This pdf is the most complete compilation of PhD Math Harvard Exams, a valuable source for the PhD candidate. Exams 1970
458 83 10MB
English Pages 623 Year 2020
Table of contents :
Cover......Page 1
Day 1......Page 2
Day 2......Page 4
Day 3......Page 6
Day 1  Solutions......Page 8
Day 2  Solutions......Page 12
Day 3  Solutions......Page 17
Day 1......Page 22
Day 2......Page 24
Day 3......Page 26
Day 1  Solutions......Page 28
Day 2  Solutions......Page 32
Day 3  Solutions......Page 36
Day 1......Page 40
Day 2......Page 42
Day 3......Page 45
Day 1......Page 48
Day 2......Page 50
Day 3......Page 52
Day 1......Page 55
Day 2......Page 56
Day 3......Page 58
Day 1  Solutions......Page 60
Day 2  Solutions......Page 63
Day 3  Solutions......Page 68
Day 1......Page 73
Day 2......Page 75
Day 3......Page 77
Day 1  Solutions......Page 79
Day 2  Solurtions......Page 85
Day 3  Solutions......Page 91
Day 1......Page 97
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Day 3......Page 101
Day 1  Solutions......Page 103
Day 2  Solutions......Page 107
Day 3  Solutions......Page 111
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Day 3......Page 118
Day 1  Solutions......Page 119
Day 2  Solutions......Page 122
Day 3  Solutions......Page 126
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Day 1  Solutions......Page 147
Day 2  Solutions......Page 155
Day 3  Solutions......Page 161
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Day 1  Solutions......Page 180
Day 2  Solutions......Page 189
Day 3  Solutions......Page 194
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Solutions  Algebraic Geometry......Page 207
Solutions  Algebraic Topology......Page 210
Solutions  Complex Analysis......Page 213
Solutions  Differential Geometry......Page 216
Solutions  Real Analysis......Page 219
Day 1......Page 221
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Solutions  AG......Page 224
Solutions  Algebra......Page 227
Solutions  Algebraic Topology......Page 228
Solutions  Complex Analysis......Page 232
Solutions  Differential Geometry......Page 234
Solutions  Real Analysis......Page 236
Day 1......Page 238
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Day 1  Solutions......Page 273
Day 2  Solutions......Page 279
Day 3  Solutions......Page 284
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Harvard University PHd MatHeMatics QUalifying exaMinations 2020 – 1970
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday January 21, 2020 (Day 1) √ 1. (A) Show Z[ p] is not a unique factorization domain for p a prime congruent to 1 mod 4. 2. (AT) Determine whether X = S 2 ∨ S 3 ∨ S 5 is homotopy equivalent to (a) a manifold, (b) a compact manifold, (c) a compact, orientable manifold. 3. (AG) We say that a curve C ⊂ P3 is a twisted cubic if it is congruent (mod the automorphism group P GL4 of P3 ) to the image of the map P1 → P3 given by φ0 : [X, Y ] 7→ [X 3 , X 2 Y, XY 2 , Y 3 ]. Now let C ⊂ P3 be any irreducible, nondegenerate curve of degree 3 over an algebraically closed field. (Here, “nondegenerate” means that C is not contained in any plane.) (a) Show that C cannot contain three collinear points. (b) Show that C is rational, that is, birational to P1 . (c) Show that C is a twisted cubic. 4. (CA) Let Ω ⊂ C be a connected open subset of the complex plane and f1 , f2 , . . . a sequence of holomorphic functions on Ω converging uniformly on compact sets to a function f . Suppose that f (z0 ) = 0 for some z0 ∈ Ω. Show that either f ≡ 0, or there exists a sequence z1 , z2 , · · · ∈ Ω converging to z0 , with fn (zn ) = 0. 5. (RA) (i) Specify the range of 1 ≤ p < ∞ for which Z
1
ϕ(f ) = 0
f (t) √ dt. t
defines a linear functional ϕ : Lp ([0, 1]) → R. (ii) For those values of p, calculate the norm of the linear functional ϕ : Lp ([0, 1]) → R. The norm of a linear functional is defined as kϕk =
ϕ(f ) f ∈Lp ([0,1]) kf kLp sup f 6=0
6. (DG) Let f : R3 → R be defined by f (x, y, z) = x2 + y 2 − 1. (i) Prove that M = f −1 (0) is a twodimensional embedded submanifold of R3 . (ii) For a, b, c ∈ R, consider the vector field X=a
∂ ∂ ∂ +b +c ∂x ∂y ∂z
For which values of a, b, c is X tangent to M at the point (1, 0, 1)?
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday January 22, 2020 (Day 2)
1. (AT) Let ∆ = {z ∈ C : z ≤ 1 be the closed unit disc in the complex plane, and let X be the space obtained by identifying z with e2πi/3 z for all z with z = 1. 1. Find the homology groups Hk (X, Z) of X with coefficients in Z. 2. Find the homology groups Hk (X, Z/3) of X with coefficients in Z/3.
2. (AG) Let C be a smooth, geometrically irreducible curve of genus 1 defined over Q, and suppose L and M are line bundles on C of degrees 3 and 5, also defined over Q. Show that C has a rational point, that is, C(Q) 6= ∅.
3. (A) Let g be an element of the finite group G. Prove that the following are equivalent: 1. g is in the center of G. 2. For every irreducible representation (V, ρ) of G, the image ρ(g) is a multiple of the identity. 3. For every irreducible representation (V, ρ) of G, the character of g has absolute value dim(G). 4. (RA) Let g ∈ L1 (R3 ) ∩ L2 (R3 ) and write gˆ for its Fourier transform defined by Z 1 gˆ(k) = e−ik·x g(x) dx 3/2 3 (2π) R For m > 0, define the function f : R3 → C by Z 1 gˆ(k) f (x) = eik·x 2 dx 3/2 k + m2 (2π) R3 Show that f solves the partial differential equation −∆f + m2 f = g in the distributional sense, i.e., show that for every test function ϕ ∈ C0∞ (R3 ), h−∆ϕ + m2 ϕ, f i = hϕ, gi. Here h·, ·i denotes the L2 (R3 )inner product.
5. (DG) Consider R2 as a Riemannian manifold equipped with the metric g = (1 + x2 )dx2 + dy 2 . (i) Compute the Christoffel symbols of the LeviCivita connection for g. (ii) Compute the parallel transport of an arbitrary vector (a, b) ∈ R2 along the curve γ(t) = (t, t) starting at t = 0. (iii) Is γ a geodesic? (iv) Are there two parallel vector fields X(t), Y (t) to the curve γ, such that g(X(t), Y (t)) = 2t? 6. (CA) Evaluate the contour integral of the following functions around the circle z = 2020 oriented counterclockwise: (i) (ii)
1 sin z ; 1 . e2z −ez
Note that
2020 π
∼ 642.98597.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday January 23, 2020 (Day 3) 1. (A) Let V be an ndimensional vector space over an arbitrary field K, and let T1 , . . . , Tn : V → V be pairwise commuting nilpotent operators on V . 1. Show that the composition T1 T2 · · · Tn = 0. 2. Does this conclusion still hold if we drop the hypothesis that the Ti are pairwise commuting? 2. (RA) (a) Let H be a Hilbert space, K ⊂ H a closed subspace, and x a point in H. Show that there exists a unique y in K that minimizes the distance kx − yk to x. (b) Give an example to show that the conclusion can fail if H is an inner product space which is not complete. 3. (AG) 1. Let the homogeneous coordinates of Pm be x0 , . . . , xm , and the homogeneous coordinates of Pn be y0 , . . . , yn , N = (m + 1)(n + 1) − 1, and the homogeneous coordinates of PN be zi,j for i = 0, . . . , m, j = 0, . . . , n. Consider the Segre embedding f : Pm × Pn → PN , given by zi,j = xi yj . Show that the degree of the Segre embedding of Pn × Pn is n+m . n 2. Let Y be a variety of dimension k in Pn , with Hilbert polynomial hY . Define the arithmetic genus of Y to be g = (−1)k (pY (0) − 1). Show that the arithmetic genus of the hypersurface H of degree d in Pn is d−1 n . 4. (CA) Find the Laurent series expansion of the meromorphic function f (z) =
1 (z − 1)(z − 2)
around the origin, valid in the annulus {z : 1 < z < 2}. 5. (DG) Define the set
1 x y 0 1 x H= : x, y ∈ R 0 0 1
(i) Equip H with a C ∞ differentiable structure so that it is diffeomorphic to R2 . (ii) Show that H is a Lie group under matrix multiplication. (iii) Show that
∂ ∂ ∂ , +x ∂y ∂x ∂y
forms a basis of leftinvariant vector fields of the associated Lie algebra. 6. (AT) Suppose that X is a space written as a union of two simply connected open subsets U1 and U2 . (a) Show that H1 X is a free abelian group. (b) Find an example in which π1 X is a nontrivial group. Why does this not contradict the the Seifertvan Kampen theorem? (c) Find an example in which π1 X is nonabelian.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday January 21, 2020 (Day 1) √ 1. (A) Show Z[ p] is not a unique factorization domain for p a prime congruent to 1 mod 4. Solution: Consider p−1=2·
p−1 √ √ = ( p + 1)( p − 1). 2
Then we claim 2 is irreducible, as it has norm 4 and there are no elements √ a + b p of norm 2, as that would imply a2 − pb2 = 2, which is impossible mod 4. But 2 clearly does not divide either factor on the right in the given ring. √ Alternatively, UFDs must be normal but (1 + p)/2 is in the normalization of the above ring.
2. (AT) Determine whether X = S 2 ∨ S 3 ∨ S 5 is homotopy equivalent to (a) a manifold, (b) a compact manifold, (c) a compact, orientable manifold. Solution: Although its Betti numbers are symmetric, its cohomology ring does not satisfy Poincar´e duality, which shows it cannot be homotopy equivalent to a compact, oriented manifold. For part (b), note that if it were homotopy equivalent to a compact, nonoriented manifold, it would have a nontrivial orientable double cover, and yet its fundamental group is trivial and so no nontrivial double covers exist. Finally, for part (a) we can embed X in a Euclidean space, such as R13 = R3 × R4 × R6 , and take a small open neighborhood which deformation retracts back to X. 3. (AG) We say that a curve C ⊂ P3 is a twisted cubic if it is congruent (mod the automorphism group P GL4 of P3 ) to the image of the map P1 → P3 given by φ0 : [X, Y ] 7→ [X 3 , X 2 Y, XY 2 , Y 3 ]. Now let C ⊂ P3 be any irreducible, nondegenerate curve of degree 3 over an algebraically closed field. (Here, “nondegenerate” means that C is not contained in any plane.)
(a) Show that C cannot contain three collinear points. (b) Show that C is rational, that is, birational to P1 . (c) Show that C is a twisted cubic. Solution: For the first part, observe that if p, q, r ∈ C are collinear, then for any fourth point s ∈ C not on the line p, q, r, the plane spanned by p, q, r and s will meet C in four points and hence contain C, contradicting nondegeneracy. To see that C is rational, choose any two distinct points p, q ∈ C; let L ⊂ P3 be the line they span and let {Hλ }λ∈P1 be the family of planes in P3 containing L. A general plane Hλ will intersect C at p, q and one other point Rλ ; conversely, a general point rλ ∈ C will lie on a unique plane Hλ . This association gives a birational isomorphism of C with P1 . Finally, given the second part we have a rational map φ from P1 to C ⊂ P3 , and since P1 is smooth this map is in fact regular. We can therefore write it as [X, Y ] 7→ [F0 (X, Y ), F1 (X, Y ), F2 (X, Y ), F3 (X, Y )] for some 4tuple [F0 , F1 , F2 , F3 ] of homogeneous cubic polynomials on P1 . Since C is nondegenerate, the Fi are linearly independent, and hence form a basis for the 4dimensional space of homogeneous cubic polynomials on P1 . If we let A ∈ GL4 be the change of basis matrix from the basis {X 3 , X 2 Y, XY 2 , Y 3 } to the basis {F0 , F1 , F2 , F3 }, then, the action of A on P3 carries the image of φ0 to C.
4. (CA) Let Ω ⊂ C be a connected open subset of the complex plane and f1 , f2 , . . . a sequence of holomorphic functions on Ω converging uniformly on compact sets to a function f . Suppose that f (z0 ) = 0 for some z0 ∈ Ω. Show that either f ≡ 0, or there exists a sequence z1 , z2 , · · · ∈ Ω converging to z0 , with fn (zn ) = 0. Solution: Suppose not. Then we can find a disc ∆ ⊂ Ω around z0 such that fn (z) 6= 0 for z ∈ ∆, and such that z0 is the sole zero of f in ∆. Now, since the functions fn and their derivatives converge uniformly on compact sets to f , we have Z Z 1 1 fn0 (z) f 0 (z) lim dz = dz. n→∞ 2πi ∂∆ fn (z) 2πi ∂∆ f (z) But by the residue theorem and the hypothesis that fn (z) 6= 0 for z ∈ ∆, the terms on the left are all zero, while the right hand side is equal to 1, a contradiction.
5. (RA)
(i) Specify the range of 1 ≤ p < ∞ for which Z
1
ϕ(f ) = 0
f (t) √ dt. t
defines a linear functional ϕ : Lp ([0, 1]) → R. (ii) For those values of p, calculate the norm of the linear functional ϕ : Lp ([0, 1]) → R. The norm of a linear functional is defined as kϕk =
sup f ∈Lp ([0,1])
ϕ(f ) kf kLp
f 6=0
Solution. (i) We use the fact that for 1 ≤ p < ∞, we can identify the dual space (Lp )∗ with Lq where q is the dual index to p, i.e., p−1 + q −1 = 1. By this identification, the claim can be rephrased as asking for which qvalues the function √1t ∈ Lq ([0, 1]). The answer is for all q ∈ [1, 2). By the relation p−1 + q −1 = 1 and the restriction to p < ∞, the answer to part (i) is the range p ∈ (2, ∞). (ii). Let p ∈ (2, ∞) or equivalently q ∈ (1, 2). We use that the identification of (Lp )∗ with Lq is in fact isometric and calculate Z kϕk = 0
1
1 √ t
1/q
q dt
=
1 1 − q/2
1/q .
6. (DG) Let f : R3 → R be defined by f (x, y, z) = x2 + y 2 − 1. (i) Prove that M = f −1 (0) is a twodimensional embedded submanifold of R3 . (ii) For a, b, c ∈ R, consider the vector field X=a
∂ ∂ ∂ +b +c ∂x ∂y ∂z
For which values of a, b, c is X tangent to M at the point (1, 0, 1)? Solution. (i) We note that f : R3 → R is smooth with derivative f∗ = ∇f = (2x, 2y, 0). This derivative has rank 1 everywhere on M . (Indeed, the points where the rank vanishes satisfy x = y = 0 and are not in M = f −1 (0).) Therefore
the inverse function theorem implies that M is an embedded submanifold of dimension 3 − 1 = 2. (ii) First, we note that (1, 0, 1) ∈ M . The vector field X is tangent to M at the point (1, 0, 1) if and only if X(f ) = 0 at (1, 0, 1). We compute X(f )(1,0,1) = (2ax + 2by)(1,0,1) = 2a which vanishes if and only if a = 0. The values of b, c ∈ R are arbitrary.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday January 22, 2020 (Day 2)
1. (AT) Let ∆ = {z ∈ C : z ≤ 1 be the closed unit disc in the complex plane, and let X be the space obtained by identifying z with e2πi/3 z for all z with z = 1. 1. Find the homology groups Hk (X, Z) of X with coefficients in Z. 2. Find the homology groups Hk (X, Z/3) of X with coefficients in Z/3.
Solution: X can be realized as a CW complex with one 0cell, one 1cell and one 2cell; the 1skeleton is just a circle S 1 and the attaching map for the 2cell is the map z 7→ z 3 . The associated cell complex is thus 0
α  Z
β  Z
 Z
 0,
with β = 0 and α given by multiplication by 3. The homology of X is then the homology of this complex, which is to say H2 (X, Z) = 0;
H1 (X, Z) = Z/3;
and H0 (X, Z) = Z.
Similarly, to find the homology with coefficients in Z/3 we tensor this complex with Z/3; now all the maps are 0 and we have H2 (X, Z/3) = H1 (X, Z/3) = H0 (X, Z/3) = Z/3.
2. (AG) Let C be a smooth, geometrically irreducible curve of genus 1 defined over Q, and suppose L and M are line bundles on C of degrees 3 and 5, also defined over Q. Show that C has a rational point, that is, C(Q) 6= ∅. Solution. Consider the line bundle N = L2 ⊗ M −1 , which has degree 1. By RiemannRoch, h0 (N ) = 1, so N has a global section σ; the zero locus of σ is then a single point p ∈ C, which is necessarily defined over Q.
3. (A) Let g be an element of the finite group G. Prove that the following are equivalent:
1. g is in the center of G. 2. For every irreducible representation (V, ρ) of G, the image ρ(g) is a multiple of the identity. 3. For every irreducible representation (V, ρ) of G, the character of g has absolute value dim(G).
Solution. (1) implies (2): ρ(g) is a Gendomorphism of V , so is a multiple of the identity by Schur’s Lemma. (2) implies (1): for any h ∈ G, ρ([g, h]) = ρ(id) for all irreducible ρ, and thus for all ρ, including the regular representation. But then [g, h] = id. (2) implies (3): Say ρ(g) = c · IV . Then ρ(g) has trace dim(V ) · c. Since some power of g (and thus of ρ(g)) is the identity, c is a root of unity. Hence c =1 and the trace has absolute value dim(V ). (3) implies (2): Since some power of ρ(g) is the identity, ρ(g) is diagonalizable and all eigenvalues are roots of unity. Hence the trace has absolute value at most dim(V ) (triangle inequality), and equals dim(V ) only when the eigenvalues are all equal to each other. But then ρ(g) is a multiple of the identity, and we’re done.
4. (RA) Let g ∈ L1 (R3 ) ∩ L2 (R3 ) and write gˆ for its Fourier transform defined by Z 1 gˆ(k) = e−ik·x g(x) dx (2π)3/2 R3 For m > 0, define the function f : R3 → C by Z 1 gˆ(k) f (x) = eik·x 2 dx 3/2 k + m2 (2π) R3 Show that f solves the partial differential equation −∆f + m2 f = g in the distributional sense, i.e., show that for every test function ϕ ∈ C0∞ (R3 ), h−∆ϕ + m2 ϕ, f i = hϕ, gi. Here h·, ·i denotes the L2 (R3 )inner product. Solution. We observe that −∆ϕ + m2 ϕ and f lie in L2 (R3 ). The former holds since −∆ϕ + m2 ϕ ∈ C0∞ (R3 ) and the latter holds because f is the inverse Fourier
transform of an L2 function. Hence, we can apply unitarity of the Fourier transform on L2 (R3 ) (Parseval’s theorem) and linearity to find \ [ + m2 ϕ, h−∆ϕ + m2 ϕ, f i = h−∆ϕ + m2 ϕ, fˆi = h−∆ϕ ˆ fˆi \ = k 2 ϕ(k) On the righthand side, we use the pointwise identity −∆ϕ(k) ˆ and ˆ the Fourier inversion theorem on f to find \ h−∆ϕ + m2 ϕ, fˆi = h(k 2 + m2 )ϕ, ˆ
k2
1 gˆi = hϕ, ˆ gˆi + m2
On the last expression, we use Parseval’s theorem again, which is allowed because ϕ, g ∈ L2 (R3 ), and we obtain hϕ, ˆ gˆi = hϕ, gi as desired.
5. (DG) Consider R2 as a Riemannian manifold equipped with the metric g = (1 + x2 )dx2 + dy 2 . (i) Compute the Christoffel symbols of the LeviCivita connection for g. (ii) Compute the parallel transport of an arbitrary vector (a, b) ∈ R2 along the curve γ(t) = (t, t) starting at t = 0. (iii) Is γ a geodesic? (iv) Are there two parallel vector fields X(t), Y (t) to the curve γ, such that g(X(t), Y (t)) = 2t?
Solution. (i). We have g
−1
=
1 1+x2
0
0 1
.
Denoting x1 = x, x2 = y, the only nonvanishing Christoffel symbol is x 1 Γ111 = (g −1 )11 ∂1 g11 = . 2 1 + x2 (ii). The equation for parallel transport ∇γ 0 (a1 , a2 ) = 0, with γ(t) = (t, t), becomes da1 t da2 1 + a = 0, = 0. dt 1 + t2 dt
The second equation is trivial and the first one can be solved by separation of variables. Implementing the initial conditions (a1 (0), a2 (0)) = (a, b) gives the a solutions a1 (t) = √1+t and a2 (t) = b. The parallel transport is therefore 2 1
2
(a (t), a (t)) =
a √ ,b . 1 + t2
(iii). By part (i), the two ODE describing the geodesic (x(t), y(t)) are given by 2 d2 x dx d2 y x = 0, + = 0. dt2 1 + x2 dt dt2 While γ(t) = (t, t) solves the second equation, it satisfies x d2 x + 2 dt 1 + x2
dx dt
2 =
t 6= 0 1 + t2
and is therefore not a geodesic. (iv). No. The scalar product of two vectors is preserved by parallel transport, since ∇ is the LeviCivita connection.
6. (CA) Evaluate the contour integral of the following functions around the circle z = 2020 oriented counterclockwise: (i) (ii)
1 sin z ; 1 . e2z −ez
Note that
2020 π
∼ 642.98597.
Solution: (i) f (z) = sin1 z is analytic in {z 6= nπ : n ∈ Z}. It has a pole of order one at nπ (reason: (sin z)0 z=nπ = cos(nπ) = (−1)n 6= 0). So Resz=nπ
1 1 = = (−1)n . sin z cos(nπ)
Therefore, Z z=2020
dz = 2πi sin z = 2πi
X
Resz=nπ
nπ≤2020
X
1 sin z
(−1)n = 2πi.
n≤642
(ii) f (z) = e2z1−ez is analytic in {e2z − ez 6= 0} = {ez 6= 1} = {z 6= 2nπi : n ∈ Z}. Since (e2z − ez )0 z=2nπi = 1 6= 0, f (z) has a pole of order one at 2nπi. So 1 1 Resz=2nπ 2z = 2z = 1. z z e −e 2e − e z=2nπ Therefore, Z z=2020
1 = 2πi 2z e − ez = 2πi
X
Resz=2nπi
2nπi≤2020
X nπ≤321
= 1286πi.
1
e2z
1 − ez
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday January 23, 2020 (Day 3)
1. (A) Let V be an ndimensional vector space over an arbitrary field K, and let T1 , . . . , Tn : V → V be pairwise commuting nilpotent operators on V . 1. Show that the composition T1 T2 · · · Tn = 0. 2. Does this conclusion still hold if we drop the hypothesis that the Ti are pairwise commuting? Solution. The key step for Part a is the Lemma 1 If S and T are commuting nilpotent operators on V then either im(ST ) ( im(T ) or T = 0. Proof Since S and T commute, im(ST ) = im(T S) ⊂ im(T ); that is, S carries the image of T to itself. If we had the equality im(ST ) = im(T ), with im(T ) 6= 0, then the restriction Sim(T ) would be invertible, and no power of S could be 0. Part (1) follows immediately. For part (2),the simplest counterexample is to 0 0 0 1 . and T = take V = K 2 , S = 1 0 0 0 2. (RA) (a) Let H be a Hilbert space, K ⊂ H a closed subspace, and x a point in H. Show that there exists a unique y in K that minimizes the distance kx − yk to x. (b) Give an example to show that the conclusion can fail if H is an inner product space which is not complete. Solution: (a): If y, y 0 ∈ K both minimize distance to x, then by the parallelogram law: kx − But
y+y 0 2
y + y0 2 y − y0 2 1 k +k k = (kx − yk2 + kx − y 0 k2 ) = kx − yk2 2 2 2
cannot be closer to x than y, by assumption, so y = y 0 .
Let C = inf y∈K kx − yk, then 0 ≤ C < ∞ because K is nonempty. We can find a sequence yn ∈ K such that kx − yn k → C, which we want to show is
m m Cauchy. The midpoints yn +y are in K by convexity, so kx − yn +y k≥C 2 2 and using the parallelogram law as above one sees that kyn − ym k → 0 as n, m → ∞. By completeness of H the sequence yn converges to a limit y, which is in K, since K is closed. Finally, continuity of the norm implies that kx − yk = C.
(b): For example choose H = C([0, 1]) ⊂ L2 ([0, 1]), K the subspace of functions with support contained in [0, 12 ], and and x = 1 the constant function. If fn is a sequence in K converging to f ∈ H in L2 norm, then Z 1 f 2 = 0 1/2
thus f vanishes on [1/2, 1], showing √ that K is closed. The distance kx − yk can be made arbitrarily close to 1/ 2 for y ∈ K by approximating χ[0,1/2] by continuous functions, but the infimum is not attained.
3. (AG) 1. Let the homogeneous coordinates of Pm be x0 , . . . , xm , and the homogeneous coordinates of Pn be y0 , . . . , yn , N = (m + 1)(n + 1) − 1, and the homogeneous coordinates of PN be zi,j for i = 0, . . . , m, j = 0, . . . , n. Consider the Segre embedding f : Pm × Pn → PN , given by zi,j = xi yj . Show that the degree of the Segre embedding of Pn × Pn is n+m . n 2. Let Y be a variety of dimension k in Pn , with Hilbert polynomial hY . Define the arithmetic genus of Y to be g = (−1)k (pY (0) − 1). Show that the arithmetic genus of the hypersurface H of degree d in Pn is d−1 n . Solution: 1. Note that the Hilbert polynomial pf (Pm ×Pn ) of the Segre embedding of Pm × Pn is the product of the Hilbert polynomials pPm , pPn , of Pm and Pn . Then m+d n+d pf (Pm ×Pn ) (d) = pPm (d) · pPn (d) = . d d (Or one can note that a homogeneous of degree d in PN pulls back to a bihomogeneous polynomial of bidgree (d, d).) Thus 1 1 m+n m n deg(f (P × P )) = (m + n)! · = . n! m! n
. Then − m−d+n 2. Note that the Hilbert polynomial is pH (m) = m+n n n n−d n d−1 . pH (0) = 1 − = 1 − (−1) n n
4. (CA) Find the Laurent series expansion of the meromorphic function f (z) =
1 (z − 1)(z − 2)
around the origin, valid in the annulus {z : 1 < z < 2}. Solution: We use partial fractions to write f as a sum of functions with only a single pole, yielding −1 1 1 = + (z − 1)(z − 2) z−1 z−2 We now take the power series expansion of the second term valid in the disc z < 2, that is, ∞ X 1 −1 1 −1 n = · = z z−2 2 1 − z/2 2n+1 n=0
and the Laurent series expansion of the first term valid in the annulus z > 1; that is, ∞ X −1 1 = −z −1 = − z −n−1 z−1 1 − z −1 n=0
The sum of these two is the Laurent series expansion. 5. (DG) Define the set
1 x y H = 0 1 x : x, y ∈ R 0 0 1
(i) Equip H with a C ∞ differentiable structure so that it is diffeomorphic to R2 . (ii) Show that H is a Lie group under matrix multiplication. (iii) Show that
∂ ∂ ∂ , +x ∂y ∂x ∂y
forms a basis of leftinvariant vector fields of the associated Lie algebra.
Solution (i). We use a single, global coordinate chart ϕ : H → R2 1 x y 0 1 x 7→ (x, y) 0 0 1
The differentiable structure defined by H and ϕ is then diffeomorphic to R2 . (ii). Let A, B ∈ H with
1 a b A = 0 1 a , 0 0 1
1 x y B = 0 1 x . 0 0 1
Elementary computation shows that 1 a + x b + ax + y ∈ H, 1 a+x AB = 0 0 0 1
1 −a a2 − b −a ∈ H, = 0 1 0 0 1
A−1
so H is a group. In coordinates, these maps can be written as (a, b) 7→ (−a, a2 − b)
(a, b, x, y) 7→ (a + x, b + ax + y), which are clearly C ∞ , so H is a Lie group.
(iii). Since dimH = 2 and the vector fields are obviously linearly independent, it suffices to show they are leftinvariant. It is convenient to identify elements A ∈ H with their coordinate vectors, say A = (a, b). As computed above, the left translation LA of B = (x, y) can be written as LA (B) = (a + x, b + ax + y). It has the Jacobian 1 0 (LA )∗ = a 1 From this we can check directly that both vector fields satisfy (LA )∗ XB = XAB . Indeed, ∂ ∂ 1 0 0 0 (LA )∗ = = = a 1 1 1 ∂y B ∂y LA (B) and (LA )∗
∂ ∂ +x ∂x ∂y
=
B
1 0 a 1
1 x
=
1 x+a
=
∂ ∂ +x ∂x ∂y
LA (B)
6. (AT) Suppose that X is a space written as a union of two simply connected open subsets U1 and U2 .
(a) Show that H1 X is a free abelian group. (b) Find an example in which π1 X is a nontrivial group. Why does this not contradict the the Seifertvan Kampen theorem? (c) Find an example in which π1 X is nonabelian. Solution: The MayerVietoris sequence H1 U ⊕ H1 V → H1 X → H0 (U ∩ V ) identifies H1 X with a subgroup of H0 (U ∩ V ) which is a free abelian group. The claim follows from the fact that a subgroup of a free abelian group is free abelian. The circle S 1 is the union of two contractible open sets and has fundamental group Z. (This doesn’t contradict Seifertvan Kampen theorem because the intersection is not connected.) Finally, if S is any discrete set, the suspension of S is the union of two cones on S, each of which is contractible. The fundamental group of the suspension of S is the free group on the points of S, so taking S to consist simply of three points (so that the suspension is a figure 8) provides an example for part (c).
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 3, 2019 (Day 1)
1. (AT) Suppose that M is a compact connected manifold of dimension 3, and that the abelianization (π1 M )ab is trivial. Determine the homology and cohomology groups of M (with integer coefficients).
2. (A) Prove that for every finite group G the number of groups homomorphisms h : Z2 → G is nG where n is the number of conjugacy classes of G.
3. (AG) Let X ⊂ Pn be a projective variety over a field K, with ideal I(X) ⊂ K[Z0 , . . . , Zn ] and homogeneous coordinate ring S(X) = K[Z0 , . . . , Zn ]/I(X). The Hilbert function hX (m) is defined to be the dimension of the mth graded piece of S(X) as a vector space over K. a. Define the Hilbert polynomial pX (m) of X. b. Prove that the degree of pX is equal to the dimension of X. c. For each m, give an example of a variety X ⊂ Pn such that hX (m) 6= pX (m).
4. (CA) Use contour integration to prove that for real numbers a and b with a > b > 0, Z π dθ π =√ . a2 − b2 0 a − b cos θ
5. (RA) Dirichlet’s function D is the function on [0, 1] ⊂ R that equals 1 at every rational number and equals 0 at every irrational number. Thomae’s function T is the function on [0, 1] whose value at irrational numbers is 0 and whose value at any given rational number r is 1/q, where r = p/q with p and q relatively prime integers, q > 0. 1. Prove that D is nowhere continuous. 2. Show that T is continuous at the irrational numbers and discontinuous at the rational numbers. 3. Show that T is nowhere differentiable.
6. (DG) Consider the Riemannian manifold (D, g) with D the unit disk in R2 and g=
1 (dx2 + dy 2 ) 1 − x2 − y 2
Find the Riemann curvature tensor of (D, g). Use this to read off the Gaussian curvature of (D, g).
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 4, 2019 (Day 2)
1. (CA) Fix a ∈ C and an integer n ≥ 2. Show that the equation az n + z + 1 = 0 has a solution with z ≤ 2.
2. (AG) Let PN be the space of nonzero homogeneous polynomials of degree d in n + 1 variables over a field K, modulo multiplication by nonzero scalars, and let U ⊂ PN be the subset of irreducible polynomials F such that the zero locus V (F ) ⊂ Pn is smooth. (a) Show that U is a Zariski open subset of PN . (b) What is the dimension of the complement D = PN \ U ? (c) Show that D is irreducible.
3. (RA) Let B denote the Banach space of continuous, real valued functions on [0, 1] ⊂ R with the sup norm. 1. State the ArzelaAscoli theorem in the context of B. 2. Define what is meant by a compact operator between two Banach spaces. 3. Prove that the operator T : B → B defined by Z x (T f )(x) = f (y) dy 0
is compact.
4. (A) Let Fq be the finite field with q elements. Show that the number of 3 × 3 nilpotent matrices over Fq is q 6 . 5. (AT) Let Symn X denote the nth symmetric power of a CW complex X, i.e. X n /Sn , where the symmetric group Sn acts by permuting coordinates. Show that for all n ≥ 2, the fundamental group of Symn X is abelian.
6. (DG) Let S 2 ⊂ R3 be the unit 2sphere, with its usual orientation. Let X be the vector field generating the flow given by cos(t) − sin(t) 0 x sin(t) cos(t) 0 · y , 0 0 1 z and let ω be the volume form induced by the embedding in R3 (so the total “volume” is 4π). Find a function f : S 2 → R satisfying df = ιX ω where ιX ω is the contraction of ω by X.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 5, 2019 (Day 3) 1. (RA) Let f : [0, 1] → R be in the Sobolev space H 1 ([0, 1]); that is, functions f such that both f and its derivative are L2 integrable. Prove that Z 1 −2πinx f (x)e dx = 0. lim n n→∞
0
2. (CA) Given that the sum X n∈Z
1 (z − n)2
converges uniformly on compact subsets of C \ Z to a meromorphic function on the entire complex plane, prove the identity X 1 π2 = . 2 (z − n)2 sin πz n∈Z
3. (AG) Let C ⊂ P3 be a smooth curve of degree 5 and genus 2. (a) By considering the restriction map ρ : H 0 (OP3 (2)) → H 0 (OC (2)), show that C must lie on a quadric surface Q. (b) Show that the quadric surface Q is unique. (c) Similarly, show that C must lie on at least one cubic surface S not containing Q. (d) Finally, deduce that there exists a line L ⊂ P3 such that the union C ∪ L is a complete intersection of a quadric and a cubic.
4. (A) Show that if p, q are distinct primes then the polynomial (xp −1)/(x−1) is irreducible mod q if an only if q is a primitive residue of p (i.e. if every integer that is not a multiple of p is congruent to q e mod p for some integer e). ii) Prove that x6 + x5 + x4 + x3 + x2 + x + 1 factors mod 23 as the product of two irreducible cubics.
5. (DG) Suppose that G is a Lie group.
(a) Consider the map ι : G → G defined by ι(g) = g −1 . Show that the derivative of ι at the identity element is multiplication by −1. (b) For g ∈ G define maps Lg , Rg : G → G by Lg (x) = gx Rg (x) = xg. Show that if ω is a kform which is biinvariant in the sense that L∗g ω = Rg∗ ω then ι∗ ω = (−1)k ω. (c) Show that biinvariant forms on G are closed.
6. (AT) Suppose that m is odd. Show that if n is odd there is a fixed point free action of Z/m on S n . What happens if n is even?
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 3, 2019 (Day 1)
1. (AT) Suppose that M is a compact connected manifold of dimension 3, and that the abelianization (π1 M )ab is trivial. Determine the homology and cohomology groups of M (with integer coefficients). Solution: Since the obstruction to orienting M is a homomorphism from π1 M to the abelian group {±1} we know that M is orientable, and we may avail ourselves of Poincar´e duality. By Poincar´e’s theorem H1 (M ) = π1 (M )ab = 0. By Poincar´e duality this gives us H 3 (M ) = Z and H 2 (M ) = 0. We now use the universal coefficient sequence 0 → Ext(Hi−1 M, Z) → H i (M ) → Hom(Hi M ; Z) → 0 to conclude that H 0 (M ) = Z and H 1 (M ) = 0, from which, say, by Poincar´e duality you can conclude that H3 (M ) = Z and H2 (M ) = 0. In short M has the homology and cohomology of S 3 .
2. (A) Prove that for every finite group G the number of groups homomorphisms h : Z2 → G is nG where n is the number of conjugacy classes of G. Solution. Z2 = ha, b  ab = bai, so the homomorphisms Z2 → G are in bijection with pairs (g, h) ∈ G such that gh = hg. Given g0 ∈ G, the group elements h such that g0 h = hg0 constitute the stabilizer of g0 under the action of G on itself by conjugation. The orbit of g0 is its conjugacy class [g0 ], so the number of solutions (g0 , h) is G/[g0 ] (orbitstabilizer theorem). Thus for each conjugacy class c the number of solutions with g ∈ c is c · G/c = G. Summing over the n conjugacy classes gives nG, Q.E.D.
3. (AG) Let X ⊂ Pn be a projective variety over a field K, with ideal I(X) ⊂ K[Z0 , . . . , Zn ] and homogeneous coordinate ring S(X) = K[Z0 , . . . , Zn ]/I(X). The Hilbert function hX (m) is defined to be the dimension of the mth graded piece of S(X) as a vector space over K. a. Define the Hilbert polynomial pX (m) of X.
b. Prove that the degree of pX is equal to the dimension of X. c. For each m, give an example of a variety X ⊂ Pn such that hX (m) 6= pX (m).
Solution: a. The Hilbert polynomial pX (m) is the unique polynomial such that pX (m) = hX (m) for all sufficiently large integers m. b. By induction on the dimension of X: if X is a finite set of d points, then the Hilbert polynomial is the constant d; and in general the Hilbert polynomial pX∩H of a general hyperplane section of a variety X is the first difference of the Hilbert polynomial pX c. Let X consist of any k distinct points of Pn . Then X is a variety of dimension 0 and degree k, so by the previous part pX (m) = k. But hX (m) is at most the dimension of the space of homogeneous degree m polynomials in n + 1 variables, so for sufficiently large k, hX (m) < k = pX (m).
4. (CA) Use contour integration to prove that for real numbers a and b with a > b > 0, Z π dθ π =√ . 2 a − b cos θ a − b2 0 Solution: Since the integrand is symmetric about θ = π, we can extend the R 2π integration interval to [0, 2π], I = 21 0 a−bdθcos θ , which can be written as an integral around an unit circle in the complex plane I I 1 1 dz 1 dz I= = 2 b 1 2 2az − bz − b i a − 2 (z + z ) iz I 1 1 =− dz. 2a 2 bi z − b z+1 Then the singular points are the zeros of the denominator, z 2 − The roots z1 , z2 are z1 =
p 1 a − a2 − b2 , b
z2 =
p 1 a + a2 − b2 . b
2a b z
+ 1 = 0.
Note that only z1 lies inside the unit circle. Thus I 1 dz = 2πiResz=z1 [ ] (z − z1 )(z − z2 ) (z − z1 )(z − z2 ) 1 = 2πi lim (z − z1 ) z→z1 (z − z1 )(z − z2 ) 1 = 2πi (z1 − z2 ) b = −2πi √ . 2 2 a − b2 Therefore I=−
π 1 b =√ . · − 2πi √ bi 2 a2 − b2 a2 − b2
5. (RA) Dirichlet’s function D is the function on [0, 1] ⊂ R that equals 1 at every rational number and equals 0 at every irrational number. Thomae’s function T is the function on [0, 1] whose value at irrational numbers is 0 and whose value at any given rational number r is 1/q, where r = p/q with p and q relatively prime integers, q > 0. 1. Prove that D is nowhere continuous. 2. Show that T is continuous at the irrational numbers and discontinuous at the rational numbers. 3. Show that T is nowhere differentiable.
Solution. For the first part: for any rational number α we can find a sequence of irrational numbers αn converging to α; since lim D(αn ) 6= D(α), D cannot be continuous at α. Similarly, if α is irrational, we can find a sequence of rational numbers αn converging to α to conclude that D cannot be continuous at α. For the second, exactly the same argument shows that T is discontinuous at rational numbers. But now if α is irrational and αn = pn /qn is a sequence of rational numbers converging to α then we must have lim qn = ∞, so lim T (αn ) = T (α) and T is continuous at α. Finally, suppose that α is irrational. We can certainly find a sequence of irrational numbers βn 6= α converging to α, from which we see that if the limit T 0 (α) exists it must be 0. But for any n we can also find a rational number αn = pn /2n with denominator 2n such that α − αn  ≤ 1/2n , from which we see that if the limit T 0 (α) exists it must be ≥ 1.
6. (DG) Consider the Riemannian manifold (D, g) with D the unit disk in R2 and g=
1 (dx2 + dy 2 ) 1 − x2 − y 2
Find the Riemann curvature tensor of (D, g). Use this to read off the Gaussian curvature of (D, g). Solution. This is a straightforward computation. We have 1 − x2 − y 2 0 −1 g = 0 1 − x2 − y 2 We set x1 = x and x2 = y and compute the Christoffel symbols x 1 − x2 − y 2 y = . 1 − x2 − y 2
Γ111 = Γ212 = Γ221 = −Γ122 = − Γ211 = Γ112 = Γ121 = Γ222
With these, we can compute the only independent component of the Riemann curvature tensor: ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ −2 R =g R = , , , , , ∂x ∂y ∂x ∂y ∂x ∂y ∂y ∂x (1 − x2 − y 2 )2 (Indeed, by antisymmetry, all other components of the Riemann curvature tensor either vanish or are equal to this up to a sign.) The Gaussian curvature K satisfies Rabcd = K(gac gbd − gad gbc ) which implies K = −2.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 4, 2019 (Day 2)
1. (CA) Fix a ∈ C and an integer n ≥ 2. Show that the equation az n + z + 1 = 0 has a solution with z ≤ 2. Solution: There are two cases. First, assume that a < 2−n . Let D denote the disk where z ≤ 2 and let ∂D denote the circle z = 2. Let f (z) = az n + z + 1 and let g(z) = z + 1. On ∂D, the function g − f obeys the inequality g(z) − f (z) = azn < 1. Since this is less than g(z) for each z ∈ ∂D, and since g has no zeros on ∂D, none of the members of the 1parameter family of functions {ft = f + t(g − f )}t∈[0,1] has a zero on ∂D. Therefore f (which is ft = 0) and g (which is ft = 1) have the same number of zeros (counting multiplicity) in D and the number is 1. Now assume that a ≥ 2−n . By the fundamental Q theorem of algebra, the function f (z) = az n + z + 1 factors as f (z) = a nk=1 (z − αk ), where αk ∈ C. In particular, n Y (−1)n a αk = 1. k=1
Thus,
Q
αk  ≤
2n .
This implies at least one of αk  ≤ 2.
2. (AG) Let PN be the space of nonzero homogeneous polynomials of degree d in n + 1 variables over a field K, modulo multiplication by nonzero scalars, and let U ⊂ PN be the subset of irreducible polynomials F such that the zero locus V (F ) ⊂ Pn is smooth. (a) Show that U is a Zariski open subset of PN . (b) What is the dimension of the complement D = PN \ U ? (c) Show that D is irreducible. Solution: In the product PN × Pn , consider the incidence correspondence Φ = {(F, p)  F (p) =
∂F (p) = 0 ∀i}. ∂Xi
This is a closed subvariety of PN × Pn and hence projective, so that its image D = π1 (Φ) ⊂ PN is closed; hence U is open. Moreover, since the projection map π2 : Φ → Pn is a projective bundle with fiber PN −n−1 , we see that Φ is
irreducible of dimension N − 1. Finally, since there exist hypersurfaces with just one singular point (e.g., cones), the general fiber of Φ over a point in its image D ⊂ PN is 0dimensional; it follows that D is again irreducible of dimension N − 1.
3. (RA) Let B denote the Banach space of continuous, real valued functions on [0, 1] ⊂ R with the sup norm. 1. State the ArzelaAscoli theorem in the context of B. 2. Define what is meant by a compact operator between two Banach spaces. 3. Prove that the operator T : B → B defined by Z x (T f )(x) = f (y) dy 0
is compact.
Solution. If B, B 0 are Banach spaces, a linear operator T : B → B 0 is said to be compact if the closure of {T v : v ∈ B, kvk ≤ 1} (that is, the closure of the image of the closed unit ball) is compact in B 0 . For our T , the image of the closed unit ball is an equicontinuous family of functions on [0, 1]. Indeed if f ∈ B with kf k ≤ 1 then Z 0 x (T f )(x0 ) − (T f )(x) = f (x) dx ≤ x0 − x x so that given > 0 the same δ (namely δ = ) works uniformly for all such T f . Moreover this image is uniformly bounded: (T f )(0) = 0 for all f , so (T f )(x) ≤ x ≤ 1 for all x ∈ [0, 1]. Hence the closure of {T v : v ∈ B, kvk ≤ 1} is compact by the Arzel` a–Ascoli theorem.
4. (A) Let Fq be the finite field with q elements. Show that the number of 3 × 3 nilpotent matrices over Fq is q 6 . Solution: Although Fq is not algebraically closed, there is still a Jordan normal form for nilpotent matrices; by CayleyHamilton, for example, we know T 3 is identically zero if T denotes our endomorphism of our vector space V , and we may consider 0 ⊂ ker T ⊂ ker T 2 ⊂ ker T 3 = V . There are only three different possibilities for the list of nontrivial dimensions, namely (3, 3), (2, 3), (1, 2),
and choosing bases appropriately, jugate, over Fq , to exactly one of 0 1 0, 0 0 0 0
we find that all nilpotent matrices are con 0 0 1 0 0 , 0 0 1 . 0 0 0 0
There is exactly one matrix in the first case. For the latter two cases, we have only to compute the stabilizer subgroup under the conjugation action for each matrix above to find the size of the conjugacy orbit. First, recall GL3 (Fq ) = (q 3 − 1)(q 3 − q)(q 3 − q 2 ) by successively picking the images of a11 a12 a13 the standard basis vectors; that a matrix a21 a22 a33 commutes with a31 a32 a33 the second matrix above is equivalent to the conditions that a11 = a22 , a21 = a31 = a23 = 0 and that the determinant vanish is that a11 = a22 , a33 be units, and so the size of this stabilizer group is (q − 1)2 q 3 . Similarly, finding the subgroup of matrices that commutes with the third matrix above is explicitly equivalent to a11 = a22 = a33 , a12 = a23 , a21 = a31 = a32 = 0 of order (q−1)q 2 . Hence, the total number of nilpotent matrices is (q 3 − 1)(q 3 − q)(q 3 − q 2 ) (q 3 − 1)(q 3 − q)(q 3 − q 2 ) + (q − 1)2 q 3 (q − 1)q 2 = 1 + (q 3 − 1)(q + 1) + (q 3 − 1)(q 3 − q)
N3 (Fq ) = 1 + = q6.
5. (AT) Let Symn X denote the nth symmetric power of a CW complex X, i.e. X n /Sn , where the symmetric group Sn acts by permuting coordinates. Show that for all n ≥ 2, the fundamental group of Symn X is abelian. Solution: This is a version of the EckmannHilton argument. If one takes as basepoint some point in the (thin) diagonal X ⊂ Symn X, then every based loop S 1 → Symn X lifts to a loop S 1 → X n , and so one directly has π1 X n π1 Symn X, but π1 X n ' (π1 X)n , and (γ1 , · · · , γn ) = (γ1 , 1, · · · 1) ◦ (1, γ2 , 1, · · · , 1) ◦ · · · ◦ (1, · · · , 1, γn ) = (γ1 , 1, · · · , 1) ◦ (γ2 , 1, · · · , 1) ◦ · · · ◦ (γn , 1, · · · , 1) = (γ1 · · · γn , 1, · · · , 1) in π1 Symn X by simply “waiting” to do each loop γi in turn and then using that as we’re in the symmetric product, it doesn’t matter in which factor we’re doing the loop. So it suffices to show (γ, 1, · · · , 1) and (σ, 1, · · · , 1) commute
but (γ, 1, · · · , 1)(σ, 1, · · · , 1) = (γ, 1, · · · , 1)(1, σ, 1, · · · , 1) = (γ, σ, 1, · · · , 1) = (σ, γ, 1, · · · , 1) = (σ, 1, · · · , 1)(γ, 1, · · · , 1) by using the above reasoning a few more times.
6. (DG) Let S 2 ⊂ R3 be the unit 2sphere, with its usual orientation. Let X be the vector field generating the flow given by x cos(t) − sin(t) 0 sin(t) cos(t) 0 · y , z 0 0 1 and let ω be the volume form induced by the embedding in R3 (so the total “volume” is 4π). Find a function f : S 2 → R satisfying df = ιX ω where ιX ω is the contraction of ω by X. Solution: The outward unit normal to S 2 is x∂x + y∂y + z∂z so the volume form of S 2 is ω = ιx∂x +y∂y +z∂z dx ∧ dy ∧ dz = x dy ∧ dz − y dx ∧ dz + z dx ∧ dy. The vector field X is −y ∂x + x ∂y. So the contraction ιX ω is y 2 dz − yz dy + x2 dz − xz dx = (x2 + y 2 ) dz − z(y dy + x dx). Using the fact that on S 2 one has x dx + y dy + z dz = 0, this becomes (x2 + y 2 + z 2 ) dz = dz. We may therefore take f (x, y, z) = z.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 5, 2019 (Day 3)
1. (RA) Let f : [0, 1] → R be in the Sobolev space H 1 ([0, 1]); that is, functions f such that both f and its derivative are L2 integrable. Prove that Z 1 −2πinx f (x)e dx = 0. lim n n→∞
0
Solution. The quantity Z
1
f (x)e−2πinx dx = fˆ(n)
0
is the nth Fourier coefficient of f . Since f ∈ H 1 ([0, 1]), it holds that nfˆ(n) is squaresummable and, in particular, forms a nullsequence.
2. (CA) Given that the sum X n∈Z
1 (z − n)2
converges uniformly on compact subsets of C \ Z to a meromorphic function on the entire complex plane, prove the identity X 1 π2 = . 2 (z − n)2 sin πz n∈Z Solution: Both sides of the desired equality are entire meromorphic functions with double poles at the integers. Moreover, they have the same polar part 1/(z − n)2 at each n ∈ Z, so that the difference g(z) =
X π2 1 − 2 sin πz n∈Z (z − n)2
is an entire holomorphic function. We claim now that g(z) is bounded, and hence by Liouville’s theorem constant. By the periodicity g(z + n) = g(z) ∀n ∈ Z, it suffices to prove that it’s bounded in the strip 0 ≤ 0 such that 2e ≡ 1 mod 7 is e = 3, so the Galois orbits on nontrivial 7th roots of unity have size 3, whence each Galoisstable factor of f7 has degree 3, Q.E.D. [The factorization is (x3 + 10x2 + 9x − 1)(x3 − 9x2 − 10x − 1).]
5. (DG) Suppose that G is a Lie group. (a) Consider the map ι : G → G defined by ι(g) = g −1 . Show that the derivative of ι at the identity element is multiplication by −1. (b) For g ∈ G define maps Lg , Rg : G → G by Lg (x) = gx Rg (x) = xg. Show that if ω is a kform which is biinvariant in the sense that L∗g ω = Rg∗ ω then ι∗ ω = (−1)k ω. (c) Show that biinvariant forms on G are closed.
Solution: For the first part note that if g(t) = exp(λt) then g(t)−1 = exp(−λt). The claim follows by taking the derivative with respect to t at t = 0. For the second part note that ι∗ ωg = ι∗ L∗g−1 ωe = Rg∗ ι∗ ωe so it suffices to show that ι∗ ωe = (−1)k ωe . But if X1 , . . . , Xk are tangent vectors at e then ι∗ (ω)(X1 , . . . , Xk ) = ω(dι(X1 ), . . . , dι Xk ) = ω(−X1 , . . . , −Xk ) = (−1)k ω(X1 , . . . , Xk ) by the first part. The third part follows from the equations (−1)(k+1) dω = ι∗ dω = dι∗ ω = (−1)k dω.
6. (AT) Suppose that m is odd. Show that if n is odd there is a fixed point free action of Z/m on S n . What happens if n is even? Solution: For the first part write n = 2k − 1 and regard S n as the unit sphere in Cn . The formula (z1 , . . . , zn ) 7→ e2πi/m (z1 , . . . , zn ) defines a free action of Z/m on S n . (This part does not require m to be odd). There can be no free action on S 2k . This follows from the Lefschetz fixed point formula. Since the automorphism group of Z is cyclic of order 2 and m is odd, there are no nontrivial actions of Z/m on Z. It follows that the Lefschetz number of any action is 2, so there must be a fixed point.
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Spring 2018 (Day 1)
Problem 1 (RA) Let X be a subset of [0, 1] with the following two properties: • For any real number, r, there is x ∈ X such that r  x is rational. •
For any two distinct x, y ∈ X, the number x  y is irrational.
Prove that X is not Lebesguemeasurable.
Problem 2 (DG) Use (x, y, z) for the Euclidean coordinate functions on R3 and let a denote the 1form a = dz +
1 2
(x dy  y dx).
a) Compute da and a ∧ da. b) Prove that the kernel of a defines a smooth, 2dimensional vector subbundle in TR3. c) Suppose that B ⊂ R3 is an open ball and that u and w are pointwise linearly independent vector fields in the kernel of a on B. Prove that the commutator of u and v is nowhere in the kernel of a.
Problem 3 (CA) How many roots of the polynomial P(z) = z4  6z + 3 occur where z < 2?
Problem 4 (T) Let X1 and X2 denote distinct copies of T2, so each is S1 × S1 with S1 denoting the circle. Define the space X to be the quotient of the disjoint union of X1 and X2 by the equivalence relation whereby any point of the form (z, 1) in X1 is identified with the corresponding (z, 1) in X2. Compute the cohomology ring of the space X. (Thus, compute H*(X; Z) and determine its cupproduct structure.)
Problem 5 (AG) Let X denote the affine curve in A2 where y2  x3 + x2 = 0. Prove that X is singular and that there is a birational morphism from A1 onto X.
Problem 6 (AN) Let G1, …, Gn denote finite groups. For each m ∈ {1, …, n}, let ρm: Gm → GL(Vm) denote a finite dimensional, complex representation of Gm. Use χm to denote the character of ρm. Set G = G1 × ··· × Gm and V = V1 ⊗ ··· ⊗ Vm. a) Define ρ: G → GL(V) by the rule ρ(g1, …, gn) = ρ1(g1) ⊗ ··· ⊗ ρn(gn). Write the character of ρ in terms of the characters {χm}1≤m≤n . b) Prove that (V, ρ) is an irreducible representation of G if and only if, for all m, each (Vm, ρm) is an irreducible representation of Gm.
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Spring 2018 (Day 2)
Problem 1 (RA) a) Let A1, A2, … be a countable collection of events in a probability space; and let A1c, A2c, … denote their respective complements. Prove the following assertion: If A1, A2, … are mutually independent, then A1c, A2c, … are also mutually independent. (A collection of events {An}n=1,2,… is said to be mutually independent when any member is independent of the mutual intersection of any finite subcollection of the other events.) b) Let A1, A2, … denote a sequence of mutually independent events in a probability space with the property that the sum of their probabilities is infinite. Prove that with probability one, the event An must occur for infinitely many values of the index n.
Problem 2 (DG) The Euclidean metric on R2 can be written using the standard rectilinear coordinates (x, y) as dx ⊗ dx + dy ⊗ dy. Let u denote a smooth function on R2 and let g denote the 2u
metric e (dx ⊗ dx + dy ⊗ dy). Let ∇ denote the corresponding LeviCivita covariant derivative for the metric g, acting on sections of T*R2. a) Write ∇(dx) and ∇(dy) in terms of u and its derivatives. b) Write the scalar curvature of the metric g in terms of u and its first and second derivatives.
Problem 3 (CA)
∞
Supposing that a is a positive number, evaluate the integral
cos2 (x)
∫ x +a 2
2
dx
using the
0
method of residues.
Problem 4 (T) Let X = T2 ∨ S2 which is the join of the torus T2 (which is S1 × S1) and the 2sphere S2. a) Describe the universal covering space of X. b) Compute π1(X). c) Compute π2(X).
Problem 5 (AG) Show that for any genus 2 curve, C, there is a divisor on C which has degree greater than zero, but is not linearly equivalent to an effective divisor. (Hint: The RiemannRoch formula states that h0(C, L) − h0(C, KC ⊗ L) = deg(L) + 1  g(C) for a line bundle L on a curve C. Here, KC denotes the canonical bundle of C and g(C) denotes the genus of C.)
Problem 6 (AN) Let k denote a finite field of 2ƒ elements for some positive integer ƒ. a) Prove that the map from k to itself given by x → x2 + x is a homomorphism of additive groups. Assuming this, then prove that exactly 2ƒ1 elements of k can be written as x2 + x for some x ∈ k. b) Prove that any given a ∈ k can be written as x2 + x for some x ∈ k if and only if ƒ1
∑ a2 i=0
i
= 0.
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Spring 2018 (Day 3) Problem 1 (RA) For ƒ: R → R a Lebesgue measurable function, let ƒ denote the norm, ƒ = ∫  ƒ dµ R
where dµ is the Lebesgue measure. Let L1(R) denote the vector space (over R) of Lebesgue measurable functions ƒ: R → R with ƒ < ∞ (we identify functions that are equal almost everywhere). If ƒ, g are functions in L1(R), define their convolution (an Rvalued function on R denoted by ƒ∗g) by the following rule: (ƒ∗g)(x) =
∫ ƒ(x  t) g(t) dt R
if
∫  ƒ(x  t) g(t) dt is finite; and (ƒ∗g)(x) = 0 otherwise. R
Prove the following: There is no function e ∈ L1(R) such that e ∗ƒ = ƒ for all ƒ ∈ L1(R). Here are two hints: First, keep in mind that a function is Lebesgue measurable when, for any real number E, the set of points in R where the function is less than E is Lebesgue measurable. Second, consider the sequence {ƒn}n=1,2,.. of functions on R which is defined as follows: For any given positive integer n, set ƒn(x) = 1 if  n1 ≤ x ≤ n1 , and set ƒn(x) = 0 otherwise.
Problem 2 (DG) View the 4dimensional sphere (denoted by S4) as the 1point compactification of R4, thus R4 ∪ ∞. A complex, rank 2 vector bundle over S4 (to be denoted by E) can be defined as follows: Cover the sphere S4 by the two open sets R4 and (R4−0) ∪ ∞. A map (to be denoted by g) from their intersection (which is R4−0) to the group SU(2) (the group of 2 × 2 unitary matrices with determinant 1) is defined by first writing the Euclidean coordinates of any given x ∈ R4 as (x1, x2, x3, x4) and using these coordinate functions to define g(x) for x ∈R4−0 by
⎛ x1 + ix 2 x 3 + ix 4 ⎞ g(x) = 1x ⎜ ⎟ . ⎝ x 3 + ix 4 x1  ix 2 ⎠ The vector bundle E is the quotient of the product C2 bundle over the R4 part of S4, and the product C2 bundle over the complement in S4 of 0 ∈ R4 by the equivalence relation that identifies pairs (x, s0) ∈ R4 × C2 and (y, s1) ∈ ((R4−0) ∪ ∞) when x and y are in R4−0 and x = y and s1 = g(x) s0. (The projection map from E to S4 sends the equivalence class of any (x, s) for x ∈ R4 to x; and it sends the equivalence class of (∞, s) to ∞.) a) Write a connection on this vector bundle. b) Compute the curvature 2form of your connection.
Problem 3 (CA) a) Prove that ∑ (z 1n)2 defines a meromorphic function on C with poles only at the n∈Z
points in Z. b) Prove that
∑ (z 1n)
2
n∈Z
2
= sinπ2 (πz) .
Problem 4 (T) a) Construct a connected, topological space X such that π1(X) is generated by two elements, denoted by a and b, subject to the relations a3 = 1 and b 3 = 1. b) For which q ≥ 1, is Hq(X; Z) independent of your choice of X?
Problem 5 (AG) The twisted cubic (to be denoted by X) is the image of the map from P1 to P3 defined using homogeneous coordinates by the rule [s : t] → [s3 : s2t : st2 : t3]. It is also the locus in P3 where the three polynomials {z0z3  z1z2, z0z2  z12, z1z3  z22} are simultaneously zero. Prove that the Hilbert polynomial of the twisted cubic, ℘X, obeys ℘X(n) = 3n + 1.
Problem 6 (AN) Prove that there is a unique positive integer n ≤ 102017 such that the last 2017 digits of n3 are 0000 ··· 00002017 (with all 2005 digits represented by ··· being zeros as well).
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday August 30, 2016 (Day 1)
1. (DG) (a) Show that if V is a C ∞ vector bundle over a compact manifold X, then there exists a vector bundle W over X such that V ⊕ W is trivializable, i.e. isomorphic to a trivial bundle. (b) Find a vector bundle W on S 2 , the 2sphere, such that T ∗ S 2 ⊕ W is trivializable. 2. (RA) Let (X, d) be a metric space. For any subset A ⊂ X, and any > 0 we set [ B (A) = B (p). p∈A
(This is the “fattening” of A.) For Y, Z bounded subsets of X define the Hausdorff distance between Y and Z by dH (Y, Z) := inf { > 0  Y ⊂ B (Z),
Z ⊂ B (Y )} . ˜ := {A ⊂ X A is closed and bounded}. Show that dH defines a metric on the set X 3. (AT) Let T n = Rn /Zn , the ntorus. Prove that any pathconnected covering space Y → T n is homeomorphic to T m × Rn−m , for some m. 4. (CA) Let f : C → C be a nonconstant holomorphic function. Show that the image of f is dense in C. 5. (A) Let F ⊃ Q be a splitting field for the polynomial f = xn − 1. (a) Let A ⊂ F × = {z ∈ F  z 6= 0} be a finite (multiplicative) subgroup. Prove that A is cyclic. (b) Prove that G = Gal(F/Q) is abelian. 6. (AG) Let C and D ⊂ P2 be two plane cubics (that is, curves of degree 3), intersecting transversely in 9 points {p1 , p2 , . . . , p9 }. Show that p1 , . . . , p6 lie on a conic (that is, a curve of degree 2) if and only if p7 , p8 and p9 are colinear.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday August 31, 2016 (Day 2)
1. (A) Let R be a commutative ring with unit. If I ⊆ R is a proper ideal, we define the radical of I to be √ I = {a ∈ R  am ∈ I for some m > 0}. Prove that
√ I=
\
p.
p⊇I p prime
2. (DG) Let c(s) = (r(s), z(s)) be a curve in the (x, z)plane which is parameterized by arc length s. We construct the corresponding rotational surface, S, with parametrization ϕ : (s, θ) 7→ (r(s) cos θ, r(s) sin θ, z(s)). Find an example of a curve c such that S has constant negative curvature −1. 3. (RA) Let f ∈ L2 (0, ∞) and consider Z ∞ F (z) = f (t)e2πizt dt 0
for z in the upper halfplane. (a) Check that the above integral converges absolutely and uniformly in any region Im(z) ≥ C > 0. (b) Show that Z sup y>0
0
∞
F (x + iy)2 dx = kf k2L2 (0,∞) .
√ 2 4. (CA) Given that 0 e−x dx =R 12 π, use contourR integration to prove that ∞ ∞ each of the improper integrals 0 sin(x2 ) dx and 0 cos(x2 ) dx converges to p π/8. R∞
5. (AT) (a) Let X = RP 3 × S 2 and Y = RP 2 × S 3 . Show that X and Y have the same homotopy groups but are not homotopy equivalent. (b) Let A = S 2 ×S 4 and B = CP 3 . Show that A and B have the same singular homology groups with Zcoefficients but are not homotopy equivalent.
6. (AG) Let C be the smooth projective curve over C with affine equation y 2 = f (x), where f ∈ C[x] is a squarefree monic polynomial of degree d = 2n. (a) Prove that the genus of C is n − 1. (b) Write down an explicit basis for the space of global differentials on C.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 1, 2016 (Day 3)
1. (AT) Model S 2n−1 as the unit sphere in Cn , and consider the inclusions ··· ···
→ S 2n−1 → S 2n+1 → · · · ↓ ↓ → Cn → Cn+1 → · · · .
Let S ∞ and C∞ denote the union of these spaces, using these inclusions. (a) Show that S ∞ is a contractible space. (b) The group S 1 appears as the unit norm elements of C× , which acts compatibly on the spaces Cn and S 2n−1 in the systems above. Calculate all the homotopy groups of the homogeneous space S ∞ /S 1 . 2. (AG) Let X ⊂ Pn be a general hypersurface of degree d. Show that if k+d > (k + 1)(n − k) k then X does not contain any kplane Λ ⊂ Pn . 3. (DG) Let H2 := {(x, y) ∈ R2 : y > 0}. Equip H2 with a metric gα :=
dx2 + dy 2 yα
where α ∈ R. (a) Show that (H2 , gα ) is incomplete if α 6= 2.
a b (b) Identify z = x + iy. For ∈ SL(2, R), consider the map c d 2 z 7→ az+b cz+d . Show that this defines an isometry of (H , g2 ). (c) Show that (H2 , g2 ) is complete. (Hint: Show that the isometry group acts transitively on the tangent space at each point.)
4. (RA) (a) Let H be a Hilbert space, K ⊂ H a closed subspace, and x a point in H. Show that there exists a unique y in K that minimizes the distance kx − yk to x. (b) Give an example to show that the conclusion can fail if H is an inner product space which is not complete. 5. (A) (a) Prove that there exists a unique (up to isomorphism) nonabelian group of order 21. (b) Let G be this group. How many conjugacy classes does G have? (c) What are the dimensions of the irreducible representations of G? 6. (CA) Find (with proof) all entire holomorphic functions f : C → C satisfying the conditions: 1. f (z + 1) = f (z) for all z ∈ C; and 2. There exists M such that f (z) ≤ M exp(10z) for all z ∈ C.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday August 30, 2016 (Day 1)
1. (DG) (a) Show that if V is a C ∞ vector bundle over a compact manifold X, then there exists a vector bundle W over X such that V ⊕ W is trivializable, i.e. isomorphic to a trivial bundle. (b) Find a vector bundle W on S 2 , the 2sphere, such that T ∗ S 2 ⊕ W is trivializable.
Solution: Since V is locally trivializable and M is compact, one can find a finite open cover Ui , i = 1, . . . , n, of M and trivializations Ti : V Ui → Rk . Thus, each Ti is a smooth map which restricts to a linear isomorphism on each fiber of V Ui . Next, choose a smooth partition of unity {fi }i=1,...,n subordinate to the cover {Ui }i=1,...,n . If p : V → M is the projection to the base, then there are maps V  Ui → R k , v 7→ fi (p(v))Ti (v) which extend (by zero) to all of V and which we denote by fi Ti . Together, the fi Ti give a map T : V → Rnk which has maximal rank k everywhere, because at each point of X at least one of the fi is nonzero. Thus V is isomorphic to a subbundle, T (V ), of the trivial bundle, Rnk . Using the standard inner product on Rnk we get an orthogonal bundle W = T (V )⊥ which has the desired property. For the second part, embed S 2 into R3 in the usual way, then T S 2 ⊕ NS 2 = T R3 S 2 where NS 2 is the normal bundle to S 2 in R3 . Dualizing we get T ∗ S 2 ⊕ (NS 2 )∗ = T ∗ R3 S 2 which solves the problem with W = (NS 2 )∗ . 2. (RA) Let (X, d) be a metric space. For any subset A ⊂ X, and any > 0 we set [ B (A) = B (p). p∈A
(This is the “fattening” of A.) For Y, Z bounded subsets of X define the Hausdorff distance between Y and Z by dH (Y, Z) := inf { > 0  Y ⊂ B (Z),
Z ⊂ B (Y )} . ˜ := {A ⊂ X A is closed and bounded}. Show that dH defines a metric on the set X ˜ dH ) is a metric space. First, since comSolution: We need to show that (X, pact sets are bounded, dH (Y, Z) is well defined for any compact sets Y, Z. Secondly, dH (Y, Z) = dH (Z, Y ) ≥ 0 is obvious from the definition. We need to prove that the distance is positive when Y 6= Z, and that dH satisfies the triangle inequality. First, let us show that dH (Y, Z) > 0 if Y 6= Z. Without loss of generality, we can assume there is a point p ∈ Y ∩ Z c . Since Z is compact, it is closed, so there exists r > 0 such that Br (p) ⊂ Z c . In particular, p is not in Br (Z). Thus Y is not contained in Br (Z) and so dH (Y, Z) ≥ r > 0. It remains to prove the triangle inequality. To this end, suppose that Y, Z, W are compact subsets of X. Fix 1 > dH (Y, Z), 2 > dH (Z, W ), then Y ⊂ B1 (Z),
Z ⊂ B1 (Y ),
Z ⊂ B2 (W ),
W ⊂ B2 (Z)
Then dH (Y, Z) < 1 , dH (Z, W ) < 2 . Let us prove that Y ⊂ B1 +2 (W ), the other containment being identical. Fix a point y ∈ Y . By our choice of 1 there exists a point z ∈ Z such that y ∈ N1 (z). By our choice of 2 there exists a point w ∈ W such that z ∈ B2 (w). Then d(y, w) ≤ d(y, z) + d(z, w) ≤ 1 + 2 so y ∈ B1 +2 (w). This proves the containment. The other containment is identical, by just swapping Y, W . Thus dH (Y, W ) ≤ 1 + 2 But this holds for all 1 , 2 as above. Taking the infimum we obtain the result. 3. (AT) Let T n = Rn /Zn , the ntorus. Prove that any pathconnected covering space Y → T n is homeomorphic to T m × Rn−m , for some m. Solution: The universal covering space of T n is Rn , so that any path connected covering space of X is of the form Rn /G, for some subgroup G ⊆ π1 (T n ). We have π1 (T n ) = π1 (S 1 ) × · · · × π1 (S 1 ) = Zn , and Zn is acting on Rn by translation. Thus, G ⊆ Zn is free. Choose a Zbasis (v1 , . . . , vm ) of G, and consider the (real!) change of basis taking (v1 , . . . , vm ) to the first m standard basis vectors (e1 , . . . , em ). Hence, G is acting on Rn by translation in the first m coordinates. Thus, Rn /G ' Rm /Zm × Rn−m ' T m × Rn−m .
4. (CA) Let f : C → C be a nonconstant holomorphic function. Show that the image of f is dense in C. Solution: Suppose that for some w0 ∈ C and some > 0, the image of f lies outside the ball B (w0 ) = {w ∈ C  w − w0  < }. Then the function g(z) =
1 f (z) − w0
is bounded and homomorphic in the entire plane, hence constant. 5. (A) Let F ⊃ Q be a splitting field for the polynomial f = xn − 1. (a) Let A ⊂ F × = {z ∈ F  z 6= 0} be a finite (multiplicative) subgroup. Prove that A is cyclic. (b) Prove that G = Gal(F/Q) is abelian.
Solution: For the first part, let m = A. Suppose that A is not cyclic, so that the order of any element in A is less than m. A is a finite abelian group so it is isomorphic to a product of cyclic groups A ' Zn1 × · · · × Znk , where ni ni+1 . In particular, the order of any element in A divides nk . Hence, for any z ∈ A, z nk = 1. However, the polynomial xnk − 1 ∈ F [x] admits at most nk < m roots in F , which is a contradiction. So, there must be some element in A with order m. For the second part, since f 0 = nxn−1 and f are relatively prime, f admits n distinct roots 1 = z0 , . . . , zn−1 . As F is a splitting field of f we can assume that F = Q(z0 , . . . , zn−1 ) ⊆ C. U = {z0 , . . . , zn−1 } ⊂ F × is a subgroup of the multiplicative group of units in F and is cyclic; moreover, Aut(U ) is isomorphic to the (multiplicative) group of units (Z/nZ)∗ . Restriction defines a homomorphism G → Aut(U ) , α 7→ αU ; this homomorphism is injective because F = Q(z0 , . . . , zn−1 ). In particular, G is isomorphic to a subgroup of the abelian group (Z/nZ)∗ . 6. (AG) Let C and D ⊂ P2 be two plane cubics (that is, curves of degree 3), intersecting transversely in 9 points {p1 , p2 , . . . , p9 }. Show that p1 , . . . , p6 lie on a conic (that is, a curve of degree 2) if and only if p7 , p8 and p9 are colinear. Solution: First, observe that we can replace C = V (F ) and D = V (G) by any two independent linear combinations C 0 = V (a0 F + a1 G) and D0 = V (b0 F + b1 G). Now suppose that p1 , . . . , p6 lie on a conic Q ⊂ P2 . Picking a seventh point q ∈ Q, we see that some linear combination C0 of C and D contains q and hence contains Q; thus C0 = Q ∪ L for some line L ⊂ P2 . Replacing C or D with C0 , we see that p7 , p8 and p9 ∈ L.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday August 31, 2016 (Day 2)
1. (A) Let R be a commutative ring with unit. If I ⊆ R is a proper ideal, we define the radical of I to be √ I = {a ∈ R  am ∈ I for some m > 0}. Prove that
√
\
I=
p.
p⊇I p prime
√ Solution: First, we prove for the case I = 0. Let f ∈ 0 so that f n = 0, and f n ∈ p, for any prime ideal p ⊆ R. Let p be a prime ideal in R. The quotient ring R/p is an integral domain and, in particular, contains no nonzero nilpotent elements. Hence, f n + p = 0 ∈ R/p so that f ∈ p. √ Now, suppose that f ∈ / 0. The set S = {1, f, f 2 , . . .} does not contain 0 so that the localisation Rf is not the zero ring. Let m ⊂ Rf be a maximal ideal. Denote the canonical homomorphism j : R → Rf . As j(f ) ∈ Rf is a unit, j(f )T ∈ / m. Then j −1 (m) ⊂ R is a prime ideal that does not contain f . Hence, f∈ / p⊆R prime p. If I ⊆ R is a proper ideal, we consider the quotient ring π : R → S = R/I. Recall the bijective correspondence {prime ideals in S} ↔ {prime ideals in R containing I} , p ↔ π −1 (p) Then, √
√ I = π −1 ( 0S ) = π −1
\ p⊆S prime
p =
\
π −1 (p) =
p⊆S prime
\
q.
q⊇I q prime
2. (DG) Let c(s) = (r(s), z(s)) be a curve in the (x, z)plane which is parameterized by arc length s. We construct the corresponding rotational surface, S, with parametrization ϕ : (s, θ) 7→ (r(s) cos θ, r(s) sin θ, z(s)). Find an example of a curve c such that S has constant negative curvature −1.
Solution: ∂ϕ (s, θ) = (r0 (s) cos θ, r0 (s) sin θ, z 0 (s)) ∂s ∂ϕ (s, θ) = (−r(s) sin θ, r(s) cos θ, 0) ∂θ The coefficients of the first fundamental form are: E = r0 (s)2 + z 0 (s)2 = 1, Curvature:
G = r(s)2
F = 0,
1 ∂2 √ r00 (s) K = −√ G = − r(s) G ∂s2
To get K = −1 we need to find r(s), z(s) such that r00 (s) = r(s), r0 (s)2 + z 0 (s)2 = 1. A possible solution is r(s) = e−s with Z p p z(s) = 1 − e−2t dt = Arcosh(r−1 ) − 1 − r2 . 3. (RA) Let f ∈ L2 (0, ∞) and consider Z ∞ F (z) = f (t)e2πizt dt 0
for z in the upper halfplane. (a) Check that the above integral converges absolutely and uniformly in any region Im(z) ≥ C > 0. (b) Show that Z sup y>0
0
∞
F (x + iy)2 dx = kf k2L2 (0,∞) .
Solution: For Im(z) ≥ C > 0 we have f (t)e2πizt  ≤ f (t)e−2Cπt thus with the Cauchy–Schwarz inequality Z
∞ 2πizt
f (t)e 0
Z dt ≤ 0
∞
1/2 Z f (t) dt 2
0
∞
−4Cπt
e
1/2 dt
which proves the claim. For the second part, Plancherel’s theorem gives Z ∞ Z ∞ 2 f (t)2 e−4πyt dt ≤ kf k2L2 (0,∞) F (x + iy) dx = 0
0
and
Z
∞
f (t) e
sup y>0
2 −4πyt
Z
∞
dt =
f (t)2 dt
0
0
by the monotone convergence theorem. R∞ √ 2 4. (CA) Given that 0 e−x dx =R 12 π, use contourR integration to prove that ∞ ∞ each of the improper integrals 0 sin(x2 ) dx and 0 cos(x2 ) dx converges to p π/8. 2
Solution: We integrate e−z dz along a triangular contour with vertices at 2 0, M , and (1 + i)M , and let M → ∞. Since e−z is holomorphic on C, RM 2 the integral vanishes. The integral from 0 to M is 0 e−x dx, which apR ∞ −x2 √ dx = 21 π. The vertical integral approaches zero, because proaches 0 e it is bounded in absolute value by Z
M
e
−(M +yi)2
0
Z
M
 dy =
y 2 −M 2
e
Z 0
M
e−M t dt
1.
To show that X and Y are not homotopy equivalent, we show that they have nonisomorphic homology groups. We make use of the following wellknown singular homology groups (with integral coefficients) H0 (S n ) = Hn (S n ) = Z,
Hi (S k ) = 0, i 6= 0, n,
H0 (RP 2 ) = H2 (RP 2 ) = Z, H1 (RP 2 ) = Z/2Z, Hi (RP 2 ) = 0, i 6= 0, 1, 2 H0 (RP 3 ) = Z, H1 (RP 3 ) = Z/2Z, Hi (RP 3 ) = 0, i 6= 0, 1 Now, the Kunneth theorem in singular homology (with Zcoefficients) gives an exact sequence
0→
M
Hi (RP 3 )⊗Z Hj (S 2 ) → H2 (X) →
i+j=2
M
T or1 (Hi (RP 3 ), Hj (S 2 )) → 0
i+j=1
Since Hk (S 2 ) is free, for every k, we have M H2 (X) ' Hi (RP 3 ) ⊗Z Hj (S 2 ) = Z i+j=2
Similarly, we compute H2 (Y ) '
M
Hi (RP 2 ) ⊗Z Hj (S 3 ) = Z/2Z.
i+j=2
In particular, X and Y are not homotopy equivalent. For the second part, B can be constructed as a cell complex with a single cell in dimensions 0, 2, 4, 6. Therefore, the homology of B is H2i (B) = Z, for i = 0, . . . , 3, and Hk (B) = 0 otherwise. The Kunneth theorem for singular cohomology (with Zcoefficients), combined with the fact that Hk (S n ) is free, for any k, gives M Hk (A) ' Hi (S 2 ) ⊗ Hj (S 4 ). i+j=k
Hence, H2i (A) = Z, for i = 0, . . . , 3, and Hk (A) = 0 otherwise. In order to show that A and B are not homotopy equivalent we will show that they have nonisomorphic homotopy groups. Consider the canonical quotient map C4 − {0} → CP 3 . This restricts to give a fiber bundle S 1 → S 7 → CP 3 . The associated long exact sequence in homotopy · · · → πk+1 (CP 3 ) → πk (S 1 ) → πk (S 7 ) → πk (CP 3 ) → · · · together with the fact that π3 (S1 ) = π4 (S 7 ), shows that π4 (CP 3 ) = 0. However, π4 (A) = π4 (S 4 ) = Z. 6. (AG) Let C be the smooth projective curve over C with affine equation y 2 = f (x), where f ∈ C[x] is a squarefree monic polynomial of degree d = 2n. (a) Prove that the genus of C is n − 1. (b) Write down an explicit basis for the space of global differentials on C. Solution: For the first part, use RiemannHurwitz: the 2 : 1 map from C to the xline is ramified above the roots of f and nowhere else (not even at infinity because deg f is even), so 2 − 2g(C) = χ(C) = 2χ(P1 ) − deg P = 4 − 2n, whence g(C) = n − 1. For the second, let ω0 = dx/y. This differential is holomorphic, with zeros of order g − 1 at the two points at infinity. (Proof by local computation around those points and the roots of P , which are the only places where holomorphy is not immediate; dx has a pole of order −2 at infinity but 1/y has zeros of order n at the points above x = ∞, while 2y dy = P 0 (x) dx takes care of the Weierstrass points.) Hence the space of holomorphic differentials contains Ω := {P (x) ω0  deg P < g}, which has dimension g. Thus Ω is the full space of differentials, with basis {ωk = xk ω0 , k = 0, . . . , g − 1}.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 1, 2016 (Day 3) 1. (AT) Model S 2n−1 as the unit sphere in Cn , and consider the inclusions ··· ···
→ S 2n−1 → S 2n+1 → · · · ↓ ↓ n n+1 → C → C → ··· .
Let S ∞ and C∞ denote the union of these spaces, using these inclusions. (a) Show that S ∞ is a contractible space. (b) The group S 1 appears as the unit norm elements of C× , which acts compatibly on the spaces Cn and S 2n−1 in the systems above. Calculate all the homotopy groups of the homogeneous space S ∞ /S 1 . Solution: The shift operator gives a normpreserving injective map T : C∞ → C∞ that sends S ∞ into the hemisphere where the first coordinate is zero. The line joining x ∈ S ∞ to T (x) cannot pass through zero, since x and T (x) cannot be scalar multiples, and hence the linear homotopy joining x to T (x) shows that T is homotopic to the identity. However, since T (S ∞ ) forms an equatorial hemisphere, there is a also a linear homotopy from T to the constant map at either of the poles. For the second part, because S 1 acts properly discontinuously on S ∞ , the quotient sequence S 1 → S ∞ → S ∞ /S 1 forms a fiber bundle. The homotopy groups of S 1 are known: π1 S 1 ∼ = Z and
π6=1 S 1 = 0 otherwise. Since S ∞ is contractible, the long exact sequence of higher homotopy groups shows that π2 (S ∞ /S 1 ) = Z and π6=2 (S ∞ /S 1 ) = 0 otherwise. 2. (AG) Let X ⊂ Pn be a general hypersurface of degree d. Show that if k+d > (k + 1)(n − k) k then X does not contain any kplane Λ ⊂ Pn . Solution: For the first, let PN be the space of all hypersurfaces of degree d in Pn , and let Γ = {(X, Λ) ∈ PN × G(k, n)  Λ ⊂ X}.
The fiber of Γ over the point [Λ] ∈ G(k, n) is just the subspace of PN corresponding to the vector space of polynomials vanishing on Λ; since the space of polynomials on Pn surjects onto space of polynomials on Λ ∼ = Pk , this the N k+d is a subspace of codimension k in P . We deduce that k+d dim Γ = (k + 1)(n − k) + N − ; k in particular, if the inequality of the problem holds, then dim Γ < N , so that Γ cannot dominate PN . 3. (DG) Let H2 := {(x, y) ∈ R2 : y > 0}. Equip H2 with a metric gα :=
dx2 + dy 2 yα
where α ∈ R. (a) Show that (H2 , gα ) is incomplete if α 6= 2. a b ∈ SL(2, R), consider the map c d 2 z→ 7 az+b cz+d . Show that this defines an isometry of (H , g2 ).
(b) Identify z = x + iy.
For
(c) Show that (H2 , g2 ) is complete. (Hint: Show that the isometry group acts transitively on the tangent space at each point.)
Solution: For the first part, consider the geodesic γ(t) with γ(0) = (0, 1), and ∂ γ 0 (0) = ∂y . In order for (H2 , gα ) to be complete, this geodesics must exist for all t ∈ (−∞, ∞). By symmetry, this geodesic must be given by x(t) = (0, y(t)). Furthermore, x(t) must have constant speed, which we may as well take to be ˙ 2 1. Thus (yy) α = 1, or in other words, y˙ = y α/2 . If α 6= 2, then the solution to this ODE is 1/(1− α ) α 2 y(t) = (1 − )t + 1 2 thus, this geodesics persists only as long as (1 − α2 )t + 1 ≥ 0. This set is always bounded from one side. Note that when α = 2, we get x(t) = (0, et ), which
exists for all time.
(b) To begin, note that dz ⊗ d¯ z = dx ⊗ dx + dy ⊗ dy, so we can write the metric as 4dz ⊗ d¯ z g2 = z − z¯2 Let A ∈ SL(2, R), we compute A∗ dz = and so A∗ d¯ z=
adz (az + b)dz dz dz −c = (ad − bc) = cz + d (cz + d)2 (cz + d)2 (cz + d)2 d¯ z . (c¯ z +d)2
It remains to compute
A∗ z − A∗ z¯ =
az + b a¯ z+b z − z¯ − = , cz + d c¯ z+d cz + d2
where we have used that A ∈ SL(2, R). Putting everything together we get A∗ g2 =
4dz ⊗ d¯ z cz + d4 · = g2 , cz + d4 z − z¯2
and so SL(2, R) acts by isometry. (c) By the computation from part (a), we know that the geodesic– let’s call it γ0 (t)– through the point (0, 1) in the direction (0, 1) exists for all time. Let z = x + iy be any point in H2 . By an isometry, we can map this point to z = iy. Without loss of generality, let us assume y = 1. It suffices to show that we can find A ∈ SL(2, R) so that A(i) = i, and A∗ V = (0, 1), where V is any unit vector in the tangent space Ti H2 , for then the geodesic through i with tangent vector V will be nothing but A−1 (γ0 (t)), and hence will exist a b . for all time. First, observe that A(i) = i, if and only if A = −b a Consider the rotation matrix cos θ − sin θ A= sin θ cos θ A straightforward computation shows that, in complex coordinates, A∗ V =
√ 1 −2 −1θ V = e V, (cos θ + i sin θ)2
that is, A∗ : Ti H2 → Ti H2 acts as a rotation. Since θ is arbitrary, and the rotations act transitively on S 2 , we’re done. 4. (RA)
(a) Let H be a Hilbert space, K ⊂ H a closed subspace, and x a point in H. Show that there exists a unique y in K that minimizes the distance kx − yk to x. (b) Give an example to show that the conclusion can fail if H is an inner product space which is not complete. Solution: (a): If y, y 0 ∈ K both minimize distance to x, then by the parallelogram law: kx − But
y+y 0 2
y + y0 2 y − y0 2 1 k +k k = (kx − yk2 + kx − y 0 k2 ) = kx − yk2 2 2 2
cannot be closer to x than y, by assumption, so y = y 0 .
Let C = inf y∈K kx − yk, then 0 ≤ C < ∞ because K is nonempty. We can find a sequence yn ∈ K such that kx − yn k → C, which we want to show is m m Cauchy. The midpoints yn +y are in K by convexity, so kx − yn +y k≥C 2 2 and using the parallelogram law as above one sees that kyn − ym k → 0 as n, m → ∞. By completeness of H the sequence yn converges to a limit y, which is in K, since K is closed. Finally, continuity of the norm implies that kx − yk = C. (b): For example choose H = C([0, 1]) ⊂ L2 ([0, 1]), K the subspace of functions with support contained in [0, 12 ], and and x = 1 the constant function. If fn is a sequence in K converging to f ∈ H in L2 norm, then Z 1 f 2 = 0 1/2
thus f vanishes on [1/2, 1], showing √ that K is closed. The distance kx − yk can be made arbitrarily close to 1/ 2 for y ∈ K by approximating χ[0,1/2] by continuous functions, but the infimum is not attained. 5. (A) (a) Prove that there exists a unique (up to isomorphism) nonabelian group of order 21. (b) Let G be this group. How many conjugacy classes does G have? (c) What are the dimensions of the irreducible representations of G?
Solution: Let G be a group of order 21, and select elements g3 and g7 of orders 3 and 7 respectively. The subgroup generated by g7 is normal — if it weren’t, then g7 and xg7 x−1 witnessing nonnormality would generate a group of order
49. In particular, we have g3 g7 g3−1 = g7j for some nonzero j ∈ Z/7. Now we use the order of g3 : g7 = g3 g3 g3 · g7 · g3−1 g3−1 g3−1 = g3 g3 (g7j )g3−1 g3−1 2
= g3 (g7j )g3−1 3
= g7j , and hence j 3 ≡ 1 (mod 7). This is nontrivially solved by j = 2 and j = 4, and these two cases coincide: if for instance g3 g7 g3−1 = g72 , then by replacing the generator g3 with g32 we instead see g32 g7 (g32 )−1 = g3 g72 g3−1 = g74 . We have the following conjugacy classes of elements: • {e} forms a class of its own. • {g7 , g74 , g72 } and {g73 , g75 , g76 } form classes by our choice of j. • Any element of order 3 generates a Sylow 3–subgroup, all of which are conjugate as subgroups. However, there cannot be an x with xg3 x−1 = g32 , since G has only elements of odd order. Hence, there are two final conjugacy classes, each of size 7: those elements conjugate to g3 and those conjugate to g32 . These five conjugacy sets give rise to five irreducible representations, which must be of dimensions 1, 1, 1, 3, and 3 (since these squaresum to G = 21). 6. (CA) Find (with proof) all entire holomorphic functions f : C → C satisfying the conditions: 1. f (z + 1) = f (z) for all z ∈ C; and 2. There exists M such that f (z) ≤ M exp(10z) for all z ∈ C.
Solution: The functions satisfying these conditions are precisely the Clinear combinations of e−2πiz , 1, and e2πiz . Indeed such f is readily seen to satisfy the two conditions. Conversely (1) means that f descends to a function of q := e2πiz ∈ C∗ , say f (z) = F (q), and then by (2) there is some M 0 such that F (q) ≤ M 0 max(q−5/π , q5/π ) for all q, whence qF and q −1 F have removable singularities at q = 0 and q = ∞ respectively, etc.
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Tuesday, January 19, 2016 (Day 1)
PROBLEM 1 (DG) Let S denote the surface in R3 where the coordinates (x, y, z) obey x2 + y2 = 1 + z2. This surface can be parametrized by coordinates t ∈ R and θ ∈ R/(2πZ) by the map (t, θ) → ψ(t, θ) = ( 1+ t 2 cos θ, 1+ t 2 sin θ, t). a) Compute the induced inner product on the tangent space to S using these coordinates. b) Compute the Gaussian curvature of the metric that you computed in Part a). c) Compute the parallel transport around the circle in S where z = 0 for the LeviCivita connection of the metric that you computed in Part a).
PROBLEM 2 (T) Let X be pathconnected and locally pathconnected, and let Y be a finite Cartesian product of circles. Show that if π1(X) is finite, then every continuous map from X to Y is nullhomotopic. (Hint: recall that there is a fiber bundle Z → R → S1.)
PROBLEM 3 (AN) Let K be the field C(z) of rational functions in an indeterminate z, and let F ⊂ K be the subfield C(u) where u = (z6 + 1)/z3. a) Show that the field extension K/F is normal, and determine its Galois group. b) Find all fields E, other than F and K themselves, such that F ⊂ E ⊂ K. For each E, determine whether the extensions E/F and K/E are normal.
PROBLEM 4 (AG) The nodal cubic is the curve in CP2 (denoted by X) given in homogeneous coordinates (x, y, z) by the locus {z y2 =x2 (x + z)}. a) Give a definition of a rational map between algebraic varieties. b) Show that there is a birational map from X to CP1. c) Explain how to resolve the singularity of X by blowing up a point in CP2.
PROBLEM 5 (RA) Let B and L denote Banach spaces, and let  · B and  · L denote their norms. a) Let L: B → L denote a continuous, invertible linear map and let m: B ⊗ B → L denote a linear map such that  m(φ ⊗ ψ) L ≤  φ B ψ B for all φ, ψ ∈ B. Prove the following assertions: • There exists a number κ > 1 depending only on L such that if a ∈ B has norm less than κ 2, then there is a unique solution to the equation Lφ + m(φ ⊗ φ) = a with  φ B < κ 1. •
The norm of the solution from the previous bullet is at most κ  a L.
b) Recall that a Banach space is defined to be a complete, normed vector space. Is the assertion of Part a) of the first bullet always true if B is normed but not complete? If not, explain where the assumption that B is complete enters your proof of Part a).
PROBLEM 6 (CA) Fix a ∈ C and an integer n ≥ 2. Show that the equation a zn + z + 1 = 0 for a complex number z necessarily has a solution with z ≤ 2.
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Wednesday, January 20, 2016 (Day 2)
PROBLEM 1 (DG) Let k denote a positive integer. A nonoptimal version of the Whitney embedding theorem states that any kdimensional manifold can be embedded into R2k+1. Using this, show that any kdimensional manifold can be immersed in R2k. (Hint: Compose the embedding with a projection onto an appropriate subspace.)
PROBLEM 2 (T) Let X be a CWcomplex with a single cell in each of the dimensions 0, 1, 2, 3, and 5 and no other cells. a) What are the possible values of H∗(X; Z)? (Note: it is not sufficient to consider Hn(X; Z) for each n independently. The value of H1(X; Z) may constrain the value of H2(X; Z), for instance.) b) Now suppose in addition that X is its own universal cover. What extra information does this provide about H∗(X; Z)?
PROBLEM 3 (AN) Let k be a finite field of characteristic p, and n a positive integer. Let G be the group of invertible linear transformations of the kvector space kn. Identify G with the group of invertible n × n matrices with entries in k (acting from the left on column vectors). n1
a) Prove that the order of G is
! (qn  qm ) where q is the number of elements of k. m=0
b) Let U be the subgroup of G consisting of uppertriangular matrices with all diagonal entries equal 1. Prove that U is a pSylow subgroup of G. c) Suppose H ⊂ G is a subgroup whose order is a power of p. Prove that there is a basis (v1, v2,..., vn) of kn such that for every h ∈H and every m ∈{1, 2, 3,..., n}, the vector h(vm)  vm is in the span of {vd: d < m}.
PROBLEM 4 (AG) Let X be a complete intersection of surfaces of degrees a and b in CP3. Compute the Hilbert polynomial of X.
PROBLEM 5 (RA) Let C 0 denote the vector space of continuous functions on the interval [0, 1]. Define a norm on C 0 as follows: If ƒ ∈ C 0, then its norm (denoted by  ƒ ) is  ƒ  = supt∈[0,1] ƒ(t) . Let C ∞ denote the space of smooth functions on [0, 1]. View C ∞ as a normed, linear space with the norm defined as follows: If ƒ ∈ C ∞, then its norm (denoted by  ƒ ∗) is  ƒ ∗ =
!
d ƒ +  ƒ) dt . ( dt
[0,1]
a) Prove that C 0 is Banach space with respect to the norm  · . In particular, prove that it is complete. b) Let ψ denote the ‘forgetful’ map from C ∞ to C 0 that sends ƒ to ƒ. Prove that ψ is a bounded map from C ∞ to C 0, but not a compact map from C ∞ to C 0.
PROBLEM 6 (CA) Let D denote the closed disk in C where z ≤ 1. Fix R > 0 and let ϕ: D → C denote a continuous map with the following properties: i) ϕ is holomorphic on the interior of D. ii) ϕ(0) = 0 and its zderivative, ϕ´, obeys ϕ´(0) = 1. iii) ϕ ≤ R for all z ∈D. Since ϕ´(0) = 1, there exists δ > 0 such that ϕ maps the z < δ disk diffeomorphically onto its image. Prove the following: a) There is a unique solution in [0, 1] to the equation 2Rδ = (1  δ)3. b) Let δ∗ denote the unique solution to this equation. If 0 < δ < δ∗, then ϕ maps the z < δ disk diffeomorphically onto its image.
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Thursday, January 21, 2016 (Day 3)
PROBLEM 1 (DG) Recall that a symplectic manifold is a pair (M, ω), where M is a smooth manifold and ω is a closed nondegenerate differential 2form on M. (The 2form ω is called the symplectic form.) a) Show that if H: M → R is a smooth function, then there exists a unique vector field, to be denoted by XH, satisfying ! XH " = dH. (Here, ι denotes the contraction operation.) b) Supposing that t > 0 is given, suppose in what follows that the flow of XH is defined for time t, and let φt denote the resulting diffeomorphism of M. Show that φt*ω = ω. c) Denote the Euclidean coordinates on R4 by (x1, y1, x2, y2) and use these to define the symplectic form ω0 = dx1 ∧ dy1 + dx2 ∧ dy2. Find a function H: R4 → R such that the diffeomorphism φt=1 that is defined by the time t = 1 flow of XH fixes the half space where x1 ≤ 0 and moves each point in the half space where x1 ≥ 1 by 1 in the y2 direction.
PROBLEM 2 (T) Let X denote a finite CW complex and let ƒ: X → X be a selfmap of X. Recall that the Lefschetz trace of ƒ, denoted by τ(ƒ), is defined by the rule "
τ(ƒ) =
# (1)
n
tr(ƒ n : H n (X; Q) ! H n (X; Q))
n=0
with ƒn denoting the induced homomorphism. Use τ(·) to answer the following: a) Does there exist a continuous map from RP2 to itself with no fixed points? If so, give an example; and if not, give a proof. b) Does there exist a continuous map from RP3 to itself with no fixed points? If so, give an example; and if not, give a proof.
PROBLEM 3 (AN) Let A be the ring Z[ 5 2016 ] = Z[X]/(X5  2016). Given that 2017 is prime in Z, determine the factorization of 2017·A into prime ideals of A.
PROBLEM 4 (AG) a) State a version of the Riemann–Roch theorem. b) Apply this theorem to show that if X is a complete nonsingular curve and P ∈ X is any point, there is a rational function on X which has a pole at P and is regular on X−{P}.
PROBLEM 5 (RA) Let ℘ denote a probability measure for a real valued random variable with mean 0. Denote this random variable by x. Suppose that the random variable x has mean equal to 2. a) Given R > 2, state a nontrivial upper bound for event that x ≥ R. (The trivial upper bound is 1.) b) Give a nonzero lower bound for the standard deviation of x. c) A function ƒ on R is Lipshitz when there exists a number c ≥ 0 such that ƒ(p)  ƒ(p´) ≤ c p  p´ for any pair p, p´ ∈ R. ˆ denote the function on R whose value at a given p ∈ R is the expectation of the Let ! random variable e ip x . (This is the characteristic function of ℘.) Give a rigorous ˆ is Lipshitz and give an upper bound for c in this case. proof that ! d) Suppose that the standard deviation of x is equal to 4. Let N denote an integer greater than 1, and let {x1, …, xN} denote a set of independent random variables each with probabilities given by ℘. Use SN to denote the random variable
1 N
(x1 + ···· + xN).
The central limit theorem gives an integral that approximates the probability of the event where SN ∈ [1, 1] when N is large. Write this integral.
PROBLEM 6 (CA) Let H ⊂ C denote the open right half plane, thus H = {z = x + iy: x > 0}. Suppose that ƒ: H → C is a bounded, analytic function such that ƒ(1/n) = 0 for each positive integer n. Prove that ƒ(z) = 0 for all z. N
(Hint: Consider the behavior of the sequence of functions {hN(z) =
! zz + 11/n/n }
N=1.2…
n =1
and, in particular, on the positive real axis.}
on H
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Tuesday, January 19, 2016 (Day 1)
PROBLEM 1 (DG) Let S denote the surface in R3 where the coordinates (x, y, z) obey x2 + y2 = 1 + z2. This surface can be parametrized by coordinates t ∈ R and θ ∈ R/(2πZ) by the map (t, θ) → ψ(t, θ) = ( 1+ t 2 cos θ, 1+ t 2 sin θ, t). a) Compute the induced inner product on the tangent space to S using these coordinates. b) Compute the Gaussian curvature of the metric that you computed in Part a). c) Compute the parallel transport around the circle in S where z = 0 for the LeviCivita connection of the metric that you computed in Part a).
PROBLEM 2 (T) Let X be pathconnected and locally pathconnected, and let Y be a finite Cartesian product of circles. Show that if π1(X) is finite, then every continuous map from X to Y is nullhomotopic. (Hint: recall that there is a fiber bundle Z → R → S1.)
PROBLEM 3 (AN) Let K be the field C(z) of rational functions in an indeterminate z, and let F ⊂ K be the subfield C(u) where u = (z6 + 1)/z3. a) Show that the field extension K/F is normal, and determine its Galois group. b) Find all fields E, other than F and K themselves, such that F ⊂ E ⊂ K. For each E, determine whether the extensions E/F and K/E are normal.
PROBLEM 4 (AG) The nodal cubic is the curve in CP2 (denoted by X) given in homogeneous coordinates (x, y, z) by the locus {z y2 =x2 (x + z)}. a) Give a definition of a rational map between algebraic varieties. b) Show that there is a birational map from X to CP1. c) Explain how to resolve the singularity of X by blowing up a point in CP2.
PROBLEM 5 (RA) Let B and L denote Banach spaces, and let  · B and  · L denote their norms. a) Let L: B → L denote a continuous, invertible linear map and let m: B ⊗ B → L denote a linear map such that  m(φ ⊗ ψ) L ≤  φ B ψ B for all φ, ψ ∈ B. Prove the following assertions: • There exists a number κ > 1 depending only on L such that if a ∈ B has norm less than κ 2, then there is a unique solution to the equation Lφ + m(φ ⊗ φ) = a with  φ B < κ 1. •
The norm of the solution from the previous bullet is at most κ  a L.
b) Recall that a Banach space is defined to be a complete, normed vector space. Is the assertion of Part a) of the first bullet always true if B is normed but not complete? If not, explain where the assumption that B is complete enters your proof of Part a).
PROBLEM 6 (CA) Fix a ∈ C and an integer n ≥ 2. Show that the equation a zn + z + 1 = 0 for a complex number z necessarily has a solution with z ≤ 2.
PROBLEM 1 SOLUTION: ! and Answer to a): The vector fields !t ! !t
! !!
along S are
! ! )+ ! = 1+tt2 (x !x + y !y !z
and
! !!
! ! . = y !x + x !y
! , ! 〉 = t 2 + 1 〈 ! , ! 〉 = 0 and 〈 ! , ! 〉 = (1 + t2), it Since their inner product is 〈 !t !! !! !t !t !! 1+t 2 follows that the square of the line element for the induced metric is 2 ds2 = 1+2t2 dt ⊗ dt + (1 + t2) dθ ⊗ dθ.
1+t
2
Answer to b): The 1forms e0 = ( 1+2 t2 )1/2 dt and e1 = (1 + t2)1/2dθ are orthonormal. Write 1+t
" 0 ! % The connection matrix of 1forms is A = $ with the 1form Γ obeying # ! 0 '& de0 = Γ ∧ e1 and de1 = Γ ∧ e0. The unique solution is Γ = 
t 1+2t 2
dθ. The Gauss curvature is denoted by κ and it is
2 1 defined by writing dΓ as κ e0 ∧ e1. Thus, κ = ( 1+2t 2 ) .
Answer to c): Since Γ = 0 on the z = 0 circle, the parallel transport is given by the ! , ! } for TS at (1, 0, 0). identity matrix when written using the orthonormal frame { !t !!
PROBLEM 2 SOLUTION: Here are two solutions: Solution 1: Let Y denote the space ×n S1. It is enough to prove that the map from X to Y factors as a map
! → Y if To prove this factorization, note that a map ƒ: X → Y lifts through a cover p: Y ! )) (they are both subgroups of π1(Y)). (See, and only if ƒ*(π1(X)) is a subgroup of p∗(π1( Y for example Proposition 1.33 in Hatcher’s book on algebraic topology.) Since π1(Rn) = 0
and ƒ∗ in this case must be the zero homomorphism, this condition is satisfied and so ƒ lifts to some ƒ! . Because Rn is contractible, this lift is nullhomotopic and any nullhomotopy pushes forward to give a nullhomotopy of ƒ. Solution 2: Recall that S1 (which is K(Z, 1)) classifies integral cohomology classes of degree 1. As a consequence, a map X → Y is (up to homotopy) determined by an ntupel of elements in H1(X; Z). The universal coefficient short exact sequence in this degree is
The two end groups are zero: The right most group is zero because H1(X; Z) is the Abelianization of π1(X) and thus it is a finite group; and finite groups have no nontrivial homomorphisms to Z. The left most group is zero because H0(X: Z) = Z and Ext(Z; Z) is trivial since Z is a free group. Thus H1(X; Z) = 0 and so all maps from X to Y are homotopic to the constant map.
PROBLEM 3 SOLUTION: Answer to a) One has [K: F] = 6 because the extension K/F is generated by the solution z of the polynomial equation z6 − uz3 + 1 = 0 which has degree 6. The Galois group contains the automorphisms α : z → 1/z and β : z → ρz, where ρ = e i2! /3 = (−1 + √−3)/2. Since α and β have orders 2 and 3 respectively, the group G generated by α and β has order at least 6. However, Gal(K/F) ≤ [K: F] = 6 with equality iff K/F is normal, so K/F must be normal with Galois group G of order 6, which is readily identified with the symmetric group the symmetric group S3 (for instance, via its permutation action on the set {1, ρ, ρ2}). Answer to b) By the fundamental theorem of Galois theory, the intermediate fields E of the Galois extension K/F correspond to subgroups H ⊂ G by E = KH (fixed subfield); K/E is always normal with Gal(K/E) = H, while E/F is normal iff H ⊴ G. Since F and K are excluded, one need not consider H = G and H = {1}. The remaining subgroups are A3 = 〈β〉, which yields the normal extension C(z3) of F, and three twoelement subgroups which yield nonnormal extensions C(z + 1/z), C(z + ρz), C(z + ρ2z). (The fact that each of these is indeed the corresponding KE can be confirmed by computing its degree as in Part a).) PROBLEM 4 SOLUTION:
Answer to a) A rational map from X to Y is an equivalence class of pairs (U, f ) where U ⊂ X is a Zariski dense open subset and f :U → Y is a regular map. Two pairs (U, f) and (V, g) are equivalent if f = g on the intersection U ∩ V. Answer to b) The projection from the point (0,0,1) ∈ CP2 to the line where z=0 restricts to a rational map p: X = {z y2 = x2(x + z)} → CP1. An inverse is given by the map given in homogeneous coordinates by the rule (u, v) → (x = (v2−u2)u, y = (v2−u2)v, z = u3). This is an inverse since x3 = (y2 − x2)z on X. It follows that p is a birational map. Answer to c) Away from the line z = 0 the blowup of CP2 at (0,0,1) is given by the locus {xt = ys} ⊂ {((x, y),(s, t))}=C2 × CP1. Consider the chart in CP1 where s ≠ 0. The blow up of X is defined here by the equations xt = y and y2 = x2(x + 1). Substituting for y gives the equation x2(t2 − x − 1) = 0 which has one irreducible component being the locus x = y = 0 (which is the exceptional curve), and the other being the locus where both t2 = x+1 and xt = y. This is the blowup of X. In the chart where t ≠ 0, the blow up of X is defined by the locus where x = ys and 1 = s2(sy + 1). By the Jacobian criterion the curve defined by these equations is nonsingular.
PROBLEM 5 SOLUTION: Answer to a) Since L is invertible, its inverse defines a bounded linear map from L to B to be denoted by L1. Using L1, one can define a map T: B → B by the rule T(φ) = L1(a  m(φ, φ)). This is relevant because φ is a fixed point of T (it obeys T(φ) = φ) if and only if φ obeys the equation Lφ + m(φ ⊗ φ) = a. Let c denote the norm of the operator L1. Then the following are computations: •
 T(φ) B ≤ c (a L +  φ B2).
•
 T(φ)  T(φ´) ≤ 4 c ( φ B +  φ´B)  φ  φ´B.
Given δ > 0, let B(δ) denote the ball of radius δ about the origin in B. If E > 0 and if  a L ≤ E then the top bullet implies that T maps B(δ) to B(cE + c δ2). Thus, if δ < (2c)1 and if E < (2c)1δ, then T maps B(δ) to itself. Meanwhile, if δ < (8c)1 then the lower bullet implies that  T(φ)  T(φ´) ≤ γ  φ  φ´B for fixed γ < 1 when φ, φ´ ∈ B(δ). This
implies in turn that T is a contraction mapping of B(δ) to itself. The contraction mapping theorem supplies a unique fixed point of T in B(δ) under these circumstances. Noting again that an element φ ∈ B is a fixed point of T if and only if φ obeys Lφ + m(φ ⊗ φ) = a, the top bullet follows if  a L ≤ (16 c)1. Take κ to be the maximum of 4c1/2 and 8c to obtaine the answer to the first bullet of Part a). The second bullet of Part a) follows directly from the fact that φ = T(φ) and  φ B2 ≤ 12  φ B because these and the inequality  T(φ) B ≤ c (a L +  φ B2) imply that
1 2
 φ B ≤ c a L.
Answer to b) The completeness of B is required. Here is an example: Take B and L to be the span of the polynomials functions on [1, 1] with the norms  ƒ B =  ƒ L = supt ƒ(t). Take the equation φ + φ2 = δ t with δ being a small, nonzero number. A solution, must be either φ =  12 + 12 (1 + 4δ2 t2)1/2 or φ =  12  12 (1 + 4δ2 t2)1/2; but neither is in B. Note that the contraction mapping theorem does not hold if the Banach space in question is not complete because the contraction mapping theorem constructs the desired solution as a limit of a Cauchy sequence in B.
PROBLEM 6 SOLUTION: There are two cases. First, assume that a < 2n. Let D denote the disk where z ≤ 2 and let ∂D denote the circle z = 2. Let f(z) = azn + z + 1 and let g(z) = z + 1. On ∂D, the function g  f obeys the inequality g(z) − f(z) = a zn < 1. Since this is less than g(z) for each z ∈ ∂D, and since g has no zeros on ∂D, none of the members of the 1parameter family of functions {fτ = f + τ (g  f)}τ∈[0,1] has a zero on ∂D. Therefore, f (which is fτ=0) and g (which is fτ=1) have the same number of zeros (counting multiplicity) in D. This number is 1 (This is Rouche’s theorem). Now assume that a ≥ 2n. By the fundamental theorem of algebra, the function f(z) = a zn + z + 1 factors as n
f(z) = a
" (z 
k
)
k =1
where the {αk}k=1….,n are complex numbers. This implies in particular the identity n
(1)n a " ! k = 1. k =1
n
hence
" ! k =1
k
 ≤ 2n. This can happen only if one or more roots αk are in D.
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Wednesday, January 20, 2016 (Day 2)
PROBLEM 1 (DG) Let k denote a positive integer. A nonoptimal version of the Whitney embedding theorem states that any kdimensional manifold can be embedded into R2k+1. Using this, show that any kdimensional manifold can be immersed in R2k. (Hint: Compose the embedding with a projection onto an appropriate subspace.)
PROBLEM 2 (T) Let X be a CWcomplex with a single cell in each of the dimensions 0, 1, 2, 3, and 5 and no other cells. a) What are the possible values of H∗(X; Z)? (Note: it is not sufficient to consider Hn(X; Z) for each n independently. The value of H1(X; Z) may constrain the value of H2(X; Z), for instance.) b) Now suppose in addition that X is its own universal cover. What extra information does this provide about H∗(X; Z)?
PROBLEM 3 (AN) Let k be a finite field of characteristic p, and n a positive integer. Let G be the group of invertible linear transformations of the kvector space kn. Identify G with the group of invertible n × n matrices with entries in k (acting from the left on column vectors). n1
a) Prove that the order of G is
! (qn  qm ) where q is the number of elements of k. m=0
b) Let U be the subgroup of G consisting of uppertriangular matrices with all diagonal entries equal 1. Prove that U is a pSylow subgroup of G. c) Suppose H ⊂ G is a subgroup whose order is a power of p. Prove that there is a basis (v1, v2,..., vn) of kn such that for every h ∈H and every m ∈{1, 2, 3,..., n}, the vector h(vm)  vm is in the span of {vd: d < m}.
PROBLEM 4 (AG) Let X be a complete intersection of surfaces of degrees a and b in CP3. Compute the Hilbert polynomial of X.
PROBLEM 5 (RA) Let C 0 denote the vector space of continuous functions on the interval [0, 1]. Define a norm on C 0 as follows: If ƒ ∈ C 0, then its norm (denoted by  ƒ ) is  ƒ  = supt∈[0,1] ƒ(t) . Let C ∞ denote the space of smooth functions on [0, 1]. View C ∞ as a normed, linear space with the norm defined as follows: If ƒ ∈ C ∞, then its norm (denoted by  ƒ ∗) is  ƒ ∗ =
!
d ƒ +  ƒ) dt . ( dt
[0,1]
a) Prove that C 0 is Banach space with respect to the norm  · . In particular, prove that it is complete. b) Let ψ denote the ‘forgetful’ map from C ∞ to C 0 that sends ƒ to ƒ. Prove that ψ is a bounded map from C ∞ to C 0, but not a compact map from C ∞ to C 0.
PROBLEM 6 (CA) Let D denote the closed disk in C where z ≤ 1. Fix R > 0 and let ϕ: D → C denote a continuous map with the following properties: i) ϕ is holomorphic on the interior of D. ii) ϕ(0) = 0 and its zderivative, ϕ´, obeys ϕ´(0) = 1. iii) ϕ ≤ R for all z ∈D. Since ϕ´(0) = 1, there exists δ > 0 such that ϕ maps the z < δ disk diffeomorphically onto its image. Prove the following: a) There is a unique solution in [0, 1] to the equation 2Rδ = (1  δ)3. b) Let δ∗ denote the unique solution to this equation. If If 0 < δ < δ∗, then ϕ maps the z < δ disk diffeomorphically onto its image.
PROBLEM 1 SOLUTION: The desired immersion will come from a projection onto the orthogonal complement of a suitably chosen, nonzero vector in R2k. To find this vector, let M denote the manifold in question and let ƒ denote the embedding of M into R2k. Let g denote the map from TM to R2k+1 that is defined as follows: Supposing that x ∈ M and v ∈ TMx set g(x, v) = ƒ∗x·v where ƒ∗ denotes the differential of ƒ. Sard’s theorem can be invoked to see that g is not surjective. Let a denote a point that is not in the image of g. (Note that a is necessarily nonzero.) Use π to denote the projection onto the orthogonal complement of a. To see that π ! ƒ is an immersion, let x denote a point in M and let v denote a nonzero vector in TMx. Suppose for the sake of argument that (π ! ƒ)∗v is zero. If this is so, then the chain rule and the fact that π is linear implies that ƒ∗x·v = ta for some nonzero t ∈ R. This implies in turn that ƒ∗x(t1v) = a which is nonsense because a is in the complement of the image of ƒ∗.
PROBLEM 2 SOLUTION: Answer to a) The cellular chain complex for X must be of the form
Since X is connected, it must have H0(X; Z) = Z, so the map c must be zero. The only other restriction is that the sequence form a complex, so b ! a = 0; but since b ! a is multiplication by some integer, either a = 0 or b = 0. In the case a = 0 and b ≠ 0, the homology groups take the form
In the case a ≠ 0 and b = 0, the homology groups take the form
In the remaing a = 0 = b case, they take the form
Answer to b) The assertion that X is its own universal cover is the same as the assertion π1(X) = 0. But, since H1(X) = π1(X)ab, this means H1(X) = 0. The only case where this is possible is when a = 0 and b ≠ 0. Moreover, since Z/b = 0 in this case, b must be a multiplicative unit: b = ±1.
PROBLEM 3 SOLUTION: Answer to a) The elements of G are in bijection with ordered bases (v1,...,vn) of kn (the map takes each matrix to its columns). For each j ∈ {0, 1, 2,..., n−1}, once vi for all i ≤ j has been chosen, then there are qn  qj choices for the index (j + 1) basis element because any of the qn elements of kn except the qm linear combinations of v1, . . . , vj will do. n1
Hence the number of possible bases is
! (qn  qm ) . m=0
Answer to b) Each factor qn  qj is qj times an integer not divisible by p because it is congruent to 1 modulo q, and q is a multiple of p. Hence the number of elements in G is n1
qd times some integer not divisible by p, where d =
! j . But qd is the order of U because j= 0
there are d entries above the diagonal, and a power of p. Hence U is a pSylow subgroup of G. Answer to c) U consists of the matrices h that satisfy the desired property with respect to the standard basis of unit vectors. Hence the matrices h that satisfy this property for the basis (v1,...,vn) constitute the subgroup of G obtained by conjugating U by the matrix with columns v1, . . . , vn. But by Sylow’s second theorem H is contained in a conjugate of U.
PROBLEM 4 SOLUTION: Let S = C[x0, x1, x2, x3] be the homogeneous coordinate ring of CP3. The coordinate ring of X is of the form S/(f, g) for some irreducible polynomials f and g of degrees a, b respectively. There is a fourterm exact sequence of graded modules
with maps given by multiplication with f and g. Hence the Hilbert polynomial of X is
PROBLEM 5 SOLUTION: Answer to a) One has to show that a Cauchy sequence {ƒn}n=1,2,… in C 0 converges to a continuous function. To do this, note that for each t ∈ [0, 1], the sequence {ƒn(t)}n∈{1,2,…} is a Cauchy sequence in R so it converges. Let ƒ(t) denote the limit. The assignment t → ƒ(t) defines a function on [0, 1]. The task is to prove that this function is continuous. This means the following: Given ε > 0, there exists δ > 0 such that ƒ(t)  ƒ(t´) < ε when t  t´ < δ. To find δ, first fix N so that ƒn(t)  ƒm(t) < 13 ε for all t ∈ [0, 1] and all pairs n, m > N. This implies that ƒn(t)  ƒ(t) ≤ 13 ε for all t. Such N exists because {ƒn}n∈{1,2,…} is a Cauchy sequence in C 0. To continue, take n > N and fix δ so that ƒn(t)  ƒn(t´) < 13 ε when t´  t < δ. It then follows by the triangle inequality that ƒ(t)  ƒ(t´) ≤ ƒ(t)  ƒn(t) + ƒ(t´)  ƒn(t´) + ƒn(t´)  ƒn(t) < ε.
Answer to b) The map ψ is bounded because for all t, one has the identity
ƒ(t) =
1
t
!0 ( ! dsd ƒ(s)ds + ƒ(r)) dr , r
and thus ƒ(t) ≤  ƒ ∗ for all t. It is not a compact map. To prove this, fix a smooth function on [0, ∞) that is equal to 1 near t = 0 and equal to 0 for t > 12 . Call this function
ƒ. Define ƒn(t) = ƒ(n t). This function is smooth on [0, 1]. The sequence {ƒn(t)} has bounded  · ∗ norm but it has no convergent sequence in C 0.
PROBLEM 6 SOLUTION: Answer to a) The function ƒ(δ) = 2Rδ/(1  δ)3 has strictly positive derivative and therefore defines a diffeomorphism from [0, 1) to [0, ∞). It follows from this that there is a single point where ƒ is equal to 1. Answer to b) To obtain the asserted lower bound for δ, note that ϕ maps the disk where z < δ diffeomorphically to its image if it is 11 on this disk and if ϕ´ > 0 on this disk. The Cauchy integral formula is used to see when this happens. Here is Cauchy’s formula: ϕ(z) =
1 2!i
" z 1 w !(w)dw . !D
Differentiating this, one sees that ϕ´´ on the z < δ disk is bounded by 2R(1  δ)3. This implies that ϕ´  1 < 2 R δ (1  δ)3 where z < δ. If ϕ´ > 0, then ϕ is a local diffeomorphism. This is the case when δ < δ∗ with δ∗ being the solution in (0, 1) to the equation 2Rδ∗(1  δ∗)3 = 1. Meanwhile, if z, z´ have norm less than δ, then ϕ(z)  ϕ(z´) ≥ (1  2 R δ (1  δ)3) z  z´ which is a positive multiple of z  z´ precisely when δ < δ∗.
Qualifying Examination HARVARD UNIVERSITY Department of Mathematics Thursday, January 21, 2016 (Day 3)
PROBLEM 1 (DG) Recall that a symplectic manifold is a pair (M, ω), where M is a smooth manifold and ω is a closed nondegenerate differential 2form on M. (The 2form ω is called the symplectic form.) a) Show that if H: M → R is a smooth function, then there exists a unique vector field, to be denoted by XH, satisfying ! XH " = dH. (Here, ι denotes the contraction operation.) b) Supposing that t > 0 is given, suppose in what follows that the flow of XH is defined for time t, and let φt denote the resulting diffeomorphism of M. Show that φt*ω = ω. c) Denote the Euclidean coordinates on R4 by (x1, y1, x2, y2) and use these to define the symplectic form ω0 = dx1 ∧ dy1 + dx2 ∧ dy2. Find a function H: R4 → R such that the diffeomorphism φt=1 that is defined by the time t = 1 flow of XH fixes the half space where x1 ≤ 0 and moves each point in the half space where x1 ≥ 1 by 1 in the y2 direction.
PROBLEM 2 (T) Let X denote a finite CW complex and let ƒ: X → X be a selfmap of X. Recall that the Lefschetz trace of ƒ, denoted by τ(ƒ), is defined by the rule "
τ(ƒ) =
# (1)
n
tr(ƒ n : H n (X; Q) ! H n (X; Q))
n=0
with ƒn denoting the induced homomorphism. Use τ(·) to answer the following: a) Does there exist a continuous map from RP2 to itself with no fixed points? If so, give an example; and if not, give a proof. b) Does there exist a continuous map from RP3 to itself with no fixed points? If so, give an example; and if not, give a proof.
PROBLEM 3 (AN) Let A be the ring Z[ 5 2016 ] = Z[X]/(X5  2016). Given that 2017 is prime in Z, determine the factorization of 2017·A into prime ideals of A.
PROBLEM 4 (AG) a) State a version of the Riemann–Roch theorem. b) Apply this theorem to show that if X is a complete nonsingular curve and P ∈ X is any point, there is a rational function on X which has a pole at P and is regular on X−{P}.
PROBLEM 5 (RA) Let ℘ denote a probability measure for a real valued random variable with mean 0. Denote this random variable by x. Suppose that the random variable x has mean equal to 2. a) Given R > 2, state a nontrivial upper bound for event that x ≥ R. (The trivial upper bound is 1.) b) Give a nonzero lower bound for the standard deviation of x. c) A function ƒ on R is Lipshitz when there exists a number c ≥ 0 such that ƒ(p)  ƒ(p´) ≤ c p  p´ for any pair p, p´ ∈ R. ˆ denote the function on R whose value at a given p ∈ R is the expectation of the Let ! random variable e ip x . (This is the characteristic function of ℘.) Give a rigorous ˆ is Lipshitz and give an upper bound for c in this case. proof that ! d) Suppose that the standard deviation of x is equal to 4. Let N denote an integer greater than 1, and let {x1, …, xN} denote a set of independent random variables each with probabilities given by ℘. Use SN to denote the random variable
1 N
(x1 + ···· + xN).
The central limit theorem gives an integral that approximates the probability of the event where SN ∈ [1, 1] when N is large. Write this integral.
PROBLEM 6 (CA) Let H ⊂ C denote the open right half plane, thus H = {z = x + iy: x > 0}. Suppose that ƒ: H → C is a bounded, analytic function such that ƒ(1/n) = 0 for each positive integer n. Prove that ƒ(z) = 0 for all z. N
(Hint: Consider the behavior of the sequence of functions {hN(z) =
! zz + 11/n/n }
N=1.2…
n =1
and, in particular, on the positive real axis.}
on H
PROBLEM 1 SOLUTION: Answer to a) To say that ω is nondegenerate is to say that the contraction operation defines a vector bundle isomorphism between TM and T*M. Answer to b) The definition of the Lie derivative is such that
! !t
(φt*ω) = φt*( L XH ! ) with
L XH ! denoting the Lie derivative of ω along the vector field XH. Cartan’s formula for L XH ! is L XH ! = d( ! XH " ) + ! XH d" and both of these terms are zero. Thus, φt*ω is independent of t and thus equal to its value at t = 0 which is ω. Answer to c) Choose a smooth function ƒ: R → [0, 1] so that ƒ(s) = 0 for s ≤ 0 and ƒ(s) = 1 for s ≥ 1. The function sending (x1, y1, x2, y2) → H(x1, y1, x2, y2) = ƒ(x1) x2 has the desired properties because XH = 0 for x1 ≤ 0 and XH = !y! for x1 ≥ 1. 2
PROBLEM 2 SOLUTION: Answer to a) The Lefschetz trace theorem states that if τ(f) ≠ 0, then f must have a fixed point. To see that τ(ƒ) is never zero, note first that the rational homology of RP2 is zero except for H0(RP2; Q), which is Q. Since ƒ0 is multiplication by 1, it τ(ƒ) is never zero. Answer to b) In this case, the nonzero rational homology is in dimensions 0 and 3, each being isomorphic to Q. As a consequence, the argument used for RP2 can not be used here. In fact, there is a selfmap with no fixed points and it is constructed momentarily. It is instructive to consider first the case of RP1 which is S1, where a rotation by angle π has no fixed points. Now viewing RP1 as (R2−0)/R∗, then this rotation through angle π is depicted using homogeneous coordinates [x1, x2] as the map [x1, x2] → [x2, x1] which can’t have a fixed point because there is no nonzero real number λ and (x1, x2) ∈ R2−0 with x2 = λx1 and x1 = λx2. To mimick this for RP3, write RP3 as (R4−0)/R* and then define the desired self map using homogeneous coordinates [x1, x2, x3, x4] by the rule whereby [x1, x2, x3, x4] → [x2, x1, x4, x3]. This has no fixed points because there is no nonzero real number λ and (x1, x2, x3, x4) ∈ R4−0 such that x2 = λx1, x1 = λx2, x4 = λx3 and x3 = λx4.
PROBLEM 3 SOLUTION:
2017A is the product of the prime ideals (2017, X + 1) and (2017, X4  X3 + X2  X + 1). In general, if the polynomial P(X) factors modulo a prime p into distinct irreducibles {Pi} then the ideal pZ[X]/(P(X)) is the product of ideals (p, Pi). In our case, p = 2017 and P = X5 − 2016 ! X5 + 1 mod p. The roots of X5 + 1 in an algebraic closure of Z/pZ are the set {1, w, w2, w3, w4} where w is a nontrivial 5th root of unity. The irreducible factors correspond to orbits of the permutation x → xp of those roots. Clearly −1 is a fixed point, and since p ! 2 mod 5 the remaining roots fall in to a single orbit w → w2 → w4 → w3 → w. Hence the irreducible factors of X5 +1 mod p are X + 1 and (X5 +1)/(X +1) which is the polynomial X4 −X3 +X2 −X +1.
PROBLEM 4 SOLUTION: Answer to a) Let X be a complete nonsingular curve of genus g. Let K denote the canonical divisor. If D is any divisor on X, let !(D) = dim(H0(X, OX(D))). The RiemannRoch theorem asserts that !(D)  !(K  D) = deg(D) + 1  g. Answer to b) Fix a point Q ≠ P and let D denote the divisor 2P  Q. Choose a positive integer n such that n > max{2g − 2, 0}. Noting that n = deg(nD) and that deg(K) = 2g  2, it follows that deg(K  nD) < 0. This implies that !(K  D) = 0. Therefore, the Riemann– Roch theorem applied to nD implies that !(nD) = n + 1  g which is greater than 1. This means that there is an effective divisor (to be denoted by D´) and a rational function on X (to be denoted by ƒ) such that n D + (ƒ) = D´. Rewriting this gives (ƒ) = D´  2n P + n Q so ƒ has poles only at P.
PROBLEM 5 SOLUTION: Answer to a) The event in question is
" !. x!R
This is no smaller than R1
" x! which x!R
in turn is no greater than R2 . Answer to b) The square of the standard deviation is the square root of the expectation of the random variable x2. Since
" x! ≤ ( "! ) ( " x ! ) 2
1/2
!
!
1/2
(∗)
!
(which is proved momentarily), and since "! = 1, it follows that ( " x 2 ! )1/2 ≥ 2. To !
!
1
2
prove (∗), note that for any t ∈ (0, ∞), the expectation of (t  t x) is the sum t 2 "!  2 " x! + t 2 " x 2 ! . !
!
!
This is nonnegative for any t ∈ (0, 1) since it is the expectation of a positive random variable. The assertion that it is nonnegative for the case t = ( " x 2 ! )1/4 ( "! )1/4 is (∗). !
!
Answer to c) Supposing that p, p´ ∈ R, then ˆ ˆ  !(p´) = " (e ix p  e ix p´ )! . !(p)
(∗∗)
!
p´
ixp
Noting that e
ixp´
e
= ix
!e
ixq
dq by the fundamental theorem of calculus, it follows
p
that eixp  eixp´ ≤ x p  p´. This understood, then (∗∗) leads to the bound ˆ ˆ !(p)  !(p´) ≤ ( " x! ) p  p´ = 2 p  p´ . !
Answer to d) The random variable SN has mean 0 and standard deviation equal to N1/2 times the standard deviation of x, thus 4 N1/2. (The expecation of SN2 is the that of N2
∑i,k=1,…,N xixk. Only the i = k terms are nonzero (because x has mean zero), there are N
of them and each is the expectation of x2 which is 16.) Denote this standard deviation of SN by σN for the moment. The central limit theorem approximates the probability in 1
question by
#
1 2! " N
ex
1
PROBLEM 6 SOLUTION
2
/ 2 " 2N
dx where σN again denotes 4 N1/2 .
This is a form of Jensen’s inequality. To elaborate, fix B so that ƒ(z) ≤ B for all z ∈ H. For each integer N, define N
/n FN(z) = ƒ(z)/hN(z) = ƒ(z) ! zz + 11/n . n =1
This function is analytic on H because the poles at z = 1, 2, 3,..., N are matched by zeros of ƒ. Moreover, the absolute value of each of the factors (z + 1/n)/(z − 1/n) approaches 1 as Re(z) → 0 (uniformly in Im(z)), and also approaches 1 as z → ∞. Hence Fn(z) ≤ B for all z ∈ H by virtue of the maximum modulus principle (the norm of an analytic function can not take on a local maximum). With the preceding understood, note that for N
any fixed, positive real z, the factor
! zz + 11/n/n
becomes unbounded as N → ∞. Hence its
n =1
product with ƒ(z) cannot remain bounded unless ƒ(z) = 0 on the real axis. But a holomorphic function on any domain has discrete zeros, so ƒ(z) must be everywhere 0.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 1, 2015 (Day 1) 1. (A) The integer 8871870642308873326043363 is the 13th power of an integer n. Find n. 2. (AG) Let C ⊂ P2 be a smooth plane curve of degree 4. (a) Describe the canonical bundle of C in terms of line bundles on P2 . What are the effective canonical divisors on C? (b) What is the genus of C? Explain how you obtain this formula. (c) Prove that C is not hyperelliptic. 3. (DG) Let M be a C ∞ manifold, T M its tangent bundle, and T C M = C⊗R T M the complexified tangent bundle. An almost complex structure on M is a C ∞ bundle map J : T M → T M such that J 2 = −1. (a) Show that an almost complex structure naturally determines, and is determined by, each of the following two structures: i) the structure of a complex C ∞ vector bundle – i.e., with fibres that are complex vector spaces – on T M compatible with its real structure. ii) a C ∞ direct sum decomposition T C M = T 1,0 M ⊕ T 0,1 M with T 0,1 M = complex conjugate of T 1,0 M . (b) Show that every almost complex manifold is orientable. (c) If S is a C ∞ , orientable, 2dimensional, Riemannian manifold, construct a natural almost complex structure on S in terms of its Riemannian structure, but one that depends only on the underlying conformal structure of S. (d) Does the almost complex structure constructed in (c) determine the conformal structure of S? You need NOT give a detailed answer to this question; a heuristic one or twosentence answer suffices. 4. (RA) In this problem V denotes a Banach space over R or C. (a) Show that any finite dimensional subspace U0 ⊂ V is closed in V . (b) Now let U1 ⊂ V a closed subspace, and U2 ⊂ V a finite dimensional subspace. Show that U1 + U2 is closed in V . 5. (AT) Consider the following three topological spaces: A = HP3 ,
B = S4 × S8,
(HP3 denotes quaternionic projective 3space.)
C = S 4 ∨ S 8 ∨ S 12 .
(a) Calculate the cohomology groups (with integer coefficients) of all three. (b) Show that A and B are not homotopy equivalent. (c) Show that C is not homotopy equivalent to any compact manifold. 6. (CA) Let f (z) be a function which is analytic in the unit disc D = {z < 1}, and assume that f (z) ≤ 1 in D. Also assume that f (z) has at least two fixed points z1 and z2 . Prove that f (z) = z for all z ∈ D.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 2, 2015 (Day 2)
1. (AT) Let CPn = (Cn+1 \ {0})/C∗ be n dimensional complex projective space. (a) Show that every map f : CP2n → CP2n has a fixed point. (Hint: Use the ring structure on cohomology.) (b) For every n ≥ 0, give an example of a map f : CP2n+1 → CP2n+1 without any fixed points and describe its induced map on cohomology. 2. (A) Let A be a commutative ring with unit. Define what it means for A to be Noetherian. Prove that the ring of continuous functions f : [0, 1] → R (with pointwise addition and multiplication) is not Noetherian. 3. (CA) Let S ⊂ C be the open halfdisc {x + iy : y > 0, x2 + y 2 < 1}. (a) Construct a surjective conformal mapping f : S → D, where D is the open unit disc {z ∈ C : z < 1}. (b) Construct a harmonic function h : S → R such that: • h(x + iy) → 0 as y → 0 from above, for all real x with x < 1, and • h(reiθ ) → 1 as r → 1 from below, for all real θ with 0 < x < π. 4. (AG) Let Q be the complex quadric surface in P3 defined by the homogeneous equation x0 x3 − x1 x2 = 0. (a) Show that Q is nonsingular. (b) Show that through each point of Q there are exactly two lines which lie on Q. (c) Show that Q is rational, but not isomorphic to P2 . 5. (DG) Let Ω be the 2form on R3 − {0} defined by Ω=
x2
1 (x dy ∧ dz + y dz ∧ dx + z dx ∧ dy). + y2 + z2
(a) Prove that Ω is closed. (b) Let f : R3 −{0} → S 2 be the map which sends (x, y, z) to ( x2 +y12 +z 2 )1/2 (x, y, z). Show that Ω is the pullback via f of a 2form on S 2 . (c) Prove that Ω is not exact.
6. (RA) Consider the linear ODE f 00 +P f 0 +Q f = 0 on the interval (a, b) ⊂ R, with P, Q denoting C ∞ real valued functions on (a, b). Recall the definition of the Wronskian W (f1 , f2 ) = f1 f20 − f10 f2 associated to any two solutions f1 , f2 of this differential equation. (a) Show that W (f1 , f2 ) either vanishes identically or is everywhere nonzero, depending on whether the two solutions f1 , f2 are linearly dependent or not. (b) Now suppose that f1 , f2 are linearly independent, real valued solutions. Show that they have at most first order zeroes, and that the zeroes occur in an alternating fashion: between any two zeroes of one of the solutions there must be a zero of the other solution.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 3, 2015 (Day 3)
1. (DG) Consider the graph S of the function F (x, y) = cosh(x) cos(y) in R3 and let Φ : R2 → S ⊂ R 3 be its parametrization: Φ(x, y) = (x, y, cosh(x) cos(y)). (a) Write down the metric on R2 that is defined by the rule that the inner product of two vectors v and w at the point (x, y) is equal to the inner product of Φ∗ (v) and Φ∗ (w) at the point Φ(x, y) in R3 . (b) Define the Gaussian curvature of a general surface embedded in R3 . (c) Compute the Gaussian curvature of the surface S at the point (0, 0, 1). 2. (RA) ∈ C(R/Z) be a continuous Cvalued function on R/Z and let P∞ Let f (x) 2πinx be its Fourier series. a e n=−∞ n (a) Show that f is C ∞ if and only if an  = O(n−k ) for all k ∈ N. (b) Prove that a sequence of functions { fn }n≥1 in C ∞ (R/Z) converges in the C ∞ topology (uniform convergence of functions and their derivatives (k) of all orders) if and only if the sequences of kth derivatives { fn }n≥1 , 2 for all k ≥ 0, converge in the L norm on R/Z. ⊗2 3. (AG) Let C be a smooth projective curve over C and ωC the square of its canonical sheaf. ⊗2 )? (a) What is the dimension of the space of sections Γ(C, ωC ⊗2 (b) Suppose g(C) ≥ 2 and s ∈ Γ(C, ωC ) is a section with simple zeros. 2 Compute the genus of Σ = {x = s} in the total space of the line bundle √ ωC , i.e. the curve defined by the 2valued 1form s.
4. (AT) Show (using the theory of covering spaces) that every subgroup of a free group is free. 5. (CA) (a) Define Euler’s Gamma function Γ(z) in the half plane Re(z) > 0 and show that it is holomorphic in this half plane. (b) Show that Γ(z) has a meromorphic continuation to the entire complex plane.
(c) Where are the poles of Γ(z)? (d) Show that these poles are all simple and determine the residue at each pole. 6. (A) Let G be a finite group, and ρ : G → GLn (C) a linear representation. Then for each integer i ≥ 0 there is a representation ∧i ρ of G on the exterior power ∧i (Cn ). Let Wi be the subspace (∧i (Cn ))G of ∧i (Cn ) fixed under this action of G. Prove that dim Wi is the T i coefficient of the polynomial 1 X det(1n + T ρ(g)) G g∈G
where 1n is the n × n identity matrix.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 1, 2015 (Day 1)
1. (A) The integer 8871870642308873326043363 is the 13th power of an integer n. Find n. Solution. Counting digits, we see that n < 100, so is determined by its residue class (mod 99).Both (mod 11) and (mod 9) raising to the 13th power is a bijection. After a little computation we find that n ≡ 2(mod 9) and n ≡ 6(mod 11). This implies, by the Chinese remainder theorem that n ≡ 83(mod 99). Hence n = 83. 2. (AG) Let C ⊂ P2 be a smooth plane curve of degree 4. (a) Describe the canonical bundle of C in terms of line bundles on P2 . What are the effective canonical divisors on C? (b) What is the genus of C? Explain how you obtain this formula. (c) Prove that C is not hyperelliptic. Solution: By the adjunction formula, the canonical divisor class of a curve of degree d is KC = OC (d − 3), that is, plane curves of degree d − 3 cut out canonical divisors on C. It follows that effective canonical divisors on C are the intersection with lines in the plane, so have degree 4. Since the degree of the canonical class is 2g − 2, the genus g = 3. Furthermore, any two points p, q ∈ C impose independent conditions on the canonical series KC ; that is, h0 (KC (−p − q)) = g − 2, so by the RiemannRoch formula h0 (OC (p + q)) = 1, i.e., C is not hyperelliptic. 3. (DG) Let M be a C ∞ manifold, T M its tangent bundle, and T C M = C⊗R T M the complexified tangent bundle. An almost complex structure on M is a C ∞ bundle map J : T M → T M such that J 2 = −1. (a) Show that an almost complex structure naturally determines, and is determined by, each of the following two structures: i) the structure of a complex C ∞ vector bundle – i.e., with fibres that are complex vector spaces – on T M compatible with its real structure. ii) a C ∞ direct sum decomposition T C M = T 1,0 M ⊕ T 0,1 M with T 0,1 M = complex conjugate of T 1,0 M . (b) Show that every almost complex manifold is orientable.
(c) If S is a C ∞ , orientable, 2dimensional, Riemannian manifold, construct a natural almost complex structure on S in terms of its Riemannian structure, but one that depends only on the underlying conformal structure of S. (d) Does the almost complex structure constructed in (c) determine the conformal structure of S? You need NOT give a detailed answer to this question; a heuristic one or twosentence answer suffices. Solution: The correspondence between J and the structure of complex vector bundle on T M is given by J ↔ multiplication by i ; this is well defined and bijective because both are C ∞ bundle maps, defined over R, of square −1. For the same reason, J ↔ bijectively corresponds to decompositions T C M = T 1,0 M ⊕ T 0,1 with T 0,1 M = complex conjugate of T 1,0 M , via T 1,0 = i − eigenspace of J , and T 0,1 = (−i) − eigenspace of J , on each fiber. That is the assertion (a). To establish (b), let {s1 , s2 , . . . , sn } denote a local C ∞ frame of T 1,0 M . The complex conjugate frame is then a frame of T 0,1 M . it follows that is a local C ∞ generator of ∧top T C M . It is defined over R, as can be checked by an easy calculation, and hence can be regarded as a local C ∞ generator of ∧top T M . Now let {t1 , t2 , . . . , tn } be another local C ∞ frame of T 1,0 M . On the overlap of the domains, the two frames are related by a C ∞ matrix valued function (ai,j ). But then i−n t1 ∧ · · · ∧ tn ∧ t1 ∧ . . . tn =  det(ai,j )2 i−n s1 ∧ · · · ∧ sn ∧ s1 ∧ . . . sn , so any two local frames induce the same orientation on M . This proves (b). On a 2dimensional Riemannian manifold S one has the notion of an angle between any two tangent vectors at a point, which depends only on the underlying conformal structure, and if S is oriented, one even has the notion of a directed angle. In this situation it makes sense to define J = rotation through an angle π/2. This is a C ∞ bundle map because the metric is smooth, and J 2 = −1 by definition. That implies (c). Finally, for (d), note that on the tangent spaces of a Riemannian surface one can make sense of a rotation though any angle if one knows the effect of a rotation about the angle π/2. 4. (RA) In this problem V denotes a Banach space over R or C. (a) Show that any finite dimensional subspace U0 ⊂ V is closed in V . (b) Now let U1 ⊂ V a closed subspace, and U2 ⊂ V a finite dimensional subspace. Show that U1 + U2 is closed in V .
Solution: For definiteness suppose V is a Banach space over R. Let {uk  1 ≤ k ≤P n} be a basis of U0 , and use this basis to identify U0 ∼ = Rn . Then, for u = k ak uk ∈ U0 ⊂ V , P kuk ≤ 0≤k≤n ak  kuk k ≤ C max0≤k≤n ak  , with C = max0≤k≤n kuk k . It follows that Rn ∼ = U0 ,→ V is bounded with respect to the sup norm on Rn (and hence with respect to any other Banach norm on Rn ). Now let {vm } be a convergent sequence in V , all of whose terms lie in the subspace U0 . But then the inverse image of this sequence in Rn must be bounded, and has a convergent subsequence. Its limit, viewed as a vector in V , must coincide with the original limit, of course. This implies (a). In establishing (b) we can replace U2 by a linear complement, in U2 , of U1 ∩ U2 . In other words, we may assume U1 ∩ U2 = 0. Any convergent sequence {vm } whose terms lie in 0 + v 00 }, with v 0 ∈ U and U1 + U2 can now be written uniquely as {vm = vm 1 m m 00 vm ∈ U2 . Let’s distinguish two cases: 00 } has a bounded subsequence, which by (a) in turn has i) The sequence {vm a subsequence that converges in U2 . But then the corresponding subsequence 0 } must converge, necessarily to a point in the closed subspace U . It of {vm 1 follows that the limit of the original series must lie in U1 + U2 . 00 k → ∞ as m → ∞. Going to an appropriate subsequence of the original ii) kvm 00 6= 0 for all m and kv 00 k−1 v 00 → v series, we may then assume that vm ˜00 ∈ U2 , m m 00 00 −1 k˜ v k = 1. Because of the hypotheses, kvm k vm → 0 ∈ V , which now implies 00 k−1 v 0 to some point v the convergence of kvm ˜0 in the closed subspace U1 . At m this point, we know that 00 k−1 v 0 = limm→∞ kvm m 00 k−1 v 0 + lim 00 −1 v 00 = v = limm→∞ kvm ˜0 + v˜00 . m→∞ kvm k m m
That is a contradiction because 0 6= v˜00 = −˜ v 0 ∈ U1 ∩ U2 = 0 .
5. (AT) Consider the following three topological spaces: A = HP3 ,
B = S4 × S8,
C = S 4 ∨ S 8 ∨ S 12 .
(HP3 denotes quaternionic projective 3space.) (a) Calculate the cohomology groups (with integer coefficients) of all three. (b) Show that A and B are not homotopy equivalent. (c) Show that C is not homotopy equivalent to any compact manifold. Solution:
1. The cohomology rings of the three spaces are as follows: H ∗ A = Z[x]/x4 , ∗
x = 4, 2
2
H B = Z[a, b]/(a , b ),
a = 4, b = 8,
H ∗ C = Z{r, s, t},
r = 4, s = 8, t = 12, with all products zero.
2. The ring structures differ: x · x = x2 6= 0, but a · a = 0. 3. If C were homotopy equivalent to a compact manifold, then it would enjoy Poincar´e duality. In particular, r could be taken to be Poincar´e dual to s and t to be the volume form, so that r · s = t. However, r · s = 0 in H ∗ C. 6. (CA) Let f (z) be a function which is analytic in the unit disc D = {z < 1}, and assume that f (z) ≤ 1 in D. Also assume that f (z) has at least two fixed points z1 and z2 . Prove that f (z) = z for all z ∈ D. Solution: First observe that we can find a fractional linear transformation S mapping D to itself and 0 to z1 . Now consider g = S −1 ◦ f ◦ S. The function g is also analytic on D, and satisfies g(z) ≤ 1 on D. One of the fixed points of g is 0, hence the function h(z) = g(z)/z is analytic; call p the other fixed point of g. We claim that h(z) ≤ 1. Before proving the claim, note that this implies the desired result, since h(p) = 1, hence h is identically 1 on D by the maximum principle. To prove the claim, we also use the maximum principle. Fix some small > 0. On {z = 1 − }, we have h(z) = g(z)/z ≤ 1/(1 − ), hence h(z) ≤ 1/(1 − ) on {z ≤ 1 − } by the maximum principle. Letting tend to 0 gives h(z) ≤ 1 on D, as desired.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 2, 2015 (Day 2)
1. (AT) Let CPn = (Cn+1 \ {0})/C∗ be n dimensional complex projective space. (a) Show that every map f : CP2n → CP2n has a fixed point. (Hint: Use the ring structure on cohomology.) (b) For every n ≥ 0, give an example of a map f : CP2n+1 → CP2n+1 without any fixed points and describe its induced map on cohomology. Solution: We have H ∗ (CP2n ; Z) = Z[x]/x2n+1 with x in degree 2. The map f induces a ring endomorphism f ∗ given by f ∗ (x) = kx for some k ∈ Z. Thus, the trace of f ∗ is 2n X Tr(f ∗ ) = 1 + k i ≡ 1 (mod 2). i=1
In particular, the trace is nonzero and hence f has a fixed point by the Lefschetz fixedpoint theorem. For the second part we can take the map f : CP2n+1 → CP2n+1 with f ([z0 , . . . , z2n+1 ]) = [−¯ z1 , z¯0 , . . . , −¯ z2n+1 , z¯2n ] since f (z) = z implies zj = −λ2 zj for some λ ∈ C∗ and all 0 ≤ j ≤ 2n + 1, a contradiction. Note that f sends the line CP1 = [z0 , z1 , 0, . . . , 0] ⊂ CP2n+1 to itself, but with reverse orientation, so f ∗ (x) = −x. 2. (A) Let A be a commutative ring with unit. Define what it means for A to be Noetherian. Prove that the ring of continuous functions f : [0, 1] → R (with pointwise addition and multiplication) is not Noetherian. Solution: A is Noetherian if it has no sequence of ideals I1 , I2 , I3 , . . . such that In ⊂ In+1 and In 6= In+1 for each n. If A is the ring of continuous functions [0, 1] → R then we get a counterexample by taking for In the ideal of functions f ∈ A such that there exists B ∈ R with f (x) ≤ Bx1/n for all x ∈ [0, 1]. The inclusions are strict because x1/n is in In but not in In+1 . [Alternatively, let In consist of the functions supported on [1/n, 1], or of the functions vanishing at 1/m for all integers m ≥ n.] 3. (CA) Let S ⊂ C be the open halfdisc {x + iy : y > 0, x2 + y 2 < 1}. (a) Construct a surjective conformal mapping f : S → D, where D is the open unit disc {z ∈ C : z < 1}.
(b) Construct a harmonic function h : S → R such that: • h(x + iy) → 0 as y → 0 from above, for all real x with x < 1, and • h(reiθ ) → 1 as r → 1 from below, for all real θ with 0 < x < π. Solution: We construct f as a composition f3 ◦ f2 ◦ f1 of conformal maps where f1 and f3 are M¨ obius transformations and f2 (z) = z 2 . Set f1 (z) = (1+z)/(1−z), which transforms D conformally into the first quadrant {(x, y) : x > 0, y > 0}, taking (−1, 1) to the positive real axis and the semicircular boundary of S to the imaginary real axis. Thus f2 ◦ f1 conformally transforms D to the upper halfplane {(x, y) : y > 0}, and finally f3 (z) := (z − i)/(z + i) takes the upper halfplane to D, whence f3 ◦ f2 ◦ f1 is a conformal map S → D as demanded. [Of course there are other variations such as f1 (z) = (z − 1)/(z + 1) etc., any of which earns full credit as long as f1 fits into f2 fits into f3 correctly.] The function h(z) = (2/π)= log f1 (z) [a.k.a. h(z) = (1/π)= log f2 (f1 (z))] is harmonic because it is the imaginary part of an analytic function, and has the requisite limiting behavior by our description of f1 in part (i) (the principal value of log(z) has imaginary part 0 for z = x > 0, and imaginary part π/2 when z = iy with y > 0). 4. (AG) Let Q be the complex quadric surface in P3 defined by the homogeneous equation x0 x3 − x1 x2 = 0. (a) Show that Q is nonsingular. (b) Show that through each point of Q there are exactly two lines which lie on Q. (c) Show that Q is rational, but not isomorphic to P2 . Solution: The partial derivatives of F (X, Y, Z, W ) = XY − ZW have no common zeroes in P3 . A line which lies on Q corresponds to an isotropic plane V in the quadratic space C4 , whereas a point on Q corresponds to an isotropic line L. The quadratic space L⊥ /L is split of dimension 2, so contains exactly two isotropic lines. These give the two isotropic planes V which contain L. Q which is the image of the Segre embedding P1 ×P1 → P3 . Since A2 ⊂ P1 ×P1 as Zariskidense subset, X is rational. To see that X P2 one can use Pic(P1 × P1 ) = Z2 Z = Pic(P2 ). 5. (DG) Let Ω be the 2form on R3 − {0} defined by Ω=
1 (x dy ∧ dz + y dz ∧ dx + z dx ∧ dy). x2 + y 2 + z 2
(a) Prove that Ω is closed.
(b) Let f : R3 −{0} → S 2 be the map which sends (x, y, z) to ( x2 +y12 +z 2 )1/2 (x, y, z). Show that Ω is the pullback via f of a 2form on S 2 . (c) Prove that Ω is not exact. Solution: Introduce spherical coordinates (r, θ, φ) by writing x = r sin(θ) cos(φ), y = r sin(θ) sin(φ) and z = r cos(θ). Written with these coordinates, Ω = sin2 (θ)dθdφ dΩ = 2 sin(θ) cos(θ)dθdθdφ + sin2 (θ)(ddθdφ − dθddφ) This is zero because d2 = 0 and because the wedge of a oneform with itself is zero. The map in these coordinates sends (r, θ, φ) to the point on (1, θ, φ). A differential form Θ is the pullback of a form on S 2 via f if and only if both Θ and dΘ annihilate the vector fields in the kernel of the differential of f . Since these vector fields are proportional ∂/∂r in this case, both of the conditions are obeyed by Ω. If Ω were exact, then its integral over S 2 would be 0, but this integral is equal to 4π. 6. (RA) Consider the linear ODE f 00 +P f 0 +Q f = 0 on the interval (a, b) ⊂ R, with P, Q denoting C ∞ real valued functions on (a, b). Recall the definition of the Wronskian W (f1 , f2 ) = f1 f20 − f10 f2 associated to any two solutions f1 , f2 of this differential equation. (a) Show that W (f1 , f2 ) either vanishes identically or is everywhere nonzero, depending on whether the two solutions f1 , f2 are linearly dependent or not. (b) Now suppose that f1 , f2 are linearly independent, real valued solutions. Show that they have at most first order zeroes, and that the zeroes occur in an alternating fashion: between any two zeroes of one of the solutions there must be a zero of the other solution. Solution: W 0 (f1 , f2 ) = f10 f20 + f1 f200 − f100 f2 − f10 f20 = = f2 (P f10 + Qf1 ) − f1 (P f20 + Qf2 ) = −P W (f1 , f2 ) , which implies W (f1 , f2 ) = c e−P . In particular, W (f1 , f2 ) either vanishes identically or not at all. The Wronskian vanishes at some x0 ∈ (a, b) if and only if the initial conditions for (f10 , f1 ) and (f20 , f2 ) are proportional at x0 , which is the case if and only if the global solutions are proportional. This implies (a). Next suppose that f1 , f2 are real valued, linearly independent solutions. Since W (f1 , f2 ) never vanishes, neither solution can have a double zero; moreover, if f1 (x0 ) = 0 at some x0 then f2 (x0 ) 6= 0, and vice versa. Finally suppose that f1 (x0 ) = 0, f1 (x1 ) = 0, with x0 < x1 and f1 (x) 6= 0 for x ∈ (x0 , x1 ). Since the zeroes are first order, the derivatives of f1 at the
two points must have opposite signs. Since the Wronskian has the same sign globally, f2 cannot have the same sign at the two points. It follows that f2 vanishes somewhere between x0 and x1 . Similarly, between any two zeros of f2 there must be a zero of f1 . That is the assertion (b).
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 3, 2015 (Day 3) 1. (DG) Consider the graph S of the function F (x, y) = cosh(x) cos(y) in R3 and let Φ : R2 → S ⊂ R 3 be its parametrization: Φ(x, y) = (x, y, cosh(x) cos(y)). (a) Write down the metric on R2 that is defined by the rule that the inner product of two vectors v and w at the point (x, y) is equal to the inner product of Φ∗ (v) and Φ∗ (w) at the point Φ(x, y) in R3 . (b) Define the Gaussian curvature of a general surface embedded in R3 . (c) Compute the Gaussian curvature of the surface S at the point (0, 0, 1). Solution: The pushforward via Φ∗ of the vectors ∂/∂x and ∂/∂y at a given point (x, y) are the vectors in R3 at (x, y, F (x, y)) given by Φ∗ (∂/∂x) = (1, 0, Fx ) and Φ∗ (∂/d∂y) = (0, 1, Fy ). The metric is g = (1 + Fx2 )dx ⊗ dx + Fx Fy (dx ⊗ dy + dy ⊗ dx) + (1 + Fy2 )dy ⊗ dy. In this problem, Fx = sinh(x) cos(y) and Fy = − cosh(x) sin(y). The Gauss curvature is the determinant of the second fundamental form as computed using an orthonormal frame for the metric whereby the inner product of any two tangent vectors is their R3 inner product. The second fundamental form is defined as follows: Let n denote a unit length normal to the surface and let (e1 , e2 ) denote an orthonormal frame at a given point. The second fundamental form has components (mab ) defined by mab = hea , ∇eb (n)i It is also defined by writing the Riemann curvature tensor R for this metric using an orthornormal frame (e1 , e2 ) for T ∗ S as R = κ(e1 ∧e2 )⊗(e1 ∧e2 ). In the case of the surface S, the normal vector is n = (−Fx , −Fy , 1)/(1 + Fx2 + Fy2 )1/2 . At the point (0, 0, 1) in S, the vectors d/dx and d/dy are orthonormal. A computation then finds that the Gauss curvature is −1. 2. (RA) ∈ C(R/Z) be a continuous Cvalued function on R/Z and let P∞ Let f (x) 2πinx be its Fourier series. a e n n=−∞ (a) Show that f is C ∞ if and only if an  = O(n−k ) for all k ∈ N. (b) Prove that a sequence of functions { fn }n≥1 in C ∞ (R/Z) converges in the C ∞ topology (uniform convergence of functions and their derivatives (k) of all orders) if and only if the sequences of kth derivatives { fn }n≥1 , for all k ≥ 0, converge in the L2 norm on R/Z.
Solution A simple integration by parts argument shows that f ∈ C 1 (R/Z) implies X∞ n an e2πinx . f 0 (x) = 2πi n=−∞
Hence for all k ∈ N and f ∈
C ∞ (R/Z),
f (k) (x) = (2πi)k
X∞ n=−∞
nk an e2πinx
is the Fourier series of a continuous, hence L2 function, with squared L2 norm X∞ n2k an 2 < ∞ . kf (k) k2L2 = (2π)2k n=−∞
It follows that for fixed k, nk an  is bounded. The topology of C ∞ (R/Z) is defined by the family of norms f 7→ kf (k) ksup , and according to (a), also by the family of seminorms f 7→ kf (k) kL2 , because X∞ nk an  kf (k) kL2 ≤ kf (k) ksup ≤ (2π)k n=−∞ X 1 P = (2π)k nk+1 an  n−1 ≤ 2π ( n6=0 n−2 )1/2 kf (k+1) kL2 . n6=0
⊗2 3. (AG) Let C be a smooth projective curve over C and ωC the square of its canonical sheaf. ⊗2 )? (a) What is the dimension of the space of sections Γ(C, ωC ⊗2 ) is a section with simple zeros. (b) Suppose g(C) ≥ 2 and s ∈ Γ(C, ωC 2 Compute the genus of Σ = {x = s} in the total space of the line bundle √ ωC , i.e. the curve defined by the 2valued 1form s. ⊗2 and g = g(C), then deg(L) = 4g − 4. For g = 0: Solution: Write L = ωC 0 h (L) = 0 since L is negative, for g = 1: h0 (L) = 1 since L is trivial, and for g ≥ 2: h0 (L) = 3g − 3 by Riemann–Roch.
For the second part note that the projection T ∗ C → C gives a natural 2:1 covering Σ → C which is ramified at the 4g − 4 zeros of s. The Riemann– Hurwitz formula gives χ(Σ) = 2χ(C) − (4g − 4), thus g(Σ) = 4g − 3. 4. (AT) Show (using the theory of covering spaces) that every subgroup of a free group is free. W Solution: For a set I of generators we let X = I S 1 , then F = π1 (X) is the free group on I. Let G ⊂ F be a subgroup, then there is a covering p : Y → X with p∗ (π1 (Y )) = G and p∗ is injective. Note that Y has the stucture of a connected 1dimensional CW complex and is thus homotopy equivalent to a wedge of S 1 ’s by contracting a maximal subtree. It follows that G ∼ = π1 (Y ) is free.
5. (CA) (a) Define Euler’s Gamma function Γ(z) in the half plane Re(z) > 0 and show that it is holomorphic in this half plane. (b) Show that Γ(z) has a meromorphic continuation to the entire complex plane. (c) Where are the poles of Γ(z)? (d) Show that these poles are all simple and determine the residue at each pole. Solution:
Z
∞
Γ(z) =
tz e−t dt/t
0
The identity Γ(z + 1) = zΓ(z) then follows from integration by parts. Rewriting this identity as Γ(z) = Γ(z + 1)/z at z = 0 with residue 1. Using this identity again, we can extend to the half plane Re(z) > −2 with a simple pole at z = −1. Continuing in this manner, we get a meromorphic continuation to the entire plane with simple poles at the negative integers. The residue at z = −n is (−1)n /n!. 6. (A) Let G be a finite group, and ρ : G → GLn (C) a linear representation. Then for each integer i ≥ 0 there is a representation ∧i ρ of G on the exterior power ∧i (Cn ). Let Wi be the subspace (∧i (Cn ))G of ∧i (Cn ) fixed under this action of G. Prove that dim Wi is the T i coefficient of the polynomial 1 X det(1n + T ρ(g)) G g∈G
where 1n is the n × n identity matrix. Solution: It is a standard consequence of Schur’s lemma that if (V, %) is any finitedimensional representation of G then
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday January 20, 2015 (Day 1) 1. (AG) Let C ⊂ P2 be a smooth plane curve of degree d. (a) Let KC be the canonical bundle of C. For what integer n is it the case that KC ∼ = OC (n)? (b) Prove that if d ≥ 4 then C is not hyperelliptic. (c) Prove that if d ≥ 5 then C is not trigonal (that is, expressible as a 3sheeted cover of P1 ). 2. (A) Let S4 be the group of automorphisms of a 4element set. Give the character table for S4 and explain how you arrived at it. 3. (DG) Let M = {(x, y, z) ∈ R3  x2 − y 2 − z 3 − z = 0}. (a) Prove that M is a smooth surface in R3 . (b) For what values of c ∈ R does the plane z = c intersect M transversely? 4. (RA) Define the Banach space L to be the completion of the space of continuous functions on the interval [−1, 1] ⊂ R using the norm Z 1 f  = f (t)dt. −1
Suppose that f ∈ L and t ∈ [−1, 1]. For h > 0, let Ih be the set of points in [−1, 1] with distance h or less from t. Prove that Z lim f (t)dt = 0 h→0 t∈Ih
5. (AT) What are the homology groups of the 5manifold RP2 × RP3 , (a) with coefficients in Z? (b) with coefficients in Z/2? (c) with coefficients in Z/3? 6. (CA) Let Ω be an open subset of the Euclidean plane R2 . A map f : Ω → R2 is said to be conformal at p ∈ Ω if its differential dfp preserves the angle between any two tangent vectors at p. Now view R2 as C and a map f : Ω → R2 as a Cvalued function on Ω.
(a) Supposing that f is a holomorphic function on Ω, prove that f is conformal where its differential is nonzero. (b) Suppose that f is a nonconstant holomorphic function on Ω, and p ∈ Ω is a point where dfp = 0. Let L1 and L2 denote distinct lines through p. Prove that the angle at f (p) between f (L1 ) and f (L2 ) is n times that between L1 and L2 , with n being an integer greater than 1.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday January 21, 2015 (Day 2) 1. (AT) Let X ⊂ R3 be the union of the unit sphere S 2 = {(x, y, z)  x2 +y 2 +z 2 = 1} and the line segment I = {(x, 0, 0)  −1 ≤ x ≤ 1}. (a) What are the homology groups of X? (b) What are the homotopy groups π1 (X) and π2 (X)? 2. (A) Let f (t) = t4 + bt2 + c ∈ Z[t]. (a) If E is the splitting field for f over Q, show that Gal(E/Q) is isomorphic to a subgroup of the dihedral group D8 . (b) Given an example of b and c for which f is irreducible, and for which the Galois group is isomorphic to Z/2Z × Z/2Z. Justify. (c) Give an example of b and c for which f is irreducible, and for which the Galois group is isomorphic to Z/4Z. Justify. (d) Give an example of b and c for which f is irreducible, and for which the Galois group is isomorphic to D8 . 3. (CA) Let a ∈ (0, 1). By using a contour integral, compute Z 2π dx . 1 − 2a cos x + a2 0 4. (AG) Let K be an algebraically closed field of characteristic 0 and let Q ⊂ Pn be a smooth quadric hypersurface over K. (a) Show that Q is rational by exhibiting a birational map π : Q → Pn−1 . (b) How does the map π factor into blowups and blowdowns? 5. (DG) Let S = {(x, y, z) ∈ R3  x2 + y 2 + z 2 = 1} be the unit sphere centered at the origin in R3 . (a) Prove that the vector field v = yz
∂ ∂ ∂ + zx − 2xy ∂x ∂y ∂z
on R3 is tangent to S at all points of S, and thus defines a section of the tangent bundle T S.
(b) Let g be the metric on S induced from the euclidean metric on R3 , and let ∇ be the associated, metric compatible, torsion free covariant derivative. The tensor ∇v is a section of T S⊗T S ∗ . Write ∇v at the point (0, 0, 1) p ∈ S using the coordinates (x1 , x2 ) given by the map (x1 , x2 ) 7→ (x1 , x2 , 1 − x21 − x22 ) from the unit disc x21 + x22 < 1 to S. 6. (RA) Let L be a positive real number. (a) Compute the Fourier expansion of the function x on the interval [−L, L] ⊂ R. (b) Prove that the Fourier transform does not converge to x pointwise on the closed interval [−L, L].
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday January 22, 2015 (Day 3) 1. (DG) The helicoid is the parametrized surface given by φ : R2 → R3 : (u, v) → (v cos u, v sin u, au) where a is a real constant. Compute its induced metric. 2. (RA) A real valued function defined on an interval (a, b) ⊂ R is said to be convex if f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y) whenever x, y ∈ (a, b) and t ∈ (0, 1). (a) Give an example of a nonconstant, nonlinear convex function. (b) Prove that if f is a nonconstant convex function on (a, b) ∈ R, then the set of local minima of f is a connected set where f is constant.
3. (AG) Let K be an algebraically closed field of characteristic 0, and let Pn be the projective space of homogeneous polynomials of degree n in two variables over K. Let X ⊂ Pn be the locus of nth powers of linear forms, and let Y ⊂ Pn be the locus of polynomials with a multiple root (that is, a repeated factor). (a) Show that X and Y ⊂ Pn are closed subvarieties. (b) What is the degree of X? (c) What is the degree of Y ? 4. (AT) Let X be a compact, connected and locally simply connected Hausdorff ˜ → X be its universal covering space. Prove that X ˜ is space, and let p : X compact if and only if the fundamental group π1 (X) is finite. 5. (CA) Prove that if f and g are entire holomorphic functions and f  ≤ g everywhere, then f = α · g for some complex number α. 6. (A) Consider the rings R = Z[x]/(x2 + 1)
and
S = Z[x]/(x2 + 5).
(a) Show that R is a principal ideal domain. (b) Show that S is not a principal ideal domain, by exhibiting a nonprincipal ideal.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday January 20, 2015 (Day 1)
1. (AG) Let C ⊂ P2 be a smooth plane curve of degree d. (a) Let KC be the canonical bundle of C. For what integer n is it the case that KC ∼ = OC (n)? (b) Prove that if d ≥ 4 then C is not hyperelliptic. (c) Prove that if d ≥ 5 then C is not trigonal (that is, expressible as a 3sheeted cover of P1 ). Solution: By the adjunction formula, the canonical divisor class is KC = OC (d − 3), that is, plane curves of degree d − 3 cut out canonical divisors on C. It follows that if d ≥ 4 then any two points p, q ∈ C impose independent conditions on the canonical series KC ; that is, h0 (KC (−p − q)) = g − 2, so by RiemannRoch h0 (OC (p + q)) = 1, i.e., C is not hyperelliptic. Similarly, if d ≥ 5 then any three points p, q, r ∈ C impose independent conditions on the canonical series KC ; by RiemannRoch it follows that h0 (OC (p + q + r)) = 1 so C is not trigonal. 2. (A) Let S4 be the group of automorphisms of a 4element set. Give the character table for S4 and explain how you arrived at it. Solution: To start with, there are five conjugacy classes in S4 : (1), (12), (123), (1234) and (12)(34). The characters of the trivial and alternating representations U and U 0 are clear. The standard representation of S4 on C4 splits as a direct sum of the trivial and a threedimensional representation V , whose character is simply the character of C4 minus one; we see that it’s irreducible because the norm of its character is 1. We get another irreducible as V 0 = V ⊗ U 0 ; its character is χV 0 = χV χU 0 . The final irreducible representation W (and its character) can be found by pulling back the standard representation of S3 via the quotient map S4 → S3 (or by the orthogonality relations). Altogether, we have
conjugacy class
e
(12)
(123)
(1234)
(12)(34)
number of elements
1
6
8
6
3
U
1
1
1
1
1
U0
1
−1
1
−1
1
V
3
1
0
−1
−1
V0
3
−1
0
1
−1
W
2
0
1
0
2
3. (DG) Let M = {(x, y, z) ∈ R3  x2 − y 2 − z 3 − z = 0}. (a) Prove that M is a smooth surface in R3 . (b) For what values of c ∈ R does the plane z = c intersect M transversely? Solution: See attached. 4. Define the Banach space L to be the completion of the space of continuous functions on the interval [−1, 1] ⊂ R using the norm Z
1
f  =
f (t)dt. −1
Suppose that f ∈ L and t ∈ [−1, 1]. For h > 0, let Ih be the set of points in [−1, 1] with distance h or less from t. Prove that Z lim f (t)dt = 0 h→0 t∈Ih
Solution: See attached. 5. (AT) What are the homology groups of the 5manifold RP2 × RP3 , (a) with coefficients in Z? (b) with coefficients in Z/2? (c) with coefficients in Z/3? Solution: RP2 and RP3 have cell complexes with sequences 0→Z→Z→Z→0
and
0→Z→Z→Z→Z→0
where the maps are alternately 0 and multiplication by 2; from this the homology groups of RP2 and RP3 can be calculated as Z, Z/2, 0 and Z, Z/2, 0, Z respectively. The rest is just Kunneth; the answers are (a): Z, (Z/2)2 , (Z/2)2 , Z, Z/2, 0; (b): Z/2, (Z/2)2 , (Z/2)3 , (Z/2)3 , (Z/2)2 , Z/2, (c): Z/3, 0, 0, Z/3, 0, 0 6. Let Ω be an open subset of the Euclidean plane R2 A map f : Ω → R2 is said to be conformal at p ∈ Ω if its differential dfp preserves the angle between any two tangent vectors at p. Now view R2 as C and a map f : Ω → R2 as a Cvalued function on Ω. (a) Supposing that f is a holomorphic function on Ω, prove that f is conformal where its differential is nonzero. (b) Suppose that f is a nonconstant holomorphic function on Ω, and p ∈ Ω is a point where dfp = 0. Let L1 and L2 denote distinct lines through p. Prove that the angle at f (p) between f (L1 ) and f (L2 ) is n times that between L1 and L2 , with n being an integer greater than 1. Solution: See attached.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday January 21, 2015 (Day 2) 1. (AT) Let X ⊂ R3 be the union of the unit sphere S 2 = {(x, y, z)  x2 +y 2 +z 2 = 1} and the line segment I = {(x, 0, 0)  −1 ≤ x ≤ 1}. (a) What are the homology groups of X? (b) What are the homotopy groups π1 (X) and π2 (X)? Solution: Under the attaching map I ,→ X, the boundary ϕ(I) is homologous to 0, so attaching I simply adds one new, nontorsion generator to H 1 ; thus H0 (X) = H 1 (X) = H 2 (X) = Z, and all other homology groups are 0. Similarly, π1 (X) = Z. For π2 (X), note that the universal cover of X is a string of spheres attached in a sequence by line segments; π2 (X) is thus the free abelian group on countably many generators. 2. (A) Let f (t) = t4 + bt2 + c ∈ Z[t]. (a) If E is the splitting field for f over Q, show that Gal(E/Q) is isomorphic to a subgroup of the dihedral group D8 . (b) Given an example of b and c for which f is irreducible, and for which the Galois group is isomorphic to Z/2Z × Z/2Z. Justify. (c) Give an example of b and c for which f is irreducible, and for which the Galois group is isomorphic to Z/4Z. Justify. (d) Give an example of b and c for which f is irreducible, and for which the Galois group is isomorphic to D8 . Solution: (a) Obviously if α is a root of f , so is −α. So let ±α1 , ±α2 be the four distinct roots of f in E. If φ is an element of the Galois group, it must permute the roots of f —moreover, φ is determined completely by its action on α1 and α2 . Also by definition of automorphism, note that φ(α1 ) cannot be a rational multiple of φ(α2 ), while φ(−α1 ) = −φ(α1 ). Hence any field automorphism must necessarily give rise to a symmetry of the following square: α1
α2
−α2
−α1
This gives the injection of Gal into D8 . (b) An obvious strategy is to find a quadratic extension of a quadratic extension, then find an element minimal is degree 4. For √ whose √ √ polynomial √ instance, the element α = 2 + 3 in E = Q( 2, 3) has a degree 4 minimal polynomial which multiplying by conjugates: √ we√can construct by repeatedly √ √ Begin√with t − 2 − 3, the multiply by (t − 2) + 3, then multiply this by (t2 + 2)2 − 3. For this choice of α, we have f (t) = t4 − 10t2 + 1. (c) Taking b = 0 and c = 1, we see that the splitting field is isomorphic to the subfield of the complex numbers generated by adjoining to Q the number α = eπi/4 . This is a degree 4 field over Q. Since we have a splitting field in characteristic zero, the Galois group has order 4. We see that the field automorphism sending α 7→ α3 has order 4, hence the Galois group is cyclic. (d) Take b = 0 and c = 2. Clearly we have roots α1 = 21/4 and α2 = i21/4 , which together lie in an extension of at least degree 8 over Q. By part (a), the Galois group must be D8 itself. 3. (CA) Let a ∈ (0, 1). By using a contour integral, compute Z 2π dx . 1 − 2a cos x + a2 0 Solution (HT): By the periodicity of cos, it suffices to compute the integral from −π to π. We note that there is a pole for the function f (z) =
1 1 − 2a cos z + a2
2
at z0 = i cosh−1 1+a 2a . Let Rt be the rectangle bordered by the lines x = ±π and y = 0, y = t. As t → ∞, the contribution from the line y = t goes to zero. On the other hand, for all values of t, the contribution to the integral from x = ±π cancel each other out. Thus the integral along the bottom edge of the rectangle (which is what we seek) is equal to 2πi times the residue of f (z) at z0 . Near z0 , we have that 1 − 2a cos z + a2 = (z − z0 )2ai sinh iz0 + . . . so we conclude the integral is given by 2πi . 2ai sinh z0 This simplifies to 2π . 1 − a2
Alternate solution (CH): Write the integral as a contour integral on the unit circle: set dx = −idz z , so that Z 0
2π
1 dx = −i 1 − 2a cos x + a2
Z z=1
1 dz. z(1 + a2 ) − az 2 − a
Factor the denominator to find the poles of the latter integrand; one is inside the unit circle and one outside. Calculate the residue at the former pole and use Cauchy’s theorem to evaluate the integral. 4. (AG) Let K be an algebraically closed field of characteristic 0 and let Q ⊂ Pn be a smooth quadric hypersurface over K. (a) Show that Q is rational by exhibiting a birational map π : Q → Pn−1 . (b) How does the map π factor into blowups and blowdowns? Solution: For the first part, we choose any point p ∈ Q and take π to be the projection from p. Since Q has degree 2, a general line in Pn through p will meet Q in one other point, so that the map π : Q → Pn−1 has degree 1; that is, it is a birational map. This map blows up the point p, and then blows down the union of the lines on Q through p. In the other direction, starting with Pn−1 we blow up the intersection Z = S ∩ H of a quadric hypersurface S ⊂ Pn−1 and a hyperplane H ⊂ Pn−1 , and then blow down the proper transform of H. 5. DG Let S = {(x, y, z) ∈ R3  x2 + y 2 + z 2 = 1} be the unit sphere centered at the origin in R3 . (a) Prove that the vector field v = yz
∂ ∂ ∂ + zx − 2xy ∂x ∂y ∂z
on R3 is tangent to S at all points of S, and thus defines a section of the tangent bundle T S. (b) Let g be the metric on S induced from the euclidean metric on R3 , and let ∇ be the associated, metric compatible, torsion free covariant derivative. The tensor ∇v is a section of T S⊗T S ∗ . Write ∇v at the point (0, 0, 1) p ∈ S using the coordinates (x1 , x2 ) given by the map (x1 , x2 ) 7→ (x1 , x2 , 1 − x21 − x22 ) from the unit disc x21 + x22 < 1 to S. Solution: See attached 6. (RA) Let L be a positive real number.
(a) Compute the Fourier expansion of the function x on the interval [−L, L] ⊂ R. (b) Prove that the Fourier transform does not converge to x pointwise on the closed interval [−L, L]. Solution: See attached. One note: the second part follows immediately from the observation that whatever the Fourier expansion converges to at −L must be the same as what it converges to at L.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday January 22, 2015 (Day 3)
1. (DG) The helicoid is the parametrized surface given by φ : R2 → R3 : (u, v) → (v cos u, v sin u, au) where a is a real constant. Compute its induced metric. Solution. Compute du + dv ⊗ dv.
∂ ∂u
and
∂ ∂v
and deduce that the metric is g = (v 2 + a2 )du ⊗
2. (RA) A real valued function defined on an interval (a, b) ⊂ R is said to be convex if f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y) whenever x, y ∈ (a, b) and t ∈ (0, 1). (a) Give an example of a nonconstant, nonlinear convex function. (b) Prove that if f is a nonconstant convex function on (a, b) ∈ R, then the set of local minima of f is a connected set where f is constant. Solution: See attached 3. (AG) Let K be an algebraically closed field of characteristic 0, and let Pn be the projective space of homogeneous polynomials of degree n in two variables over K. Let X ⊂ Pn be the locus of nth powers of linear forms, and let Y ⊂ Pn be the locus of polynomials with a multiple root (that is, a repeated factor). (a) Show that X and Y ⊂ Pn are closed subvarieties. (b) What is the degree of X? (c) What is the degree of Y ? Solution: First, X is the image of the map P1 → Pn sending [a, b] ∈ P1 to (ax + by)n ∈ Pn . This is projectively equivalent (in characteristic 0!) to the degree n Veronese map, whose image is a closed curve of degree n. Second, Y is the zero locus of the discriminant, which is a polynomial of degree 2n − 2 in the coefficients of a polynomial of degree n (this number can be deduced from the RiemannHurwitz formula, which says that a degree n map from P1 to P1 has 2n − 2 branch points; that is, a general line in Pn meets Y in 2n − 2 points).
4. (AT) Let X be a compact, connected and locally simply connected Hausdorff ˜ → X be its universal covering space. Prove that X ˜ is space, and let p : X compact if and only if the fundamental group π1 (X) is finite. Solution: See attached 5. (CA) Prove that if f and g are entire holomorphic functions and f  ≤ g everywhere, then f = α · g for some complex number α. Solution: The conclusion trivially holds in the case g = 0; from now on, assume that g is not the zero function. The identity theorem implies that the zeros of g are isolated, so h := f /g is meromorphic. The function h is bounded by hypothesis, so Riemann’s theorem implies that h can be extended to an entire bounded function. Liouville’s theorem implies that h is constant, which implies the conclusion. 6. (A) Consider the rings R = Z[x]/(x2 + 1)
and
S = Z[x]/(x2 + 5).
(a) Show that R is a principal ideal domain. (b) Show that S is not a principal ideal domain, by exhibiting a nonprincipal ideal. Solution: For the first, the fact that R is a principal ideal domain follows from the fact that it’s a Euclidean domain, with size function z2 : for any a, b ∈ R we can write b = ma + r with r < a; carrying this out repeatedly shows that the ideal generated by two elements of R can be generated by one. For the second, the ideal (2, 1 + x) ⊂ S is not principal.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 2, 2014 (Day 1) 1. (AG) For any 0 < k < m ≤ n ∈ Z, let M ∼ = Pmn−1 be the space of nonzero m × n matrices mod scalars, and let Mk ⊂ M be the subset of matrices of rank k or less. (a) (b) (c) (d)
Show that Mk is closed in M (in the Zariski topology). Show that Mk is irreducible. What is the dimension of Mk ? What is the degree of M1 ?
Solution: For the first, Mk is the zero locus of the (k + 1) × (k + 1) minors, which are homogeneous polynomials of degree k + 1 on M ∼ = Pmn−1 . For the second and third, we introduce the incidence correspondence Φ = {(Λ, A) ∈ G(n − k, n)  Λ ⊂ ker(A)}. Since Φ is fibered over G(n − k, n) with fibers Pkm−1 , it is irreducible of dimension k(n−k)+km−1 = mn−1−(m−k)(n−k); since it is generically onetoone over Mk , we conclude that Mk is likewise irreducible of that dimension. Finally, M1 is the Segre variety Pm−1 × Pn−1 ⊂ Pmm−1 , which has degree m+n−2 m−1 . 2. (A) Let S3 be the group of automorphisms of a 3element set. (a) Classify the conjugacy classes of S3 . (b) Classify the irreducible representations of S3 . (c) Write the character table for S3 . Solution: (a) Conjugacy classes of symmetric groups are given by the types of cycles one can write on the set of n elements. For n = 3, we have shapes given by (1), (12), (123) so we have three conjugacy classes. (b) By the orthogonality relations, the number of irreps are equal to the number of conjugacy classes. On the other hand, we can produce three irreps: The trivial, the sign, and the geometric representation corresponding to S3 ∼ = D6 ; i.e., the symmetries 2 of an equilateral triangle embedded in R . (c) Compute the traces of each conjugacy class. End up with the table (1) (12) (123) triv 1 1 1 sign 1 1 1 geom 2 0 1
3. (DG) Let x, y, z be the standard coordinates on R3 . Consider the unit sphere S2 ⊂ R3 . 1. Compute the critical points of the function xS2 . Show that they are isolated and nondegenerate. 2. Equip S2 with the standard metric induced from R3 . Compute the gradient vector field of xS2 . Compute the integral curves of this vector field. Solution: 1. The unit sphere is defined by x2 + y 2 + z 2 = 1. Regarding y, z as independent variables and x as dependent variable, we have the equations 2x∂y x + 2y = 0 and 2x∂z x + 2z = 0. For a critical point, ∂y x = ∂z x = 0, and hence y = z = 0. Then x = ±1. Hence the critical points are (1, 0, 0) and (−1, 0, 0). They are isolated in S2 . Differentiating once more and put x = ±1 and y = z = 0 for computing the Hessians, we get ∂y ∂z x = 0 and ∂y2 x = ∂z2 x = ∓1. Hence the Hessians at the critical points are nondegenerate. 2. The gradient vector field is V (x, y, z) = (1, 0, 0) − h(1, 0, 0), (x, y, z)i(x, y, z) = (1 − x2 , −xy, −xz). The integral curves are great arcs connecting (−1, 0, 0) to (1, 0, 0). To get their parametrized forms, we need to solve the equation x0 = 1 − x2 , y 0 = −xy, z 0 = −xz with the boundary condition that x(t → −∞) = −1, x(t → ∞) = 1, y(t → −∞) = y(t → ∞) = z(t → −∞) = z(t → ∞) = 0. The first equation gives e2λt − 1 x = 2λt e +1 where λ can be taken to be any positive real constant (which just corresponds to scale of time). We fix λ = 1. Subsituting to the second and third equations, we get y=
C1 et C2 e t , z = . 1 + e2t 1 + e2t
Since x2 + y 2 + z 2 = 1, we get C12 + C22 = 2. Hence the solutions are 2t e − 1 2et cos θ 2et sin θ (x, y, z) = , , e2t + 1 1 + e2t 1 + e2t where θ is a real constant.
4. (RA) Find a solution for the heat equation ∂ ∂2 u(x, t) − 2 u(x, t) = 0, ∂t ∂x
(t > 0,
0 < x < 1),
with the initial condition u(x, 0) = A where A is a constant and the boundary conditions u(0, t) = u(1, t) = 0, t > 0. Solution: In view of the boundary conditions (Dirichlet), using linearity and separation of variables, we can write a solution of the form u(x, t) =
∞ X
2 π2 t
Bn sin(nπx)e−n
.
n=1
The coefficients Bn can be computed using a Fourier decomposition of the function Rf (x) = u(x, 0) given by the initial condition. A quick calculation 1 (Bn = 2 0 sin(nπx)f (x)dx) gives: B2n = 0 B2n−1 =
4A , (2n − 1)π
n = 1, 2, 3, · · ·
5. (AT) (a) Show that a continuous map f : X → RPn factors through S n → RPn if and only if the induced map f ∗ : H 1 (RPn ; Z/2) → H 1 (X, Z/2) is zero. (b) Show that a continuous map f : X → CPn factors through S 2n+1 → CPn if and only if the induced map f ∗ : H 2 (CPn ; Z) → H 2 (X, Z) is zero. 6. (CA) Let f be a meromorphic function on a contractible region U ⊂ C, and let γ be a simple closed curve inside that region. Recall that the argument principle for a meromorphic function says that the integral Z 0 1 f 2πi γ f is equal to the number of zeroes minus the number of poles of f inside γ. (a) Prove Rouch´e’s Theorem. That is, assume (1) f and g are holomorphic in U , (2) γ is a simple, smooth, closed curve in U , and (3) f  > g on γ. Then the number of zeroes of f + g inside γ is equal to the number of zeroes of f inside γ. You may assume the Argument Principle. (b) Show that for any n, the roots of the polynomial n X i=0
all have absolute value less than 2.
zi
Solution: (a) Apply the argument principle to f + g. Take note that the derivative of log(1 + fg ) shows up. (b) Let f = z n and g be the summation of z i from i = 0 to n − 1. Apply Rouch´e’s theorem, noting that z n has the same number of roots as f + g (since they are polynomials of equal degree).
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 3, 2014 (Day 2)
1. (AT) (a) Let X and Y be compact, oriented manifolds of the same dimension n. Define the degree of a continuous map f : X → Y . (b) What are all possible degrees of continuous maps f : CP3 → CP3 ? ∼ Z → H n (X, Z) ∼ Solution: For the first part, the induced map f ∗ : H n (Y, Z) = = Z is multiplication by some integer d; this is the degree of f . For the second part, note that H ∗ (CP3 , Z) ∼ = Z[ζ]/(ζ 4 ) and that f ∗ is a ring homomorphism. If f ∗ (ζ) = mζ, then f ∗ (ζ 3 ) = m3 ζ 3 and so the degree must be a cube. To see that all cubes occur, just consider the map [X, Y, Z, W ] 7→ [X m , Y m , Z m , W m ] for positive d = m3 ; take complex conjugates to exhibit maps with negative degrees.
2. (A) (a) Show that every finite extension of a finite field is simple (i.e., generated by attaching a single element). (b) Fix a prime p ≥ 2 and let Fp be the field of cardinality p. For any n ≥ 1, show that any two fields of degree n over Fp are isomorphic as fields. Solution: (a) If E/F is the field extension, then E × is cyclic. Taking a generator x, we see that E = F (x). (b) Any field extension of degree n is a n splitting field for the polynomial X p − X, hence is unique. 3. (CA) Fix two positive real numbers a, b > 0. Calculate the value of the integral Z ∞ cos(ax) − cos(bx) dx. x2 −∞ Solution: We compute the keyhole integral over a simple closed curve C = Cr ∪ CR ∪ [−R, −r] ∪ [r, R], where the closed intervals are on the yaxis of the complex plane. The curve Cr is a semicircle in the upper halfplane of radius 0 < r < R, oriented so as to agree with the positive orientation on the real axis. Likewise CR is on the upper halfplane. We integrate the function F (z) =
exp(iaz) − exp(ibz) . z2
R In the interior of C, F has no singularities, so C F = 0. Along Cr , we can use the power series expansion of exp to see that exp(iaz) − exp(ibz) 1 − 1 + iaz − ibz + (iaz)2 /2 − (ibz)2 /2 + . . . i(a − b) = = +h(z) 2 2 z z z for some holomorphic function h(z). So Z Z Z i(a − b) lim F (z)dz = lim dz + lim h(z)dz r→0 Cr r→0 Cr r→0 Cr z Z i(a − b) dz + lim h(r) − h(−r) = lim r→0 r→0 Cr z Z 0 i = lim (a − b) ireit dt + 0 it r→0 π re = (a − b)(−i2 )π = (a − b)π.
(1) (2) (3) (4) (5)
On the other hand, we utilize the following estimate as R → ∞: Since y > 0 and a > 0, eiaz = e−ay eix has a modulus less than 1. Likewise for eibz . This means that on CR , 2 exp(iaz) − exp(ibz)  ≤ 2.  2 z R R Hence the integral CR F (z)dz is bounded by 2π/R, which tends to zero as R → ∞. So we obtain that Z F (z)dz (6) 0 = lim r→0,R→∞ C Z Z Z +∞ = lim F dz + lim F dz + F dz (7) r→0 Cr
R→0 CR
Z
−∞
+∞
= π(a − b) +
F (z)dz.
(8)
−∞
Looking at the real part of this equality we arrive at the conclusion that Z +∞ π(b − a) = Real F (z)dz −∞
which is what we seek. 4. (AG) Let C ⊂ P2 be the smooth plane curve of degree d > 1 defined by the homogeneous polynomial F (X, Y, Z) = 0 (a) If p ∈ C, find the homogeneous linear equation of the tangent line Tp C ⊂ P2 to C at p. (b) Let P2∗ be the dual projective plane, whose points correspond to lines in P2 . Show that the Gauss map g : C → P2∗ sending each point p ∈ C to its tangent line Tp C ∈ P2∗ is a regular map.
(c) Let C ∗ ⊂ P2∗ be the dual curve of C; that is, the image of the Gauss map. Assuming that the Gauss map is birational onto its image, what is the degree of C ∗ ⊂ P2∗ ? Solution: For the first part, the tangent line Tp C is given by the equation ∂F ∂F ∂F (p) · X + (p) · Y + (p) · Z = 0. ∂X ∂Y ∂Z For the second, the Gauss map is given by ∂F ∂F ∂F g : p 7→ (p), (p), (p) . ∂X ∂Y ∂Z Since these have no common zeroes, the map is regular. For the third, since the partial derivatives of F are homogeneous of degree d − 1, the preimage of a general line in P2∗ —that is, the zero locus of a general linear combination— will consist of d(d − 1) points (since the partials have no common zeroes, by Bertini a general linear combination will have only simple zeroes); thus deg(C ∗ ) = d(d − 1). 5. (DG) Let U the be upper half plane U = {(x, y) ∈ R2 y > 0} and introduce the Poincar´e metric g = y −2 (dx ⊗ dx + dy ⊗ dy). Write the geodesic equations. Solution: A direct calculation gives x00 − y2 x0 y 0 = y 00 − y1 [(x0 )2 + 3(y 0 )2 ] = 0. 6. (RA) (a) Define what is meant by an equicontinuous sequence of functions on the closed interval [−1, 1] ⊂ R. (b) Prove the ArzelaAscoli theorem: that if {fn }n=1,2,... is a bounded, equicontinuous sequence of functions on [−1, 1], then there exists a continuous function f on [−1, 1] and an infinite subsequence Λ ⊂ {1, 2, . . . } such that ! lim
n∈Λ and n→∞
sup fn (t) − f (t)
=0
t∈[−1,1]
. Solution: First, a sequence {fn } of functions is equicontinuous if ∀ > 0 there exists a δ > 0 such that if t − t0  < δ then fn (t) − fn (t0 ) < for all n. For the second part, here is a fourstep proof: 1. We first show there exists a subsequence Λ ⊂ N such that ∀r ∈ Q∩[−1, 1], the sequence {fn (r)}n∈Λ converges. We do this by first ordering Q ∩ [−1, 1], choosing a subsequence Λ1 ⊂ N such that {fn (r1 )}n∈Λ1 converges
(we can do this because {fn } is bounded); then choosing a subsequence Λ2 ⊂ Λ1 of that such that {fn (r2 )}n∈Λ2 converges, and so on; we can do this such that if nk is the smallest integer in Λk then nk ∈ / Λk+1 . We take Λ = {n1 , n2 , n3 , . . . }; since all but finitely many elements of Λ are in Λk , it follows that {fn (rk )}n∈Λ converges. Denote the limit limn∈Λ fn (rk ) by f (rk ). 2. Second, we claim that the function f on Q ∩ [−1, 1] defined in the first part satisfies the condition that ∀ > 0 there exists a δ > 0 such that for r, r0 ∈ Q ∩ [−1, 1], r − r0  < δ =⇒ f (r) − f (r0 ) < . This follows from the “up, over and down” argument: we have f (r) − f (r0 ) ≤ f (r) − fn (r) + fn (r) − fn (r0 ) + fn (r0 ) − f (r0 ) and we can bound each term on the right by /3 (the middle term by equicontinuity). It follows that for any t ∈ [−1, 1] and any sequence {q1 , q2 , . . . } ⊂ Q ∩ [−1, 1] converging to t, the sequence f (qn ) is Cauchy; denote the limit by f (t). 3. We claim that the function defined in the second part is continuous. This is again an up, over and down argument: if t, t0 ∈ [−1, 1] and r, r0 ∈ Q ∩ [−1, 1], we have f (t) − f (t0 ) ≤ f (t) − f (r) + f (r) − f (r0 ) + f (r0 ) − f (t0 ) and again if we require t, t0 , r, r0 to all lie in a sufficiently small interval we can bound each term by /3. 4. We repeat the argument one more time. Choose N large, and consider the rational numbers r ∈ Q ∩ [−1, 1] with denominator N ; that is, {k/N }k=−N,−N +1,...,N . For any t ∈ [−1, 1], we choose r = k/N close to t and write fn (t) − f (t) ≤ fn (t) − fn (r) + fn (r) − f (r) + f (r) − f (t) and once more we can bound each term by /3 by choosing n sufficiently large and r sufficiently close to t.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 4, 2014 (Day 3)
1. (DG) The symplectic group Sp(2n, R) is defined as the subgroup of Gl(2n, R) that preserves the matrix 0 In Ω= −In 0 where In is the n × n identify matrix. That is, it is composed of elements of Gl(2n, R) that satisfy the relation M T ΩM = Ω. (a) Show that every symplectic matrix is invertible with inverse M −1 = Ω−1 M T Ω. (b) Show that the square of the determinant of a symplectic metric is 1. (In fact, the determinant of a symplectic matrix is always 1, but you don’t need to show this.) (c) Compute the dimension of the symplectic group. Solution: (a) Direct consequence from the definition since we can write (Ω−1 M T Ω)M = I2n (b) Take the determinant on both side of the defining equation M T ΩM = Ω and use the fact that det M T = det M . (c) Using the exponential map, describe the tangent space at the identity to be defined T my the matrices m such that m Ω + Ωm = 0 which can also be written as a b in terms of n × n blocks (a, b, c and d) and ΩmT Ω = m. Writing m = c d deduce the condition that these blocks have to satisfy (d = −aT , bT = b, cT = c) to have ΩmT Ω = m. It follows that the dimension is n(2n + 1). 2. (RA) Suppose that σ is a positive number and f is a nonnegative function on R such that Z Z Z f (x)dx = 1; xf (x)dx = 0 and x2 f (x)dx = σ 2 . R
R
R
Let P denote the probability measure on R with density function f . (a) Supposing that ρ is a positive number, give a nontrivial upper bound in terms of σ for the probability as measured by P of the subset [ρ, ∞).
(b) Given a positive integer N , let {X1 , . . . , XN } denote N independent random variables on R, each with the same probability measure P. Let SN be the random variable on RN given by SN =
N 1 X Xi N i=1
What are the mean and standard deviation of SN ? (c) Let {X1 , X2 , . . . , XN } be independent random variables on R, each with the same probability measure P, and let PN (x) denote the function on R given by the probability that N 1 X √ Xk < x. N k=1
Given x ∈ R, what is the limit as N → ∞ of the sequence {PN (x)}? R∞ Solution: For the first part, the probability assigned to the interval is ρ f (x)dx. An bound is derived by noting that the probability is no greater than R ∞ upper 2 x2 f (x)dx and this in turn is at most σρ2 This is Chebyshevs inequality. ρ ρ2 For the second part, the mean is 0 and the standard deviation is
√1 σ. N
Finally, the central limit theorem says that Z x 1 2 2 √ e−x /2σ . lim PN (x) = N →∞ −∞ σ 2π 3. (AG) Let X be the blowup of P2 at a point. (a) Show that the surfaces P2 , P1 × P1 and X are all birational. (b) Prove that no two of the surfaces P2 , P1 × P1 and X are isomorphic. Solution: For the first part, we can simply observe that all three surfaces contain the affine plane A2 as a Zariski open subset. For the second, there are many invariants that we can use to distinguish P2 from P1 × P1 : the topological Euler characteristic; the selfintersection of the canonical bundle, or the rank of the Picard group all work. To see that X is not isomorphic to either, note that X contains a curve of negative selfintersection (the exceptional divisor), while P2 and P1 × P1 do not. 4. (AT) Suppose that G is a finite group whose abelianization is trivial. Suppose also that G acts freely on S 3 . Compute the homology groups (with integer coefficients) of the orbit space M = S 3 /G. Solution: Note that M is a smooth manifold, and that π1 M = G. By Poincare’s theorem H1 S 3 /G = 0, as is H 1 (S 3 /G; A) = hom(π1 M, A) for
any abelian group A. This implies that M is orientable. It then follows from Poincare duality that H2 (M ; A) = 0 for any abelian group A and that H3 (M ; A) = A. 5. (CA) Recall that a function u : R2 → R is called harmonic if ∆u := ∂x2 u + ∂y2 u = 0. Prove the following statements using harmonic conjugates and standard complex analysis. (a) Show that the average value of a harmonic function along a circle is equal to the value of the harmonic function at the center of the circle. (b) Show that the maximum value of a harmonic function on a closed disk occurs only on the boundary, unless u is constant. Solution: Cauchy Integral Formula, and maximum principle. In detail: (a) Let v be the harmonic conjugate for u so u + iv = f is an analytic function on R2 . By Cauchy’s integral formula, Z 1 f (z) f (a) = dz 2πi γ z − a for any point a and any closed, simple curve surrounding a. Taking γ to be a circle of radius R centered at a, and parametrizing z(t) = Reit + a, we thus have Z 1 u + iv u(a) + iv(a) = iReit dt. 2πi γ Reit Equating real and imaginary parts, we obtain Z 1 udt. u(a) = 2πR γ (b) If u is harmonic, let v be a harmonic conjugate so f = u+iv is holomorphic. By the maximum modulus principle, ef  must obtain a maximum only along the boundary (unless f is constant). But ef  = eu  = eu , and since exp is strictly monotone and continuous, eu obtains a maximum if and only if u does. 6. (A) Let G be a finite group. (a) Let V be any Crepresentation of G. Show that V admits a Hermitian, Ginvariant inner product. (b) Let N be a C[G]module which is finitedimensional over C, and let M ⊂ N a submodule. Show that the inclusion splits. (c) Consider the action of S3 on C3 given by permuting the axes. Decompose C3 into irreducible S3 representations.
Solution: (a) Put an arbitrary inner product h, i on V , then define (v, w) :=
1 X hgv, gwi. G g∈G
(b) Take the orthogonal complement to M under a Ginvariant inner product. (c) Clearly the diagonal z1 = z2 = z3 is an invariant subspace. By using the methods above (or by writing an invariant linear equation) we deduce that an orthogonal complement is given by the plane z1 + z2 + z3 = 0. This twodimensional representation has no subrepresentations, so must be the unique 2dimensional irreducible representation.
Qualifying Exams I, 2014 Spring 1. (Algebra) Let k = Fq be a ﬁnite ﬁeld with q elements. Count the number of monic irreducible polynomials of degree 12 over k. 2. (Algebraic Geometry) (a) Show that the set of lines L ⊂ P3C may be identiﬁed with a quadric hypersurface in P5C . (b) Let L0 ⊂ P3C be a given line. Show that the set of lines not meeting L0 is isomorphic to the aﬃne space A4C . 3. (Complex Analysis) (a) Compute ∫ z 31 dz z 2 + 3)2 (¯ z 4 + 2)3 z=1 (2¯ Note that the integrand is not a meromorphic function. (b) Evaluate the integral
∫
∞
x=0
(
sin x x
)3 dx
by using the theory of residues. Justify carefully all the limiting processes in your computation. 4. (Algebraic Topology) Suppose that X is a ﬁnite connected CW complex such that π1 (X) is ﬁnite and nontrivial. Prove that the univer˜ of X cannot be contractible. (Hint: Lefschetz ﬁxed point sal covering X theorem.) 5. (Differential Geometry) Let P2 = (C3 − {0})/C× , which is called the complex projective plane. 1. Show that P2 is a complex manifold by writing down its local coordinate charts and transitions. 2. Deﬁne L ⊂ P2 × C3 to be the subset containing elements of the form ([x], λx), where x ∈ C3 − {0} and λ ∈ C. Show that L is the total space of a holomorphic line bundle over P2 by writing down its local trivializations and transitions. It is called the tautological line bundle. 3. Using the standard Hermitian metric on C3 or otherwise, construct a Hermitian metric on the tautological line bundle. Express the metric in terms of local trivializations. 1
6. (Real Analysis) (Schwartz’s Theorem on Perturbation of Surjective Maps by Compact Maps Between Hilbert Spaces). Let E, F be Hilbert spaces over C, S : E → F be a compact Clinear map, and T : E → F be a continuous surjective Clinear map. Prove that the cokernel of S+T : E → F is ﬁnitedimensional and the image of S + T : E → F is a closed subspace of F. Here the compactness of the Clinear map S : E → F means that for any sequence {xn }∞ n=1 in E with ∥xn ∥E ≤ 1 for all n ∈ N there exists some ∞ subsequence {xnk }∞ k=1 of {xn }n=1 such that S (xnk ) converges in F to some element of F as k → ∞. Hint: Verify ﬁrst that the conclusion is equivalent to the following equivalent statement for the adjoints T ∗ , S ∗ : F 7→ E for T, S. The kernel of T ∗ + S ∗ is ﬁnitedimensional and the image of T ∗ + S ∗ is closed. Then prove the equivalent statement.
2
Qualifying Exams II, 2014 Spring 1. (Algebra) Let A be a ﬁnite group of order n, and let V1 , · · · , Vk be its irreducible representations. (a) the dimensions of the vector spaces Vi satisfy the equality ∑k Show that 2 (dim V ) = n. i i=1 (b) What are the dimensions of the irreducible representations of the symmetric group S6 of six elements. 2. (Algebraic Geometry) Let C ⊂ P2 be a smooth plane curve of degree ≥ 3. (a) Show that C admits a regular map f : C → P1 of degree d − 1. (b) Show that C does not admit a regular map f : C → P1 of degree e with 0 < e < d − 1. 3. (Complex Analysis) Suppose that f is holomorphic in an open set containing the closed unit disk {z ≤ 1} in C, except for a pole at z0 on the unit circle {z = 1}. Show that if ∞ ∑
an z n
n=0
denote the power series expansion of f in the open unit disk {z < 1} , then an = z0 . n→∞ an+1 lim
4. (Algebraic Topology) Show that if n > 1, then every map from the real projective space RPn to the ntorus T n is nullhomotopic. 5. (Differential Geometry) Let S2 := {x2 + y 2 + z 2 = 1} ⊂ R3 be the unit sphere in the Euclidean space. Let C = {(r cos t, r sin t, h) : t ∈ R} be a circle in S2 , where r, h > 0 are constants with r2 + h2 = 1. 1. Compute the holonomy of the sphere S2 (with the standard induced metric) around the circle C.
3
2. By using GaussBonnet theorem or otherwise, compute the total curvature ∫ κ dA D
where D = S ∩ {z ≥ h} is the disc bounded by the circle C, and dA is the area form of S2 . 2
6. (Real Analysis) (Commutation of Diﬀerentiation and Summation of Integrals). Let Ω be an open subset of Rd and a < b be real numbers. For any positive integer n let fn (x, y) be a complexvalued measurable function on Ω × (a, b). Let a < c < b. Assume that the following three conditions are satisﬁed. (i) For each n and for almost all x ∈ Ω the function fn (x, y) as a function of y is absolutely continuous in y for y ∈ (a, b). (ii) The function function
∂ ∂y
fn (x, y) is measurable on Ω × (a, b) for each n and the ∞ ∑ ∂ fn (x, y) ∂y n=1
is integrable on Ω × (a, b). (iii) The ∑∞ function fn (x, c) is measurable on Ω for each n and the function n=1 fn (x, c) is integrable on Ω. Prove that the function ∫ y 7→
∞ ∑
fn (x, y)dx
x∈Ω n=1
is a welldeﬁned function for almost all points y of (a, b) and that ( ) ∫ ∞ ∞ ∫ ∑ ∑ ∂ d fn (x, y)dx = fn (x, y) dx dy x∈Ω n=1 ∂y x∈Ω n=1 for almost all y ∈ (a, b). Hint: Use Fubini’s theorem to exchange the order of integration and use convergence theorems for integrals of sequences of functions to exchange the order of summation and integration. 4
Qualifying Exams III, 2014 Spring 1. (Algebra) Prove or disprove: There exists a prime number√p such √ that the principal ideal (p) in the ring of integers OK in K = Q( 2, 3) is a prime ideal 2. (Algebraic Geometry) Let Γ = {p1 , · · · , p5 } ⊂ P2 be a conﬁguration of 5 points in the plane. (a) What is the smallest Hilbert function Γ can have? (b) What is the largest Hilbert function Γ can have? (c) Find all the Hilbert functions Γ can have. 3. (Complex Analysis) (Cauchy’s Integral Formula for Smooth Functions and Solution of ∂¯ Equation). (a) Let Ω be a bounded domain in C with smooth boundary ∂Ω. Let f be a C ∞ complexvalued function on some open ¯ of Ω in C. neighborhood U of the topological closure Ω (i) Show that for a ∈ Ω, 1 f (a) = 2πi where
with z = x +
√
∫ z∈∂Ω
∂f 1 = ∂ z¯ 2
f (z)dz 1 + z−a 2πi (
∧ d¯ z , z−a
∂f dz ∂ z¯ Ω
)
∂f √ ∂f + −1 ∂x ∂y
−1 y and x, y real.
(ii) Show that a ∈ Ω, 1 f (a) = − 2πi where
∫ z∈∂Ω
∂f 1 = ∂z 2
1 h(z) = 2πi ∂h (z) ∂ z¯
f (z)d¯ z 1 + z¯ − a ¯ 2πi
(
(iii) For z ∈ Ω deﬁne
Show that
∫
∫
∂f √ ∂f − −1 ∂x ∂y ∫ ζ∈Ω
= f (z) on Ω. 5
∧ d¯ z , z¯ − a ¯
∂f dz ∂z
Ω
) .
f (ζ) dζ ∧ dζ¯ . ζ −z
( ) dz Hint: For (i), apply Stokes’s theorem to d f (z) z−a on Ω minus a closed disk of radius ε > 0 centered at a and then let ε → 0. ( For the proof of) (iii), for any ﬁxed z ∈ Ω, apply Stokes’s theorem to d f (ζ) log ζ − zdζ¯ (with variable ζ) on Ω minus a closed disk of radius ε > 0 centered at z and then let ε → 0. Then apply ∂∂z¯ and use (ii). (b) Let Dr be the open disk of radius r > 0 in C centered at 0. Prove that for any C ∞ complexvalued function g on D1 there exists some C ∞ complexvalued function h on D1 such that ∂h = g on D1 . ∂ z¯ Hint: First use (a)(iii) to show that for 0 < r < 1 there exists some C ∞ r complexvalued function hr on D1 such that ∂h = g on Dr . Then use some ∂ z¯ approximation and limiting process to construct h. 4. (Algebraic Topology) Suppose that X is contractible and that some point a of X has a neighborhood homeomorphic to Rk . Prove that Hn (X \ {a}) ≃ Hn (S k−1 ) for all n. 5. (Differential Geometry) Let U+ = R2 − (R≤0 × {0}), U− = R2 − (R≥0 × {0}), and U0 = R2 − (R × {0}). Let B be obtained by gluing U+ and U− over U0 by the map ψ : U0 → U0 deﬁned by ψ(x, y) = (x, y) when y < 0, and ψ(x, y) = (x + y, y) when y > 0. 1. Show that B is a manifold. 2. Show that the trivial connections on the tangent bundles of U+ and U− glue together and give a global connection on the tangent bundle T B. Compute the curvature of this connection. 3. Compute the holonomy of the above connection around the loop γ : [0, 2π] → B determined by γU+ (θ) = (cos θ, sin θ) for θ ∈ (0, 2π). 6. (Real Analysis) (Bernstein’s Theorem on Approximation of Continuous Functions by Polynomials). Use the probabilistic argument outlined in 6
the two steps below to prove the following theorem of Bernstein. Let f be a realvalued continuous function on [0, 1]. For any positive integer n let ( )( ) n ∑ j n j x (1 − x)n−j Bn (f ; x) = f j n j=0 be the Bernstein polynomial. Then Bn (f ; x) converges to f uniformly on [0, 1] as n → ∞. Step One. For 0 < x < 1 consider the binomial distribution ( ) n j b(n, x, j) = x (1 − x)n−j j for 0 ≤ j ≤ n, which is the probability of getting j heads and n − j tails in tossing a coin n times if the probability of getting a head is x. Verify that the mean √ µ of this probability distribution is nx and its standard deviation σ is nx(1 − x). Step Two. Let X be the random variable which assumes the value j with probability ( b(n, ) x, j) for 0 ≤ j ≤ n. Consider (the) random variable Y = f (x) − f Xn which assumes the value f (x) − f nj with probability b(n, x, j) for 0 ≤ j ≤ n. Prove Bernstein’s theorem by bounding, for an arbitrary positive number ε, the sum which deﬁnes the expected value E(Y ) of the random variable Y , after breaking the sum up into two parts deﬁned respectively by j − µ ≥ ησ and j − µ < ησ for some appropriate positive number η depending on ε and the uniform bound of f .
7
Solutions of Qualifying Exams I, 2014 Spring 1. (Algebra) Let k = Fq be a ﬁnite ﬁeld with q elements. Count the number of monic irreducible polynomials of degree 12 over k. Solution. Let G := Gal(Fq12 /Fq ) act naturally on Fq12 . The set of monic irreducible polynomials of degree 12 are in onetoone correspondence with the set of Gorbits of order 12 in Fq12 . An orbit Gα has order 12 exactly when the subﬁeld Fq (α) coincides with Fq12 , i.e., exactly when ∪ K α ∈ Fq12 \ Fq ≤KFq12
The maximal proper subﬁelds of Fq12 are Fq6 and Fq4 . By inclusionexclusion principle, the number of the polynomials sought is equal to q 12 − q 6 − q 4 + q 2 . 12 2. (Algebraic Geometry) (a) Show that the set of lines L ⊂ P3C may be identiﬁed with a quadric hypersurface in P5C . (b) Let L0 ⊂ P3C be a given line. Show that the set of lines not meeting L0 is isomorphic to the aﬃne space A4C . Solution. (a) If P3 = PV is the projective space of onedimensional subspaces of a 4dimensional vector space V , then we associate to∧the line L 2 spanned by two vectors V ∼ = P5 . ∧2 v, w ∈ V the wedge product v ∧ w ∈ P Since a 2form η ∈ V is decomposable if and only if Q(η) = η ∧ η = 0 ∈ ∧ 4 ∼ V = C, this identiﬁes the set of lines with the zeroes of the quadratic form Q. (b) Choose 2 planes Λ, Λ′ ⊂ P3 containing L0 . Any line not meeting L0 is determined by its points of intersection with the two planes, giving an isomorphism between the set of lines not meeting L0 and (Λ \ L0 ) × (Λ′ \ L0 ) ∼ = A2 × A2 ∼ = A4 . 3. (Complex Analysis) (a) Compute ∫ z 31 dz z 2 + 3)2 (¯ z 4 + 2)3 z=1 (2¯ Note that the integrand is not a meromorphic function. 1
(b) Evaluate the integral ∫
∞
x=0
(
sin x x
)3 dx
by using the theory of residues. Justify carefully all the limiting processes in your computation. Solution. (a) Since z¯ =
1 z
for z = 1, it follows that
∫
z 31 dz z 2 + 3)2 (¯ z 4 + 2)3 z=1 (2¯ ∫ z 31 = )2 (( ) )3 dz. ( ( ) 1 4 z=1 2 1 2 + 3 +2 z z Use the change of variables z = ∫ −
w=1
1 w
to transform the integral to
( ) dw − 2 . w (2w2 + 3) (w4 + 2)3 1 w31 2
The negative sign in front of the integral comes from the change of orientation when the parametrization z = eiθ for 0 ≤ θ ≤ 2π is transformed to the parametrization w = e−iθ for 0 ≤ θ ≤ 2π. This new integral can be rewritten as ∫ dw , 33 2 2 4 3 w=1 w (3 + 2w ) (2 + w ) which is equal to 2πi times the residue of the meromorphic function 1 w33 (3 + 2w2 )2 (2 + w4 )3 at w = 0. We have the power series expansion of the factor 1 1 1 = ( ) 2 9 1 + 2 w2 2 (3 + 2w2 ) 3 )k ( ∞ 1 ∑ (−2)(−3) · · · (−2 − k + 1) 2 2 = w 9 k=0 k! 3 2
at w = 0 and the power series expansion of the factor 1 1 1 ) ( 3 = 4 8 1 + 1 w4 3 (2 + w ) 2 )ℓ ( ∞ 1 ∑ (−3)(−4) · · · (−3 − ℓ + 1) 1 4 = w 8 ℓ=0 ℓ! 2 at w = 0. Contributions to the residue in question from the two power series expansions come from 2k + 4ℓ = 32, which means that k must be divisible by 2 and there are only 9 choices for ℓ from 0 to 8 inclusively (with the corresponding value k = 32−4ℓ = 16 − 2ℓ). Hence the residue in question is 2 equal to the following sum 1 ∑ (−2)(−3) · · · (−2 − (16 − 2ℓ) + 1) 72 (16 − 2ℓ)! 8
ℓ=0
( )16−2ℓ ( )ℓ 2 (−3)(−4) · · · (−3 − ℓ + 1) 1 3 ℓ! 2
of 9 terms. The ﬁnal answer is that ∫ z=1
z 31 dz (2¯ z 2 + 3)2 (¯ z 4 + 2)3
is equal to 2πi ∑ (−2)(−3) · · · (−2 − (16 − 2ℓ) + 1) 72 (16 − 2ℓ)! 8
ℓ=0
( )16−2ℓ ( )ℓ 2 (−3)(−4) · · · (−3 − ℓ + 1) 1 . 3 ℓ! 2
(b) By Euler’s formula we have sin x = ( 3
sin x =
eix −e−ix 2i
eix − e−ix 2i
and )3
) 1 ( 3ix e − 3eix + 3e−ix − e−3ix −8i ( ix ) 1 e − e−ix e3ix − e−3ix = − 3 . 4 2i 2i
=
Thus sin3 x is the imaginary part of 3 ix 1 3ix e − e . 4 4 3
The power series expansion of 3 iz 1 3iz e − e 4 4 is
( )) 1 ( ( )) 1 ( ) 3( 1 + iz + O z 2 − 1 + 3iz + O z 2 = + O z 2 . 4 4 2 The Rlinear combination 3 iz 1 3iz 1 e − e − 4 4 2 vanishes to order 2 at z = 0 and its imaginary part for z = x real is equal to sin3 x. Let − 1 e3iz + 34 eiz − 12 f (z) = 4 . z3 Its behavior near z = 0 is given by 2
f (z) =
− 14 (3iz) + 2
3 (iz)2 4 2 z3
+ O (z 3 )
=
( ) 31 + O z3 4z
and we have a simple pole for f at z = 0 whose residue Res0 f is 34 . Integrating f (z) dz over the boundary of the set which is equal to the upper halfdisk of radius R > 0 minus the upper halfdisk of radius r with 0 < r < R and letting R → ∞ and r → 0, we get )3 ( ) ∫ ∞ ( sin x 3 3π dx = Im (πi Res0 f ) = Im πi = x 4 4 x=−∞ and
∫
∞
x=0
(
sin x x
)3 dx =
3π . 8
To justify the limiting process, we have to show that the integral ∫ f (z) dz CR
4
over the upper halfcircle of radius R centered at the origin 0 approaches 0 as R → ∞. This is a consequence of the fact that both e3iz  and eiz  are ≤ 1 for Im z ≥ 0 so that ∫ 1 f (z) dz ≤ 3 πR → 0 as R → ∞. R CR We need also to compute the integral ∫ f (z) dz Cr
over the upper halfcircle of radius r centered at the origin 0 in the counterclockwise sense as r → 0+. This is done by using f (z) =
( ) 31 + O z3 4z
and the parametrization z = reiθ for 0 ≤ θ ≤ π so that ∫ ∫ 31 lim f (z) dz = lim dz r→0+ C r→0+ C 4 z r r ∫ π 3 1 3 = ireiθ dθ = πi . iθ 4 θ=0 4 re 4. (Algebraic Topology) Suppose that X is a ﬁnite connected CW complex such that π1 (X) is ﬁnite and nontrivial. Prove that the univer˜ of X cannot be contractible. (Hint: Lefschetz ﬁxed point sal covering X theorem.) ˜ is also a ﬁnite CW complex. Solution. Since X is a ﬁnite CW complex, X ˜ ˜ Suppose X is contractible. Then X has the same homology as a point, i.e. ˜ = Z and Hi (X) ˜ = 0 for i ̸= 0. Then by the Lefschetz ﬁxed point H0 (X) ˜ → X ˜ has a ﬁxed point. On the other theorem any continuous map f : X ˜ is isomorphic to π1 (X), hand, the group of covering transformations of X hence is nontrivial. Since a nonidentity covering transformation does not ˜ cannot be contractible. have ﬁxed points, we obtain a contradiction. Thus X 5. (Differential Geometry) Let P2 = (C3 − {0})/C× , which is called the complex projective plane. 5
1. Show that P2 is a complex manifold by writing down its local coordinate charts and transitions. 2. Deﬁne L ⊂ P2 × C3 to be the subset containing elements of the form ([x], λx), where x ∈ C3 − {0} and λ ∈ C. Show that L is the total space of a holomorphic line bundle over P2 by writing down its local trivializations and transitions. It is called the tautological line bundle. 3. Using the standard Hermitian metric on C3 or otherwise, construct a Hermitian metric on the tautological line bundle. Express the metric in terms of local trivializations. Sketched Solution. 1. The charts are ϕ0 : U0 = {[x, y, z] : z ̸= 0} → C2 = V0 by [x, y, z] 7→ (x/z, y/z), and ϕ1 , ϕ2 are deﬁned similarly. The transition from V0 to V1 is (X, Y ) 7→ [X, Y, 1] 7→ (1, Y /X, 1/X) for X ̸= 0, and other transitions are computed in a similar way. 2. The local trivialization over U0 is ([x, y, 1], λ(x, y, 1)) 7→ ([x, y, 1], λ), and that over U1 and U2 are deﬁned in a similar way. The transition over U0 ∩ U1 is ([x, y, 1], λ) 7→ ([x, y, 1], λ(x, y, 1)) = ([1, y/x, 1/x], λx(1, y/x, 1/x)) 7→ ([x, y, 1], xλ). The transition over U12 and U02 are similarly deﬁned. 3. Deﬁne a metric by ([x], λx) 7→ ∥λx∥. Over U0 , it is given by ([x, y, 1], λ) 7→ ∥λ(x, y, 1)∥. It is similar for the other trivializations U1 , U2 . 6. (Real Analysis) (Schwartz’s Theorem on Perturbation of Surjective Maps by Compact Maps Between Hilbert Spaces). Let E, F be Hilbert spaces over C, S : E → F be a compact Clinear map, and T : E → F be a continuous surjective Clinear map. Prove that the cokernel of S+T : E → F is ﬁnitedimensional and the image of S + T : E → F is a closed subspace of F. Here the compactness of the Clinear map S : E → F means that for any sequence {xn }∞ n=1 in E with ∥xn ∥E ≤ 1 for all n ∈ N there exists some ∞ subsequence {xnk }k=1 of {xn }∞ n=1 such that S (xnk ) converges in F to some element of F as k → ∞. 6
Hint: Verify ﬁrst that the conclusion is equivalent to the following equivalent statement for the adjoints T ∗ , S ∗ : F → E of T, S. The kernel of T ∗ + S ∗ is ﬁnitedimensional and the image of T ∗ + S ∗ is closed. Then prove the equivalent statement. Solution. We prove ﬁrst the equivalent statement for the adjoints T ∗ , S ∗ : F → E for T, S and then at the end obtain from it the original statement for T, S : E → F . The adjoint S ∗ of the compact operator S is again compact (see e.g., p.189 of Stein and Shakarchi’s Real Analysis). Since T is surjective, by the open mapping theorem for Banach spaces and in particular for Hilbert spaces, the map T : E → F is open. It implies that F is the quotient of E by the kernel of T . Thus T ∗ is the isometry between F and the orthogonal complement of the kernel of T in E, when a Hilbert space is naturally identiﬁed with its dual by using its inner product according to the Riesz representation theorem (see e.g., Theorem 5.3 on p.182 of Stein and Shakarchi’s Real Analysis). Now we verify that the kernel of T ∗ + S ∗ is ﬁnitedimensional by showing that its closed unit ball is compact. Take a sequence of points {yn }n∈N in the kernel of T ∗ + S ∗ with ∥yn ∥F ≤ 1 for n ∈ N. Then T ∗ yn + S ∗ yn = 0 for n ∈ N. Since S ∗ is compact, there exists a subsequence {ynk }k∈N of {yn }n∈N such that S ∗ ynk → z in E for some z ∈ E. From T ∗ ynk → −z as k → ∞ and the fact that T ∗ is the isometry between F and the orthogonal complement of kernel of T in E, it follows that ynk converges to the unique element zˆ in F such that T ∗ zˆ = −z. Since z = limk→∞ S ∗ ynk = S ∗ zˆ and −z = T ∗ zˆ, it follows that T ∗ zˆ + S ∗ zˆ = 0 and z is in the kernel of T ∗ + S ∗ . Thus the closed unit ball of the kernel of T ∗ + S ∗ is compact. Since every locally compact Hilbert space is ﬁnite dimensional, it follows that the kernel of T ∗ + S ∗ is ﬁnitedimensional. Now we verify that the image of T ∗ + S ∗ is closed. Suppose for some sequence of points {yn }n∈N in F we have the convergence of T ∗ yn + S ∗ yn in E to some element z in E. We have to show that z belongs to the image of T ∗ + S ∗ . By replacing yn by its projection onto the orthogonal complement (Ker (T ∗ + S ∗ ))⊥ of the kernel of T ∗ + S ∗ in F , we can assume without loss of generality that each yn belongs to (Ker (T ∗ + S ∗ ))⊥ . We claim that the sequence of points {yn }n∈N in (Ker (T ∗ + S ∗ ))⊥ is bounded in the norm ∥·∥F of F , otherwise we can deﬁne yˆn = ∥yynn∥ so F
7
that T ∗ yˆn + S ∗ yˆn → 0 as n → ∞ with ∥ˆ yn ∥F = 1 for all n ∈ N. Since S ∗ is compact, there is a subsequence {ˆ ynk }k∈N of {ˆ yn }n∈N with S ∗ yˆnk converging to some element u in E. Thus T ∗ yˆnk = (T ∗ yˆnk + S ∗ yˆnk ) − S ∗ yˆnk converges to the element −u in E. Since T ∗ is the isometry between F and the orthogonal complement of kernel of T in E, it follows that yˆnk converges to the unique element v in F such that T ∗ v = −u. This means that (T ∗ + S ∗ ) (v) = 0 and v ∈ Ker (T ∗ + S ∗ ). On the other hand, v being the limit of the sequence yˆnk in (Ker (T ∗ + S ∗ ))⊥ must be in (Ker (T ∗ + S ∗ ))⊥ also. Thus, v = 0, which contradicts the fact that it is the limit of yˆnk with ∥ˆ ynk ∥F = 1 for all k ∈ N. This ﬁnishes the proof of the claim that sequence of points {yn }n∈N in (Ker (T ∗ + S ∗ ))⊥ . Since the sequence of points {yn }n∈N in (Ker (T ∗ + S ∗ ))⊥ is bounded in the norm ∥·∥F of F , by the compactness of S ∗ we can select a a subsequence {ynk }k∈N of {yn }n∈N with S ∗ ynk converging to some element w in E. Thus T ∗ ynk = (T ∗ ynk + S ∗ ynk ) − S ∗ ynk converges to the element z − w in E. Since T ∗ is the isometry between F and the orthogonal complement of kernel of T in E, it follows that ynk converges to the unique element t in F such that T ∗ t = z−w. With w = S ∗ t = limk→∞ ynk , this implies that (T ∗ + S ∗ ) (t) = z. This ﬁnishes the proof that the image of T ∗ + S ∗ is closed. Since we now know that the image of T ∗ + S ∗ is closed, it follows from the Riesz representation theorem that the map S +T maps E onto the orthogonal complement Ker(T ∗ +S ∗ )⊥ of the kernel Ker(T ∗ +S ∗ ) of T ∗ +S ∗ in F . Hence the image of T + S is closed and the cokernel of T + S is ﬁnitedimensional.
8
Solutions of Qualifying Exams II, 2014 Spring 1. (Algebra) Let A be a ﬁnite group of order n, and let V1 , · · · , Vk be its irreducible representations. (a) the dimensions of the vector spaces Vi satisfy the equality ∑k Show that 2 (dim V ) = n. i i=1 (b) What are the dimensions of the irreducible representations of the symmetric group S6 of six elements. Solution. (a) Use the character theory and show that Vi appears (dim Vi ) times in the regular representation C[A]. (b) Irreducible representations of S6 correspond to conjugacy classes in S6 , and then to partitions of 6, of which there are p(6) = 11. Then use the “hooklength formula”, d! . dim Vλ = ∏ ( hook lengths ) They are: 16, 10 (twice), 9 (twice), 5 (four times) and 1 (twice). 2. (Algebraic Geometry) Let C ⊂ P2 be a smooth plane curve of degree ≥ 3. (a) Show that C admits a regular map f : C → P1 of degree d − 1. (b) Show that C does not admit a regular map f : C → P1 of degree e with 0 < e < d − 1. Solution. (a) Solution: Simply project from any point p ∈ C to a complementary line. (b) Since the canonical series of C is cut on C by plane curves of degree d − 3, by RiemannRoch the general ﬁber of any map f : C → P1 of degree e must consist of e points of C that fail to impose independent conditions on curves of degree d − 3. But any set d − 2 or fewer points in the plane impose independent conditions on curves of degree d − 3.
9
3. (Complex Analysis) Suppose that f is holomorphic in an open set containing the closed unit disk {z ≤ 1} in C, except for a pole at z0 on the unit circle {z = 1}. Show that if ∞ ∑
an z n
n=0
denote the power series expansion of f in the open unit disk {z < 1} , then an = z0 . n→∞ an+1 lim
Solution. Since z0 is the only pole of the meromorphic function f on an open set containing the closed unit disk in C, we can express f (z) in the form m ∑
Ak
k=1
(z − z0 )k
+ g(z)
with A∑ 1 , · · · , Am ∈ C, where m ≥ 1 and Am = h (z0 ) ̸= 0 and g(z) is a power n series ∞ n=0 bn z with radius of convergence R > 1. For any positive number r with z0  < r < R we can ﬁnd a positive number B such that bn  ≤
B rn
for all nonnegative integer n. By using the binomial expansion of diﬀerentiating the geometric series (n+k−1) (n+k−1) = , we have k−1 n a n = bn +
m ∑
(−1)k Ak
1 z−z0
1 (z−z0 )k
in z (k − 1)times) and noting that
(n + k − 1)(n + k − 2) · · · (n + 2)(n + 1) (k − 1)! (z0 )n+k
k=1
(or
.
In the computation of the limit an , n→∞ an+1 < z10 and bn  ≤ lim
B since Am = h (z0 ) ̸= 0 and 1r , the dominant term from rn an is (n + m − 1)(n + m − 2) · · · (n + 2)(n + 1) (−1)m Am (m − 1)! (z0 )n+m
10
and we get (−1)m Am (n+m−1)(n+m−2)···(n+2)(n+1) an (m−1)!(z0 )n+m lim = z0 . = lim (n+m)(n+m−1)···(n+3)(n+2) n→∞ an+1 n→∞ (−1)m Am n+1+m (m−1)!(z ) 0
The dominant term from an means that an minus the dominant term and then divided by the dominant term would have limit zero when n → ∞. 4. (Algebraic Topology) Show that if n > 1, then every map from the real projective space RPn to the ntorus T n is nullhomotopic. Solution. Recall that π1 (RPn ) = Z/2Z and π1 (T n ) = π1 (S 1 × · · · × S 1 ) = Zn . Now if f : RPn → T n is any map, then the induced homomorphism f∗ : π1 (RPn ) → π1 (T n ) must be trivial because Zn has no nontrivial elements of ﬁnite order. Let p : Rn → T n be the standard covering map. Then, by the general lifting lemma, we obtain a continuous map f˜: RPn → Rn such that f = p ◦ f˜. Since Rn is contractible, we obtain that f˜ is nullhomotopic, from which it follows that f is nullhomotopic. 5. (Differential Geometry) Let S2 := {x2 + y 2 + z 2 = 1} ⊂ R3 be the unit sphere in the Euclidean space. Let C = {(r cos t, r sin t, h) : t ∈ R} be a circle in S2 , where r, h > 0 are constants with r2 + h2 = 1. 1. Compute the holonomy of the sphere S2 (with the standard induced metric) around the circle C. 2. By using GaussBonnet theorem or otherwise, compute the total curvature ∫ κ dA D
where D = S ∩ {z ≥ h} is the disc bounded by the circle C, and dA is the area form of S2 . 2
Sketched Solution. 1. The holonomy is rotation by 2πh. √ 2. The total curvature is 2π − 2π r2 h2 + (1 − r2 )2 . 11
6. (Real Analysis) (Commutation of Diﬀerentiation and Summation of Integrals). Let Ω be an open subset of Rd and a < b be real numbers. For any positive integer n let fn (x, y) be a complexvalued measurable function on Ω × (a, b). Let a < c < b. Assume that the following three conditions are satisﬁed. (i) For each n and for almost all x ∈ Ω the function fn (x, y) as a function of y is absolutely continuous in y for y ∈ (a, b). (ii) The function function
∂ ∂y
fn (x, y) is measurable on Ω × (a, b) for each n and the ∞ ∑ ∂ fn (x, y) ∂y n=1
is integrable on Ω × (a, b). (iii) The ∑∞ function fn (x, c) is measurable on Ω for each n and the function n=1 fn (x, c) is integrable on Ω. Prove that the function ∫ y 7→
∞ ∑
fn (x, y)dx
x∈Ω n=1
is a welldeﬁned function for almost all points y of (a, b) and that d dy
∫
∞ ∑
fn (x, y)dx =
x∈Ω n=1
∞ ∫ ∑ n=1
x∈Ω
(
) ∂ fn (x, y) dx ∂y
for almost all y ∈ (a, b). Hint: Use Fubini’s theorem to exchange the order of integration and use convergence theorems for integrals of sequences of functions to exchange the order of summation and integration. Solution. The theorem of Fubini which we will use states that if F (x, y) on Ω1 × Ω2 (with Ωj open in Rdj for j = 1, 2) and if ∫ F (x, y) < ∞, (x,y)∈Ω1 ×Ω2
12
then
∫
(∫
)
(∫
∫
F (x, y)dy dx = x∈Ω1
y∈Ω2
) F (x, y)dx dy.
y∈Ω2
x∈Ω1
One consequence of the theorem of dominated convergence which we will use is the folloing exchange of integration and summation. If Fn (x) is a sequence ˜ of Rd˜ such that of measurable functions on an open subset Ω ∫
∞ ∑
˜ x∈Ω n=1
then
∫
∞ ∑ ˜ x∈Ω n=1
Fn (x) < ∞,
Fn (x) =
∞ ∫ ∑ n=1
˜ x∈Ω
Fn (x).
These two results make it possible for us to both exchange the order of integration and the order of summation and integration in the following equation for a < η < b, (†) ( (∞ ∫ ( ( ) ) ) ) ∫ η ∑ ∫ ∞ ∫ η ∑ ∂ ∂ fn (x, y) dx dy = fn (x, y) dy dx, ∂y ∂y y=c x∈Ω n=1 x∈Ω n=1 y=c because the function
∞ ∑ ∂ fn (x, y) ∂y n=1
is integrable on Ω × (a, b). Since for almost all x ∈ Ω the function fn (x, y) as a function of y is absolutely continuous in y, it follows that ) ∫ η ( ∂ fn (x, y) dy = fn (x, η) − fn (x, c) ∂y y=c for almost all x ∈ Ω, which implies that (∞ ∫ ( (∞ ) ) ) ∫ ∫ ∑ η ∑ ∂ fn (x, y) dy dx = (fn (x, η) − fn (x, c)) dx ∂y x∈Ω y=c x∈Ω n=1 (∞ ) n=1 ∫ (∞ ) ∫ ∑ ∑ = fn (x, η) dx − fn (x, c) dx, x∈Ω
n=1
13
x∈Ω
n=1
∑ because ∞ n=1 fn (x, c) is integrable on Ω. Putting this together with (†) yields (‡) ( (∞ ) (∞ ) ( ) ) ∫ η ∑ ∫ ∫ ∞ ∫ ∑ ∑ ∂ fn (x, y) dx dy = fn (x, η) dx− fn (x, c) dx. ∂y y=c x∈Ω x∈Ω x∈Ω n=1 n=1 n=1 Diﬀerentiating both sides of (‡) with respect to η and applying the fundamental theorem of calculus in the theory of Lebesgue and then replacing η by y, we obtain (∞ ) ( ) ∫ ∞ ∫ ∑ ∑ ∂ ∂ fn (x, y) dx = fn (x, y) dx ∂y ∂y x∈Ω n=1 n=1 x∈Ω for almost all y ∈ (a, b).
14
Solutions of Qualifying Exams III, 2014 Spring 1. (Algebra) Prove or disprove: There exists a prime number√p such √ that the principal ideal (p) in the ring of integers OK in K = Q( 2, 3) is a prime ideal. Solution. If there were, the decomposition group and the inertia group at (p) would be isomorphic to the whole Gal (K/Q) ≃ Z/2 × Z/2 and to the trivial group, respectively, and the quotient would not be cyclic. 2. (Algebraic Geometry) Let Γ = {p1 , · · · , p5 } ⊂ P2 be a conﬁguration of 5 points in the plane. (a) What is the smallest Hilbert function Γ can have? (b) What is the largest Hilbert function Γ can have? (c) Find all the Hilbert functions Γ can have. Solution. (a) The smallest Hilbert function Γ can have occurs if Γ consists of 5 collinear points; the Hilbert function in this case is (hΓ (0), hΓ (1), hΓ (2), . . . ) = (1, 2, 3, 4, 5, 5, . . . ). (b) The largest Hilbert function Γ can have occurs if Γ consists of 5 general points; the Hilbert function in this case is (1, 3, 5, 5, . . . ). (c) The only other Hilbert function Γ can have occurs when Γ consists of four collinear points and one point not collinear with those; the Hilbert function in this case is (1, 3, 4, 5, 5, . . . ). 3. (Complex Analysis) (Cauchy’s Integral Formula for Smooth Functions and Solution of ∂¯ Equation). (a) Let Ω be a bounded domain in C with smooth boundary ∂Ω. Let f be a C ∞ complexvalued function on some open ¯ of Ω in C. neighborhood U of the topological closure Ω (i) Show that for a ∈ Ω, 1 f (a) = 2πi where
with z = x +
√
∫ z∈∂Ω
1 ∂f = ∂ z¯ 2
1 f (z)dz + z−a 2πi (
∫ Ω
∂f ∂f √ + −1 ∂x ∂y
−1 y and x, y real. 15
∧ d¯ z , z−a
∂f dz ∂ z¯
)
(ii) Show that a ∈ Ω, 1 f (a) = − 2πi where
∫ z∈∂Ω
∂f 1 = ∂z 2
f (z)d¯ z 1 + z¯ − a ¯ 2πi
(
∫
∧ d¯ z , z¯ − a ¯
∂f dz ∂z
Ω
∂f ∂f √ − −1 ∂x ∂y
) .
(iii) For z ∈ Ω deﬁne 1 h(z) = 2πi Show that
∂h (z) ∂ z¯
∫ ζ∈Ω
f (ζ) dζ ∧ dζ¯ . ζ −z
= f (z) on Ω.
) ( dz on Ω minus a closed Hint: For (i), apply Stokes’s theorem to d f (z) z−a disk of radius ε > 0 centered at a and then let ε → 0. ( For the proof of) (iii), for any ﬁxed z ∈ Ω, apply Stokes’s theorem to d f (ζ) log ζ − zdζ¯ (with variable ζ) on Ω minus a closed disk of radius ε > 0 centered at z and then let ε → 0. Then apply ∂∂z¯ and use (ii). (b) Let Dr be the open disk of radius r > 0 in C centered at 0. Prove that for any C ∞ complexvalued function g on D1 there exists some C ∞ complexvalued function h on D1 such that ∂h = g on D1 . ∂ z¯ Hint: First use (a)(iii) to show that for 0 < r < 1 there exists some C ∞ r = g on Dr . Then use some complexvalued function hr on D1 such that ∂h ∂ z¯ approximation and limiting process to construct h. Solution. (a) Take an arbitrary positive number ε less than the distance from a to the boundary of Ω. Let Bε be the closed disk of radius ε > 0 centered at a. Application of Stokes’s theorem to ( ) ∂f d¯ z ∧ dz dz d f (z) = ∂ z¯ z−a z−a on Ω − Bε yields ∫
∂f d¯ z∧ ∂ z¯ Ω−Bε
dz = z−a
∫
dz f (z) − z−a ∂Ω 16
∫ f (z) z−a=ε
dz . z−a
iθ We ( use the )parametrization z = a + εe to evaluate the last integral and use iθ f a + εe − f (a) → 0 as ε → 0 from the continuous diﬀerentiability of f at 0 to conclude that ∫ dz lim+ f (z) = 2πif (a). ε→0 z−a z−a=ε
Hence 1 f (a) = 2πi
∫ z∈∂Ω
1 f (z)dz + z−a 2πi
∫
∧ d¯ z . z−a
∂f dz ∂ z¯ Ω
This ﬁnishes the proof of the formula in (i). For the proof of the formula in (ii) we apply (i) to f (z) to get 1 f (a) = 2πi
∫ z∈∂Ω
f (z)dz 1 + z−a 2πi
∫
∧ d¯ z . z−a
∂f dz ∂ z¯ Ω
and then we take the complexconjugates of both sides to get 1 f (a) = − 2πi
∫ z∈∂Ω
f (z)d¯ z 1 + z¯ − a ¯ 2πi
∫ Ω
∧ d¯ z , z¯ − a ¯
∂f dz ∂z
which is the formula in (ii). For the proof of (iii) we apply Stokes’s theorem to ( ) ∂f f (ζ)dζ ∧ dζ¯ d f (ζ) log ζ − z2 dζ¯ = log ζ − z2 dζ ∧ dζ¯ + ∂ζ ζ −z on Ω − Bε yields ∫ ∫
Ω−Bε
=
∂f log ζ − z2 dζ ∧ dζ¯ + ∂ζ ∫ 2 ¯ f (ζ) log ζ − z dζ −
∫ Ω−Bε
z−a=ε
∂Ω
f (ζ)dζ ∧ dζ¯ ζ −z
¯ f (ζ) log ζ − z2 dζ.
With its evaluation by the parametrization z = a + εeiθ , the last integral ∫ f (ζ) log ζ − z2 dζ¯ z−a=ε
17
approaches 0 as ε → 0+ so that ∫ ∫ ∫ ∂f f (ζ)dζ ∧ dζ¯ 2 ¯ ¯ log ζ − z dζ ∧ dζ + = f (ζ) log ζ − z2 dζ. ∂ζ ζ − z Ω Ω ∂Ω ∂ We apply ∂∂z¯ to both sides and separately justify the commutation of ∂x ∂ with integration and the commutation of ∂y with integration, because on the righthand sides of the following two formulae both integrals over Ω after diﬀerentiation are absolutely convergent. ( ) ∫ ∫ ∂ ∂f ∂f ∂ 2 2 log ζ − z dζ ∧ dζ¯ = log ζ − z dζ ∧ dζ¯ ∂x Ω ∂ζ ∂x Ω ∂ζ
and ∂ ∂y
∫ Ω
∂f log ζ − z2 dζ ∧ dζ¯ = ∂ζ
∫ Ω
∂f ∂ζ
(
) ∂ 2 ¯ log ζ − z dζ ∧ dζ. ∂y
We get ∫
∂f dζ ∂ζ
−
∧ dζ¯
ζ − z¯
Ω
∂ + ∂z
∫ Ω
∫ f (ζ)dζ ∧ dζ¯ f (ζ) dζ¯ =− , ζ −z ¯ ∂Ω ζ − z
or ∂ ∂z
(
1 2πi
∫ Ω
) ∫ ∫ f (ζ)dζ ∧ dζ¯ f (ζ) dζ¯ 1 1 =− + ζ −z 2πi ∂Ω ζ − z¯ 2πi Ω
∂f dζ ∂ζ
∧ dζ¯
ζ − z¯
,
which by the formula in (ii) is equal to f (z). This ﬁnishes the proof of the formula in (iii). For use in (b) we also observe that ∂ ∂ z¯
(
1 2πi
∫ Ω
) ∫ ∫ f (ζ)dζ ∧ dζ¯ 1 f (ζ) dζ¯ 1 =− + ζ −z 2πi ∂Ω ζ − z 2πi Ω
This implies that ∂ ∂ z¯
(
1 2πi
∫ Ω
∂f dζ ∂ζ
∧ dζ¯
ζ −z
.
) f (ζ)dζ ∧ dζ¯ ζ −z
is uniformly bounded on compact subsets of Ω. By induction on k and by ¯ in U in going from the applying the argument to ∂f on a neighborhood of Ω ∂ζ 18
kth step to the (k + 1)st step in the induction process, we conclude that all the kth partial derivatives of ∫ f (ζ)dζ ∧ dζ¯ ζ −z Ω with respect to x and y (i.e., with respect to z and z¯) are uniformly bounded on compact subsets of Ω. Hence ∫ f (ζ)dζ ∧ dζ¯ ζ −z Ω is C ∞ on Ω as a function of z. (b) Choose rn = 1 −
1 . 2n
We can set
1 hrn (z) = 2πi
∫ ζ∈Drn+1
g(ζ)dζ ∧ dζ¯ ζ −z
on Drn to get ∂z¯hrn = g on Drn from (a)(iii). As observed above, hrn (z) is an inﬁnitely diﬀerentiable function on Drn . We now look at the approximation and limiting process to construct h on all of D1 such that ∂z¯h = g on D1 . For n ≥ 3 the function hrn − hrn−1 is holomorphic on Drn−1 . By using the Taylor polynomial Pn of hrn − hrn−1 centered at 0 of degree Nn for Nn suﬃciently large, we have
ˆ rn = hrn on Drn−2 . Let h ˆ rn − h ˆr = h k
( ) hrn − hrn−1 − Pn ≤ 1 2n ∑ − nk=3 Pn on Drn . Then for any n > k ≥ 3 from
n ( ∑
n ) ∑ ) ( ˆr ˆhr − h = hrℓ − hrℓ−1 − Pℓ ℓ−1 ℓ ℓ=k+1
ℓ=k+1
it follows that n n ∑ ∑ 1 1 ˆ ˆ hrℓ − hrℓ−1 − Pℓ ≤ ≤ k hrn − hrk ≤ 2ℓ 2 ℓ=k+1 ℓ=k+1
19
on Drk−1 . Thus, for any ﬁxed k ≥ 3 the sequence {hrn − hrk }∞ n=k+1 is a Cauchy sequence of holomorphic functions on Drk−1 and we can deﬁne h = ˆ rn on D with ∂h = g on D, because h − hr is holomorphic on Dr limn→∞ h k k−1 ∂ z¯ ∂hrk−1 ˆ and ∂ z¯ = g on Drk−1 . Since hrn is inﬁnitely diﬀerentiable on Drk−1 , it follows that h is inﬁnitely diﬀerentiable on each Drk−1 and hence is inﬁnitely diﬀerentiable on all of D1 . 4. (Algebraic Topology) Suppose that X is contractible and that some point a of X has a neighborhood homeomorphic to Rk . Prove that Hn (X \ {a}) ≃ Hn (S k−1 ) for all n. Solution. We have the following piece of the long exact homology sequence: Hk (X) → Hk (X, X \ {a}) → Hk−1 (X \ {a}) → Hk−1 (X). Now for k > 1, the outer two groups are 0, hence Hk (X, X \ {a}) ≃ Hk−1 (X \ {a}). Let U be a neighborhood of a homeomorphic to Rm and let C = X \ U. Then C ⊂ X \ {a}, which is open. Hence, by excision, Hk (X, X \ {a}) ≃ Hk (U, U \ {a}) ≃ Hk (Rm , Rm \ {a}). On the other hand, we have the same piece of exact sequence of (Rm , Rm \ {a}) : ψ(x, y) = (x, y) when y < 0, and Hk (Rm ) → Hk (Rm , Rm \ {a}) → Hk−1 (Rm \ {a}) → Hk−1 (Rm ), and the outer two groups are 0 for k > 1. Since Rm \ {a} deformation retracts onto S m−1 , putting everything together we obtain that for k > 1, Hk (S m−1 ) ≃ Hk (X \ {a}). 5. (Differential Geometry) Let U+ = R2 − (R≤0 × {0}), U− = R2 − (R≥0 × {0}), and U0 = R2 − (R × {0}). Let B be obtained by gluing U+ and U− over U0 by the map ψ : U0 → U0 deﬁned by ψ(x, y) = (x, y) when y < 0, and ψ(x, y) = (x + y, y) when y > 0. 20
1. Show that B is a manifold. 2. Show that the trivial connections on the tangent bundles of U+ and U− glue together and give a global connection on the tangent bundle T B. Compute the curvature of this connection. 3. Compute the holonomy of the above connection around the loop γ : [0, 2π] → B determined by γU+ (θ) = (cos θ, sin θ) for θ ∈ (0, 2π). Sketched Solution. 1. U+ and U− already serve as charts of B, and the transition between them is aﬃne. 2. Since the transition is aﬃne, the diﬀerential d is preserved by the transition. The curvature is just zero. 3. The holonomy is given by the matrix ( ) 1 1 . 0 1 6. (Real Analysis) (Bernstein’s Theorem on Approximation of Continuous Functions by Polynomials). Use the probabilistic argument outlined in the two steps below to prove the following theorem of Bernstein. Let f be a realvalued continuous function on [0, 1]. For any positive integer n let ( )( ) n ∑ j n j Bn (f ; x) = f x (1 − x)n−j n j j=0 be the Bernstein polynomial. Then Bn (f ; x) converges to f uniformly on [0, 1] as n → ∞. Step One. For 0 < x < 1 consider the binomial distribution ( ) n j x (1 − x)n−j b(n, x, j) = j for 0 ≤ j ≤ n, which is the probability of getting j heads and n − j tails in tossing a coin n times if the probability of getting a head is x. Verify that the mean √ µ of this probability distribution is nx and its standard deviation σ is nx(1 − x). 21
Step Two. Let X be the random variable which assumes the value j with probability ( b(n, ) x, j) for 0 ≤ j ≤ n. Consider (the) random variable Y = f (x) − f Xn which assumes the value f (x) − f nj with probability b(n, x, j) for 0 ≤ j ≤ n. Prove Bernstein’s theorem by bounding, for an arbitrary positive number ε, the sum which deﬁnes the expected value E(Y ) of the random variable Y , after breaking the sum up into two parts deﬁned respectively by j − µ ≥ ησ and j − µ < ησ for some appropriate positive number η depending on ε and the uniform bound of f . Solution. Step One. From ( ) ( ) n (n − 1)(n − 2) · · · (n − j + 1) n−1 j =n =n j (j − 1)! j−1 it follows that µ=
n ∑
j b(n, x, j)
j=0
( ) n ∑ n j x (1 − x)n−j = j j j=0 ( ) n ∑ n−1 = n x xj−1 (1 − x)n−j j − 1 j=1 = nx (x + (1 − x))n−1 = nx. From
( ) ( ) ( ) n n−1 n−2 j(j − 1) = n(j − 1) = n(n − 1) j j−1 j−2
22
and ( ) n j x (1 − x)n−j E(X(X − 1)) = j (j − 1) b(n, x, j) = j (j − 1) j j=0 j=0 ( ) n ∑ 2 n−2 = n(n − 1)x xj−2 (1 − x)n−j j − 2 j=2 ) n−2 ( ∑ n−2 j 2 = n(n − 1)x x (1 − x)n−2−j j j=0 n ∑
n ∑
= n(n − 1)x2 (1 + (1 − x))n−2 = n(n − 1)x2 it follows that
and σ =
√
( ) σ 2 = E (X − µ)2 = E(X 2 ) − 2µE(X) + µ2 = E(X 2 ) − µ2 = n(n − 1)x2 + nx − (nx)2 = nx ((n − 1)x + 1 − nx) = nx(1 − x).
nx(1 − x).
Step Two. Given any ε > 0. By the uniform continuity of f on [0, 1] there exists some δ > 0 such that f (x1 ) − f (x2 ) < ε for x1 − x2  < δ. Choose a positive number η suﬃciently large so that 1 ε 2 sup f  < 2 η [0,1] 2 and then choose a positive N with η √ < δ. N We are going to prove that f − Bn (f ; x) < ε on [0, 1] for n ≥ N by bounding the sum which deﬁnes the expected value E(Y ) of the random variable Y , after breaking the sum up into two parts deﬁned respectively by j − µ ≥ ησ and j − µ < ησ. 23
First of all, for any ﬁxed x ∈ [0, 1], ( )) ( ) n ( ∑ j n j x (1 − x)n−j f (x) − Bn (f ; x) = f (x) − f n j j=0 ( )) n ( ∑ j f (x) − f = b(n, x, j) n j=0 ( ) n ∑ f (x) − f j b(n, x, j), ≤ n j=0 which is the expected value E(Y ) of the random variable Y , because ( ) n ( ) ∑ n j n j n−j f (x) x (1 − x) = f (x) x (1 − x)n−j j j j=0 j=0
n ∑
= f (x) (x + (1 − x))n = f (x). For the estimation of the part ( ) ∑ f (x) − f j b(n, x, j) n j−nx 2. Show that BG (the classifying space of G) cannot have homology groups whose direct sum has ﬁnite rank. 5. (Differential Geometry) Let H = {(x, y) ∈ R2 : y > 0} be the upper half plane. Let g be the Riemannian metric on H given by g=
(dx)2 + (dy)2 . y2
(H, g) is known as the halfplane model of the hyperbolic plane. (a) Let γ(θ) = (cos θ, sin θ) and η(θ) = (cos θ + 1, sin θ) for θ ∈ (0, π) be two paths in H. Compute the angle A at their intersection point shown in Figure 1, measured by the metric g.
Figure 1: Angle A between the two curves γ and η in the upper half plane H.
(b) By computing the LeviCivita connection ∑ ∂ ∂ = Γkij ∇ ∂ ∂xi ∂x ∂xk j k=1 2
of g or otherwise (where (x1 , x2 ) = (x, y)), show that the path γ, after arclength reparametrization, is a geodesic with respect to the metric g. 2
6. (Real Analysis) any positive integer n let Mn be a positive number ∑For ∞ such that the series n=1 Mn of positive numbers is convergent and its limit is M . Let a < b be real numbers and fn (x) be a realvalued continuous function on [a, b] for any positive integer n such that its derivative fn′ (x) exists ∑ for every a < x < b with fn′ (x) ≤ Mn for a < x < b. Assume that the series ∞ n=1 fn (a) of real numbers converges. Prove that ∑ (a) the series ∞ n=1 fn (x) converges to some realvalued function f (x) for every a ≤ x ≤ b, (b) f ′ (x) exists for every a < x < b , and (c) f ′ (x) ≤ M for a < x < b. Hint for (b): For ﬁxed x ∈ (a, b) consider the series of functions ∞ ∑ fn (y) − fn (x) n=1
y−x
of the variable y and its uniform convergence.
3
Qualifying Exams II, 2013 Fall 1. (Algebra) Find all the ﬁeld automorphisms of the real numbers R. Hint: Show that any automorphism maps a positive number to a positive number, and deduce from this that it is continuous. 2. (Algebraic Geometry) What is the maximum number of ramiﬁcation points that a mapping of ﬁnite degree from one smooth projective curve over C of genus 1 to another (smooth projective curve of genus 1) can have? Give an explanation for your answer. 3. ( (Complex Analysis) Let ω and η be two complex numbers such that ) ω Im η > 0. Let G be the closed parallelogram consisting of all z ∈ C such that z = λω + ρη for some 0 ≤ λ, ρ ≤ 1. Let ∂G be the boundary of G and Let G0 = G − ∂G be the interior of G. Let P1 , · · · , Pk , Q1 , · · · , Qℓ be points in G0 and let m1 , · · · , mk , n1 , · · · , nℓ be positive integers. Let f be a function on G such that ∏ f (z) ℓj=1 (z − Qj )nj ∏k mp p=1 (z − Pp ) is continuous and nowhere zero on G and is holomorphic on G0 . Let φ(z) and ψ(z) be two polynomials on C. Assume that f (z + ω) = eφ(z) f (z) if both ψ(z) z and z + ω are in G. Assume ∑k also that ∑ℓ f (z + η) = e f (z) if both z and z + η are in G. Express p=1 mp − j=1 nj in terms of ω and η and the coeﬃcients of φ(z) and ψ(z). 4. (Algebraic Topology) (a) Fix a basis for H1 of the twotorus (with integer coeﬃcients). Show that for every element x ∈ SL(2, Z), there is an automorphism of the twotorus such that the induced map on H1 acts by x. Hint: SL(2, Z) also acts on the universal cover of the torus. (b) Fix an embedding j : D2 × S 1 → S 3 . Remove its interior from S 3 to obtain a manifold X with boundary T 2 . Let f be an automorphism of the twotorus and consider the glued space Xf := (D2 × S 1 ) ∪f X. If X is homotopy equivalent to D2 × S 1 , compute the homology groups of Xf . 4
5. (Differential Geometry) Let M = U (n)/O(n) for n ≥ 1, where U (n) is the group of n×n unitary matrices and O(n) is the group of n×n orthogonal matrices. M is a real manifold called the Lagrangian Grassmannian. (a) Compute and state the dimension of M . (b) Construct a Riemannian metric which is invariant under the left action of U (n) on M . (c) Let ∇ be the corresponding LeviCivita connection on the tangent bundle T M , and X, Y, Z be any U (n)invariant vector ﬁelds on M . Using the given identity (which you are not required to prove) 1 ∇X Y = [X, Y ], 2 show that the Riemannian curvature tensor R of ∇ satisﬁes the formula 1 R(X, Y )Z = [Z, [X, Y ]]. 4 6. (Real Analysis) Show that there is no function f : R → R whose set of continuous points is precisely the set Q of all rational numbers.
5
Qualifying Exams III, 2013 Fall 1. (Algebra) Consider the function ﬁelds K = C(x) and L = C(y) of one variable, and regard L as a ﬁnite extension of K via the Calgebra inclusion x 7→
−(y 5 − 1)2 4y 5
Show that the extension L/K is Galois and determine its Galois group. 2. (Algebraic Geometry) Is every smooth projective curve of genus 0 deﬁned over the ﬁeld of complex numbers isomorphic to a conic in the projective plane? Give an explanation for your answer. 3. (Complex Analysis) Let f (z) = z +e−z for z ∈ C and let λ ∈ R, λ > 1. Prove or disprove the statement that f (z) takes the value λ exactly once in the open right halfplane Hr = {z ∈ C : Re z > 0}. 4. (Algebraic Topology) (a) Let X and Y be locally contractible, connected spaces with ﬁxed basepoints. Let X ∨ Y be the wedge sum at the basepoints. Show that π1 (X ∨ Y ) is the free product of π1 X with π1 Y . (b) Show that π1 (X × Y ) is the direct product of π1 X with π1 Y . (c) Note the canonical inclusion f : X ∨ Y → X × Y . Assume that X and Y have abelian fundamental groups. Show that the map f∗ on fundamental groups exhibits π1 (X × Y ) as the abelianization of π1 (X ∨ Y ). Hint: The Hurewicz map is natural. 5. (Differential Geometry) (a) Let S1 = R/Z be a circle and consider the connection √ ∇ := d + π −1dθ deﬁned on the trivial complex line bundle over S1 , where θ is the standard coordinate on S1 = R/Z descended from R. By solving the diﬀerential equation for ﬂat sections f (θ) √ ∇f = df + π −1f dθ = 0 or otherwise, show that there does not exist global ﬂat sections with respect to ∇ over S1 . 6
(b) Let T = V /Λ be a torus, where Λ is a lattice and V = Λ ⊗ R is the real vector space containing Λ. Let L be the trivial complex line bundle equipped with the standard Hermitian metric. By identifying ﬂat U (1) connections with U (1) representations of the fundamental group π1 (T ) or otherwise, show that the space of ﬂat unitary connections on L is the dual torus T ∗ = V ∗ /Λ∗ , where Λ∗ := Hom(Λ, Z) is the dual lattice and V ∗ := Hom(V, R) is the dual vector space. 6. (Real Analysis) (Fundamental Solutions of Linear Partial Diﬀerential Equations with Constant Coeﬃcients). Let Ω be an open (−M, M ) in ∑ interval dν R with M > 0. Let n be a positive integer and L = nν=0 aν dx ν be a linear diﬀerential operator of order n on R with constant coeﬃcients, where the coeﬃcients a0 , · · · ,∑ an−1 , an ̸= 0 are complex numbers and x is the coordinate n dν ν of R. Let L∗ = ν=0 (−1) aν dxν . Prove, by using Plancherel’s identity, that there exists a constant c > 0 which depends only on M and an and is independent of a0 , a1 , · · · , an−1 such that for any f ∈ L2 (Ω) a weak solution u of Lu = f exists with ∥u∥L2 (Ω) ≤ c ∥f ∥L2 (Ω) . Give one explicit expression for c as a function of M and an . Hint: A weak solution u of Lu = f means that (f, ψ)L2 (Ω) = (u, L∗ ψ)L2 (Ω) for every inﬁnitely diﬀerentiable function ψ on Ω with compact support. For the solution of this problem you can consider as known and given the following three statements. (I) If there exists a positive number c > 0 such that ∥ψ∥L2 (Ω) ≤ c ∥L∗ ψ∥L2 (Ω) for all inﬁnitely diﬀerentiable complexvalued functions ψ on Ω with compact support, then for any f ∈ L2 (Ω) a weak solution u of Lu = f exists with ∥u∥L2 (Ω) ≤ c ∥f ∥L2 (Ω) . ∑ k (II) Let P (z) = z m + m−1 k=0 bk z be a polynomial with leading coeﬃcient 1. If F is a holomorphic function on C, then ∫ 2π ( iθ ) ( iθ ) 2 1 2 P e F e dθ. F (0) ≤ 2π θ=0 (III) For an L2 function f on R which is zero outside Ω = (−M, M ) its Fourier transform ∫ M fˆ(ξ) = f (x)e−2πixξ dx −M
7
as a function of ξ ∈ R can be extended to a holomorphic function ∫ M fˆ (ξ + iη) = f (x)e−2πix(ξ+iη) dx −M
on C as a function of ξ + iη.
8
Solutions of Qualifying Exams I, 2013 Fall 1. (Algebra) Consider the algebra M2 (k) of 2 × 2 matrices over a field k. Recall that an idempotent in an algebra is an element e such that e2 = e. (a) Show that an idempotent e ∈ M2 (k) different from 0 and 1 is conjugate to 1 0 e1 := 0 0 by an element of GL2 (k). (b) Find the stabilizer in GL2 (k) of e1 ∈ M2 (k) under the conjugation action. (c) In case k = Fp is the prime field with p elements, compute the number of idempotents in M2 (k). (Count 0 and 1 in.) Solution. (a) Since e 6= 0, 1, the image and the kernel of e are both onedimensional. Let v1 be a nonzero element in the image, so v1 = e(v0 ) for some v0 ∈ k ⊕2 . Then e(v1 ) = e(e(v0 )) = e2 (v0 ) = e(v0 ) = v1 . Pick a nonzero element v2 in the kernel of e, and we get a basis of k ⊕2 in which e takes the form e1 . (b) For a general element g=
a b c d
to be in the stabilizer, it must satisfy ge1 = e1 g. Writing the equation in four entries out, one sees that it means b = c = 0 (and a, d arbitrary). So the centralizer is the subgroup of diagonal matrices. (c) By (a) and (b), the set of rank 1 idempotents is in bijection with GL2 (Fp )/T (Fp ), whose cardinality is (p2 − 1)(p2 − p) = (p + 1)p. (p − 1)(p − 1) So the total number of idempotents is equal to p2 + p + 2.
1
2. (Algebraic Geometry) (a) Find an everywhere regular differential nform on the affine nspace An . (b) Prove that the canonical bundle of the projective ndimensional space Pn is O(−n − 1). Solution (Sketch). Part (a) is really a hint for Part (b). Letting x1 , x2 , . . . , xn be affine ( An ) coordinates, put ω := dx1 ∧ dx2 · · · ∧ dxn giving (a) . Denoting the corresponding homogenous Pn coordinates t0 , t1 , . . . , tn , with xi := ti /t0 for i = 1, 2, . . . , n extend ω to Pn writing dxi = dti /t0 − ti /t20 dt0 and wedging to discover that the divisor of poles of ω is (n + 1)H where H is the hyperplane at infinity (t0 = 0) and then conclude (appropriately). ˜ be a domain in 3. (Complex Analysis) (Bol’s Theorem of 1949). Let W ˜ . Let ε > 0 and C and W be a relatively compact nonempty subdomain of W Gε be the set of all (a, b, c, d) ∈ C such that max (a − 1, b, c, d − 1) < ε. ˜ for z ∈ W and (a, b, c, d) ∈ Gε . Let Assume that cz + d 6= 0 and az+b ∈W cz+d m ≥ 2 be an integer. Prove that there exists a positive integer ` (depending ˜ such on m) with the property that for any holomorphic function ϕ on W that az + b (cz + d)2m ϕ(z) = ϕ cz + d (ad − bc)m for z ∈ W and (a, b, c, d) ∈ Gε , the `th derivative ψ(z) = ϕ(`) (z) of ϕ(z) on ˜ satisfies the equation W az + b (ad − bc)`−m ψ(z) = ψ cz + d (cz + d)2(`−m) for z ∈ W and (a, b, c, d) ∈ Gε . Express ` in terms of m. Hint: Use Cauchy’s integral formula for derivatives. Solution. Let
az + b cz + d for A ∈ Gε . We take a positive integer ` which we will determine later as a function of n. We use Cauchy’s integral formula for derivatives to take ˜ we use U (z) to denote an open the `th derivative ψ(z) of ϕ(z). For z ∈ W Az =
2
˜ and use ∂U (z) to denote its boundary. The `th neighborhood of z in W ˜ is given by the formula derivative ψ of ϕ at z ∈ W Z ϕ(ζ)dζ `! ψ(z) = √ 2π −1 ζ∈∂U (z) (ζ − z)`+1 and
`! ψ(Az) = √ 2π −1
It follows from
Z ζ∈∂U (Az)
ϕ(ζ)dζ (ζ − Az)`+1
˜. when Az ∈ W
ζ ∈ U (z) ⇐⇒ Aζ ∈ U (Az), ζ ∈ ∂U (z) ⇐⇒ Aζ ∈ ∂U (Az),
with the change of variable ζ 7→ Aζ, that Z Z ϕ(Aζ)d(Aζ) ϕ(ζ)dζ = . `+1 `+1 Aζ∈∂U (Az) (Aζ − Az) ζ∈∂U (Az) (ζ − Az) From the following straightforward direct computation of the discrete version of the formula for the derivative of fractional linear transformation aζ + b az + b − cζ + d cz + d (aζ + b)(cz + d) − (az + b)(cζ + d) = (cζ + d)(cz + d) (acζz + bcz + adζ + bd) − (acζz + adz + bcζ + bd) = (cζ + d)(cz + d) (ad − bc)(ζ − z) = (cζ + d)(cz + d)
Aζ − Az =
we obtain Z Aζ∈∂U (Az)
ϕ(Aζ)d(Aζ) = (Aζ − Az)`+1
Z ζ∈∂U (z)
ϕ
aζ+b cζ+d
ad−bc dζ (cζ+d)2
(ad−bc)`+1 (ζ−z)`+1 (cζ+d)`+1 (cz+d)`+1 m
Z =
ad−bc ϕ(ζ) (ad−bc) dζ (cζ+d)2m (cζ+d)2 (ad−bc)`+1 (ζ−z)`+1 (cζ+d)`+1 (cz+d)`+1
ζ∈∂U (z)
(cz + d)`+1 = (ad − bc)`−m 3
Z ζ∈∂U (z)
ϕ(ζ)dζ (cζ + d)`−1−2m . (ζ − z)`+1
The extra factor (cζ + d)`−1−2m inside the integrand on the extreme righthand side becomes 1 and can be dropped if ` − 1 − 2m = 0, that is, if ` = 2m + 1. Thus, if ` = 2m + 1, then ψ(Az) =
(cz + d)`+1 ψ(z). (ad − bc)`−m
That is, ψ(z) = ψ
az + b cz + d
(ad − bc)`−m , (cz + d)2(`−m)
because ` = 2m + 1 implies ` + 1 = 2(` − m). 4. (Algebraic Topology) (a) Show that the Euler characteristic of any contractible space is 1. (b) Let B be a connected CW complex made of finitely many cells so that its Euler characteristic is defined. Let E → B be a covering map whose fibers are discrete, finite sets of cardinality N . Show the Euler characteristic of E is N times the Euler characteristic of B. (c) Let G be a finite group with cardinality > 2. Show that BG (the classifying space of G) cannot have homology groups whose direct sum has finite rank. Solution. (a) The homology of a point with coefficients P in a ifield k is H0 = k, Hi = 0 for i > 0. Hence its Euler characteristic is (−1) dim Hi = 1. All contractible spaces are homotopy equivalent so their Euler characteristic is that of the point. (b) For any open cover {Ui }, we know that the chain complex of singular chains living in Ui for some i has equivalent homology to the chain complex of all chains. Taking the cover of B by trivializing neighborhoods Ui , the chain complex of chains living in Ui receives a map from chains in E living in π −1 (Ui ). The latter is simply G direct sums of the former, and the chain map between them is the “add every component” map. This shows the ranks of homology of E is N times the rank of homology of B. (c) Strictly speaking, this problem cannot be solved based on easy machinery (as far as I know). A much more reasonable problem would be: Prove BG is not homotopy equivalent to anything made up of only finitely many cells. I did not take off points for people not distinguishing between this condition, 4
and the condition stated in the problem itself. We know BG = EG/G, but EG is contractible. So χ(EG) = 1. If BG has finite homology, χ(BG) = 1/G, which cannot be an integer unless G = 1. 5. (Differential Geometry) Let H = {(x, y) ∈ R2 : y > 0} be the upper half plane. Let g be the Riemannian metric on H given by g=
(dx)2 + (dy)2 . y2
(H, g) is known as the halfplane model of the hyperbolic plane. (a) Let γ(θ) = (cos θ, sin θ) and η(θ) = (cos θ + 1, sin θ) for θ ∈ (0, π) be two paths in H. Compute the angle A at their intersection point shown in Figure 1, measured by the metric g.
Figure 1: Angle A between the two curves γ and η in the upper half plane H.
(b) By computing the LeviCivita connection 2
X ∂ ∂ ∇ ∂ = Γkij ∂xi ∂x ∂xk j k=1 of g or otherwise (where (x1 , x2 ) = (x, y)), show that the path γ, after arclength reparametrization, is a geodesic with respect to the metric g. √ Solution. (a) The intersection point is (1/2, 3/2): solving for γ(θ) = (cos θ, sin θ) = (cos φ + 1, sin φ) = η(φ) we obtain θ = π/3, φ = 2π/3. 5
The angle A satisfies hγ 0 (π/3), −η 0 (2π/3)ig γ 0 (π/3)g  − η 0 (2π/3)g √ √ h(− 3/2, 1/2), ( 3/2, 1/2)ig √ √ = (− 3/2, 1/2)g ( 3/2, 1/2)g − 21 y12 = 1
cos A =
y2
=−
1 2
and so A = 2π/3. (b) Using the formula 1 Γijk = g il (gjl,k + gjl,j − gjk,l ) 2 one obtains Γijk =
−1 (δij δk,2 + δki δj,2 − δjk δi,2 ). y
After arclength reparametrization, the tangent vectors of the path are v(θ) =
γ 0 (θ) = (− sin2 θ, sin θ cos θ). γ 0 (θ)g
Then
0
∇v(θ) v(θ) = v (θ) +
Γ11 Γ12 Γ21 Γ22
· v(θ)
where Γ11 Γ12 Γ21 Γ22
= (− sin θ)Γ111 + (cos θ)Γ121 = (− sin θ)Γ112 + (cos θ)Γ122 = (− sin θ)Γ211 + (cos θ)Γ221 = (− sin θ)Γ212 + (cos θ)Γ222
Thus one has ∇v(θ) v(θ) = 0.
6
= − cot θ; = 1; = −1; = − cot θ.
6. (Real Analysis) any positive integer n let Mn be a positive number PFor ∞ such that the series n=1 Mn of positive numbers is convergent and its limit is M . Let a < b be real numbers and fn (x) be a realvalued continuous function on [a, b] for any positive integer n such that its derivative fn0 (x) exists P for every a < x < b with fn0 (x) ≤ Mn for a < x < b. Assume that the series ∞ n=1 fn (a) of real numbers converges. Prove that P (a) the series ∞ n=1 fn (x) converges to some realvalued function f (x) for every a ≤ x ≤ b, (b) f 0 (x) exists for every a < x < b , and (c) f 0 (x) ≤ M for a < x < b. Hint for (b): For fixed x ∈ (a, b) consider the series of functions ∞ X fn (y) − fn (x) n=1
y−x
of the variable y and its uniform convergence. Solution. (a) Fix x ∈P (a, b]. For q > p ≥ 1, by the Mean Value Theorem applied to the function qn=p fn on [a, x] we can find a < ξp,q < x such that q X
fn (x) −
n=p
q X
fn (a) = (x − a)
n=p
q X
fn0 (ξp,q ) ,
n=p
which implies that q q q X X X fn (x) ≤ fn (a) + (x − a) fn0 (ξp,q ) n=p n=p n=p q q X X ≤ fn (a) + (x − a) Mn . n=p
n=p
P∞
P∞
Since both series n=1 fn (a) and n=1 Mn are convergent and therefore Cauchy, for any ε > 0 we can find a positive integer N1 such that q X ε fn (a) < 2 n=p
7
for q > p ≥ N1 and we can find a positive integer N2 such that q X ε Mn < 2(x − a) n=p
for q > p ≥ N2 . Thus for n ≥ max(N1 , N2 ) we have q X fn (x) < ε n=p P P∞ and the series ∞ n=1 fn (x) is Cauchy. Hence the series n=1 fn (x) converges to some realvalued function f (x) for every a ≤ x ≤ b. (b) Before the proof of the statement in (b), we would like to state that the uniform limit of continuous functions is continuous. That is, if hn (x) is a sequence of functions on a metric space E which converges to a function h(x) on E uniformly on E and if for some x0 ∈ E and for every n the function hn (x) is continuous at x = x0 , then h(x) is continuous at x0 . This results from the socalled 3ε argument as follows. Given any ε > 0. The uniform convergence of hn → h on E implies that there exists some positive integer N such that hN (x) − h(x) < ε for all x ∈ E. Since hN is continuous at x = x0 , there exists some δ > 0 such that hN (x) − hN (x0 ) < ε for dE (x, x0 ) < δ (where dE (·, ·) is the metric of the metric space E). Thus for dE (x, x0 ) < δ we have h(x) − h (x0 ) ≤ h(x) − hN (x)+hN (x) − hN (x0 )+hN (x0 ) − h (x0 ) < 3ε, which implies the continuity of h at x = x0 . We now prove the statement in (b). Take x0 ∈ (a, b). We introduce the function gn,x0 (x) on [a, b] which is defined by f (x)−f (x ) gn,x0 (x) = n x−xn0 0 for x 6= x0
gn,x0 (x0 ) = fn0 (x0 ) .
It follows from the continuity of fn on [a, b] and the existence of fn0 (x0 ) that gn,x0 is a continuous function on [a, b]. 8
When x ∈ [a, b] with x 6= x0 , by the Mean Value Theorem fn (x) − fn (x0 ) = fn0 (ξx ) x − x0 for some ξx strictly between x0 and x and as a consequence gn,x0 (x) = fn0 (x0 ) ≤ Mn . When x = x0 , gn,x0 (x) = fn0 (x0 ) ≤ Mn . P Thus gn,x0 (x)P ≤ Mn for x ∈ [a, b]. From ∞ n=1 Mn ≤ M < ∞ it follows ∞ that the seriesP n=1 gn,x0 is uniformly convergent on [a, b]. It follows that the uniform limit ∞ n=1 gn,x0 is a continuous function on [a, b] by the 3ε argument given above. For x 6= x0 ∞ X
gn,x0 (x) =
x − x0
n=1
n=1
The continuity of
∞ X fn (x) − fn (x0 )
P∞
n=1
=
f (x) − f (x0 ) . x − x0
gn,x0 (x) at x = x0 means that the limit of f (x) − f (x0 ) x − x0
exists as x → x0 , which implies that f 0 (x0 ) exists and is equal to ∞ X
gn,x0 (x0 ) =
∞ X
n=1
fn0 (x0 ) .
n=1
(c) From 0
f (x0 ) =
∞ X
gn,x0 (x0 ) =
n=1
and
fn0
∞ X
fn0 (x0 )
n=1
(x0 ) ≤ Mn , it follows that 0
f (x0 ) ≤
∞ X n=1
9
Mn = M.
Solutions of Qualifying Exams II, 2013 Fall 1. (Algebra) Find all the field automorphisms of the real numbers R. Hint: Show that any automorphism maps a positive number to a positive number, and deduce from this that it is continuous. Solution. If t > 0, there exists an element s 6= 0 such that t = s2 . If ϕ is any field automorphism of R, then ϕ(t) = ϕ(s2 ) = (ϕ(s))2 > 0. It follows that ϕ preserves the order on R: If t < t0 , then ϕ(t0 ) = ϕ(t + (t0 − t)) = ϕ(t) + ϕ(t0 − t) > ϕ(t). Any real number α is determined by the set (Dedekind’s cut) of rational numbers that are less than α, and any field automorphism fixes each rational number. Therefore ϕ is the identity automorphism. 2. (Algebraic Geometry) What is the maximum number of ramification points that a mapping of finite degree from one smooth projective curve over C of genus 1 to another (smooth projective curve of genus 1) can have? Give an explanation for your answer. Solution (Sketch). By the RiemannHurwitz formula, if we have a mapping f of finite degree d from one smooth projective (irreducible, say) curve onto another the Euler characteristic of the source curve is d times the Euler characteristic of the target minus a certain nonnegative number e, and moreover e is zero if and only if the mapping is unramified. Now compute: the Euler characterstic of our source and target curves is, by hypothesis, 0 and so this e is zero, and therefore the mapping is unramified. 3. (Complex Analysis) Let ω and η be two complex numbers such that Im ωη > 0. Let G be the closed parallelogram consisting of all z ∈ C such that z = λω + ρη for some 0 ≤ λ, ρ ≤ 1. Let ∂G be the boundary of G and Let G0 = G − ∂G be the interior of G. Let P1 , · · · , Pk , Q1 , · · · , Q` be points in G0 and let m1 , · · · , mk , n1 , · · · , n` be positive integers. Let f be a function on G such that Q f (z) `j=1 (z − Qj )nj Qk mp p=1 (z − Pp ) 10
is continuous and nowhere zero on G and is holomorphic on G0 . Let ϕ(z) and ψ(z) be two polynomials on C. Assume that f (z + ω) = eϕ(z) f (z) if both ψ(z) z and z + ω are in G. Assume Pk also that P` f (z + η) = e f (z) if both z and z + η are in G. Express p=1 mp − j=1 nj in terms of ω and η and the coefficients of ϕ(z) and ψ(z). Solution. Let A = 0, B = η, C = η + ω, and D = ω. Since Im ωη > 0, it follows that going from A to B, to C, to D and then back to A is in the counterclockwise direction. By the argument principle I k ` X X 1 d log f mp − nj = √ 2π −1 ∂G p=1 j=1 Z Z Z Z 1 = √ d log f + −−→ d log f + −−→ d log f + −−→ d log f −→ 2π −1 AB Z ZBC ZCD ZDA 1 d log f − −−→ d log f + −−→ d log f − −−→ d log f = √ −→ 2π −1 AB CD AD Z BC Z 1 = √ − −→ dϕ(z) + −−→ dψ(z) 2π −1 AB AD 1 (−ϕ(η) + ϕ(0) + ψ(ω) − ψ(0)) . = √ 2π −1 Thus, the answer is k X p=1
mp −
` X j=1
nj =
1 √
2π −1
(−ϕ(η) + ϕ(0) + ψ(ω) − ψ(0)) .
4. (Algebraic Topology) (a) Fix a basis for H1 of the twotorus (with integer coefficients). Show that for every element x ∈ SL(2, Z), there is an automorphism of the twotorus such that the induced map on H1 acts by x. Hint: SL(2, Z) also acts on the universal cover of the torus. (b) Fix an embedding j : D2 × S 1 → S 3 . Remove its interior from S 3 to obtain a manifold X with boundary T 2 . Let f be an automorphism of the twotorus and consider the glued space Xf := (D2 × S 1 ) ∪f X. If X is homotopy equivalent to D2 × S 1 , compute the homology groups of Xf . 11
Solution. (a) Given g ∈ SL(2, Z) ⊂ SL(2, R) let x : R2 → R2 be the induced action. Since g is in SL(2, Z) it respects the relationship of whether two vectors in R2 differ by integer coordinates. So the map on the torus [(x1 , x2 )] 7→ [g(x1 , x2 )] is welldefined. This clearly sends a homology generating pair given by the curves (x1 , 0) and (0, x2 ) to the expected images via g. (b) There is an ambiguity in the problem about how f glues X and D2 × S 1 together; so I gave full credit regardless of whether you identified this ambiguity or not. Note Xf = (D2 × S 1 ) ∪S 1 ×S 1 X. Write U = D2 × S 1 and V = X. The MayerVietoris sequence gives H0 (U ∩ V )
/
H0 (U ) ⊕ H0 (V )
/
H0 (U ∪ V )
/
H1 (U ∩ V )
/
H1 (U ) ⊕ H1 (V )
/
H1 (U ∪ V )
/
H2 (U ∩ V )
/
H2 (U ) ⊕ H2 (V )
/
H2 (U ∪ V )
/
H3 (U ∩ V )
/
H3 (U ) ⊕ H3 (V )
/
H3 (U ∪ V )
/
but because we know the homology of D2 × S 1 ' S 1 and S 1 × S 1 , we can fill in various groups in the long exact sequence: /
Z⊕Z
/
H0 (U ∪ V )
/
Z⊕Z
/
H1 (U ∪ V )
Zj Z2 j
g
Zj
/
0⊕0
/
H2 (U ∪ V )
0
/
0⊕0
/
H3 (U ∪ V )
Since g is an isomorphism, we know H1 must inject into Z, but the inclusion map H0 (U ∩ V ) → H0 (U ) ⊕ H0 (V ) is an injection, so H1 (U ∪ V ) = 0. 12
We know H0 is either equal to Z from the long exact sequence above, or by observing that Xf is pathconnected. If f induces an isomorphism, we see H2 must be zero; this was the intent of the problem, but you can get a different answer based on how you interpreted the ”gluing” by f . Finally, H3 is also isomorphic to Z by the exactness of the above sequence. 5. (Differential Geometry) Let M = U (n)/O(n) for n ≥ 1, where U (n) is the group of n×n unitary matrices and O(n) is the group of n×n orthogonal matrices. M is a real manifold called the Lagrangian Grassmannian. (a) Compute and state the dimension of M . (b) Construct a Riemannian metric which is invariant under the left action of U (n) on M . (c) Let ∇ be the corresponding LeviCivita connection on the tangent bundle T M , and X, Y, Z be any U (n)invariant vector fields on M . Using the given identity (which you are not required to prove) 1 ∇X Y = [X, Y ], 2 show that the Riemannian curvature tensor R of ∇ satisfies the formula 1 R(X, Y )Z = [Z, [X, Y ]]. 4 Solution. (a) T[I] M ∼ = u(n)/o(n) ∼ = Sym2 (Rn ) where Sym2 (Rn ) denotes the space of real n × n symmetric matrices. Thus dim M =
n(n + 1) . 2
(b) Define a metric on Sym2 (Rn ) by hA, Bi = tr(AB t ) = tr(AB). g ∈ O(n) acts on T[I] M ∼ = Sym2 (Rn ) by g · A = gAg −1 . Then hg · A, g · Bi = tr(g · ABg −1 ) = hA, Bi. Hence this metric is invariant under the action of O(n). By translating the metric to tangent spaces at other points by the action of U (n), this gives a welldefined invariant metric on U (n)/O(n). 13
(c) 1 ∇X Y = [X, Y ]. 2 Then R(X, Y )Z = ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z 1 1 = ([X, [Y, Z]] − [Y, [X, Z]]) − [[X, Y ], Z] 4 2 1 = [Z, [X, Y ]] 4 where the last equality follows from Jacobi identity. 6. (Real Analysis) Show that there is no function f : R → R whose set of continuous points is precisely the set Q of all rational numbers. Solution. For fixed δ > 0 let C(δ) be the set of points x ∈ R such that for some ε > 0 we have f (x0 ) − f (x00 ) < δ for all x0 , x00 ∈ (x − ε, x + ε). Clearly C(δ) is open since for every x ∈ C(δ), we have (x − ε, x + ε) ⊂ C(δ). Now let C denote the set of continuous points of f. From the definitions, we have that ∞ \ C= C(1/n). n=1
Now suppose that C = Q. Then R−Q=
∞ [
Xn ,
n=1
where Xn = R − C(1/n). Since C(1/n) is open, Xn is closed. Also Q is countable, say Q = {q1 , q2 , . . . }. Let Yn = {qn }. Then ! ! ∞ ∞ [ [ R= Xn ∪ Yn , n=1
n=1
i.e. we have written R as a countable union of closed sets. Then by Baire’s theorem, some Xn or Yn has nonempty interior. Clearly it cannot be one of the Yn . So there exists Xn containing an interval (a, b). But this is impossible because Xn ⊂ R − Q and every interval contains a rational number. Thus, we obtain a contradiction, which shows that C 6= Q. 14
Solutions of Qualifying Exams III, 2013 Fall 1. (Algebra) Consider the function fields K = C(x) and L = C(y) of one variable, and regard L as a finite extension of K via the Calgebra inclusion x 7→
−(y 5 − 1)2 4y 5
Show that the extension L/K is Galois and determine its Galois group. Solution. Consider the intermediate extension K 0 = C(y 5 ). Then clearly [L : K 0 ] = 5 and [K 0 : K] = 2, therefore [L : K] = 10. Thus, to prove that L/K is Galois it is enough to find 10 field automorphisms of L over K. Choose a primitive 5th root of 1, say ζ = e2πi/5 . For i ∈ Z/5 and s ∈ {±1}, the Calgebra automorphism σi,s of L defined by y 7→ ζ i y s leaves x, hence K, fixed. There can be many ways to determine the group, here’s one. Looking at the law of composition of these automorphisms, one sees that the subgroup Gal(L/K 0 ) ' Z/5, (which is necessarily normal, being of index 2) is not central, for conjugation by σ0,−1 acts as −1 on it. So the group is the dihedral group of 10 elements. 2. (Algebraic Geometry) Is every smooth projective curve of genus 0 defined over the field of complex numbers isomorphic to a conic in the projective plane? Give an explanation for your answer. Solution (Sketch). Yes. Apply the RiemannRoch theorem which guarantees the existence of a nonconstant meromorphic function with a simple pole at exactly one point. Argue that this meromorphic function identifies the curve with P1 , and using that fact, embed the curve as a conic in the plane in any convenient way, e.g., If t0 , t1 are projective (P1 ) coordinates, let z0 = t20 , z1 = t0 t1 z2 = t21 be the map to P2 . The conic, then, would be z0 z2 = z12 . (Alternatively: one can consider the complete linear system attached to the anticanonical divisor.) 15
3. (Complex Analysis) Let f (z) = z +e−z for z ∈ C and let λ ∈ R, λ > 1. Prove or disprove the statement that f (z) takes the value λ exactly once in the open right halfplane Hr = {z ∈ C : Re z > 0}. Solution. First, let us consider the real function f (x) = x + e−x . Since f is continuous, f (0) = 1 and limx→∞ f (x) = ∞, by the intermediate value theorem, there exists u ∈ R such that f (u) = λ. Now let us show that such u is unique. Let R > 2λ and let Γ be the closed right half disk of radius R centered at the origin n πo π . {z = x + iy ∈ C : x = 0, y ≤ R} ∪ z ∈ C : z = R, − ≤ arg(z) ≤ 2 2 Let F (z) = λ − z and G(z) = −e−z . Then for z ∈ Γ, we have G(z) = e−Rez  ≤ 1 since Re z ≥ 0, while F (z) > 1 by construction. Hence by Rouch´e’s theorem, λ − f (z) = F (z) + G(z) has the same number of zeros inside Γ as F (z), namely 1. Since this is true for all R large enough, we conclude that the point u is unique. 4. (Algebraic Topology) (a) Let X and Y be locally contractible, connected spaces with fixed basepoints. Let X ∨ Y be the wedge sum at the basepoints. Show that π1 (X ∨ Y ) is the free product of π1 X with π1 Y . (b) Show that π1 (X × Y ) is the direct product of π1 X with π1 Y . (c) Note the canonical inclusion f : X ∨ Y → X × Y . Assume that X and Y have abelian fundamental groups. Show that the map f∗ on fundamental groups exhibits π1 (X × Y ) as the abelianization of π1 (X ∨ Y ). Hint: The Hurewicz map is natural. Solution. (a) This follows form the Van Kampen theorem: Writing X ∨ Y as the union X ∪∗ Y we have that π1 (X ∨ Y ) ∼ = π1 (X) ∗π (∗) π1 (Y ) = π1 (X) ∗ π1 Y . 1
(b) There is the obvious continuous map M aps∗ (S 1 , X) × M aps∗ (S 1 , Y ) → M aps∗ (S 1 , X × Y ) given by sending (t 7→ γX (t), t 7→ γY (t)) 7→ (t 7→ (γX (t), γY (t))). This map is a continuous so it induces a map π0 (M aps∗ (S 1 , X) × M aps∗ (S 1 , Y )) → π0 M aps∗ (S 1 , X × Y ) 16
where the lefthand side is isomorphic to π0 M aps∗ (S 1 , X)×π0 M aps∗ (S 1 , Y )). Further, the above map is clearly a bijection, so it induces an injection and a surjection on π0 . (c) The Hurewicz map is natural so we have a commutative diagram π1 (X ∨ Y )
f∗
/ π1 (X
×Y)
q
H1 (X ∨ Y )
f∗
/
H1 (X × Y )
where the vertical maps are abelianizations by the Hurewicz theorem. But the lowerright corner is equal to H1 (X) × H1 (Y ) by the Kunneth theorem (since X and Y are connected), and the bottom copy of f∗ is the obvious isomorphism on H1 . Since q is an abelianization by definition, but the bottom arrow and rightmost arrow are both isomorphisms, the top arrow must also be an abelianization. 5. (Differential Geometry) (a) Let S1 = R/Z be a circle and consider the connection √ ∇ := d + π −1dθ defined on the trivial complex line bundle over S1 , where θ is the standard coordinate on S1 = R/Z descended from R. By solving the differential equation for flat sections f (θ) √ ∇f = df + π −1f dθ = 0 or otherwise, show that there does not exist global flat sections with respect to ∇ over S1 . (b) Let T = V /Λ be a torus, where Λ is a lattice and V = Λ ⊗ R is the real vector space containing Λ. Let L be the trivial complex line bundle equipped with the standard Hermitian metric. By identifying flat U (1) connections with U (1) representations of the fundamental group π1 (T ) or otherwise, show that the space of flat unitary connections on L is the dual torus T ∗ = V ∗ /Λ∗ , where Λ∗ := Hom(Λ, Z) is the dual lattice and V ∗ := Hom(V, R) is the dual vector space.
17
Solution. (a) The differential equation √ f 0 (θ) + π −1f (θ) = 0 has a unique solution f (θ) = Ae−π
√
−1θ
up to a constant A ∈ C. This is not a welldefined function over S1 because f (0) 6= f (1). (b) The space of flat Gconnections over T can be identified as Hom(π1 (T ), G)/AdG. Since π1 (T ) = Λ and for the abelian group G = U (1) the adjoint action is trivial, we have Hom(π1 (T ), G)/AdG = Hom(Λ, U (1)) = T ∗ . 6. (Real Analysis) (Fundamental Solutions of Linear Partial Differential Equations with Constant Coefficients). Let Ω be an open (−M, M ) in Pninterval dν R with M > 0. Let n be a positive integer and L = ν=0 aν dxν be a linear differential operator of order n on R with constant coefficients, where the coefficients a0 , · · · ,P an−1 , an 6= 0 are complex numbers and x is the coordinate n dν ν ∗ of R. Let L = ν=0 (−1) aν dxν . Prove, by using Plancherel’s identity, that there exists a constant c > 0 which depends only on M and an and is independent of a0 , a1 , · · · , an−1 such that for any f ∈ L2 (Ω) a weak solution u of Lu = f exists with kukL2 (Ω) ≤ c kf kL2 (Ω) . Give one explicit expression for c as a function of M and an . Hint: A weak solution u of Lu = f means that (f, ψ)L2 (Ω) = (u, L∗ ψ)L2 (Ω) for every infinitely differentiable function ψ on Ω with compact support. For the solution of this problem you can consider as known and given the following three statements. (I) If there exists a positive number c > 0 such that kψkL2 (Ω) ≤ c kL∗ ψkL2 (Ω) for all infinitely differentiable complexvalued functions ψ on Ω with compact support, then for any f ∈ L2 (Ω) a weak solution u of Lu = f exists with kukL2 (Ω) ≤ c kf kL2 (Ω) .
18
P k (II) Let P (z) = z m + m−1 k=0 bk z be a polynomial with leading coefficient 1. If F is a holomorphic function on C, then Z 2π 1 2 P eiθ F eiθ 2 dθ. F (0) ≤ 2π θ=0 (III) For an L2 function f on R which is zero outside Ω = (−M, M ) its Fourier transform Z M ˆ f (x)e−2πixξ dx f (ξ) = −M
as a function of ξ ∈ R can be extended to a holomorphic function Z M f (x)e−2πix(ξ+iη) dx fˆ (ξ + iη) = −M
on C as a function of ξ + iη. Solution. This problem is to compute the constant c in Lemma 3.3 on p.225 of the book of Stein and Shakarchi on Real Analysis by going over its arguments and keeping track of the constants involved in each step. Introduce the polynomial Q(ζ) =
n X
(−1)k ak (2πζ)k
k=0
so that
(#)
∗ ψ (ζ) = Q(ζ)ψ(ζ) d ˆ L
any ψ ∈ C0∞ (R), where b denotes taking the Fourier transform. Consider 1 n in the first the special case where an = (2πi) n so that the coefficient of ξ √ polynomial Q(ζ) of degree n in ζ is 1. Writing ζ = ξ + −1η (with both ξ and η real) and taking the L2 of both sides of (#) over R as functions of η. Then Z ∞ Z ∞ 2 2 d ˆ ([) Q (ξ + iη) ψ (ξ + iη) dξ = L∗ ψ (ξ + iη) dξ. −∞
−∞
Since from the definition of Fourier transform Z ∞ Z ∗ −2πi(ξ+iη)x ∗ d (L ψ) (x)e dx = L ψ (ξ + iη) = x=−∞
∞
x=−∞
19
(L∗ ψ) (x)e2πηx e−2πiξx dx,
∗ ψ (ξ + iη) is equal to the value at ξ of the Fourier transd it follows that L form of the function (L∗ ψ) (x)e2πηx . Thus, by applying Plancherel’s identity to the function (L∗ ψ) (x)e2πηx , we get Z ∞ 2 d L∗ ψ (ξ + iη) dξ ξ=−∞ Z ∞ Z ∞ ∗ 2 4πηM 2πηx (L ψ) (x)e dx ≤ e (L∗ ψ) (x)2 dx, = −∞
x=−∞
because the support of ψ(x) (as well as the support of (L∗ ψ) (x)) is in the interval Ω = (−M, M ). Thus from ([) it follows that Z ∞ Z ∞ 2 4πηM (L∗ ψ) (x)2 dx. (]) Q (ξ + iη) ψˆ (ξ + iη) dξ ≤ e −∞
−∞
Setting η = sin θ in (]), we get from η ≤ 1 that Z Z ∞ 2 4πM ˆ (†) Q (ξ + i sin θ) ψ (ξ + i sin θ) dξ ≤ e
∞
(L∗ ψ) (x)2 dx.
−∞
−∞
Replacing ξ by ξ + cos θ in the integrand on the lefthand side of (†), we get Z ∞ 2 ˆ (‡) Q (ξ + cos θ + i sin θ) ψ (ξ + cos θ + i sin θ) dξ −∞ Z ∞ 4πM ≤e (L∗ ψ) (x)2 dx. −∞
By Statement (III) given above the function ψˆ (ξ + iη) as a function of ξ+iη ∈ C is holomorphic on C. Since Q (ξ + iη) as a function of ξ + iη ∈ C is a polynomial of degree n with leading coefficient 1, it follows from Statement ˆ + z) and P (z) = Q(ξ + z) that (II) applied to F (z) = ψ(ξ Z 2π 2 1 ˆ 2 ψ (ξ) ≤ Q (ξ + cos θ + i sin θ) ψˆ (ξ + cos θ + i sin θ) dθ. 2π θ=0 Integrating both sides over ξ ∈ (−∞, ∞) and using (‡), we get Z ∞ Z ∞ Z 2π 2 1 ˆ 2 ˆ ψ (ξ) ≤ Q (ξ + cos θ + i sin θ) ψ (ξ + cos θ + i sin θ) dθ dξ 2π θ=0 ξ=−∞ ξ=−∞ Z 2π Z ∞ 2 1 = Q (ξ + cos θ + i sin θ) ψˆ (ξ + cos θ + i sin θ) dξ dθ 2π θ=0 ξ=−∞ Z ∞ Z 2π Z ∞ 1 2 ∗ 4πM 4πM (L ψ) (x) dx dθ = e (L∗ ψ) (x)2 dx. ≤ e 2π θ=0 −∞ −∞ 20
By applying Plancherel’s formula to ψ, we conclude that kψ (ξ)k2L2 (Ω) ≤ e4πM k(L∗ ψ) (x)k2L2 (Ω) 1 under the additional assumption that an = (2πi) When this additional n. assumption is not satisfied, we can apply the argument for the special case to 1 L an (2πi)n
instead of to L to conclude that kψ (ξ)k2L2 (Ω) ≤
e4πM n 2
an (2π) 
k(L∗ ψ) (x)k2L2 (Ω) ,
or kψ (ξ)kL2 (Ω) ≤ c k(L∗ ψ) (x)kL2 (Ω) , with c=
e2πM . an  (2π)n
By Statement (I) given above, when we set c=
e2πM , an  (2π)n
we can conclude that for any f ∈ L2 (Ω) a weak solution u of Lu = f exists with kukL2 (Ω) ≤ c kf kL2 (Ω) .
21
Qualifying Exams I, Jan. 2013 1. (Real Analysis) Suppose fj , j = 1, 2, . . . and f are real functions on [0, 1]. Define fj → f in measure if and only if for any ε > 0 we have lim µ{x ∈ [0, 1] : fj (x) − f (x) > ε} = 0
j→∞
where µ is the Lebesgue measure on [0, 1]. In this problems, all functions are assumed to be in L1 [0, 1]. (a) Suppose that fj → f in measure. Does it implies that Z lim fj (x) − f (x)dx = 0. j→∞
Prove it or give a counterexample. (b) Suppose that fj → f in measure. Does this imply that fj (x) → f (x) almost everywhere in [0, 1]? Prove it or give a counter example. (c) Suppose that fj (x) → f (x) almost everywhere in [0, 1]. Does it implies that fj → f in measure? Prove it or give a counter example. 2. (Complex analysis) The following questions are independent. a) For any a ∈ (−1, 1), compute
b) For any p > 1, compute
Z
2π 0
Z
dt . 1 + a cos t
∞ 0
dx . xp + 1
3. (Differential Geometry) The Heisenberg group is the subgroup of Sl(3, R) composed of the 3 × 3, upper triangular matrices with 1 on the diagonal, this being the set of matrices of the form: 1 x z 0 1 y , with (x, y, z) ∈ R3 . 0 0 1 This group is observably diffeomorphic to R3 . (a) Compute the Lie algebra of the Heisenburg group. (b) Exhibit a leftinvariant Riemannian metric on the Heisenberg group. 4. (Algebraic Topology) Let H be the space of quaternions, and denote by S3 the unit sphere inside H. The quaternion group G = {±1, ±i, ±j, ±k} acts on H by left multiplication, and the action preserves the unit sphere S3 . Let X be the quotient space S3 /G. Compute its fundamental group π1 (X) and its first homology group H1 (X, Z). (H is spanned by four independent unit vectors 1, i, j, k as a real normed vector space. The multiplication is associative and, between two elements of H, it is bilinear and determined by the rules i2 = j 2 = k 2 = ijk = −1, where 1 is the multiplicative identity.)
1/22/2013 1
5. (Algebra) Let k be a field, and let G be a finite group acting on a vector space V . a) If k = C, prove that any subrepresentation U ⊆ V has a Gstable complement, that is, subrepresentation U ′ ⊆ V such that V = U ⊕ U ′ .
b) Now suppose that k = Z/pZ for some prime p, and that G acts doubly transitively on a set X of size n (that is, if x, y, x′ , y ′ ∈ X with x 6= y and x′ 6= y ′ then there exists g ∈ G such that g(x) = x′ and g(y) = y ′ ). Let V be the tracezero subspace of the corresponding permutation representation over k (recall that the permutation representation is the vector space k S with the natural action of G, so that V is the subspace consisting of vectors whose n coordinates sum to 0). Prove that if n ≡ 0 mod p then the trivial subrepresentation generated by (1, 1, . . . , 1) has no Gstable complement, except for one choice of (p, n, G), and determine that one choice. 6. (Algebraic Geometry) Prove that the following complex algebraic varieties are pairwise nonisomorphic. (a) X1 = Spec C[x, y]/(y 2 − x3 ), X2 = Spec C[x, y]/(y 2 − x3 − x) and X3 = Spec C[x, y]/(y 2 − x3 − x2 ).
(b) X1 = Spec C[x, y]/(xy 2 + x2 y) and X2 = Spec C[x, y, z]/(xy, yz, zx).
(c) X1 = P1C × P1C , X2 = P2C and X3 = the blowing up of X2 at the point [0 : 0 : 1].
2
1/22/2013
Qualifying Exams II, Jan. 2013 (1) (Real Analysis) (a) For any bounded positive function f define Z 1 Z A(f ) := f (x) log f (x)dx, B(f ) := 0
1 0
Z f (x)dx log
1 0
f (y)dy .
There are three possibilities: (i) A(f ) ≥ B(f ) for all bounded positive functions, (ii) B(f ) ≥ A(f ) for all bounded positive functions, and (iii) A(f ) ≥ B(f ) for some functions while B(f ) ≥ A(f ) for some functions. Decide which possibility is correct and prove your answer. If you use any inequality, state all assumptions of the inequality precisely and clearly. (b) Let fˆ denote the Fourier transform of the function f on R. Suppose that f ∈ C ∞ (R) and kfˆ(ξ)kL2 (R) ≤ α, kξ1+ε fˆ(ξ)kL2 (R) ≤ β for some ε > 0. Find a bound on kf kL∞ (R) in terms of α, β and ε.
(2) (Complex analysis) Is there a conformal map between the following domains? If the answer is yes, give such a conformal map. If it is no, prove it. a) From D = {z ∈ C : z < 1} to H = {z ∈ C : Im(z) > 0}. b) From the intersection of the open disks D((0, 0), 3) and D((0, 3), 2) to C. c) From H\(0, i] to H. d) From D to C\(−∞, − 14 ]. (3) (Differential Geometry) View R2 × C as the product complex line bundle over R2 and let θ0 denote the connection on this line bundle whose covariant derivative acts on a given section s as ds with d being the exterior derivative. Let A denote the connection i (xdy − ydx). θ0 + 1 + x2 + y 2 (a) Compute as a function of r ∈ (0, ∞) the linear map from C to C that is obtained by using A to parallel transport a given nonzero vector in C in the clockwise direction on the circle where x2 + y 2 = r2 from the point (r, 0) to itself. (b) Give a formula for the curvature 2form of the connection A. (4) (Algebra) a) Let K/F be a field extension of degree 2n + 1 generated by t. Prove that for every c ∈ K there exists a unique rational function f ∈ F [T ] such that deg(f ) ≤ n and c = f (t). [The degree of a rational function f is the smallest d such that f = P/Q for polynomials P, Q each of degree at most d.] b) Deduce that if [K : F ] = 3 then PGL2 (F ) acts simply transitively by fractional linear transformations on K\F (the complement of F in K). If F  = q < ∞, compute PGL2 (F ) directly, and verify that it equals K − F .
1/23/2013 1
2
(5) (Algebraic Topology) Use Z to denote the subset of R2 that is given using standard polar coordinates (r, θ) by the equation r = cos2 (2θ). The set Z is depicted in Figure 1.
Figure 1. The set Z. (a) Compute the fundamental group of Z. (b) Let D denote the closed unit disk in R2 centered at the origin. The boundary of D is the unit circle, this denoted here by ∂D. Parametrize ∂D by the angle φ ∈ [0, 2π) and let f denote the map from the boundary of ∂D to Z that sends the angle φ to the point in Z with polar coordinates (r = cos2 (2φ), θ = φ). Let X denote the space that is obtained from the disjoint union of D and Z by identifying φ ∈ ∂D with f (φ) ∈ Z. Give a set of generators and relations for the fundamental group of X. (6) (Algebraic Geometry) Let f and g be irreducible homogeneous polynomials in S = C[X0 , X1 , X2 , X3 ] of degrees 2 and 3, respectively. For parts (a) and (b), combinatorial polynomials (such as T2 = T (T − 1)/2) are acceptable in the final answer. (a) Compute the Hilbert polynomial of X = Proj(S/(g)) embedded in P = P3C = Proj(S). (b) Compute the Hilbert polynomial of Y = Proj(S/(f, g)) embedded in P . (c) Assuming in addition that Y is nonsingular, use your answer for part (b) to compute its geometric genus dimC Γ(Y, Ω1Y /C ).
1/23/2013
Qualifying Exams III, Jan. 2013 (1) (Real Analysis) Assume that X1 , X2 , . . . are independent random variables uniformly distributed on [0, 1]. Let Y (n) = n inf{Xi , 1 ≤ i ≤ n}. Prove that Y (n) converges weakly to an exponential random variable, i.e. for any continuous bounded function f : R+ → R, Z (n) E f (Y ) −→ f (u)e−u du. n→∞
R+
(2) (Complex Analysis) The following questions are independent. a) Describe all harmonic functions on the plane R2 that are bounded from above. b) Let h : H = {z ∈ C : Im(z) > 0} → C be holomorphic. Assume that h(z) ≤ 1 for any z ∈ H and i is a zero of h of order m ≥ 1. Prove that, for any z ∈ H, z − i m . h(z) ≤ z + i
(3) (Differential Geometry) Use (t, x, y, z) to denote the Euclidean coordinates for R4 . Let t 7→ a(t) denote a strictly positive function on R. A Riemannian metric on R4 is given by the quadratic form: g = dt ⊗ dt + a(t)2 dx ⊗ dx + dy ⊗ dy + dz ⊗ dz . Compute the components of the Riemann curvature tensor of g using the orthonormal basis {dt, a dx, a dy, a dz} for T⋆ R4 . (4) (Algebraic Topology) Let K ⊂ R3 denote a knot, this being a compact, connected, dimension 1 submanifold. (a) Compute the homology of the complement in R3 of any given knot K. (b) Figure 1 shows a picture of the trefoil knot.
Figure 1. The trefoil knot. Sketch on this picture a curve or curves in the complement of K that generate(s) the first homology of R3 − K. (c) A Seifert surface for a knot in R3 is a connected, embedded surface with boundary, with the knot being the boundary (we do not impose orientability here). By way of an example, view the unit circle in the xy plane as a knot in R3 . This is called the unknot. The unit disk in the xy plane is a Seifert surface for the unknot. (i) Compute the second homology of the complement in R3 of any given Seifert surface for the unknot. (ii) Sketch a Seifert surface for the unknot whose complement is not simply connected.
1/24/2013 1
2
(5) (Algebra) Let k be a finite field of cardinality q, and let L = k(t), the field of rational functions over k in an indeterminate t. Set x = tq − t, K = k(x), and F = k(xq−1 ). a) Prove that L/K is a Galois extension with Gal(L/K) = (k, +) (the additive group of k). b) Prove that L/F is Galois. What is Gal(L/F ), and how does Gal(L/F ) act on t? (6) (Algebraic Geometry) Let X0 be the affine plane curve defined by the equation y 3 − 3y = x5
over the complex numbers, and let X be the projective smooth model of X0 . (a) Show that X0 is nonsingular. (b) Find all a ∈ C for which the polynomial Pa (y) = y 3 − 3y − a has repeated roots. For each such a, factor the polynomial Pa (y). (c) Let π : X −→ P1C be the unique extension of the coordinate map x : X0 −→ A1C . Describe the ramification divisor of π and compute its degree. (d) Compute the genus of X by applying Hurwitz’s theorem to π : X −→ P1 .
1/24/2013
Questions in AG for the qualifying exam, Spring 2013 (draft version by Suh). P1. Prove that the following complex algebraic varieties are pairwise nonisomorphic. (a) X1 = Spec C[x, y]/(y 2 − x3 ), X2 = Spec C[x, y]/(y 2 − x3 − x) and X3 = Spec C[x, y]/(y 2 − x3 − x2 ). (b) X1 = Spec C[x, y]/(xy 2 + x2 y) and X2 = Spec C[x, y, z]/(xy, yz, zx). (c) X1 = P1C × P1C , X2 = P2C and X3 = the blowing up of X2 at the point [0 : 0 : 1]. P2. Let f and g be irreducible homogeneous polynomials in S = C[X0 , X1 , X2 , X3 ] of degrees 2 and 3, respectively. For parts (a) and (b), combinatorial polynomials (such as T2 = T (T − 1)/2) are acceptable in the final answer. (a) Compute the Hilbert polynomial of X = Proj(S/(g)) embedded in P = P3C = Proj(S). (b) Compute the Hilbert polynomial of Y = Proj(S/(f, g)) embedded in P . (c) Assuming in addition that Y is nonsingular, use your answer for part (b) to compute its geometric genus dimC Γ(Y, Ω1Y /C ). P3. Let X0 be the affine plane curve defined by the equation y 3 − 3y = x5 over the complex numbers, and let X be the projective smooth model of X0 . (a) Show that X0 is nonsingular. (b) Find all a ∈ C for which the polynomial Pa (y) = y 3 − 3y − a has repeated roots. For each such a, factor the polynomial Pa (y). (c) Let π : X −→ P1C be the unique extension of the coordinate map x : X0 −→ A1C . Describe the ramification divisor of π and compute its degree. (d) Compute the genus of X by applying Hurwitz’s theorem to π : X −→ P1 .
1
A solution. A1. (a) Only X2 among the three is nonsingular. The normalization map is a settheoretic bijection in the case of X1 , but not in the case of X3 . (b) Since X1 embeds into the affine plane, the Zariski cotangent space at every Cvalued point of X1 has dimension at most 2. At (0, 0, 0) ∈ X2 , the Zariski cotangent space has dimension 3. (c) By B´ezout’s theorem, any two distinct irreducible curves on X2 intersect; this is not the case of X1 , nor of X3 . The exceptional divisor on X3 has selfintersection −1, while no prime divisor on X1 has strictly negative selfintersection (one can translate any prime divisor into a different prime divisor, using the action of P GL2 ×P GL2 by fractional linear transformations). A2. (a) One has a short exact sequence of sheaves on P : 0 −→ OP (−3) −→ OP −→ OX −→ 0, hence the Hilbert polynomial is T +3 T +3−3 3 3 hX (T ) = − = T 2 + T + 1. 3 3 2 2 (b) By the assumptions on f and g, they are relatively prime, and we have an exact sequence 0 −→ OP (−5) −→ OP (−2) ⊕ OP (−3) −→ OP −→ OY −→ 0, hence the Hilbert polynomial is T +3 T +1 T T −2 hY (T ) = − − + = 6T − 3 3 3 3 3 (c) By Serre duality it is equal to the arithmetic genus, or 1 − χ(OY ) = 1 − hY (0) = 4. A3. (a) By the Jacobian criterion of smoothness, at a singular point we have y 3 − 3y = x5 , 0 = 5x4 and 3y 2 − 3 = 0, which has no solution. (b) We have Pa′ (y) = 3(y 2 − 1), and Pa (y) =
y ′ P (y) − 2y − a, 3 a
so Pa (y) is separable exactly when a 6= ±2. When a = 2, Pa (y) = (y + 1)2 (y − 2) and when a = −2, Pa (y) = (y − 1)2 (y + 2). 2
(c) For x ∈ A1 (C), the fiber π −1 (x) consists of three distinct points when x5 6= ±2. When x5 = ±2, the fiber consists of two points, one point Px with multiplicity two and the other with multiplicity one. Since 3 is prime to 5, π −1 (∞) consists of one point P∞ with multiplicity three. In this notation, the ramification divisor is X R= 1 · [Px ] + 2[P∞ ], x5 =±2
and has degree 12. (d) Let g be the genus of X, and Hurwitz formula reads 2g − 2 = 3 · (2 · 0 − 2) + 12, so g = 4.
3
PROPOSED QUESTIONS FOR QUALIFYING EXAMINATION IN ALGEBRAIC TOPOLOGY (2013 SPRING)
(1) Let H be the space of quaternions, and denote by S3 the unit sphere inside H. The quaternion group G = {±1, ±i, ±j, ±k} acts on H by left multiplication, and the action preserves the unit sphere S3 . Let X be the quotient space S3 /G. Compute its fundamental group π1 (X) and its first homology group H1 (X, Z). (H is spanned by four independent unit vectors 1, i, j, k as a real normed vector space. The multiplication between two elements of H is bilinear and is determined by the rules i2 = j 2 = k 2 = ijk = −1, and 1 is the multiplicative identity.) (2) Use Z to denote the subset of R2 that is given using standard polar coordinates (r, θ) by the equation r = cos2 (2θ). The set Z is depicted in Figure 1.
Figure 1. The set Z. (a) Compute the fundamental group of Z. (b) Let D denote the closed unit disk in R2 centered at the origin. The boundary of D is the unit circle, this denoted here by ∂D. Parametrize ∂D by the angle φ ∈ [0, 2π) and let f denote the map from the boundary of ∂D to Z that sends the angle φ to the point in Z with polar coordinates (r = cos2 (2φ), θ = φ). Let X denote the space that is obtained from the disjoint union of D and Z by identifying φ ∈ ∂D with f (φ) ∈ Z. Give a set of generators and relations for the fundamental group of X. 1
2 PROPOSED QUESTIONS FOR QUALIFYING EXAMINATION IN ALGEBRAIC TOPOLOGY (2013 SPRING)
(3) Let K ⊂ R3 denote a knot, this being a compact, connected, dimension 1 submanifold. (a) Compute the homology of the complement in R3 of any given knot K. (b) Figure 2 shows a picture of the trefoil knot.
Figure 2. The trefoil knot. Sketch on this picture a curve or curves in the complement of K that generate(s) the first homology of R3 − K. (c) A Seifert surface for a knot in R3 is a connected, embedded surface with boundary, with the knot being the boundary (we do not impose orientability here). By way of an example, view the unit circle in the xy plane as a knot in R3 . This is called the unknot. The unit disk in the xy plane is a Seifert surface for the unknot. (i) Compute the second homology of the complement in R3 of any given Seifert surface for the unknot. (ii) Sketch a Seifert surface for the unknot whose complement is not simply connected.
PROPOSED QUESTIONS FOR QUALIFYING EXAMINATION IN ALGEBRAIC TOPOLOGY (2013 SPRING) 3
Proposed Answer (1) Since G is a finite group and it acts on S3 freely, the quotient map S3 → X is a covering map. S3 is simply connected and hence it is the universal cover of X, where G acts as Deck transformations. Thus π1 (X) = G. The first homology group is the abelianization of the fundamental group. Thus H1 (X) = π1 (X)/[π1 (X), π1 (X)] = G/[G, G] = G/{±1} (2) (a) (b) (3) (a) (b) (c)
= Z2 ⊕ Z2 .
The fundamental group is Z ∗ Z ∗ Z ∗ Z. The fundamental group is generated by a, b, c, d with abcd = 1. H0 = Z, H1 = Z, H2 = Z, H3 = 0 by Mayer Vietoris sequence. A circle surrounding a segment of the knot. (i) H2 = Z. (ii) A Mobius strip.
Exercise 1. The following questions are independent. a) For any a ∈ (−1, 1), compute Z 2π
dt . 1 + a cos t
0
b) For any p > 1, compute
Z
∞ 0
dx . xp + 1
Exercise 2. Is there a conformal map between the following domains? If the answer is yes, give such a conformal map. If it is no, prove it. a) From D = {z ∈ C : z < 1} to H = {z ∈ C : Im(z) > 0}. b) From the intersection of the open disks D((0, 0), 3) and D((0, 3), 2) to C. c) From H\(0, i] to H. d) From D to C\(−∞, − 41 ]. Exercise 3. The following questions are independent. a) Describe all harmonic functions on the plane R2 that are bounded from above. b) Let h : H = {z ∈ C : Im(z) > 0} → C be holomorphic. Assume that h(z) ≤ 1 for any z ∈ H and i is a zero of h of order m ≥ 1. Prove that, for any z ∈ H, z − i m . h(z) ≤ z + i
1
Solution of exercise 1. a) The case a = 0 is obvious, we assume a 6= 0 from now. We have Z Z Z Z 2π 1 2dz 2dz dz 1 dt = = = . 1 a 2 + 2z + a 1 + a cos t i az i a(z − z z + iz 1 + 1 )(z − z2 ) T T T 0 2 z √
√
2
2
where z1 = −1− a1−a , z2 = −1+ a1−a . The point z1 is outside the closed unit disk while the point z2 is inside the open unit disk; thus the residue theorem leads to Z 2π 2 2π dt . = 2πRes , z2 = √ 1 + a cos t a(z − z1 )(z − z2 ) 1 − a2 0 1 b) For z = reiθ , r > 0, −π < θ < π, let log z = log r + iθ The function 1+x p is the restriction to (0, +∞) of 1 f (z) = 1+exp(p log z) , which is analytic in C − (−∞, 0]. For 0 < ǫ < R < ∞ we consider the positively oriented contour Γ = Γ(ǫ, R) made of the following four arcs:
• the closed interval [ǫ, R], • the arc of circle of radius R and angle from 0 to • the interval joining Rei
2π p
and ǫei
2πi p .
2π p
• the arc of circle of radius ǫ and angle from
2πi p
to 0.
For R > 1 there is only one 0 of f inside the contour, zp = ei p . The residue at zp can be computed to be π
ei p . p π
Res(f, zp ) = −
The residue theorem applied to Γ for R → ∞ and ǫ → 0 allows to conclude that Z ∞ π dx 1 = . p+1 x p sin(π/p) 0
Solution of exercise 2. 1+z a) Yes, i 1−z . b) No. The inverse map would be a conformal map from the plane to a bounded set. It would therefore constant, √ by Liouville’s theorem. This is absurd. c) Yes, z 2 + 1. To see this, z 7→ z 2 sends the domain to C/[−1, ∞). Therefore z 2 + 1 sends the domain to C/[0, ∞). Taking the square root maps it to the upper half plane. d) Yes, it can be deduced from questions a) and c), going from the disk to the upper half plane and then z the complement of a ray. It is the Koebe function (1−z) 2. Solution of exercise 3. a) Let v be the harmonic conjugate of u. Then the function H(z) = exp(u(x, y) + iv(x, y)), z = x + iy, 2
is entire and bounded in the complex plane. By Liouville’s theorem it is constant, which implies that u is constant. 1+z b) The map Φ : z 7→ i 1−z is conformal from D to H. Let f (z) = h(Φ(z)). The function g(z) = f (z)/z m is analytic on D and g(z) ≤ 1 on ∂D. The maximum principle implies that g(z) ≤ 1 on D and therefore f (z) ≤ zm for any z in D. This implies the result.
3
Differential Geometry (Qualifying exams, Spring 2013) (1) The Heisenberg group is the subgroup of Sl(3, R) composed of the 3 × 3, upper triangular matrices with 1 on the diagonal, this being the set of matrices of the form: 1 x z 0 1 y , with (x, y, z) ∈ R3 . 0 0 1 This group is observably diffeomorphic to R3 . (a) Compute the Lie algebra of the Heisenburg group. (b) Exhibit a leftinvariant Riemannian metric on the Heisenberg group. (2) View R2 × C as the product complex line bundle over R2 and let θ0 denote the connection on this line bundle whose covariant derivative acts on a given section s as ds with d being the exterior derivative. Let A denote the connection i (xdy − ydx). θ0 + 1 + x2 + y 2 (a) Compute as a function of r ∈ (0, ∞) the linear map from C to C that is obtained by using A to parallel transport a given nonzero vector in C in the clockwise direction on the circle where x2 + y 2 = r2 from the point (r, 0) to itself. (b) Give a formula for the curvature 2form of the connection A. (3) Use (t, x, y, z) to denote the Euclidean coordinates for R4 . Let t 7→ a(t) denote a strictly positive function on R. A Riemannian metric on R4 is given by the quadratic form: g = dt ⊗ dt + a(t)2 dx ⊗ dx + dy ⊗ dy + dz ⊗ dz .
Compute the components of the Riemann curvature tensor of g using the orthonormal basis {dt, adx, ady, adz} for T⋆ R4 .
1
2
Solutions Question 1. We denote an element of the Heisenberg group as 1 x z M = 0 1 y . 0 0 1 (1.a). To identify the generators of the Lie algebra, we compute the left MaurerCartan oneform: θL = M −1 dM = dxX + dyY + (dz − xdy)Z, where
0 1 0 X = 0 0 0 , 0 0 0
0 0 0 Y = 0 0 1 , 0 0 0
0 0 1 Z = 0 0 0 , 0 0 0
are the generators of the Lie algebra. The Lie bracket is given by the commutators of these matrices: [X, Z] = [Y, Z] = 0,
[X, Y ] = Z.
(1.b). Denoting the transpose of a matrix U as U t and its trace as Tr U , the following is trivially a leftinvariant Riemannian metric: gL = Tr(θL · θLt ) = dx2 + dy 2 + (dz − xdy)2 . Question 2 (2.a). The parametric equation of the curve is γ : [0, 2π] → R2 : t 7→ γt = (r cos t, −r sin t). Pulling back the connection to the curve gives ir2 dt . 1 + r2 To find the parallel transport of s0 ∈ C along the curve γt , we solve the first order differential equation with initial condition s(0) = s0 : γt⋆ A = γt⋆ θ0 −
∇γ˙ t s(t) = 0, where γ˙ t is dγt /dt and the covariant derivative on the curve is computed with respect to the connection oneform γt⋆ A: d ir2 dt − . dt 1 + r2 With the initial condition s(0) = s0 , the solution is t r2 i . s(t) = s0 exp 1 + r2 The parallel transport of s0 is then given by evaluating s(t) at the end point t = 2π. This defines the following linear map: 2πr2 C → C : s0 7→ s0 exp i . 1 + r2 ∇γ˙ t =
3
(2.b). The curvature of the connection A is : 2i dx ∧ dy. Ω(A) = dA + A ∧ A = (1 + x2 + y 2 )2 Question 3 We use the following notation: e0 = dt,
e1 = adx,
e2 = ady,
e3 = adz,
a˙ =
da , dt
a ¨=
d2 a . dt2
The metric can then be rewritten as 3 X g= em ⊗ em . m=0
We can compute the connection from the Cartan’s structure equations with zero torsion: 3 X m ω m n ∧ en = 0, m = 0, 1, 2, 3. de + n=0
Since the connection is compatible with the metric, ω m n has to be antisymmetric in m and n. we can then uniquely determine its components. A direct calculation gives: a˙ de0 = 0, dei = e0 ∧ ei , i = 1, 2, 3, a from which we get: 0 −e1 −e2 −e3 a˙ e1 0 0 0 . ω= 2 0 0 0 a e e3 0 0 0
The Riemann curvature is then: 0 −a¨ a e0 ∧ e1 −a¨ a e0 ∧ e2 −a¨ a e 0 ∧ e1 1 a¨ a e 0 ∧ e1 0 −a˙ 2 e1 ∧ e2 −a˙ 2 e1 ∧ e2 . R(ω) = dω + ω ∧ ω = 2 0 2 2 1 2 a e ∧e a˙ e ∧ e 0 −a˙ 2 e1 ∧ e3 a a¨ a¨ a e0 ∧ e3 a˙ 2 e1 ∧ e3 a˙ 2 e2 ∧ e3 0
1. Suppose fj , j = 1, 2, . . . and f are real functions on [0, 1]. Define fj → f in measure if and only if for any ε > 0 we have lim µ{x ∈ [0, 1] : fj (x) − f (x) > ε} = 0 j→∞
where µ is the Lebesgue measure on [0, 1]. In this problems, all functions are assumed to be in L1 [0, 1]. (a) Suppose that fj → f in measure. Does it implies that Z lim fj (x) − f (x)dx = 0. j→∞
Prove it or give a counterexample. (b) Suppose that fj → f in measure. Does this imply that fj (x) → f (x) almost everywhere in [0, 1]? Prove it or give a counter example. (c) Suppose that fj (x) → f (x) almost everywhere in [0, 1]. Does it implies that fj → f in measure? Prove it or give a counter example. 2. (a) For any bounded positive function f define A(f ) :=
Z
1
f (x) log f (x)dx,
B(f ) :=
0
Z
1
f (x)dx log 0
Z
1 0
f (y)dy .
There are three possibilities: (i) A(f ) ≥ B(f ) for all bounded positive functions, (ii) B(f ) ≥ A(f ) for all bounded positive functions, and (iii) A(f ) ≥ B(f ) for some functions while B(f ) ≥ A(f ) for some functions. Decide which possibility is correct and prove your answer. If you use any inequality, state all assumptions of the inequality precisely and clearly. (b) Let fˆ denote the Fourier transform of the function f on R. Suppose that f ∈ C ∞ (R) and kfˆ(ξ)kL2 (R) ≤ α,
kξ1+ε fˆ(ξ)kL2 (R) ≤ β
for some ε > 0. Find a bound on kf kL∞ (R) in terms of α, β and ε. 3. Assume that X1 , X2 , . . . are independent random variables uniformly distributed on [0, 1]. Let Y (n) = n inf{Xi , 1 ≤ i ≤ n}. Prove that Y (n) converges weakly to an exponential random variable, i.e. for any continuous bounded function f : R+ → R, Z f (u)e−u du. E f (Y (n) ) −→ n→∞
1
R+
Solutions: 1a. no. let f = 0 and fj (x) = j, 0 ≤ x ≤ 1/j;
= 0 otherwise
1b. no. let f = 0 and fj are the jth Haar functions. 1c. yes. Fix any ε > 0. Let Ej = {x ∈ [0, 1] : fj (x) − f (x) > ε},
Fj = ∪k≥j Ej
By changing x on a set of measure zero, we have fj (x) → f (x) for all x. Thus lim Fj = ∅
j→∞
Hence µ(Fj ) → 0. Alternatively, WLOG, we assume that f = 0. Let gj = min(1, fj ). Then gj → 0. Thus µ(Ej ) ≤ ε−1
Z
1 0
gj (x)dx → 0.
2a. Since the function x → x log x is convex, by Jensen inequality, we have A(f ) ≥ B(f ). 2b. From the Fourier inversion formula and Schwarz inequality, Z Z Z 2πixξ 2 1+ε ˆ ˆ f (x) ≤ f (ξ)e dξ ≤ f (ξ) (1 + ξ) dξ (1 + ξ)−1−ε dξ ≤ ε−1 C[α2 + β 2 ]
for some constant C.
3. Up to a permutation, we can assume that x1 ≤ x2 . . . ≤ xn . In this case, Y (n) = nx1 . Thus Z 1 Z 1 Z 1 dxn f (nx1 ) dx2 . . . dx1 Ef (Y (n) ) = n! 0
= n! =n
Z
Z
1
dx1 0
Z
1
dx2 . . . x1
1 0
dx1 f (nx1 )(1 − x1 )n−1 =
xn−1
x1
Z
Z
1 xn−2
dxn−1 (1 − xn−1 )f (nx1 )
n 0
dx f (x)(1 − x/n)n−1 →
where we have used dominated convergence in the last step.
2
Z
f (u)e−u du. R+
Tuesday, September 4, 2012 Qualifying Exams I, 2012 Fall 1. (Real Analysis) Suppose fj , (j = 1, 2, . . .) and f are real functions on [0, 1]. Suppose that fj (x) → f (x) almost everywhere for x ∈ [0, 1]. Furthermore, we assume that sup kfj kL2 [0,1] ≤ 1, j≥1
kf kL2 [0,1] ≤ 1.
(a) Is it always true that lim kfj − f kL2 [0,1] = 0 ?
j→∞
Prove it or give a counterexample. (b) Is it always true that lim kfj − f kL1 [0,1] = 0 ?
j→∞
Prove it or give a counterexample. 2. (Complex analysis) Evaluate the following path integral Z 1 z 5 sin( 2 )dz, z C where C is the union of two positively oriented unit circles z = 1 and z − 1/2 = 1. Find also the Laurent series in the domain 0 < z < ∞. You need to prove your answers or state precisely the theorems you use. 3. (Differential Geometry) (a) Prove that SUN (the set of N × N unitary matrices with determinant 1) is a submanifold of MN (C) (the set of N × N matrices with entries in C).
(b) Find the dimension of SUN and prove that its tangent space at the identity is the space of N × N antihermitian and trace zero matrices.
(c) Prove that the submanifolds SLN (the set of N × N matrices with determinant 1) and UN (the set of N × N unitary matrices) of MN (C) do not intersect transversally. 4. (Algebraic Topology) (Third Homotopy Group of 2Sphere). Let Φ : C2 − {0} → CP1 be defined by mapping the inhomogeneous coordinates (z1 , z2 ) of C2 to the homogeneous coordinates [z1 , z2 ] of the complex projective line CP1 . Let f : S 3 → S 2 be the map from the 3sphere S 3 to the 2sphere S 2 defined by restricting Φ to the unit 3sphere in C2 . Let g be the element in the third homotopy group π3 S 2 of the 2sphere S 2 which is represented by f . Let γ : Z → π3 S 2 be the group homomorphism which maps the element 1 of the additive group Z to the element g of π3 S 2 . Compute the kernel and the cokernel of the group homomorphism γ. Justify each step of your computation. 5. (Algebra) Let p be a prime number, n > 1 an integer, and K = Fpn the field with pn elements. Considering the field automorphism of K defined by F (x) = xp as an Fp linear endomorphism of K, compute its characteristic polynomial. 6. (Algebraic Geometry) (a) Let X be a closed subvariety of Pn . Show that degree of X is one if and only if X is a linear subspace of Pn . (b) Write down the statement of B´ezout’s theorem for Pn . (c) Let Y ⊂ Pn be a closed subvariety of dimension k and degree a, and Z := vd (Y ) ⊂ PN where vd : Pn → PN is the Veronese map of degree d. Compute the degree of Z.
1
Wednesday, September 5, 2012 Qualifying Exams II, 2012 Fall 1. (Real Analysis) Let u(t, x) be the solution to the equation 1 uxx (t, x), 2 Suppose u(t, x) is smooth in both t and x and ∂t u(t, x) =
u(0, x) = f (x).
u(t, x) + uxx (t, x) ≤ e−x . (a) Prove that there are constants C1 and C2 such that Z C1 (x − y)2 u(t, x) = √ f (y)dy exp −C2 t t R You can use the formula that the inverse Fourier transform (in p) of e−tp
2
/2
is
2 C √1 e−C2 x /t . t
(b) From this formula and Holder (or Jensen) inequality, prove the following inequality ku(t, ·)k2L2 (R) ≤ Ct−1/2 kf k2L1 (R) . 2. (Complex analysis) Let n be a positive integer, what is the residue of the Gamma function at the pole z = −n? 3. (Differential Geometry) A metric on R2 written in polar coordinates has the form dr2 + f (r, θ)2 dθ2 . 2 Prove that its Gaussian curvature is K = −f −1 ∂∂ 2fr .
4. (Algebraic Topology) (Fundamental Group of Space Obtained by Glueing). Denote by RP2 the real projective plane (which is the quotient of the 2sphere with antipodal points identified). Denote by T 2 the real 2dimensional torus (which is the quotient of a closed rectangle with opposite sides identified). Let D be the interior of a closed disk in T 2 whose boundary is C. Let G be the interior of a closed disk in RP2 whose boundary is E. Let X be the space obtained by glueing T 2 − D to RP2 − G along a homeomorphism between the two circles C and E. Compute the fundamental group of X by describing a presentation of it. Then compute the first homology group H1 (X, Z) of X with coefficients in Z. 5. (Algebra) Consider the principal ideal I = (y 2 − x3 + x) in the polynomial algebra A = Q[x, y], and let R := A/I. (a) Show that the ring automorphism σ e of A that sends x to x and y to −y descends to an automorphism σ of R. (b) For any associative and commutative ring S with unit and any unital ring automorphism τ of S, show that the image of a principal ideal (resp. a prime ideal) of S under τ is still a principal ideal (resp. a prime ideal). (c) Show that the ideal p = (¯ x, y¯)R generated by the image of x and that of y is a prime ideal. (d) Show that while p2 is a principal ideal, p is not. (Hint. Use (b) and the fact that p is left invariant by σ.) 6. (Algebraic Geometry) Let X be a closed subvariety in PN over an algebraically closed field. Let hX : N → N be the Hilbert function of X. (a) Prove that when X = {x1 , . . . , xd } ⊂ PN (xi 6= xj whenever i 6= j), hX (m) = d for m sufficiently large. (b) Compute hX for the rational normal curve X, which is the image of P1 ֒→ PN defined by sending [Z0 , Z1 ] ∈ P1 to [Z0N , Z0N −1 Z1 , . . . , Z1N ] ∈ PN . (c) Prove that there exists a polynomial pX such that hX (m) = pX (m) for all sufficiently large natural numbers m.
1
Thursday, September 6, 2012 Qualifying Exams III, 2012 Fall 1. (Real Analysis) Suppose that (Xj )j≥1 is a sequence of random variables on the same probability space with mean EXj = 1 for all j. Suppose we know that EXj Xk − EXj EXk  ≤ f (k − j) for some sequence f (m) with
P∞
m=1
f (m) < A. Prove that
P(n−1
n X
Xj ≥ 2) ≤
j=1
B n
and find a relation between B and A. 2. (Complex Analysis) Denote by U the open unit ball in C. Suppose that fn is a Cauchy sequence of analytic functions in L2 (U ). Prove that fn converges uniformly on every compact subsets of U to an analytic function. 3. (Differential Geometry) Consider Rn with a coordinate system (x1 , · · · , xn ). Compute the LeviCevita connection of the Riemann manifold (M, g) where M ⊂ Rn is defined by xn > 0 and the metric g is given by: g=
dx1 ⊗ dx1 + · · · + dxn ⊗ dxn . x2n
4. (Algebraic Topology) (Universal Cover of OnePoint Union of Two Real Projective Planes). Let RP2 denote the real projective plane (which is the quotient of the 2sphere with antipodal points identified). Let X be the onepoint union RP2 ∨ RP2 (or wedge sum) of two real projective planes (i.e., the result obtained by identifying, in the disjoint union of two real projective planes, one single given point on one with one single given point on the other). Find the universal cover of X. 5. (Algebra) Let p be a prime number and denote by Fp the field with p elements. Consider the groups G = GL2 (Fp ) and G′ = PGL2 (Fp ) = G/F× p , where the group of units in Fp embeds diagonally into G. (a) Compute G and G′ . (b) Find a pSylow subgroup of G. (c) Let X be the set of onedimensional subspaces of the twodimensional Fp vectorspace F2p . Show that the natural action of G on X descends to a faithful action of G′ on X. (d) Show that PGL2 (F3 ) is isomorphic to S4 , the symmetric group on four letters. 6. (Algebraic Geometry) Let X be a smooth projective curve of genus g. (a) Show that the canonical divisor KX of X has degree 2g − 2. (b) When g = 1, show that KX is linearly equivalent to the zero divisor. (c) Let D be an effective divisor on X, and denote by D the complete linear system associated to D. Show that dim D ≤ deg D and equality holds if and only if D = 0 or g = 0.
1
FALL 2012  Solutions for the Algebraic Geometry Questions
PROPOSED ANSWERS IN AG
Day 1
(1) (a) If X is a linear subspace of Pn , it intersects transversely with a generic linear subspace at a single point. Thus deg X = 1. Now consider the converse. Let k = dim X. Consider the projection π : X → Pk+1 from a general (n − k − 2)plane in Pn such that X ¯ ⊂ Pk+1 . Since X is of degree one, X ¯ is birational onto its image X k+1 ¯ is simply a hyperplane in P . Then the inverse image of X under the projection is a hyperplane in Pn . That means X is contained in a hyperplane P n−1 in Pn . Doing the same thing for X ⊂ P n−1 and continue the inductive process, we obtain X ⊂ P k+1 ⊂ P k+2 . . . ⊂ P n−1 ⊂ Pn where one sits in the next one as a hyperplane. Thus X is a linear subspace of Pn . (b) B´ezout’s theorem states that if X and Y are two subvarieties of Pn which intersect generically transversely, then deg(X ∩ Y ) = (deg X)(deg Y ).
Day 2
(c) Recall that the Veronese map vd : Pn → PN of degree d is defined by sending [Z0 , . . . , Zn ] to [P0 (Z0 , . . . , Zn ), . . . , PN (Z0 , . . . , Zn )], where Pi , i = 0, . . ! . , N are all " the monomials of degree d in n + 1 variables, n+d where N = − 1. d The degree of Z equals to the number of intersection points of Z with a generic linear subspace L of PN of dimension N −k, which is the zero set of k linear functions l1 , . . . , lk on PN . Since vd is an embedding, it suffices to compute the number of intersection points in the inverse image. The inverse image of the intersection between Z and L equals to the intersection of Y and the zero set of vd∗ l1 , . . . , vd∗ lk , which are degree d polynomials. By B´ezout’s theorem, the number of intersection points equals to (deg Y )(deg vd∗ l1 ) . . . (deg vd∗ lk ) = dk a. (2) (a) Recall that the Hilbert function hX : N → N is defined by # $ hX (m) = dim (K[z0 , . . . , zN ]/IX )m where z0 , . . . , zN are the homogeneous coordinates of PN , IX is the defining ideal of X, and (K[z0 , . . . , zN ]/IX )m denotes its mth graded piece. Consider the linear map ev : (K[z0 , . . . , zN ]/IX )m → K d defined by sending f ∈ (K[z0 , . . . , zN ]/IX )m to (f (v1 ), . . . , f (vd )), where vi ∈ K N +1 are chosen representatives of xi ∈ PN for all i = 1, . . . , d. It is welldefined and oneone since f ∈ IX if and only if f (vi ) = 0 for all i = 1, . . . , d. When m ≥ d − 1, It is also surjective: Let {ej }dj=1 be the standard basis of K N +1 . For every j = 1, . . . , d, there exists P ∈ K[z0 , . . . , zN ]m such that ev([P ]) = ej ∈ K d defined 1
2
PROPOSED ANSWERS IN AG
as follows. For each i ∈ {1, . . . , d} − {j}, choose a linear function li on K N +1 such that li (vi ) = 0 and li (vj ) = 1. (This is possible since vi and vj cannot be linearly dependent, or otherwise pi = pj .) For i = d + % 1, . . . , m + 1, choose a linear function li with li (vj ) = 1. Then P = i#=j li has degree m and satisfies P (vk ) = 0 for k ∈ {1, . . . , d} − {j} and P (vj ) = 1. Thus (K[z0 , . . . , zN ]/IX )m ∼ = K d as vector spaces, and hence hX (m) = dim(K[z0 , . . . , zN ]/IX )m = d. (b) Consider the pullback K[z0 , . . . , zN ]m → K[Z0 , Z1 ]N m by the embedding P1 "→ PN . It is surjective: Every monomial in Z0 and Z1 of degree N m can be written as Z0jN Z1kN Z0p Z1q for some j, k, p, q ∈ Z≥0 k with p, q < N . Then it is the image of z0j zN zq when q (= 0, or j k z0 zN when q = 0. Moreover, the kernel is exactly those polynomials in K[z0 , . . . , zN ]m which vanish on the rational normal curve. Thus (K[z0 , . . . , zN ]/IX )m ∼ = K[Z0 , Z1 ]N m as vector spaces, which has dimension N m + 1. (c) Let n = dim X. Let P be a linear subspace of dimension N − n which intersects X transversely. P is the zero set of n linear functions l1 , . . . , ln . The number of intersection points between X and P is d = deg X. Let X (i) = X ∩ {l1 = . . . = li = 0} for i = 0, . . . , n. Then for all i = 0, . . . , n − 1, we have a homomorphism A(X (i) )m → A(X (i) )m+1
given by multiplication by li+1 . Here A(X (i) ) := K[z0 , . . . , zN ]/IX (i) denotes the coordinate ring of X (i) . Since the intersection is transverse, this homomorphism is injective. Moreover A(X (i) )m+1 /Im(A(X (i) )m ) ∼ = A(X (i+1) )m+1 given by restriction. Thus hX (i) (m + 1) − hX (i) (m) = hX (i+1) (m + 1)
Day 3
for all m ∈ N. By (b), hX (n) (m) = d for all sufficiently large m. We conclude that hX = hX (0) is a polynomial. (3) (a) By RiemannRoch, h0 (KX ) − h0 (OX ) = deg(KX ) − g + 1. Since h0 (OX ) = 1 and h0 (KX ) = g, deg(KX ) = 2g − 2. (b) deg KX = 0 while h0 (KX ) = 1. Thus there exists a meromorphic function f such that KX + (f ) is an effective divisor. KX + (f ) is of degree 0 because (f ) and KX are. This forces KX + (f ) = 0. It follows that KX ∼ 0. (c) Since D is effective, dim D = h0 (D) − 1
= h0 (K − D) + deg D − g ≤ h0 (K) + deg D − g = deg D.
Equality holds if and only if h0 (K − D) = h0 (K). Obviously D = 0 implies this equality. When g = h0 (K) = 0, 0 ≤ h0 (K −D) ≤ h0 (K) = 0. Thus h0 (K − D) = h0 (K) = 0 and equality holds.
PROPOSED ANSWERS IN AG
3
Conversely, suppose equality holds, yet g (= 0. Then D ∼ 0, and since D is effective, D = 0.
FALL 2012  Qualifying Exams Solutions for Algebra
Day 1
A1. The characteristic polynomial is T n − 1. The Galois group of K = Fpn over Fp is the cyclic group of order n generated by F . By the normal basis theorem there is an element v ∈ K such that v, F v, · · · , F n−1 v
forms an Fp basis of K. So det(T · In − F ) = T n − 1.
Day 2
A2. (a) It fixes the generator of I, hence preserves the ideal I. (b) If P is a prime ideal, and if s, t ∈ S satisfy s · t ∈ τ (P ), then τ −1 (s) · τ −1 (t) ∈ P and either τ −1 (s) ∈ P or τ −1 (t) ∈ P , hence s ∈ τ (P ) or t ∈ τ (P ). In addition, by definition 1 #∈ P , which implies 1 #∈ τ (P ).
The image under τ of the principal ideal generated by α is that generated by τ (α).
(c) We have R/p = Q[x, y]/(x, y) = Q, which is an integral domain. (d) First, we have p2 = (¯ x). For ⊆, note y¯2 = x ¯(¯ x2 − 1). For ⊇, note that 2 2 2 3 p contains both x ¯ and −¯ y =x ¯−x ¯ .
Suppose p = (α), and write α = P1 (x) + P2 (x)¯ y . Since p is fixed by σ, we have p2 = pσ(p) = (P1 (x)+P2 (x)¯ y )(P1 (x)−P2 (x)¯ y ) = (P1 (x)2 −P2 (x)2 (x3 −x)).
Because this ideal contains x, there are Q1 (x) and Q2 (x) such that x = (P12 − P22 · (x3 − x))(Q1 + y¯Q2 ) in R. First, by taking half the trace ((1 + σ)/2), we may assume Q2 = 0. Then the degree consideration in Q[x] leads to P2 = 0 and P1 (x) ∈ Q× , which is absurd.
Day 3
A3. (a) G = (p2 − 1)(p2 − p) = p(p − 1)2 (p + 1), G#  = p(p − 1)(p + 1). (b) The upper unitriangular matrices. (c) The units act trivially on X. Conversely, any g ∈ G fixing every line fixes both [1 : 0] and [0 : 1], hence g is diagonal. If the entries were distinct, it would fail to fix [1 : 1]. (d) From (c) we get an injection PGL2 (F3 ) $→ S4 , since when p = 3, X has 4 elements. The two groups have the same order, by (a). 2
FALL 2012 Solutions of Qualifying Exam Problems in Algebraic Topology in September 2012
Day 1
Problem 1 (Third Homotopy Group of 2Sphere). Let Φ : C2 − {0} → CP1 be defined by mapping the inhomogneous coordinates (z1 , z2 ) of C2 to the homogeneous coordinates [z1 , z2 ] of the complex projective line CP1 . Let f : S 3 → S 2 be defined by restricting Φ to the unit 3sphere in C2 . Define the group homomorphism γ : Z → π3 (S 2 ) by setting γ(1) to be the element of π3 (S 2 ) defined by f . Compute the kernel and the cokernel of the group homomorphism γ : Z → π3 (S 2 ). Justify each step of your computation. Solution. The action of C − {0} on C2 − {0} defined by scalar multiplication of vectors in C2 makes Φ : C2 − {0} → CP1 a principal bundle with structure group C − {0}. The map f : S 3 → S 2 defined by restricting Φ to the unit 3sphere in C2 is the principal bundle with circle group S 1 as the structure group and is, in particular, a fiber bundle over S 2 with fiber S 1 . The exact sequence of homotopy groups for the fiber bundle f : S 3 → S 2 is ! " ! " ! " ! " · · · → πi S 1 → πi S 3 → πi S 2 → πi−1 S 1 → · · · and, in particular, the sequence ! " ! " ! " ! " π3 S 1 → π3 S 3 → π3 S 2 → π2 S 1
is exact. Since the universal cover of S 1 is R which is contractible and since any continuous map from the simplyconnected 3sphere S 3 or 2sphere S 2 to S 1 can be lifted to the contractible universal cover R of S 1 , it follows that both π3 (S 1 ) and π2 (S 1 ) vanish and from the above exact sequence of four terms the map π3 (S 3 ) → π3 (S 2 ) induced by f : S 3 → S 2 is an isomorphism.
Since πj (S 3 ) is trivial for j = 1, 2 (as every element of πj (S 3 ) for j = 1, 2 can be represented by a continuous map S j → S 3 whose image misses some point of S 3 ), it follows from Hurewicz’s theorem (relating homotopy groups to homology groups) that the map π3 (S 3 ) → H3 (S 3 ) is an isomorphism and, in particular, the homotopy group π3 (S 3 ) is the cyclic group Z whose generator is represented by the identity map of S 3 . Hence the group homomorphism γ : Z → π3 (S 2 ) is an isomorphism and both its kernel and cokernel are trivial.
1
Day 2
Problem 2 (Fundamental Groups of Spaces Obtained by Glueing). Denote by RP2 the real projective plane (which is the quotient of the 2sphere with antipodal points identified). Denote by T 2 the real 2dimensional torus (which is the quotient of a closed rectangle with opposite sides identified). Let D be the interior of a closed disk in T 2 whose boundary is C. Let G be the interior of a closed disk in RP2 whose boundary is E. Let X be the space obtained by glueing T 2 − D to RP2 − G along a homeomorphism between the two circles C and E. Compute the fundamental group of X by describing a presentation of it. Then compute H1 (X, Z). Solution. Let f : C → E be the homeomorphism used to glue together T 2 −D and RP2 − G to construct X. The fundamental group π1 (X) of X will be computed by applying the theorem of van Kampen to X = (T 2 − D) ∪f (RP2 − G). The fundamental group π1 (T 2 − D) of T 2 − D is the free group generated by two elements a and b which are represented by two standard basis loops of T 2 avoiding the topological closure of D. The removal of D from T 2 makes the relation aba−1 b−1 = 1 in π1 (T 2 − D) disappear, because when T 2 is represented by identifying the opposite sides of a rectangle, the removal of a disk in the center of the rectangle makes the relation obtained by going around the boundary of the rectangle impossible. From this picture of removing a disk in the center of rectangle, we know that going around the boundary of the rectangle shows that aba−1 b−1 is homotopic in T 2 − D to a loop going once around the circle C (or E under identification by f ). The space RP2 − G is the same as the M¨obius band, as one can easily see by considering the map from S 2 minus two antipodal disks to RP2 − G defined by identifying antipodal points. The generator c of the fundamental group π1 (RP2 − G) of the M¨obius band RP2 − G is the loop represented by going around the center line of the M¨obius band once. The loop c2 is homotopic in RP2 − G to going once around the circle E (or C identified by f ), because going around the edge of the M¨obius band once is the same as going around the the centerline of the M¨obius band twice. By van Kampen’s theorem, c2 needs to be identified with aba−1 b−1 , because both represent going around C or E once (which are identified by the glueing homeomorphism f ). Hence the fundamental group π1 (X) of X is equal to the free group generated by three elements a, b, c subject to one single relation c−2 aba−1 b−1 = 1. We can compute H1 (X, Z) by abelianizing π1 (X). In the abelianization of π1 (X) the element aba−1 b−1 of π1 (X) becomes 1 and the single relation c−2 aba−1 b−1 = 1
2
in π1 (X) becomes the single relation c−2 = 1. Hence H1 (X, Z) ≈ Z ⊕ Z ⊕ (Z /2Z ) .
Day 3
Problem 3 (Universal Cover of OnePoint Union of Two Real Projective Planes). Let RP2 denote the real projective plane (which is the quotient of the 2sphere with antipodal points identified). Let X be the one point union RP2 ∨ RP2 (or wedge sum) of two real projective planes (i.e., the result obtained by identifying, in the disjoint union of two real projective planes, one point on one identified with one point on the other). Find the universal cover of X. 2 2 ˜ =# Solution. Let X n∈Z (S + (2n + 1, 0, 0)), where S is the unit 2sphere (centered at the origin of radius 1) in R3 and S 2 + (2n + 1, 0, 0) means the translate of S 2 by the vector (2n + 1, 0, 0) so that S 2 + (2n + 1, 0, 0) ˜ is is the 2sphere in R3 centered at the origin of radius 1. The space X an infinite string of touching 2spheres of radius 1 centered at (2n + 1, 0, 0) touching the two adjacent 2spheres. Let ϕ : R3 → R3 be the map defined by ϕ(x, y, z) = (−x + 2, −y, −z) and ψ : R3 → R3 be the map with ψ(x, y, z) = (−x − 2, −y, −z). The map ϕ, when restricted to the 2sphere of radius 1 centered at (1, 0, 0), is simply the antipodal map on that 2sphere. The map ψ, when restricted to the 2sphere of radius 1 centered at (−1, 0, 0), is simply the antipodal map on that 2sphere. The group G of transformations in R3 ˜ to give the quotient X. Since X ˜ is generated by ϕ and ψ acts free on X ˜ is the universal cover of X. The argument given up to simply connected, X this point is already the complete rigorous solution of the problem of finding ˜ of X. the universal cover X ˜ as the universal If one wants to know how one arrives at the candidate X cover of X, one can do it either geometrically or algebraically. ˜ is to find all liftings to X ˜ The geometric way to arrive at the candidate X 2 2 2 2 2 of the universal cover S → RP of the first summand P of RP ∨ RP (with ˜ and also all the liftings the image for each individual lifting a 2sphere in X) 2 2 of the universal cover S → RP of the second summand P2 of RP2 ∨ RP2 ˜ The first (again with the image for each individual lifting a 2sphere in X). sequence of 2spheres alternates between the second sequence of 2spheres with points of touching being the inverse images of the common point of the two copies of P2 in RP2 ∨ RP2 . The universal cover is the union of the two sequences of 2spheres alternatingly touching each other to form an infinite 3
sting of touching 2spheres. For the rigorous proof that such a construction ˜ is indeed the universal cover of X, one goes back to the argument in of X the preceding paragraph. ˜ is to use the theorem of The algebraic way to arrive at the candidate X van Kampen to conclude that the fundamental group of RP2 ∨ RP2 is equal to the group (Z /2Z ) ∗ (Z /2Z ) amalgamated from the fundamental group Z /2Z of the first summand of RP2 ∨ RP2 and the fundamental group Z /2Z of the second summand of RP2 ∨ RP2 . The map ϕ : R3 → R3 defined by ϕ(x, y, z) = (−x + 2, −y, −z) can be used to represent the generator of the fundamental group Z /2Z of the right summand of RP2 ∨ RP2 , because it represents the antipodal map of the 2sphere in R3 of radius 1 centered at (1, 0, 0). The map ψ : R3 → R3 defined by ψ(x, y, z) = (−x − 2, −y, −z) can be used to represent the generator of the fundamental group Z /2Z of the left summand of RP2 ∨ RP2 , because it represents the antipodal map of the 2sphere in R3 of radius 1 centered at (−1, 0, 0). The group G generated by ϕ and ψ is the fundamental group (Z /2Z ) ∗ (Z /2Z ) of RP2 ∨ RP2 . The orbit of the union of the two unit 2spheres centered respectively at (1, 0, 0) ˜ of X. Again, for and (0, 0, 1) under the group G is the universal cover X ˜ is indeed the universal cover of X, the rigorous proof that such an orbit X one goes back to the argument in the first paragraph of this solution of the problem.
4
FALL 2012  Qualifying Exam Solutions for Complex Analysis 1. Differential Geometry
Day 1
Exam I, Complex analysis. The integrand z 5 sin( z12 ) is an analytic function on the punctured complex plane (0 < z < ∞). The Taylor series for sin(u) is sin(u) =
∞ ! n=0
(−)n u2n+1 . (2n + 1)!
As we are in the domain 0 < z < ∞, we can substitute with u = z −2 : ! (−)n 1 z −4n+3 , z sin( 2 ) = z (2n + 1)! k=0 ∞
5
The pole for z = 0 is at n = 1. It follows that the residue is − 3!1 . For the integral, we see that the function is analytic everywhere within the first circle z < 1, with the exception of the z = 0. There are no singularities on the boundary. The second contour is a deformation of the first one without meeting a singularity. So it has the same value. It follows that the residue at z = 0 will contribute twice and the final answer is −2 2πi = − 2πi . 3! 3
Day 2
Exam II, Q2, Complex analysis. The residue of the Γ function at n z = −n is (−1) . n! Defining the Γ function by the integral " ∞ tz−1 e−t dt, Γ(z) = 0
we have " Γ(z) = =
"
∞
t
z−1 −t
e dt +
1 ∞ 1
tz−1 e−t dt +
"
1 −t z−1
dte t 0
∞ ! (−1)n n=0
n!
=
"
∞ 1
t
z−1 −t
e dt +
"
1
dtt 0
z−1
∞ ! (−1)n n=0
1 . z+n
The first integral is an analytic function of z and the second terms n . shows that the residue at z = −n is (−1) n! Alternatively, you can the functional relation Γ(z + 1) = zΓ(z) ti show that the residue at z = 0 is Γ(1) = 1. You then reduce the residue of Γ(z) at z = −n to the residue at z = 0 using recursively the functional relation Γ(z) = Γ(z+1) . z
Day 3
Exam III, Q2, Complex analysis. We consider analytic functions such that " 2 f (z)f (z)dz ≤ ∞. f 2 := U
1
n!
tn
2
Using Cauchy integral formula, an analytic function f of L2 (U ) admits the following estimate for every compact K strictly included in U : supz∈K f (z) ≤ CK f 2 ,
where CK is a constant depending on the compact space K. You can use this estimate together with CauchySchwarz inequality to prove the uniform convergence of the sequence (fn ). To show that the limit is analytic, you can use Moreva theorem.
FALL 2012  Solutions for Differential Geometry QUALIFYING EXAMINATION Harvard University Department of Mathematics Differential geometry, Paul
Day
1. (a) Prove that SUN (the set of N × N unitary matrices with determinant 1) is a submanifold of MN (C) (the set of N × N matrices with entries in C). (b) Precise the dimension of SUN and its tangent space at identity. (c) Prove that the submanifolds SLN (the set of N × N matrices with determinant 1) and UN (the set of N × N unitary matrices) of MN (C) do not intersect transversally. Solution. (a) Assume first that SUN is a submanifold in a neighborhood of Id: it is locally the zero set of a submersion F . Let M ∈ SUN . Let LM : MN (C) → MN (C) be the left multiplication by M . It is a diffeomorphism with inverse LM −1 . Then SUN is locally, in a neighborhood of M , the zero set for the submersion F ◦ LM −1 .
Hence we just need to consider the case of a neighborhood of Id. Let E = {M ∈ MN (C) : M t = M }, and consider the map ! MN (C) → E×R . Φ: M %→ (M t M , &(det(M )))
We have Φ(−1) (Id, 0) = {M ∈ MN (C) : det M = ±1}, so in a neighborhood of identity Φ(M ) = (Id, 0) is an equation for SUN : we need to check that Φ is a submersion at Id. A calculation yields dΦId (H) = (H t +H, &(T r(H))). For any (M, λ) ∈ E ×R, one therefore can find H such that dΦId (H) = (M, λ) (choose for example H0 with a real trace such that M = H0 + H0 , and H to be H0 with iλ added in the upper left entry). This proves that dΦId is surjective, concluding the proof. (b) The tangent space at identity for SUN is the kernel of dΦId (H) = (H t + H, &(T r(H))), that is to say matrices with trace 0 and equal to the opposite of transpose of their conjugate. The dimension of SU (N ) is therefore 2N (N − 1)/2 + N − 1 = N 2 − 1. (c) SLN and UN do not intersect transversally at Id. Indeed the tangent spaces of these two varieties at Id are both included in the subvector space of MN (C) consisting in matrices with purely imaginary trace.
Day
2. Let M be a dimension 2 Riemannian manifold. We write its metric in polar coordinates as dr2 + f (r, θ)2 dθ2 . Prove that its Gaussian curvature is
2
K = −f −1 ∂∂rf2 . Solution. Let f " = ∂f /∂r and f "" = ∂ 2 f /∂r2 . Then (Rer ,eθ er ) · eθ = ∇er ∇eθ er · eθ − ∇eθ ∇er er · eθ − ∇[er ,eθ ] er · eθ " " # f = ∇e r eθ · eθ − 0 − ∇− f " e er · eθ f f θ "" "2 " ff − f f f" = + ∇ e · e + ∇e θ e θ e θ θ r f2 f f f f "" − f "2 f "2 = + 2 eθ · eθ f2 f "" f , = f ""
so K = − ff .
Day
3. Exercise about calculating the LeviCivita connection for the ndimensional hyperbolic space. $ Solution. Remember that ∇∂i ∂j = k Γkij ∂k , where Γkij =
1 1 % kn g (∂i gnj + ∂j gni − ∂n gij ) = g kk (∂i gkj + ∂j gki − ∂k gij ), 2 n 2
the last equality because g is diagonal. A calculation then yields, on {i = k, j = n} ∪ {j = k, i = n}, Γkij = −x−1 n , and on {i = j, k = n, i *= n}, k −1 k Γij = xn . For all other indices, Γij = 0.
FALL 2012  Qualifying Exam Solutions for Real Analysis
Day
Solution: 1. (a) Let f = 0 and
√
fn (x) =
n,
0 ≤ x ≤ 1/n
and fn (x) = 0 otherwise. Then fn (x) → 0 a.e. but $fn $2 = 1 for all n. This is a counterexample. (b) Let gn = fn − f and for M > 0 rewrite gn = hn + kn where hn (x) = gn (x)1(gn (x) > M ). Then by the dominated convergence theorem, we have ! lim kn (x)dx = 0. n→∞
Also,
!
hn (x)dx ≤ M −1
!
hn (x)2 dx ≤ M −1
!
gn (x)2 dx ≤ M −1 ($fn $2 + $f $2 )2 ≤ 4/M
Since M can be arbitrary large, this proves that ! lim gn (x)dx = 0. n→∞
Day
2. Define the Fourier transform by f (p) =
!
e−ixp f (x)dx
Take the Fourier transform in x to get ∂t u ˆ(t, p) = −
p2 u ˆ(t, p) 2
Here the assumptions on u(t, x) make sure that the Fourier transforms can be taken on both sides of the equation. Hence 2 2 u ˆ(t, p) = e−tp /2 u ˆ(0, p) = e−tp /2 fˆ(p) Take the inverse Fourier transform to get u(t, x) = [g(t, ·) ∗ f ](x) where g(t, ·) is the inverse transform of e−tp Gaussian, i.e.,
2
/2
. Using the wellknown formula of the Fourier transform of
2 C1 g(t, x) = √ e−C2 x /t t
we prove (a). To prove (b), we have $u(t, ·)$2L2 (R)
≤C
!
−1
dx t
"!
#
(x − y)2 exp −C t R
1
$
f (y)dy
%2
Since f ∈ L1 , we can use Holder (or Jensen) inequality to have "!
#
(x − y)2 exp −C t R
$
f (y)dy
%2
#
!
(x − y)2 ≤ exp −2C t R
$
f (y)dy
!
f (y)dy R
Combining these two inequalities, we have # $ ! ! (x − y)2 2 −1 f (y)dy$f $L1 (R) ≤ Ct−1/2 $f $2L1 (R) . $u(t, ·)$L2 (R) ≤ C dx t exp −2C t R
Day
This proves (b). 3. Let Yj = Xj − 1. We have n n n & (2 ' ' ' E n−1 Yj = n−2 E Yi Yj ≤ n−2 E f (i − j) ≤ n−1 A j=1
i,j=1
i,j=1
By the Chebyshev’s inequality, we have P(n−1
n ' j=1
Yj ≥ 1) ≤ A/n
Hence A = B.
2
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday August 30, 2011 (Day 1)
1. Let f be a differentiable function on R whose Fourier transform is bounded and has compact support. (a) Prove that there exists a constant C ∈ R such that the k th derivative of f is bounded by C k+1 for all k ≥ 0.
(b) Prove that f does not have compact support unless it is identically zero. 2. Let F ⊂ K ⊂ L be fields. (a) Show that [L : F ] = [L : K][K : F ]. √ √ (b) Compute [Q( 3, 2) : Q]. √ √ (c) Show that x3 − 2 is irreducible over Q( 2). 3. Consider the rational map ϕ : P2 → P2 given by ϕ(X, Y, Z) = (XY, Y Z, XZ). (a) Show that ϕ is birational. (b) Find open subsets U, V ⊂ P2 such that ϕ : U → V is an isomorphism.
(c) Let Γ ⊂ P2 × P2 be the graph of ϕ (that is, the closure in P2 × P2 of the graph of the map on any open set where it’s regular). Describe the projection π1 : Γ → P2 as a blowup of P2 .
4. For any positive integer n, evaluate Z
∞ 0
x1/n dx 1 + x2
5. Let M be a closed manifold (compact, without boundary). Let f : M → R be a smooth function. For t ∈ R let Xt = f −1 (t). If there is no critical value of f in [a, b] show that Xa and Xb are submanifolds, and Xb is diffeomorphic to Xa . 6. A covering space is abelian if it is normal and its group of deck transformations is abelian. Determine all connected abelian covering spaces of S 1 ∨ S 1 (the figure 8). (Hint: one way to do this might be to consider their relation to covering spaces of S 1 × S 1 .)
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday August 31, 2011 (Day 2)
1. Find the Laurent expansion of the meromorphic function f (z) =
1 (z − 1)(z − 2)
(a) valid in the open unit disc z < 1;
(b) valid in the annulus 1 < z < 2; and
(c) valid in the complement z > 2 of the closed disc of radius 2 around 0.
2. For any closed, connected, compact, oriented nmanifolds X and Y , write X#Y for their oriented connected sum. (a) Show that Hi (X#Y ; Z) ∼ = Hi (X; Z) ⊕ Hi (Y ; Z) for all 0 < i < n.
(b) Compute the cohomology ring H ∗ ((S 2 × S 8 )#(S 4 × S 6 ); Z) and show in particular that it satisfies Poincar´e duality. 3. Let S 3 = {(x, y, z, t) ∈ R4  x2 + y 2 + z 2 + t2 = 1},
and let α be the 1form on R4 given by
α = xdy − ydx + zdt − tdz. (a) Prove that the restriction of the form α ∧ dα to S 3 is nowhere 0.
(b) Compute the integral of α ∧ dα over S 3 .
(c) Let U ⊂ S 3 be an open subset, and let v and w be everywhere independent vector fields on U with α(v) ≡ α(w) ≡ 0. If [v, w] is the Lie bracket of v and w, show that α([v, w]) is nowhere zero on U . (Hint: use polar coordinates on R4 = R2 × R2 .)
4. Let Z[i] = Z[x]/(x2 + 1). (a) What are the units in the ring Z[i]? (b) What are the primes in Z[i]? (c) Factorize 11 + 7i into primes in Z[i]. 5. Let Y ⊂ Pn be an irreducible variety of dimension r and degree d > 1, and let p ∈ Y be a nonsingular point. Define X to be the closure of the union of all lines p, q, with q ∈ Y and q 6= p.
(a) Show that X is a variety of dimension r + 1. (b) Show that the degree of X is strictly less than d. (c) Give an example where the degree of X is strictly less than d − 1. 6. Let L2 (R) denote the completion of the Banach space of smooth functions with compact support using the norm whose square is Z 2 f 2; kf k = R
and let L21 (R) be the completion of the Banach space of smooth functions with compact support using the norm whose square is Z df kf k21 = ( )2 + f 2 . dx R df (a) Prove that the map f 7→ dx from the space of smooth, compactly supported functions to itself extends to a bounded, linear map φ from L21 (R) to L2 (R).
(b) Prove that this extended map φ does not have closed image. df − f from the space of smooth, compactly (c) Prove that the map f 7→ dx supported functions to itself extends to an isometry from L21 (R) to L2 (R).
(d) Prove that the map f 7→ cokernel.
df √ x dx − 1+x2 f
has closed image and 1dimensional
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 1, 2011 (Day 3)
1. Let Λ1 , Λ2 and Λ3 ⊂ P2n+1 be pairwise skew (that is, disjoint) nplanes, and let X ⊂ P2n+1 be the union of all lines L ⊂ P2n+1 that meet all three. (a) Show that through every point p ∈ Λ1 there is a unique line meeting both Λ2 and Λ3 . (b) Show that X ⊂ P2n+1 is a closed subvariety. (c) What is the dimension of X?
2. (a) Define the degree of a map f : S n → S n
(b) Show that the degree of f is zero if f is not surjective. (c) Show that if f has no fixed points, it has the same degree as the antipodal map. What is this degree? (d) Show that Z/2 is the only group that can act freely on S 2n .
3. Let p be a prime and G = GL2 (Fp ). (a) Find the order of G. (b) Show that the order of every element of G divides either (p2 − 1) or p(p − 1). 4. Let H = {z = x + iy : y > 0} ⊂ C be the upper half plane, with the metric ds2 =
dxdy . y2
(a) What are the equations for the geodesics? Prove that they are either straight lines or semicircles. (b) Compute the scalar curvature. 5. Let X be a Banach space. Assume that the dual X ∗ of X is separable. Show that X is separable. 6. Let f : C → C be continuous on all of C and analytic on C \ [−1, 1]. Show that f is entire, that is, analytic on all of C.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday August 31, 2010 (Day 1)
1. (CA) Evaluate
Z
∞ −∞
sin2 x dx. x2
2. (A) Let b be any integer with (7, b) = 1 and consider the polynomial fb (x) = x3 − 21x + 35b. (a) Show that fb is irreducible over Q. (b) Let P denote the set of b ∈ Z such that (7, b) = 1 and the Galois group of fb is the alternating group A3 . Find P . 3. (T) Let X be the Klein bottle.1 (a) Compute the homology groups Hn (X, Z). (b) Compute the homology groups Hn (X, Z/2). (c) Compute the homology groups Hn (X × X, Z/2). 4. (RA) Let f be a Lebesgue integrable function on the closed interval [0, 1] ⊂ R. (a) Suppose that g is a continuous function on [0, 1] such that the integral of f − g is less than ǫ2 . Prove that the set where f − g > ǫ has measure less than ǫ. (b) Show that for every ǫ > 0, there is a continuous function g on [0, 1] such that the integral of f − g is less than ǫ2 . 5. (DG) Let v denote a vector field on a smooth manifold M and let p ∈ M be a point. An integral curve of v through p is a smooth map γ : U → M from a neighborhood U of 0 ∈ R to M such that γ(0) = p and the differential dγ carries the tangent vector ∂/∂t to v(γ(t)) for all t ∈ U . (a) Prove that for any p ∈ M there is an integral curve of v through p.
(b) Prove that any two integral curves of v through any given point p agree on some neighborhood of 0 ∈ R. 1
The Klein bottle is obtained from the square I 2 = {(x, y) : 0 ≤ x, y ≤ 1} ⊂ R2 by the equivalence relation (0, y) ∼ (1, y) and (x, 0) ∼ (1 − x, 1)
(c) A complete integral curve of v through p is one whose associated map has domain the whole of R. Give an example of a nowhere zero vector field on R2 that has a complete integral curve through any given point. Then, give an example of a nowhere zero vector field on R2 and a point which has no complete integral curve through it. 6. (AG) Show that a general hypersurface X ⊂ Pn of degree d > 2n − 3 contains no lines L ⊂ Pn .
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 1, 2010 (Day 2)
1. (T) If Mg denotes the closed orientable surface of genus g, show that continuous maps Mg → Mh of degree 1 exist if and only if g ≥ h. 2. (RA) Let f ∈ C(S 1 ) be a continuous function with a continuous P first derivative f ′ (x). Let {an } be the Fourier coefficients of f . Prove that n an  < ∞.
3. (DG) Let S ⊂ R3 be the surface given as a graph z = ax2 + 2bxy + cy 2 where a, b and c are constants.
(a) Give a formula for the curvature at (x, y, z) = (0, 0, 0) of the induced Riemannian metric on S. (b) Give a formula for the second fundamental form at (x, y, z) = (0, 0, 0). (c) Give necessary and sufficient conditions on the constants a, b and c that any curve in S whose image under projection to the (x, y)plane is a straight line through (0, 0) is a geodesic on S. 4. (AG) Let V and W be complex vector spaces of dimensions m and n respectively, and A ⊂ V a subspace of dimension l. Let PHom(V, W ) ∼ = Pmn−1 be the projective space of nonzero linear maps φ : V → W mod scalars, and for any integer k ≤ l let Ψk = {φ : V → W : rank(φA ) ≤ k} ⊂ Pmn−1 . Show that Ψk is an irreducible subvariety of Pmn−1 , and find its dimension. 5. (CA) Find a conformal map from the region Ω = {z : z − 1 > 1 and z − 2 < 2} ⊂ C between the two circles z − 1 = 1 and z − 2 = 2 onto the upperhalf plane. 6. (A) Let G be a finite group with an automorphism σ : G → G. If σ 2 = id and the only element fixed by σ is the identity of G, show that G is abelian.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 2, 2010 (Day 3) 1. (DG) Let D ⊂ R2 be the closed unit disk, with boundary ∂D ∼ = S 1 . For any 2 smooth map γ : D → R , let A(γ) denote the integral over D of the pullback γ ∗ (dx ∧ dy) of the area 2form dx ∧ dy on R2 . (a) Prove that A(γ) = A(γ ′ ) if γ = γ ′ on the boundary of D. (b) Let α : ∂D → R2 denote a smooth map, and let γ : D → R2 denote a smooth map such that γ∂D = α. Give an expression for A(γ) as an integral over ∂D of a function that is expressed only in terms of α and its derivatives to various orders. (c) Give an example of a map γ such that γ ∗ (dx ∧ dy) is a positive multiple of dx ∧ dy at some points and a negative multiple at others. 2. (T) Compute the fundamental group of the space X obtained from two tori S 1 × S 1 by identifying a circle S 1 × {x0 } in one torus with the corresponding circle S 1 × {x0 } in the other torus. 3. (CA) Let u be a positive harmonic function on C. Show that u is constant. √ 4. (A) Let R = Z[ −5]. Express the ideal (6) = 6R ⊂ R as a product of prime ideals in R. 5. (AG) Let Q ⊂ P5 be a smooth quadric hypersurface, and L ⊂ Q a line. Show that there are exactly two 2planes Λ ∼ = P2 ⊂ P5 contained in Q and containing L. 6. (RA) Let C ∞ denote the space of smooth, real valued functions on the closed interval I = [0, 1]. Let H denote the completion of C ∞ using the norm whose square is the functional Z df 2 2 ( ) + f dt. f 7→ dt I (a) Prove that the map of C ∞ to itself given by f 7→ T (f ) with T (f )(t) =
Z
t
f (s)ds 0
extends to give a bounded map from H to H, and prove that the norm of T is 1.
(b) Prove that T is a compact mapping from H to H (c) Let C 1/2 be the Banach space obtained by completing C ∞ using the norm given by f (t) − f (t′ ) f 7→ sup + sup f (t). t − t′ 1/2 t t6=t′ Prove that the inclusion of C ∞ into H and into C 1/2 extends to give a bounded, linear map from H to C 1/2 .
(d) Give an example of a sequence in H such that all elements have norm 1 and such that there are no convergent subsequences in C 1/2 .
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday January 19, 2010 (Day 1) 1. Let (X, µ) be a measure space with µ(X) < ∞. For q > 0, let Lq = Lq (X, µ) denote the Banach space completion of the space of bounded functions on X with the norm Z 1 q q f q = f  µ . X
Now suppose that 0 < p ≤ q. Prove that all functions in Lq are in Lp , and that the inclusion map Lq ֒→ Lp is continuous. 2. Let X ⊂ Pn be an irreducible projective variety of dimension k, G(ℓ, n) the Grassmannian of ℓplanes in Pn for some ℓ < n − k, and C(X) ⊂ G(ℓ, n) the variety of ℓplanes meeting X. Prove that C(X) is irreducible, and find its dimension. 3. Let λ be real number greater than 1. Show that the equation zeλ−z = 1 has exactly one solution z with z < 1, and that this solution z is real. (Hint: use Rouch´e’s theorem.) 4. Let k be a finite field, with algebraic closure k. (a) For each integer n ≥ 1, show that there is a unique subfield kn ⊂ k containing k and having degree n over k. (b) Show that kn is a Galois extension of k, with cyclic Galois group. (c) Show that the norm map kn× → k × (sending a nonzero element of kn to the product of its Galois conjugates) is a surjective homomorphism. 5. Suppose ω is a closed 2form on a manifold M . For every point p ∈ M , let Rp (ω) = {v ∈ Tp M : ωp (v, u) = 0 for all u ∈ Tp M }. Suppose that the dimension of Rp is the same for all p. Show that the assignment p 7→ Rp as p varies in M defines an integrable subbundle of the tangent bundle T M . 6. Let X be a topological space. We say that two covering spaces f : Y → X and g : Z → X are isomorphic if there exists a homeomorphism h : Y → Z such that g ◦ h = f . If X is a compact oriented surface of genus g (that is, a gholed torus), how many connected 2sheeted covering spaces does X have, up to isomorphism?
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday January 20, 2010 (Day 2)
1. Let a be an arbitrary real number and b a positive real number. Evaluate the integral Z ∞ cos(ax) dx cosh(bx) 0 (Recall that cosh(x) = cos(ix) = 21 (ex + e−x ) is the hyperbolic cosine.) 2. For any irreducible plane curve C ⊂ P2 of degree d > 1, we define the Gauss ∗ map g : C → P2 to be the rational map sending a smooth point p ∈ C to its ∗ tangent line; we define the dual curve C ∗ ⊂ P2 of C to be the image of g. (a) Show that the dual of the dual of C is C itself. (b) Show that two irreducible conic curves C, C ′ ⊂ P2 are tangent if and only if their duals are. 3. Let Λ1 and Λ2 ⊂ R4 be complementary 2planes, and let X = R4 \ (Λ1 ∪ Λ2 ) be the complement of their union. Find the homology and cohomology groups of X with integer coefficients. 4. Let X = {(x, y, z) : x2 + y 2 = 1} ⊂ R3 be a cylinder. Show that the geodesics on X are helices, that is, curves such that the angle between their tangent lines and the vertical is constant. 5. (a) Show that if every closed and bounded subspace of a Hilbert space E is compact, then E is finite dimensional. (b) Show that any decreasing sequence of nonempty, closed, convex, and bounded subsets of a Hilbert space has a nonempty intersection. (c) Show that any closed, convex, and bounded subset of a Hilbert space is the intersection of the closed balls that contain it. (d) Deduce that any closed, convex, and bounded subset of a Hilbert space is compact in the weak topology. 6. Let p be a prime, and let G be the group Z/p2 Z ⊕ Z/p2 Z. (a) How many subgroups of order p does G have? (b) How many subgroups of order p2 does G have? How many of these are cyclic?
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday January 21, 2010 (Day 3)
1. Consider the ring A = Z[x]/(f )
where
f = x4 − x3 + x2 − 2x + 4.
Find all prime ideals of A that contain the ideal (3). 2. Let f be a holomorphic function on a domain containing the closed disc {z : z ≤ 3}, and suppose that f (1) = f (i) = f (−1) = f (−i) = 0. Show that f (0) ≤
1 max f (z) 80 z=3
and find all such functions for which equality holds in this inequality. 3. Let f : R → R+ be a differentiable, positive real function. Find the Gaussian curvature and mean curvature of the surface of revolution S = {(x, y, z) : y 2 + z 2 = f (x)}. 4. Show that for any given natural number n, there exists a finite Borel measure on the interval [0, 1] ⊂ R such that for any real polynomial of degree at most n, we have Z 1
p dµ = p′ (0).
0
Show, on the other hand, that there does not exist a finite Borel measure on the interval [0, 1] ⊂ R such that for any real polynomial we have Z
1
p dµ = p′ (0). 0
5. Let X = RP2 × RP4 . (a) Find the homology groups H∗ (X, Z/2) (b) Find the homology groups H∗ (X, Z) (c) Find the cohomology groups H ∗ (X, Z)
6. By a twisted cubic curve we mean the image of the map P1 → P3 given by [X, Y ] 7→ F0 (X, Y ), F1 (X, Y ), F2 (X, Y ), F3 (X, Y )
where F0 , . . . , F3 form a basis for the space of homogeneous cubic polynomials in X and Y . (a) Show that if C ⊂ P3 is a twisted cubic curve, then there is a 3dimensional vector space of homogeneous quadratic polynomials on P3 vanishing on C. (b) Show that C is the common zero locus of the homogeneous quadratic polynomials vanishing on it. (c) Suppose now that Q, Q′ ⊂ P3 are two distinct quadric surfaces containing C. Describe the intersection Q ∩ Q′ .
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 1, 2009 (Day 1)
1. (RA) Let H be a Hilbert space and {ui } an orthonormal basis for H. Assume that {xi } is a sequence of vectors such that X xn − un 2 < 1. Prove that the linear span of {xi } is dense in H. 2. (T) Let CPn be complex projective nspace. (a) Describe the cohomology ring H ∗ (CPn , Z) and, using the Kunneth formula, the cohomology ring H ∗ (CPn × CPn , Z). (b) Let ∆ ⊂ CPn × CPn be the diagonal, and δ = i∗ [∆] ∈ H2n (CPn × CPn , Z) the image of the fundamental class of ∆ under the inclusion i : ∆ → CPn × CPn . In terms of your description of H ∗ (CPn × CPn , Z) above, find the Poincar´e dual δ ∗ ∈ H 2n (CPn × CPn , Z) of δ. 3. (AG) Let X ⊂ Pn be an irreducible projective variety, G(1, n) the Grassmannian of lines in Pn , and F ⊂ G(1, n) the variety of lines contained in X. (a) If X has dimension k, show that dim F ≤ 2k − 2, with equality holding if and only if X ⊂ Pn is a kplane. (b) Find an example of a projective variety X ⊂ Pn with dim X = dim F = 3. 4. (CA) Let Ω ⊂ C be the open set Ω = {z : z < 2 and z − 1 > 1}. Give a conformal isomorphism between Ω and the unit disc ∆ = {z : z < 1}. 5. (A) Suppose φ is an endomorphism of a 10dimensional vector space over Q with the following properties. 1. The characteristic polynomial is (x − 2)4 (x2 − 3)3 . 2. The minimal polynomial is (x − 2)2 (x2 − 3)2 . 3. The endomorphism φ − 2I, where I is the identity map, is of rank 8. Find the Jordan canonical form for φ.
6. (DG) Let γ : (0, 1) → R3 be a smooth arc, with γ ′ 6= 0 everywhere. (a) Define the curvature and torsion of the arc. (b) Characterize all such arcs for which the curvature and torsion are constant.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 2, 2009 (Day 2) 1. (CA) Let ∆ = {z : z < 1} ⊂ C be the unit disc, and ∆∗ = ∆ \ {0} the punctured disc. A holomorphic function f on ∆∗ is said to have an essential singularity at 0 if z n f (z) does not extend to a holomorphic function on ∆ for any n. Show that if f has an essential singularity at 0, then f assumes values arbitrarily close to every complex number in any neighborhood of 0—that is, for any w ∈ C and ∀ǫ and δ > 0, there exists z ∈ ∆∗ with z < δ
and f (z) − w < ǫ. ∗
2. (AG) Let S ⊂ P3 be a smooth algebraic surface of degree d, and S ∗ ⊂ P3 the dual surface, that is, the locus of tangent planes to S. (a) Show that no plane H ⊂ P3 is tangent to S everywhere along a curve, and deduce that S ∗ is indeed a surface. (b) Assuming that a general tangent plane to S is tangent at only one point (this is true in characteristic 0), find the degree of S ∗ . 3. (T) Let X = S 1 ∨ S 1 be a figure 8, p ∈ X the point of attachment, and let α and β : [0, 1] → X be loops with base point p (that is, such that α(0) = α(1) = β(0) = β(1) = p) tracing out the two halves of X. Let Y be the CW complex formed by attaching two 2discs to X, with attaching maps homotopic to α2 β and αβ 2 . (a) Find the homology groups Hi (Y, Z). (b) Find the homology groups Hi (Y, Z/3). 4. (DG) Let f = f (x, y) : R2 → R be smooth, and let S ⊂ R3 be the graph of f , with the Riemannian metric ds2 induced by the standard metric on R3 . Denote the volume form on S by ω. (a) Show that ω=
s
∂f ∂x
2
+
∂f ∂y
(b) Find the curvature of the metric ds2 on S
2
+ 1.
5. (RA) Suppose that O ⊂ R2 is an open set with finite Lebesgue measure. Prove that the boundary of the closure of O has Lebesgue measure 0. √ 6. (A) Let R be the ring√of integers in the field Q( −5), and S the ring of integers in the field Q( −19). (a) Show that R is not a principal ideal domain (b) Show that S is a principal ideal domain
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday September 3, 2009 (Day 3)
1. (A) Let c ∈ Z be an integer not divisible by 3. (a) Show that the polynomial f (x) = x3 − x + c ∈ Q[x] is irreducible over Q. (b) Show that the Galois group of f is the symmetric group S3 . 2. (CA) Let τ1 and τ2 ∈ C be a pair of complex numbers, independent over R, and Λ = Zhτ1 , τ2 i ⊂ C the lattice of integral linear combinations of τ1 and τ2 . An entire meromorphic function f is said to be doubly periodic with respect to Λ if f (z + τ1 ) = f (z + τ2 ) = f (z) ∀z ∈ C. (a) Show that an entire holomorphic function doubly periodic with respect to Λ is constant. (b) Suppose now that f is an entire meromorphic function doubly periodic with respect to Λ, and that f is either holomorphic or has one simple pole in the closed parallelogram {aτ1 + bτ2 : a, b ∈ [0, 1] ⊂ R}. Show that f is constant. 3. (DG) Let M and N be smooth manifolds, and let π : M × N → N be the projection; let α be a differential kform on M × N . Show that α has the form π ∗ ω for some kform ω on N if and only if the contraction ιX (α) = 0 and the derivative LX (α) = 0 for any vector field X on M × N whose value at every point is in the kernel of the differential dπ. 4. (RA) Show that the Banach space ℓp can be embedded as a summand in Lp (0, 1)—in other words, that Lp (0, 1) is isomorphic as a Banach space to the direct sum of ℓp and another Banach space. 5. (T) Find the fundamental groups of the following spaces: (a) SL2 (R) (b) SL2 (C) (c) SO3 (C) 6. (AG) Let X ⊂ An be an affine algebraic variety of pure dimension r over a field K of characteristic 0.
(a) Show that the locus Xsing ⊂ X of singular points of X is a closed subvariety. (b) Show that Xsing is a proper subvariety of X.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday January 27, 2009 (Day 1) 2
1. Let Pn −1 be the space of nonzero n × n matrices mod scalars, and consider the subset 2 2 Σ = {(A, B) : AB = 0} ⊂ Pn −1 × Pn −1 . 2 −1
(a) Prove that Σ is a Zariski closed subset of Pn
2 −1
× Pn
.
(b) Is Σ irreducible? (c) What is the dimension of Σ? 2. Consider the integral Z
∞
sin x · xa−1 dx.
0
(a) For which real values of a does the integral converge absolutely? For which does it converge conditionally? (b) Evaluate the integral for those values of a for which it does converge. 3. (a) Let p be a prime number. Show that a group G of order pn (n > 1) has a nontrivial normal subgroup, that is, G is not a simple group. (b) Let p and q be primes, p > q. Show that a group G of order pq has a normal Sylow psubgroup. If G has also a normal Sylow qsubgroup, show that G is cyclic. (c) Give a necessary and sufficient condition on p and q for the existence of a nonabelian group of order pq. Justify your answer. 4. Let X = S 1 ∨ S 1 be a figure 8. (a) Exhibit two threesheeted covering spaces f : Y → X and g : Z → X such that Y and Z are not homeomorphic. (b) Exhibit two threesheeted covering spaces f : Y → X and g : Z → X such that Y and Z are homeomorphic, but not as covering spaces of X (i.e., there is no homeomorphism φ : Y → Z such that g ◦ φ = f ). (c) Exhibit a normal (that is, Galois) threesheeted covering space of X. (d) Exhibit a nonnormal threesheeted covering space of X. (e) Which of the above would still be possible if we were considering twosheeted covering spaces instead of threesheeted?.
5. Suppose T is a bounded operator in a Hilbert space V and there exist a basis {ek } for V such that T ek = λk ek . Prove that T is compact if λk → 0 as k → ∞. 6. Let Σ ⊂ R3 be a smooth 2dimensional submanifold, and n : Σ → R3 a smooth map such that n(p) is a unit length normal to Σ at p. Identify the tangent bundle T Σ as the subspace of pairs (p, v) ∈ Σ × R3 such that v · n(p) = 0, where · designates the Euclidean inner product. Suppose now that t → p(t) is a smoothly parametrized curve in R3 that lies on Σ. Prove that this curve is a geodesic if and only if p′′ (t) · (n(p(t)) × p′ (t)) = 0 ∀t Here, p′ is the derivative of the map t → p(t) and p′′ is the second derivative.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday January 28, 2009 (Day 2) 1. Let C ⊂ Pn be a smooth algebraic curve. (a) Let Λ ⊂ Pn be a general (n − 4)plane. Show that the projection map πΛ : C → P3 is an embedding. (b) Now let Λ ⊂ Pn be a general (n − 3)plane. Show that the projection map πΛ : C → P2 is birational onto its image, and that the image curve has only nodes (ordinary double points) as singularities. 2. Show that the function defined by f (z) =
∞ X
n
z2
n=0
is analytic in the open disc z < 1, but has no analytic continuation to any larger domain. 3. (a) Let K be the splitting field of the polynomial f (x) = x3 − 2 over Q. Find the Galois group G of K/Q and describe its action on the roots of f . (b) Let K be the splitting field of the polynomial X 4 + aX 2 + b (where a, b ∈ Q) over the rationals. Assuming that the polynomial is irreducible, prove that the Galois group G of the extension K/Q is either C4 , or C2 × C2 , or the dihedral group D8 . 4. Let {fn } be a sequence of functions on the interval X = (0, 1) ⊂ R, and suppose fn → f in Lp (X) for all p : 1 ≤ p < ∞. Does it imply that fn → f almost everywhere? Does it imply that there is a subsequence of fn converging to f almost everywhere? Prove your answer or give a counterexample. 5. View S 2n+1 as the unit sphere in Cn+1 , and in particular S 1 as the unit circle in C. Define an action of S 1 on S 2n+1 by λ : (z1 , . . . , zn+1 7→ (λz1 , . . . , λzn+1 ) The quotient is the space CPn . View the projection map π : S 2n+1 → CPn as a principal S 1 bundle. (a) Explain why the restriction to S 2n+1 of the 1form 1 X A= (z k dzk − zk dz k ) 2 1≤k≤n=1
defines a connection on this bundle.
(b) What is the pullback to S 2n+1 of the curvature 2form of this connection? 6. Let X = S 2 × RP3 and Y = S 3 × RP2 (a) Find the homology groups Hn (X, Z) and Hn (Y, Z) for all n. (b) Find the homology groups Hn (X, Z/2) and Hn (Y, Z/2) for all n. (c) Find the homotopy groups π1 (X) and π1 (Y ).
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday January 29, 2009 (Day 3)
1. Let C ⊂ P1 × P1 be an algebraic curve of bidegree (a, b) (that is, the zero locus of a bihomogeneous polynomial of bidegree (a, b)), and let C ′ ⊂ P3 be the image of C under the Segre embedding σ : P1 × P1 → P3 . (a) What is the degree of C ′ ? (b) Assume now that max(a, b) ≥ 3. Show that C ′ lies on one and only one quadric surface Q ⊂ P3 (namely, the quadric surface σ(P1 × P1 )). 2. Find the Laurent expansion f (z) =
X
an z n
n∈Z
around 0 of the function f (z) =
z2
1 +z+1
(a) valid in the open unit disc {z : z < 1}, and (b) valid in the complement {z : z > 1} of the closed unit disc in C. 3. Let A be a commutative ring. Show that an element a ∈ A belongs to the intersection of all prime ideals in A if and only if it’s nilpotent. 4. Let f be a given realvalued function on X = (0, 1) ⊂ R, and define a function φ : [1, ∞) → R by φ(p) = kf kpLp (X) . Prove that φ is convex. 5. Let X ⊂ R2 be a connected onedimensional real analytic submanifold, not contained in a line. Prove that not every tangent line to X is bitangent—that is, it is not the case that for all p ∈ X there exists q 6= p ∈ X such that the tangent line to X at p equals the tangent line to X at q as lines in R2 . 6. Let X and Y be two CW complexes. (a) Show that χ(X × Y ) = χ(X)χ(Y ). (b) Let A and B be two subcomplexes of X such that X = A ∪ B. Show that χ(X) = χ(A) + χ(B) − χ(A ∩ B).
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 16 2008 (Day 1)
1. (a) Prove that the Galois group G of the polynomial X 6 + 3 over Q is of order 6. (b) Show that in fact G is isomorphic to the symmetric group S3 . (c) Is there a prime number p such that X 6 + 3 is irreducible over the finite field of order p? 2. Evaluate the integral Z
∞ 0
√
t dt. (1 + t)2
3. For X ⊂ R3 a smooth oriented surface, we define the Gauss map g : X → S 2 to be the map sending each point p ∈ X to the unit normal vector to X at p. We say that a point p ∈ X is parabolic if the differential dgp : Tp (X) → Tg(p) (S 2 ) of the map g at p is singular. (a) Find an example of a surface X such that every point of X is parabolic. (b) Suppose now that the locus of parabolic points is a smooth curve C ⊂ X, and that at every point p ∈ C the tangent line Tp (C) ⊂ Tp (X) coincides with the kernel of the map dgp . Show that C is a planar curve, that is, each connected component lies entirely in some plane in R3 . 4. Let X = (S 1 × S 1 ) \ {p} be a oncepunctured torus. (a) How many connected, 3sheeted covering spaces f : Y → X are there? (b) Show that for any of these covering spaces, Y is either a 3times punctured torus or a oncepunctured surface of genus 2. 5. Let X be a complete metric space with metric ρ. A map f : X → X is said to be contracting if for any two distinct points x, y ∈ X, ρ(f (x), f (y)) < ρ(x, y). The map f is said to be uniformly contracting if there exists a constant c < 1 such that for any two distinct points x, y ∈ X, ρ(f (x), f (y)) < c · ρ(x, y).
(a) Suppose that f is uniformly contracting. Prove that there exists a unique point x ∈ X such that f (x) = x. (b) Give an example of a contracting map f : [0, ∞) → [0, ∞) such that f (x) 6= x for all x ∈ [0, ∞). 6. Let K be an algebraically closed field of characteristic other than 2, and let Q ⊂ P3 be the surface defined by the equation X 2 + Y 2 + Z 2 + W 2 = 0. (a) Find equations of all lines L ⊂ P3 contained in Q. (b) Let G = G(1, 3) ⊂ P5 be the Grassmannian of lines in P3 , and F ⊂ G the set of lines contained in Q. Show that F ⊂ G is a closed subvariety.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 17 2008 (Day 2)
1. (a) Show that the ring Z[i] is Euclidean. (b) What are the units in Z[i]? (c) What are the primes in Z[i]? (d) Factorize 11 + 7i into primes in Z[i]. 2. Let U ⊂ C be the open region U = {z : z − 1 < 1 and z − i < 1} . Find a conformal map f : U → ∆ of U onto the unit disc ∆ = {z : z < 1}. 3. Let n be a positive integer, A a symmetric n × n matrix and Q the quadratic form X Q(x) = Ai,j xi xj . 1≤i,j≤n
Define a metric on Rn using the line element whose square is X ds2 = eQ(x) dxi ⊗ dxi . 1≤i≤n
(a) Write down the differential equation satisfied by the geodesics of this metric (b) Write down the Riemannian curvature tensor of this metric at the origin in Rn . 4. Let H be a separable Hilbert space and b : H → H a bounded linear operator. (a) Prove that there exists r > 0 such that b + r is invertible. (b) Suppose that H is infinite dimensional and that b is compact. Prove that b is not invertible. 5. Let X ⊂ Pn be a projective variety. (a) Define the Hilbert function hX (m) and the Hilbert polynomial pX (m) of X. (b) What is the significance of the degree of pX ? Of the coefficient of its leading term?
(c) For each m, give an example of a variety X ⊂ Pn such that hX (m) 6= pX (m). 6. Let X = S 2 ∨ RP2 be the wedge of the 2sphere and the real projective plane. (This is the space obtained from the disjoint union of the 2sphere and the real projective plane by the equivalence relation that identifies a given point in S 2 with a given point in RP2 , with the quotient topology.) (a) Find the homology groups Hn (X, Z) for all n. (b) Describe the universal covering space of X. (c) Find the fundamental group π1 (X).
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday January 31 2008 (Day 3)
1. For z ∈ C \ Z, set f (z) = lim
N →∞
N X
n=−N
1 z+n
!
(a) Show that this limit exists, and that the function f defined in this way is meromorphic. (b) Show that f (z) = π cot πz. 2. Let p be an odd prime. (a) What is the order of GL2 (Fp )? (b) Classify the finite groups of order p2 . (c) Classify the finite groups G of order p3 such that every element has order p. 3. Let X and Y be compact, connected, oriented 3manifolds, with π1 (X) = (Z/3Z) ⊕ Z ⊕ Z
and
π1 (Y ) = (Z/6Z) ⊕ Z ⊕ Z ⊕ Z.
(a) Find Hn (X, Z) and Hn (Y, Z) for all n. (b) Find Hn (X × Y, Q) for all n. 4. Let Cc∞ (R) be the space of differentiable functions on R with compact support, and let L1 (R) be the completion of Cc∞ (R) with respect to the L1 norm. Let f ∈ L1 (R). Prove that Z 1 lim f (y) − f (x)dy = 0 h→0 h y−x 0 be small and R large, and let γ be the contour which starts at ǫi, travels along the ray z = [0, ∞) + ǫi until it reaches the circle z = R, traverses most of that circle counterclockwise stopping at the ray z = [0, ∞) − ǫi, then travels along that ray backwards, and finally traverses the semicircle Rz = ǫ in the left halfplane to get back to ǫi. Consider the contour integral contribution from the first ray is approximately the desired γ f (z) dz. The R∞√ integral I = 0 t/(1 + t2 ) dt; the contribution from the large circle is small, √ because when z = R,  z/(1 + z)2  is about R−3/2 , and the perimeter of the circle is only about 2πR; the contribution from the second ray is about I again, because the sign from traveling in the opposite direction cancels the √ sign coming from the branch cut in z; and the contribution from the small circle is small because f (z) is bounded in a neighborhood of 0. So √ √ Z z d z 1 2I = lim dz = 2πi =π = 2πi √ ǫ→0,R→∞ γ (1 + z)2 dz z=−1 2 −1 and thus I = π.
3. For X ⊂ R3 a smooth oriented surface, we define the Gauss map g : X → S 2 to be the map sending each point p ∈ X to the unit normal vector to X at p. We say that a point p ∈ X is parabolic if the differential dgp : Tp (X) → Tg(p) (S 2 ) of the map g at p is singular. (a) Find an example of a surface X such that every point of X is parabolic. (b) Suppose now that the locus of parabolic points is a smooth curve C ⊂ X, and that at every point p ∈ C the tangent line Tp (C) ⊂ Tp (X) coincides with the kernel of the map dgp . Show that C is a planar curve, that is, each connected component lies entirely in some plane in R3 . Solution.
(a) Let X be the xyplane; then the Gauss map g : X → S 2 is constant, so its differential is everywhere zero and hence singular. (b) Consider the Gauss map of X restricted to C, gC : C → S 2 . Then for any point p ∈ C, d(gC )p = (dgp )Tp (C) , which is 0 by assumption. Hence gC is locally constant on C. That is, on each connected component C0 of C there is a fixed vector (the value of gC at any point of the component) normal to all of C0 . Hence C0 lies in a plane in R3 normal to this vector. 4. Let X = (S 1 × S 1 ) \ {p} be a oncepunctured torus. (a) How many connected, 3sheeted covering spaces f : Y → X are there? (b) Show that for any of these covering spaces, Y is either a 3times punctured torus or a oncepunctured surface of genus 2. Solution. (a) By covering space theory, the number of connected, 3sheeted covering spaces of a space Z is the number of conjugacy classes of subgroups of index 3 in the fundamental group π1 (Z). (We consider two covering spaces of Z isomorphic only when they are related by an homeomorphism over the identity on Z, not one over any homeomorphism of Z.) So we may replace X by the homotopy equivalent space X ′ = S 1 ∨ S 1 . If we view this new space X ′ as a graph with one vertex and two directed loops labeled a and b, then a connected 3sheeted cover of X ′ is a connected graph with three vertices and some directed edges labeled a or b such that each vertex has exactly one incoming and one outgoing edge with each of the labels a and b. Temporarily treating the three vertices as having distinct labels x, y, z, we find six ways the a edges can be placed: loops at x, y and z; a loop at x and edges from y to z and from z to y; similarly but with the loop at y; similarly but with the loop at z; edges from x to y, y to z, and z to x; and edges from x to z, z to y, and y to x. Analogously there are six possible placements for the b edges. Considering all possible combinations, throwing out the disconnected ones, and then treating two graphs as the same if they differ only in the labels x, y, z, we arrive at seven distinct possibilities. (b) Let C be a small loop in S 1 × S 1 around the removed point p, and let X0 ⊂ X be the torus with the interior of C removed, so that X0 is a compact manifold with boundary C = S 1 . Now let Y be any connected, 3sheeted covering space of X. Pull back the covering map Y → X along the inclusion X0 → X to obtain a 3sheeted covering space Y0 of X0 . Since X0 → X is a homotopy equivalence, so is Y0 → Y and in particular Y0 is still connected. We can recover Y from Y0 by gluing a strip D×[0, 1) along the preimage D of C in Y0 . So, it will suffice to show that Y is
either a torus with three small disks removed, or a surface of genus two with one small disk removed. Since Y0 is a 3sheeted cover of X0 , it is a compact oriented surface with boundary. By the classification of compact oriented surfaces with boundary, Y0 can be formed by taking a surface of some genus g and removing some number d of small disks. The boundary of Y0 is D, the preimage of C, which is a (not necessarily connected) 3sheeted cover of C. So Y0 has either one or three boundary circles, i.e., d = 1 or d = 3. Moreover, we can compute using the Euler characteristic that 2 − 2g − d = χ(Y0 ) = 3χ(X0 ) = −3. If d = 3, then g = 1; if d = 1, then g = 2. So Y is correspondingly either a 3times punctured torus or a oncepunctured surface of genus two. 5. Let X be a complete metric space with metric ρ. A map f : X → X is said to be contracting if for any two distinct points x, y ∈ X, ρ(f (x), f (y)) < ρ(x, y). The map f is said to be uniformly contracting if there exists a constant c < 1 such that for any two distinct points x, y ∈ X, ρ(f (x), f (y)) < c · ρ(x, y). (a) Suppose that f is uniformly contracting. Prove that there exists a unique point x ∈ X such that f (x) = x. (b) Give an example of a contracting map f : [0, ∞) → [0, ∞) such that f (x) 6= x for all x ∈ [0, ∞). Solution. (a) We first show there exists at least one fixed point of f . Let x0 ∈ X be arbitrary and define a sequence x1 , x2 , . . . , by xn = f (xn−1 ). Let d = ρ(x0 , x1 ). By the uniformly contracting property of f , ρ(xn , xn+1 ) ≤ dcn for every n. Now observe ρ(xn , xn+k ) ≤ ρ(xn , xn+1 ) + · · · + ρ(xn+k−1 , xn+k ) ≤ dcn + · · · + dcn+k−1 ≤ dcn /(1 − c). This expression tends to 0 as n increases, so (xn ) is a Cauchy sequence and thus has a limit x by the completeness of X. Now f is continuous, because it is uniformly contracting, so f (x) = lim f (xn ) = lim xn+1 = x, n→∞
n→∞
and x is a fixed point of f , as desired. To show that f has at most one fixed point, suppose x and y were distinct points of X with f (x) = x and f (y) = y. Then ρ(x, y) = ρ(f (x), f (y)) < cρ(x, y), which is impossible since ρ(x, y) > 0 and c < 1. (b) Let f (x) = x + e−x . Then f ′ (x) = 1 − e−x ∈ [0, 1) for all x ≥ 0, so by the Mean Value Theorem 0 ≤ f (x) − f (y) < x − y for any x > y ≥ 0. Thus f is contracting. But f has no fixed points, because x + e−x/2 6= x for all x. 6. Let K be an algebraically closed field of characteristic other than 2, and let Q ⊂ P3 be the surface defined by the equation X 2 + Y 2 + Z 2 + W 2 = 0. (a) Find equations of all lines L ⊂ P3 contained in Q. (b) Let G = G(1, 3) ⊂ P5 be the Grassmannian of lines in P3 , and F ⊂ G the set of lines contained in Q. Show that F ⊂ G is a closed subvariety. Solution. (a) Since K is algebraically closed and of characteristic other than 2, we may replace the quadratic form X 2 + Y 2 + Z 2 + W 2 with any other nondegenerate one, set A = X + √ √ such as AB + √ CD. More explicitly, √ −1Y , B = X − −1Y , C = Z + −1W , D = −Z + −1W ; this change of coordinates is invertible because we can divide by 2, and AB − CD = X 2 + Y 2 + Z 2 + W 2. A line contained in the surface in P3 defined by AB −CD = 0 is the same as a plane in the subset of the vector space K 4 defined by v1 v2 −v3 v4 = 0. Define a bilinear form (·, ·) on K 4 by (v, w) = v1 w2 + v2 w1 − v3 w4 − v4 w3 . Then we want to find all the planes V ⊂ K 4 such that (v, v) = 0 for every v ∈ V . Observe that (v + w, v + w) − (v, v) − (w, w) = (v, w) + (w, v) = 2(v, w), so it is equivalent to require that (v, w) = 0 for all v and w ∈ V . Suppose now that V is such a plane inside K 4 . Then V has nontrivial intersection with the subspace {v1 = 0}; let v ∈ V be a nonzero vector with v1 = 0. Since v1 v2 − v3 v4 = 0, we must have either v3 = 0 or v4 = 0. Assume without loss of generality that v3 = 0. Write u = v2 , t = v4 ; then (u, t) 6= (0, 0). Now consider any vector w ∈ V ; then 0 = (w, v) = w1 v2 + w2 v1 − w3 v4 − w4 v3 = uw1 − tw3 .
So there exists r ∈ K such that w1 = rt and w3 = ru. We also have 1 0 = (w, w) = w1 w2 − w3 w4 = rtw2 − ruw4 . 2 Hence either r = 0 or there exists s ∈ K such that w2 = su and w4 = st. So V ⊂ { (w1 , 0, w3 , 0)  w1 , w3 ∈ K } ∪ { (rt, su, ru, st)  r, s ∈ K }. Since V has dimension 2, we conclude that V must be equal to one of these two planes. This discussion was under the assumption that v3 = 0 rather than v4 = 0; in the second case, we find that V is of one of the forms { (w1 , 0, 0, w4 )  w1 , w4 ∈ K } or { (rt, su, st, ru)  r, s ∈ K } for (u, t) 6= (0, 0). But we obtain { (w1 , 0, w3 , 0)  w1 , w3 ∈ K } by setting (u, t) = (0, 1) in { (rt, su, st, ru)  r, s ∈ K } and { (w1 , 0, 0, w4 )  w1 , w4 ∈ K } by setting (u, t) = (0, 1) in { (rt, su, ru, st)  r, s ∈ K }. Hence all such planes V are of one of the forms (1)
Vu,t = { (rt, su, ru, st)  r, s ∈ K } or
(2)
Vu,t = { (rt, su, st, ru)  r, s ∈ K } for some (u, t) 6= (0, 0). And it is easy to see conversely that each of these subspaces is twodimensional and lies in the subset of K 4 determined by (v, v) = 0. Translating this back into equations for the lines on the surface Q, we obtain two families of lines: rt + su rt − su ru − st ru + st (1) : Lu,t =  r, s ∈ K , : √ : √ 2 2 2 −1 2 −1 rt + su rt − su st − ru st + ru (2) Lu,t = : √ : √ :  r, s ∈ K , 2 2 2 −1 2 −1 (1)
(2)
where (u, t) ranges over K 2 \ {(0, 0)}. The families L∗,∗ and L∗,∗ are disjoint, and two pairs (u, t) and (u′ , t′ ) yield the same line in a given family if and only if one pair is a nonzero scalar multiple of the other. (b) By the result of the previous part, F is the image of a regular map P1 ∐ P1 → G, so F is a closed subvariety of G.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday September 17 2008 (Day 2)
1. (a) Show that the ring Z[i] is Euclidean. (b) What are the units in Z[i]? (c) What are the primes in Z[i]? (d) Factorize 11 + 7i into primes in Z[i]. Solution. √ (a) We define a norm on Z[i] in the usual way, a + bi = a2 + b2 . Then we must show that for any a and b in Z[i] with b 6= 0, there exist q and r in Z[i] with a = qb + r and r < b. Let q0 = a/b ∈ C and let q ∈ Z[i] be one of the Gaussian integers closest to q0 ; the real and √ imaginary parts of q differ by at most 12 from those of q0 , so q − q0  ≤ 2/2 < 1. Now let r = a − qb. Then r = a − qb = (q0 − q)b = q0 − qb < b as desired. (b) If u ∈ Z[i] is a unit, then there exists u′ ∈ Z[i] such that uu′ = 1, so uu′  = 1 and hence u = 1 (since √ z > 0 for every z ∈ Z[i]). Writing u = a + bi, we obtain 1 = u = a2 + b2 so either a = ±1 and b = 0 or a = 0 and b = ±1. The four possibilities u = 1, −1, i, −i are all clearly units. (c) Since Z[i] is Euclidean, it contains a greatest common divisor of any two elements, and it follows that irreducibles and primes are the same: if z is irreducible and z ∤ x and n ∤ y, then gcd(x, z) = gcd(y, z) = 1, so 1 ∈ (x, z) and 1 ∈ (y, z); hence 1 ∈ (xy, z), so z ∤ xy. Let z ∈ Z[i]. If z ≤ 1, then z is either zero or a unit so is not prime. If √ z = p, p ∈ Z a prime, then u must be a prime in Z[i], because  ·  is multiplicative and z2 ∈ Z for all z ∈ Z[i]. It remains to consider z for which z2 is composite. √ Write N = z, and factor N = p1 p2 · · · pr in Z. Note that z  z z¯ = N = p1 p2 · · · pr so if z is prime, then z divides one of the primes p = pi in Z[i]. Moreover z¯ also divides p so N = z z¯ divides p2 ; since N is composite we must have N = p2 . That is, z z¯ = p2 ; by assumption the left side is a factorization
into irreducibles, so up to units each p on the right hand side must be a product of some terms on the left; the only possibility is z = pu, z¯ = p¯ u for some unit u. Now when p ≡ 3 (mod 4), p is indeed a prime in Z[i], because then p  a2 + b2 =⇒ p  a, b =⇒ p2  a2 + b2 , so there are no √ elements of Z[i] with norm p. If p ≡ 1 (mod 4), then p can be written in the form p = a2 + b2 , so p = (a + bi)(a − bi) and p is not in fact a prime. In conclusion, the primes of Z[i] are √ • elements z ∈ Z[i] with z = p, p ∈ Z prime (necessarily congruent to 1 mod 4); • elements of the form pu with p ∈ Z a prime congruent to 3 mod 4 and u ∈ Z[i] a unit. √ √ + 49 = 170; so 11 + 7i will be (d) We first compute 11 + 7i = 121 √ √ √ a product of primes is only one √ with norms 2, 5 and 17. There √ prime with norm 2 up to units and only two with a norm 5; a quick calculation yields 11 + 7i = (1 + i)(1 + 2i)(1 − 4i). 2. Let U ⊂ C be the open region U = {z : z − 1 < 1 and z − i < 1} . Find a conformal map f : U → ∆ of U onto the unit disc ∆ = {z : z < 1}. Solution. The map z 7→ 1/z takes the open discs { z : z − 1 < 1 } and { z : z − i < 1 } holomorphically to the open halfplanes { z : ℜz ≥ 21 } and { z : ℑz ≤ − 12 } respectively, so it takes U to their intersection. So we can define a conformal isomorphism f0 from U to the interior U ′ of the fourth quadrant by 1 1−i . f0 (z) = − z 2 Now we can send U ′ to the lower half plane by the squaring map, and that to 1 ∆ by the M¨ obius transformation z 7→ z−i/2 − i. Thus the composite 1 ( z1
−
1−i 2 2 )
+
i 2
−i
is actually a conformal isomorphism from U to ∆. 3. Let n be a positive integer, A a symmetric n × n matrix and Q the quadratic form X Q(x) = Ai,j xi xj . 1≤i,j≤n
Define a metric on Rn using the line element whose square is X ds2 = eQ(x) dxi ⊗ dxi . 1≤i≤n
(a) Write down the differential equation satisfied by the geodesics of this metric (b) Write down the Riemannian curvature tensor of this metric at the origin in Rn . Solution. We first compute the Christoffel symbols Γm ij with respect to the standard basis for the tangent space (∂/∂xk . The metric tensor in these coordinates is with inverse g ij = δij e−Q(x) .
gij = δij eQ(x) Its partial derivatives are
X ∂ ∂ Alk xl . gij = δij eQ(x) Q(x) = 2δij eQ(x) ∂xk ∂xk l
Then (using implicit summation notation) 1 km ∂ ∂ ∂ m Γ ij = g gkj + gik − gij 2 ∂xi ∂xj ∂xk 1 δkm e−Q(x) (2δkj eQ(x) Ali xl + 2δik eQ(x) Alj xl − 2δij eQ(x) Alk xl ) = 2 = (δmj Ali + δim Alj − δij Alm )xl . (a) The geodesic equation is 0 = = =
d2 xm dxi dxj + Γm ij 2 dt dt dt d2 xm dxi dxj + (δmj Ali + δim Alj − δij Alm )xl 2 dt dt dt X X dxi 2 dxi dxm X d2 xm +2 Ali xl − Alm xl dt2 dt dt dt i,l
l
i
(where we have written summations explicitly on the last line). (b) The Riemannian curvature tensor is given by ∂ l ∂ l Γ ik − Γ ij + Γl js Γs ik − Γl ks Γsij ∂xj ∂xk = (δlk Ari + δil Ark − δik Arl ) − (δlj Ari + δil Arj − δij Arl )
Rl ijk =
+ (δls Atj + δjl Ats − δjs Atl )xt (δsk Aui + δis Auk − δik Aus )xu − (δls Atk + δkl Ats − δks Atl )xt (δsj Aui + δis Auj − δij Aus )xu . 4. Let H be a separable Hilbert space and b : H → H a bounded linear operator. (a) Prove that there exists r > 0 such that b + r is invertible.
(b) Suppose that H is infinite dimensional and that b is compact. Prove that b is not invertible. Solution. (a) It is equivalent to show that there exists ǫ > 0 such that 1−ǫb is invertible. Since b is bounded there is a constant C such that bv ≤ Cv for all v ∈ H. Choose ǫ < 1/C and consider the series a = 1 + ǫb + ǫ2 b2 + · · · . For any v the sequence v + ǫbv + ǫ2 b2 v + · · · converges by comparison to a geometric series. So this series converges to a linear operator a and a(1 − ǫb) = (1 − ǫb)a = 1, that is, a = (1 − ǫb)−1 . (b) Suppose for the sake of contradiction that b is invertible. Then the open mapping theorem applies to b, so if U ⊂ H is the unit ball, then b(U ) contains the ball around 0 of radius ε for some ε > 0. By the definition of a compact operator, the closure V of b(U ) is a compact subset of H. But H is infinite dimensional, so there is an infinite orthonormal set v1 , v2 , . . . , and the sequence εv1 , εv2 , . . . is contained in V but has no limit point, a contradiction. Hence b cannot be invertible. 5. Let X ⊂ Pn be a projective variety. (a) Define the Hilbert function hX (m) and the Hilbert polynomial pX (m) of X. (b) What is the significance of the degree of pX ? Of the coefficient of its leading term? (c) For each m, give an example of a variety X ⊂ Pn such that hX (m) 6= pX (m). Solution. (a) The homogeneous coordinate ring S(X) is the graded ring S(Pn )/I, where S(Pn ) is the ring of polynomials in n+1 variables and I is the ideal generated by those homogeneous polynomials which vanish on X. Then hX (m) is the dimension of the mth graded piece of this ring. The Hilbert polynomial pX (m) is the unique polynomial such that pX (m) = hX (m) for all sufficiently large integers m. (b) The degree of pX is the dimension d of the variety X ⊂ Pn , and its leading term is deg X/d!. (c) Let X consist of any k distinct points of Pn . Then X is a variety of dimension 0 and degree k, so by the previous part pX (m) = k. But hX (m) is at most the dimension of the space of homogeneous degree m polynomials in n + 1 variables, so for sufficiently large k, hX (m) < k = pX (m).
6. Let X = S 2 ∨ RP2 be the wedge of the 2sphere and the real projective plane. (This is the space obtained from the disjoint union of the 2sphere and the real projective plane by the equivalence relation that identifies a given point in S 2 with a given point in RP2 , with the quotient topology.) (a) Find the homology groups Hn (X, Z) for all n. (b) Describe the universal covering space of X. (c) Find the fundamental group π1 (X). Solution. ˜ n (A ∨ B, Z) = H ˜ n (A, Z) ⊕ (a) The wedge A ∨ B of two spaces satisfies H ˜ Hn (B, Z) for all n, so H0 (X, Z) = Z,
H1 (X, Z) = Z/2Z,
H2 (X, Z) = Z.
˜ of X can be constructed as the union (b) The universal covering space X of the unit spheres centered at (−2, 0, 0), (0, 0, 0) and (2, 0, 0) in R3 ; the ˜ by sending x to −x, and the quotient is X. group Z/2Z acts freely on X ˜ Topologically, X is the wedge sum S 2 ∨ S 2 ∨ S 2 . ˜ by a free action (c) Since X is the quotient of the simply connected space X of the group Z/2Z, we have π1 (X) = Z/2Z.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday January 31 2008 (Day 3) 1. For z ∈ C \ Z, set f (z) = lim
N →∞
N X
n=−N
1 z+n
!
(a) Show that this limit exists, and that the function f defined in this way is meromorphic. (b) Show that f (z) = π cot πz. Solution. (a) We can rewrite f as N X 1 1 1 + f (z) = + lim z N →∞ z+n z−n n=1
!
∞
1 X 2z . = + z z 2 − n2 n=1
For any z ∈ C \ R, the terms of this sum are uniformly bounded near z by a convergent series. So this sum of analytic functions converges uniformly near z and thus f is analytic near z. We can apply a similar 1 to conclude that f has a simple pole at each argument to f (z) − z−n integer n (with residue 1). (b) The meromorphic function π cot πz also has a simple pole at each integer n with residue limz→n (z − n)(π cot πz) = 1, so f (z) − π cot πz is a global analytic function. Moreover ! N X 1 1 f (z + 1) − f (z) = lim − N →∞ z+1+n z+n n=−N 1 1 − = lim N →∞ z + 1 + N z−N = 0 for all z ∈ C \ Z, and cot π(z + 1) = cot πz, so f (z) − π cot πz is periodic with period 1. Its derivative is ∞ X d 1 1 1 f ′ (z) − π cot πz = − 2 + − − + π 2 sin2 πz. dz z (z + n)2 (z − n)2 n=1
This is again an analytic function with period 1, and it approaches 0 as the imaginary part of z goes to ∞, so it must be identically 0. So
f (z) − π cot πz is constant; since it is an odd function, that constant must be 0. 2. Let p be an odd prime. (a) What is the order of GL2 (Fp )? (b) Classify the finite groups of order p2 . (c) Classify the finite groups G of order p3 such that every element has order p. Solution. (a) To choose an invertible 2 × 2 matrix over Fp , we first choose its first column to be any nonzero vector in p2 − 1, then its second column to be any vector not a multiple of the first in p2 − p ways. So GL2 (Fp ) has (p2 − 1)(p2 − p) elements. (b) Let G be a group with p2 elements. As a pgroup, G must have nontrivial center Z. If Z = G, then G is abelian and so G = (Z/pZ)2 or G = Z/p2 Z. Otherwise Z has order p. So there is a short exact sequence 1 → Z → G → Z/pZ → 1. The sequence splits, because we can pick a generator for Z/pZ and choose a preimage for it in G; this preimage has order p (G cannot contain an element of order p2 or it would be cyclic) so it determines a splitting Z/pZ → G. Hence G is the direct product of Z and Z/pZ (because Z is central in G). So there are no new groups in this case. (c) Let G be a group with p3 elements in which every element has order p, and let Z be the center of G; again Z is nontrivial. If Z has order p3 , then G is abelian, and since every element has order p, G must be (Z/pZ)3 . If Z has order p2 , then Z must be isomorphic to (Z/pZ)2 , and there is a short exact sequence 1 → Z → G → Z/pZ → 1. Again, we can split this sequence by choosing a preimage of a generator of Z/pZ, so G is the direct product Z × Z/pZ. Hence Z is not really the center of G, and there are no groups in this case. Finally, suppose Z has order p; then there is a short exact sequence 1 → Z → G → (Z/pZ)2 → 1. Let a and b be elements of G whose images together generate (Z/pZ)2 . Then the image of c = bab−1 a−1 is 0 ∈ (Z/pZ)2 , so c lies in Z. If a and b commuted, we could split this sequence which would lead to a
contradiction as before. Hence c is a generator of Z. We can write every element of G uniquely in the form ai bj ck with 0 ≤ i, j, k < p, and we know the commutation relations between a, b and c; it’s easy to see that G is isomorphic to the group of uppertriangular 3 × 3 matrices over Fp with ones on the diagonal via the isomorphism 1 j k ai bj ck ↔ 0 1 i . 0 0 1 It remains to check that in this group every element really has order p. But one can check by induction that n 1 j k 1 nj nk + n(n−1) ij 2 0 1 i = 0 1 ni 0 0 1 0 0 1 and setting n = p, the right hand side is the identity because p is odd. 3. Let X and Y be compact, connected, oriented 3manifolds, with π1 (X) = (Z/3Z) ⊕ Z ⊕ Z
and
π1 (Y ) = (Z/6Z) ⊕ Z ⊕ Z ⊕ Z.
(a) Find Hn (X, Z) and Hn (Y, Z) for all n. (b) Find Hn (X × Y, Q) for all n. Solution. (a) (We omit the coefficient group Z from the notation in this part.) By the Hurewicz theorem, H1 (X) is the abelianization of π1 (X), so H1 (X) = (Z/3Z) ⊕ Z ⊕ Z. By Poincar´e duality, H 2 (X) = (Z/3Z) ⊕ Z ⊕ Z as well. Now by the universal coefficient theorem for cohomology, H 1 (X) is (noncanonically isomorphic to) the free part of H1 (X). So H 1 (X) = Z ⊕ Z, and by Poincar´e duality again H2 (X) = Z⊕Z too. Of course, H3 (X) = Z because X is a connected oriented 3manifold. So the homology groups of X are H0 (X) = Z,
H1 (X) = (Z/3Z) ⊕ Z2 ,
H2 (X) = Z2 ,
H3 (X) = Z.
H2 (Y ) = Z3 ,
H3 (Y ) = Z.
Entirely analogous arguments for Y yield H0 (Y ) = Z,
H1 (Y ) = (Z/6Z) ⊕ Z3 ,
(b) The module Q is flat over Z (TorZn (Q, −) = 0 for n > 0) so for any space A, Hn (A, Q) = Q ⊗ Hn (A, Z). In particular, H0 (X, Q) = Q,
H1 (X, Q) = Q2 ,
H2 (X, Q) = Q2 ,
H3 (X, Q) = Q,
H0 (Y, Q) = Q,
H1 (Y, Q) = Q3 ,
H2 (Y, Q) = Q3 ,
H3 (Y, Q) = Q.
The K¨ unneth theorem over a field k states that H∗ (A×B, k) = H∗ (A, k)⊗ H∗ (B, k) for any spaces A and B. So the homology groups Hn (X × Y, Q) for n = 0, . . . , 6 are Q,
Q5 ,
Q11 ,
Q14 ,
Q11 ,
Q5 ,
Q.
Note. Actually, there are no compact connected 3manifolds M with π1 (M ) = (Z/3Z)⊕Z⊕Z or π1 (M ) = (Z/6Z)⊕Z⊕Z⊕Z. The only abelian groups which are the fundamental groups of compact connected 3manifolds are Z/nZ, Z, Z ⊕ Z ⊕ Z, and (Z/2Z) ⊕ Z. 4. Let Cc∞ (R) be the space of differentiable functions on R with compact support, and let L1 (R) be the completion of Cc∞ (R) with respect to the L1 norm. Let f ∈ L1 (R). Prove that Z 1 lim f (y) − f (x)dy = 0 h→0 h y−x . lim sup k h→0 h y−x 0. By the given definition of L1 (R), there is a differentiable function g on R with compact support such that f − g1 ≤ ε/4k. Write f1 = f − g. I claim that Z Z 1 1 lim sup f (y) − f (x) dy = lim sup f1 (y) − f1 (x) dy, h→0 h y−x n, show that the induced map f∗ : Hk (CPm , Z) → Hk (CPn , Z) is zero for all k > 0. (b) If m = n, the induced map f∗ : H2m (CPm , Z) ∼ = Z → H2m (CPm , Z) ∼ =Z is multiplication by some integer d, called the degree of the map f . What integers d occur as degrees of continuous maps f : CPm → CPm ? Justify your answer. 2. Let a1 , a2 , . . ., an be complex numbers. Prove there exists a real x ∈ [0, 1] such that n X 2πikx 1 − a e ≥ 1. k k=1
3. Suppose that ∇ is a connection on a Riemannian manifold M . Define the torsion tensor τ via τ (X, Y ) = ∇X Y − ∇Y X − [X, Y ], where X, Y are vector fields on M . ∇ is called symmetric if the torsion tensor vanishes. Show that ∇ is symmetric if and only if the Christoffel symbols with respect to any coordinate frame are symmetric, i.e. Γkij = Γkji . Remember that if {Ei } is a coordinate frame, and ∇ is a connection, the Christoffel symbols are defined via X ∇Ei Ej = Γkij Ek . k
4. Recall that a commutative ring is called Artinian if every strictly descending chain of ideals is finite. Let A be a commutative Artinian ring. (a) Show that any quotient of A is Artinian. (b) Show that any prime ideal in A is maximal. (c) Show that A has only finitely many prime ideals.
5. Let X ⊂ Pn be a smooth hypersurface of degree d > 1, and let Λ ∼ = Pk ⊂ X a kdimensional linear subspace of Pn contained in X. Show that k ≤ (n − 1)/2. 6. Let l∞ (R) denote the space of bounded real sequences {xn }, n = 1, 2, . . . . Show that there exists a continuous linear functional L ∈ l∞ (R)∗ with the following properties: a) inf xn ≤ L({xn }) ≤ sup xn , b) If limn→∞ xn = a then L({xn }) = a, c) L({xn }) = L({xn+1 }). Hint: Consider subspace V ⊂ l∞ (R) generated by sequences {xn+1 − xn }. Show that {1, 1, . . . } 6∈ V¯ and apply HahnBanach.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday 25 January 2005 (Day 1)
1. (a) Show that, up to isomorphism, there is a unique group of order 15. (b) Show that, up to isomorphism, there are exactly two groups of order 10. 2. Let m and n be positive integers, and k another positive integer less than m and n. Let N = mn − 1, and realize the complex projective space PN as the space of nonzero m × n complex matrices modulo scalars. Let Xk ⊂ PN be the subset of matrices of rank k or less. Show that Xk is an irreducible closed algebraic subset of PN , and compute its dimension. 3. (a) Consider f (x) ∈ L1 (Rn ), Rn equipped with Lebesgue measure. Show that the function Φf : Rn → L1 (Rn ) given by Φf (y)(x) = f (x + y) is continuous. (b) Let f (x) ∈ L1 (R) and Z lim h→0
also absolutely continuous. Prove that f (x + h) − f (x) − f 0 (x) dx = 0. h
4. Let −1 < a < 1 and 0 < β < π. Compute Z ∞ xa dx 1 + 2x cos β + x2 0
and express the answer as a rational function of π, sin aβ, sin aπ, sin β with rational coefficients. 5. Let f : [a, b] → R+ be smooth, and let X ⊂ R3 be the surface of revolution formed by revolving z = f (x) about the xaxis. (a) For which functions f is the Gaussian curvature of X • always positive? • always negative? • identically zero?
(b) Characterise by a differential equation those functions f such that X is a minimal surface.
6. Illustrate how a Klein bottle, K, may be decomposed as the union of two M¨obius bands, joined along their common boundary. Using this decomposition and van Kampen’s theorem, obtain a presentation of π1 (K), and hence show that there is surjection from π1 (K) to the dihedral group of order 2n, for all n. Prove that, for all n ≥ 3, there exists a connected nsheeted covering space, ˜ n → K, that is not a normal covering. What is the topology of the covering K ˜ 5 in your example? space K
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday 26 January 2005 (Day 2) 1. (a) Show that every group of order pn , p prime, has a nontrivial center. (b) Let G be a group of order pn , let k be a (possibly infinite) field of characteristic p, and let M be a finitedimensional kvector space on which G acts. Show that if σ ∈ G satisfies σ p = 1, then σ fixes (pointwise) a nonzero subspace of M . (c) Under the same assumptions as in the previous part, show that there is a nonzero vector of M fixed by G. 2. Let H1 and H2 be two distinct hyperplanes in Pn . Show that any regular function on Pn − (H1 ∩ H2 ) is constant. P 3. (a) A (formal) sum ∞ n=1 an of complex numbers an ∈ C is Cesaro summable provided the limit m 1 X Si lim m→∞ m i=1 Pi exists, where Si := a . Given a sequence of complex numbers P∞ j=1 j2 P∞ {an }n≥1 , assume na  < ∞, and n n=1 n=1 an is Cesaro summable. P∞ Show that n=1 an converges. (b) Let f (θ) P ∈ C 0 (S 1 ) (a continuous function on the unit circle) with the 2 ˆ property ∞ n=−∞ f (n) n < ∞, where Z 2π 1 ˆ f (n) = f (θ)e−inθ dθ. 2π 0 Show that Sn f → f uniformly
as n → ∞ where Sn f are the partial Fourier sums of f , i.e. Sn f (θ) =
n X
fˆ(k)eikθ .
k=−n
4. Let f (z) be a holomorphic function on {z < 1} and continuous up to {z ≤ 1}. Let M be thesupremum of f  on {z ≤ 1}. Let L be the intersection of {z ≤ 1} and Re z = 21 . Let m be the supremum of f  on L. Show that f (0)3 ≤ m M 2 . (Hint: Consider the product of some functions obtained from f .)
5. Let C1 and C2 be smooth curves in R3 . Suppose that there exist unitspeed parametrizations ρ1 : R → R3 , ρ2 : R → R3 of C1 , C2 such that: • the curvatures κ1 , κ2 of ρ1 , ρ2 coincide and are never zero;
• the torsions τ1 , τ2 of ρ1 , ρ2 coincide.
Show that there exists an isometry I of R3 taking C1 to C2 . 6. Let S n denote the unit sphere in Rn+1 , and let f : S m → S n be a map satisfying f (−x) = −f (x) for all x. Assuming that m and n are both at least 1, show that the resulting map f¯ : RPm → RPn induces a nonzero map f¯∗ : π1 (RPm ) → π1 (RPn ). Use Z/2 cohomology to show that m ≤ n. In the case that n is even and m = n, show that f¯ must have a fixed point.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday 27 January 2005 (Day 3)
1. Let p be a prime, and let K be a field of characteristic not equal to p that contains the p th roots of unity. Show that every cyclic extension L of K of degree p can be obtained by adjoining a root of the polynomial xp −a for some a ∈ K. 2. Let X be the Veronese surface, i.e. the image of the 2uple embedding of P2 in P5 . If C ⊂ X is a closed irreducible curve, show that there exists a hypersurface H ⊂ P5 such that H ∩ X = C, where the intersection is considered settheoretically. 3. Show that kf ∗ gk2L2 (R) ≤ kf ∗ f kL2 (R) kg ∗ gkL2 (R) for all f, g ∈ L2 (R) (with the understanding that either side may be infinite). Can there be such an inequality with L1 (R) instead of L2 (R)? (Hint: use delta functions.) 4. By applying the Argument Principle to the domain which is the intersection of a quadrant and the disk centered at the origin of radius R with R → ∞, find out how many roots of the following equation lie in each of the four quadrants: z 4 + z 3 + 4z 2 + 2z + 3 = 0. (Hint: First observe that there are no zeroes on the nonnegative real axis and the imaginary axis. Then verify that there are no zeroes on the negative real axis by separately grouping certain terms together in the case Re x ≤ −1 and in the case Re x > −1. Then consider the change of arguments along the positive imaginary axis and a large quartercircle.) 5. Let ω be a closed nondegenerate 2form on a compact smooth manifold M , and let f : M → R be smooth. (a) Show that there is a unique vector field X on M such that ιX ω = df . (b) For each t ∈ R, let ρt : M → M be the timet flow of the vector field X. Show that ρ?t ω = ω and ρ?t f = f for all t ∈ R.
(c) Let M be the unit 2sphere in R3 and let ω be the standard volume form on M . Find a function f : M → R so that the corresponding map ρt is rotation about the zaxis by the angle t.
6. Give an example of a pair of CW complexes, (X, A), satisfying all of the following three conditions: (i) there exists an n such that X is obtained from A by adding a single ncell, X = A ∪ φ en ; (ii) the attaching map φ : S n−1 → A for this ncell induces the zero map Hn−1 (S n−1 ) → Hn−1 (A); and (iii) the space A is not a retract of X. Justify your answer.
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 21, 2004 (Day 1) Each of the six questions is worth 10 points. 1) Let H be a (real or complex) Hilbert space. We say that a set of vectors {φ n } ⊂ H, n = 1, 2, . . ., has “property D” provided H is the closure of the space of all finite linear combinations of the φn . Now let {φn }, n = 1, 2, . . ., be an orthonormal set having property D, and {ψn } a set of vectors satisfying ∞ X n=1
kφn − ψn k2 < 1 ,
where k k refers to the Hilbert space norm. Show that {ψn } also has property D. 2) Let K be the splitting field of the polynomial x4 − x2 − 1. Show that the Galois group of K over Q is isomorphic to the dihedral group D8 , and compute the lattice of subfields of K. 3) Let S be a smooth surface in R3 defined by r(u, v), where r is the radius vector of R3 and (u, v) are curvilinear coordinates on S. Let H and K be respectively the mean curvature and the Gaussian curvature of S. Let A and B be respectively the supremum of the absolute value of H and K on S. Let a be a positive number and n be the unit normal vector of S. Consider the surface S˜ defined by ρ~(u, v) = r(u, v) + a n(u, v). Let C be a curve in S defined by u = u(t) and v = v(t). Let C˜ be the curve in S˜ defined by t 7→ ρ~(u(t), v(t)). √Show that the length of C˜ is no less than the length of C multiplied by 1 − a A + A2 + 4B . (Hint: compare the first fundamental form of S˜ with the difference of the first fundamental form of S and 2a times the second fundamental form of S.) 4) Compute the integral Z
∞ 0
xa−1 dx 1 + x4
0 x − y a) Show that Hf (x) := lim→0 g (x) exists for each x ∈ R, and that Hf ∈ C ∞ (R). b) Exhibit a universal constant C such that
kHf kL2 = C kf kL2 . Show how to extend the operator H to an isomorphism from L 2 (R) to itself.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday September 23, 2004 (Day 3) Each of the six questions is worth 10 points. 1) a) Let G be a group of order n, acting on a finite set S. Show that the number of orbits of this action equals 1X #{x ∈ S  gx = x}. g∈G n
b) Let S be the set of integer points in the rectangle [0, 3] × [0, 2]. We consider two subsets of S equivalent if one can be transformed into the other by a series of reflections around the horizontal and vertical axes of symmetry of the rectangle. How many equivalence classes of fourelement subsets of S are there? 2) Let U ⊂ C be a connected open subset. Carefully define the topology of locally uniform convergence on O(U ), the space of holomorphic functions on U . Show that O(U ), equipped with this topology, is a Fr´echet space. 3) Consider the two dimensional torus T2 = S 1 × S 1 , where S 1 = R /2π Z. For any fixed α ∈ R, find all functions f ∈ L2 (T2 ) with the property f (x1 + α, x1 + x2 ) = f (x1 , x2 ) .
4) Let Γ be a set of seven points in CP3 , no four of them lying in a plane. What is the dimension of the subspace of homogeneous quadratic polynomials in C[X0 , X1 , X2 , X3 ] vanishing along any subset {p1 , . . . , pm } ⊂ Γ, m ≤ 7 ? 5) (Smooth Version of Michael Artin’s Generalization of the Implicit Function Theorem.) Let a and b be positive numbers. Let R be the ring of all Rvalued infinitely differentiable functions on the open interval (−a, a). For elements F , G, H in R we say that F is congruent to G modulo H in R if there exists some element Q of R such that F − G = QH as functions on (−a, a). Let f (x, y) be an Rvalued infinitely differentiable function on {x < a, y < b} with f (0, 0) = 0. Denote by 5
fy (x, y) the firstorder partial derivative of f (x, y) with respect to y. Let g(x) be an element of R such that g(0) = 0 and supx a and f (b) < b. Let x1 ∈ [a, b], and define a sequence via xn = f (xn−1 ). Show that limn→∞ xn exists. If we call this number x∗ , show that f (x∗ ) = x∗ . 2c. Describe all the irreducible complex representations of the group S 4 (the symmetric group on four letters). 3c. Suppose f is a biholomorphism between two closed annuli in C A(R) = {z ∈ C  1 ≤ z ≤ R}
and
A(S) = {z ∈ C  1 ≤ z ≤ S},
with R, S > 1. (i) Show that f can be extended to a biholomorphic map from C \ {0} to C \ {0}.
(ii) Prove that R = S.
4c. Use homotopy groups to show that there is no retraction r : RP n → RPk if n > k > 0. (Here RPn is real projective space of dimension n.) 5c. Let α : I → R3 be a regular curve with nonzero curvature everywhere. Show that if the torsion τ (t) = 0 for all t ∈ I, then α(t) is a plane curve (i.e., the image of α lies entirely in a plane). 6c. Let X be a kdimensional irreducible subvariety of Pn . In the Grassmannian G(1, n) of lines in Pn , let S(X) be the set of lines which are secant to X, i.e., which meet X in at least two distinct points. Consider also the union C(X) of all these secant lines, which is a subset of Pn . (i) Prove that if X is not a linear subspace of Pn , then the closure of S(X) is an irreducible subvariety of G(1, n) of dimension 2k. (ii) Prove that the closure of C(X) is an irreducible subvariety of P n of dimension at most 2k + 1. 5
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday 30 September 2003 (Day 1)
There are six problems. Each question is worth 10 points, and parts of questions are of equal weight unless otherwise specified. ~ 1a. Let S be an embedded closed surface in R3 with the position vector X(p) and ~ the unit outward normal vector N (p) for p ∈ S. For a fixed (small) t, define a surface St to be the set o n ~ ~ (p) ∈ R3  p ∈ S . + tN St = X(p) Let κ1 , κ2 be the principal curvatures of S at the point p with respect to the outward normal vector. Let Ht be the mean curvature of St at the point ~ ~ (p) with respect to the outward normal vector (mean curvature is X(p) + tN defined to be the sum of the two principal curvatures). Show that Ht =
κ2 κ1 + . 1 − tκ1 1 − tκ2
¯ = {z ∈ C : 2a. Let D = {z ∈ C : z < 1} be the open unit disk in C and D ¯ →D ¯ is analytic, onetoone z ≤ 1} be the closed unit disk. Suppose f : D ¯ . Also suppose g : D ¯ → D ¯ is analytic in D and in D and continuous in D ¯ continuous in D, with g(0) = f (0) and g(D) ⊂ f (D). Prove g 0 (0) ≤ f 0 (0). 3a. Use the RiemannHurwitz (or any other) method to compute the genus of the Fermat curve, which is given in CP2 with homogeneous coordinates (x : y : z) by the equation xn + y n = z n (assume that the base field is C). 4a. Let k be a finite field with q elements and let Γ = GL(2, k) denote the group of invertible 2 × 2 matrices over k. (i) How many elements are there in Γ? (ii) How many complex irreducible representations does Γ have? (iii) Consider the representation of Γ on the space of complexvalued functions on P1 over k (induced by the natural action of Γ on P1 ). Let V be the quotient of this space by the subspace of constant functions. Prove that V is an irreducible representation of Γ. 1
5a. Let V be a Hilbert space, and W a vector subspace of V . Show that V = W ⊕ W ⊥. 6a. What is π1 (RP3 )? Show that any continuous map f : RP3 → S 1 is nullhomotopic.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday 1 October 2003 (Day 2)
There are six problems. Each question is worth 10 points, and parts of questions are of equal weight unless otherwise specified. 1b. Let (H2 , g) be the twodimensional hyperbolic space, where H2 = {(x, y) ∈ R2 : y > 0} is the upper half plane of R2 = C and the metric g is given by g=
dx2 + dy 2 . y2
(i) Suppose a, b, c and d are real numbers such that ad − bc = 1. Define ϕ(z) = for any z = x +
az + b cz + d
√ −1y. Prove that ϕ is an isometry for (H2 , g).
(ii) Prove that (H2 , g) has constant Gaussian curvature. 2b. Prove the open mapping theorem for analytic functions of one complex variable: “if U is a connected open subset of C and f : U → C is holomorphic and nonconstant, then f (U ) is open.” You may assume that a holomorphic function that is constant on an open subset of U is constant on U . 3b. Prove that if k is a field of characteristic p and f (x) ∈ k[x] is a polynomial, then the map from the curve y p + y = f (x) to A1k sending (x, y) to x is everywhere unramified. 4b.
(i) Let k be an algebraically closed field. Assume that k is uncountable. Let now V be a vector space over k of at most countable dimension and A : V → V be a linear operator. Prove that there exists λ ∈ k such that the operator A − λ idV is not invertible. (Hint: show first that in the field 1 k(t) of rational functions over k the elements t−λ are linearly independent (for different values of λ) and then use this fact.) (ii) Show that (i) is not necessarily true if k is countable. 3
(iii) Use (i) to show that for k uncountable every maximal ideal in the ring k[x1 , . . . , xn ] is generated by (x−λ1 , . . . , x−λn ) for some (λ1 , . . . , λn ) ∈ k n . 5b. Give an example or show that none exist. (i) A function f : R → R whose set of discontinuities is precisely the set Q of rational numbers. (ii) A function f : R → R whose set of discontinuities is precisely the set R \ Q of irrational numbers. 6b. Let X be the manifoldwithboundary D 2 × S 2 . Calculate H2 (X; Z), H 2 (X; Z) and H 2 (X, ∂X; Z), using any techniques you choose. Calculate the map j ∗ : H 2 (X, ∂X; Z) → H 2 (X; Z) that arises from the inclusion j : (X, ∅) → (X, ∂X).
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday 2 October 2003 (Day 3)
There are six problems. Each question is worth 10 points, and parts of questions are of equal weight unless otherwise specified. 1c. Let Σ be an embedded, compact surface without boundary in R 3 . Show that there exists at least one point p in Σ which has strictly positive Gaussian curvature. P n 2c. Determine for which x ∈ Qp the exponential power x /n! converges. P n series Do the same for the logarithmic power series x /n. 3c. Let V be a variety over an algebraically closed field k, and suppose V is also a group, i.e., there are morphisms ϕ : V × V → V (multiplication or addition), and ψ : V → V (inverse) that satisfy the group axioms. Then V is called an algebraic group.
(i) Suppose that V is a nonsingular plane cubic. Describe a way to put a group structure on V . You do not have to prove that the maps you define are morphisms, but you do have to prove that they satisfy the axioms of a group. (ii) Let V be defined by y 2 z = x3 in P2 . Prove that V − {(0, 0, 1)} can be equipped with the structure of algebraic group. (iii) Let V be defined by x3 + y 3 = xyz in P2 . Prove that V − {(0, 0, 1)} can be equipped with the structure of algebraic group. 4c. Compute the following integral: Z ∞ 0
5c.
log x dx. (1 + x2 )2
(i) Let a, b be nonnegative numbers, and p, q such that 1 < p < ∞ and 1/p + 1/q = 1. Establish Young’s inequality: ab ≤
5
a p bq + . p q
(ii) Using Young’s inequality, prove the H¨older inequality: If f ∈ L p [0, 1] and g ∈ Lq [0, 1], where p and q are as above, then f g ∈ L1 [0, 1], and Z f g ≤ kf kp · kgkq . (iii) For 1 < p < ∞, and g ∈ Lq , consider the linear functional F on Lp given by Z F (f ) = f g. Show that kF k = kgkq . (Recall that kF k = sup{F (f )/kf k : f ∈ Lp }.)
(iv) Establish similar results for p = 1 and p = ∞. 6c.
(i) Prove that every continuous map f : CP6 → CP6 has a fixed point.
(ii) Exhibit a continuous map f : CP3 → CP3 without a fixed point. (Hint: Try the case of CP1 first and write your answer in terms of homogeneous coordinates.)
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday 4 February 2003 (Day 1)
There are six problems. Each question is worth 10 points, and parts of questions are of equal weight. 1a. Let k be any field. Show that the ring k[x] has infinitely many maximal ideals. 2a. Let f be an entire function. Suppose that f vanishes to even order at every zero of f . Prove there exists a holomorphic function g such that g 2 = f . 3a. Let k be a field. Let a, b be relatively prime positive integers. Is there an element in the field of fractions of the kalgebra A = k[X, Y ]/(Y a − X b ) that generates the integral closure of A (i.e., generates it as kalgebra)? If so, find such an element; if not, prove not. 4a. Let (f (v) cos(u), f (v) sin(u), g(v)) be a parametrization of a surface of revolution S ⊂ R3 where (u, v) ∈ (0, 2π) × (a, b). If S is given the induced metric from R3 , prove that the following map from S to R2 is locally conformal where R2 is given the standard Euclidean metric: ! Z p 0 (f (v))2 + (g 0 (v))2 dv . (u, v) → u, f (v) 5a. Let X be the space obtained by identifying the three edges of a triangle using the same orientation on each edge, as shown below.
Compute π1 (X), H∗ (X), and H∗ (X × X). 6a. New Real Analysis Problem 1.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday 5 February 2003 (Day 2)
There are six problems. Each question is worth 10 points, and parts of questions are of equal weight. 1b. Let X ⊂ C2 be the curve defined by x2 (y 2 − 1) = 1, and let X ⊂ P2 be its closure. (i) Find the singularities of X and classify them into nodes, cusps, and so on. (ii) Find the genus of the smooth completion of X. 2b. Let 0 < s < 1. Evaluate the integral Z ∞ 0
3b.
xs−1 dx. 1+x
(i) Consider Rn with the standard Euclidean metric and let p ∈ Rn be an arbitrary point. For any x ∈ Rn let ρp (x) be the distance from p to x. Viewing ρp (x) as a smooth function of x away from p, verify that  grad(ρp (x))2 = 1 and that the integral curves of grad(ρp (x)) are straight lines. (Here grad(ρp (x)) refers to the usual gradient vector field of the function ρp (x).) (ii) More generally, given a smooth function f on a Riemannian manifold (M, gij ), define grad(f ) to be the vector field given locally by X df ∂ g ij . dxi ∂xj i,j Show that if  grad(f )2 = 1 then the integral curves of the vector field grad(f ) are geodesics.
4b. A mechanical linkage is a collection of points (some fixed, some not) in the plane connected by rigid struts, each with a fixed length. Its configuration space is the set of all solutions to the constraints that the struts have a fixed
2
length, with the topology induced from the product of the plane with itself. For instance, this mechanical linkage x1 x2
(0, 0) x0
denotes a fixed vertex) can be described by the equations
(in which
{x0 , x1 , x2 ∈ R2 x0 = (0, 0), x0 − x1  = 1, x2 − x1  = 1} The configuration space of this linkage is the torus S 1 × S 1 .
Identify topologically the configuration space of the linkages (0, 2)
(0, 0)
(3, 0)
(2, 2)
and (0, 0)
(2, 0)
All edges have length 1, and the fixed vertices are at the indicated locations. Hint: Consider the position of the central point, and compute the Euler characteristic. 5b. Let Hd be the space of degree d curves in P2 , where d > 1. We identify Hd with the projectivization of the vector space of degree d homogeneous polynomials in three variables, so Hd = PN for some N . (i) Find N , the dimension of Hd . (ii) For a fixed point p ∈ P2 find the dimension of the set Σp ⊂ Hd of curves that have a singularity at p. (iii) Find the dimension of the set Σ ⊂ Hd of singular curves.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday 6 February 2003 (Day 3) There are six problems. Each question is worth 10 points, and parts of questions are of equal weight. 1c. Let Z ⊂ Pn be a variety of degree d. Choose a point P ∈ / Z and let P Z be the union of lines containing the points P and Q, where the union is taken over all points Q ∈ Z. Prove that the degree of P Z is at most d. (Hint: Intersect with a suitable hyperplane and use induction on dimension.) 2c. Let p, q, and r be nonconstant nonvanishing entire holomorphic functions that satisfy the equation p + q + r = 0. Prove there exists an entire function h such that p, q and r are constant multiples of h. 3c. Let M be a smooth manifold with a connection ∇ on the tangent bundle. Recall the following definitions of the torsion tensor T and curvature tensor R: For arbitrary vector fields X, Y and Z on M we have and
T (X, Y ) := ∇Y X − ∇X Y − [X, Y ] R(X, Y )Z := ∇Y ∇X Z − ∇X ∇Y Z − ∇[Y,X] Z.
Assuming we have a torsionfree connection (T = 0), verify the following identity: R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = 0. (Hint: Begin by assuming that X, Y, Z are coordinate vector fields, then justify that there is no loss in generality in doing this.) 4c.
(i) What is the symmetry group G of the following pattern? What is the topological space R2 modulo G?
4
(ii) What is the commutator subgroup of G? Draw generators for the commutator subgroup on a copy of the pattern (see Page 6). 5c. Let ρ be a twodimensional (complex) representation of a finite group G such that ρ(g) has 1 as an eigenvalue for every g ∈ G. Prove that ρ is the sum of two onedimensional representations. 6c. Let k be a field. Let f, g be polynomials in k[x, y] with no common factor. Show that the quotient ring k[x, y]/(f, g) is a finite dimensional vector space over k.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday October 1, 2002 (Day 1)
There are six problems. Each question is worth 10 points, and parts of questions are of equal weight. 1a. Exhibit a polynomial of degree three with rational coefficients whose Galois group over the field of rational numbers is cyclic of order three. 2a. The Catenoid C is the surface of revolution in R3 of the curve x = cosh(z) about the z axis. The Helicoid H is the surface in R3 generated by straight lines parallel to the xy plane that meet both the z axis and the helix t 7−→ [cos(t), sin(t), t]. (Recall that sinh(x) =
ex + e−x ex − e−x and cosh(x) = .) 2 2
(i) Show that both C and H are manifolds by exhibiting natural coordinates on each. (ii) In the coordinates above, write the local expressions for the metrics g C and gH , induced by R3 , on C and H, respectively. (iii) Is there a covering map from H to C that is a local isometry? 3a. In Rn , consider the Laplace equation u11 + u22 + · · · + unn = 0. Show that the equation is invariant under orthonormal transformations. Find all rotationally symmetric solutions to this equation. (Here u ii denotes the second derivative in the ith coordinate of a function u.) 4a. Let C denote the unit circle in C. Evaluate I e1/z C 1 − 2z 5a. Let G(1, 3) be the Grassmannian variety of lines in CP 3 .
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(i) Show that the subset I ⊂ G(1, 3)2 I = {(l1 , l2 )  l1 ∩ l2 6= ∅} is irreducible in the Zariski topology. (Hint: Consider the space of triples (l1 , l2 , p) ∈ G(1, 3)2 × CP 3 such that p ∈ l1 ∩ l2 , and consider two appropriate projections.) (ii) Show that the subset J ⊂ G(1, 3)3 J = {(l1 , l2 , l3 )  l1 ∩ l2 6= ∅, l2 ∩ l3 6= ∅, l3 ∩ l1 6= ∅} is reducible. How many irreducible components does it have? 6a. For the purposes of this problem, a manifold is a CW complex which is locally homeomorphic to Rn . (In particular, it has no boundary.) (i) Show that a connected simplyconnected compact 2manifold is homotopy equivalent to S 2 . (Do not use the classification of surfaces.) (ii) Let M be a connected simplyconnected compact orientable 3manifold. Compute π3 (M ). (iii) Show that a connected simplyconnected compact orientable 3manifold is homotopy equivalent to S 3 . (iv) Find a simplyconnected compact 4manifold that is not homotopy equivalent to S 4 .
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday October 2, 2002 (Day 2)
There are six problems. Each question is worth 10 points, and parts of questions are of equal weight. 1b. Let C[S4 ] be the complex group ring of the symmetric group S4 . For n ≥ 1 let Mn (C) be the algebra of all n × n matrices with complex entries. Prove that the algebra C[S4 ] is isomorphic to a direct sum M Mni (C) i=1,...,t
and calculate the ni ’s. 2b.
(i) Show that the 2 dimensional sphere S 2 is an analytic manifold by exhibiting an atlas for which the change of coordinate functions are analytic functions. Write the local expression of the standard metric on S 2 in the above coordinates. (ii) Put a metric on R2 such that the corresponding curvature is equal to 1. Is this metric complete?
3b. Let C ∈ CP 2 be a smooth projective curve of degree d ≥ 2. Let CP 2∗ be the dual space of lines in CP 2 and C ∗ ⊂ CP 2∗ the dual curve of lines tangent to C. Find the degree of C ∗ . (Hint: Project from a point.) 4b. Let f : R → R be any function. Prove that the set of points x ∈ R where f is continuous is a countable intersection of open sets. 5b. Prove that the only meromorphic functions f (z) on C ∪ {∞} are rational functions. 6b.
(i) Show that the fundamental group of a Lie group is abelian. (ii) Find π1 (SL2 (R)).
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday October 3, 2002 (Day 3)
There are six problems. Each question is worth 10 points, and parts of questions are of equal weight. 1c. Let H = {(u, v) ∈ R2  v > 0} and B = {(x, y) ∈ R2  x2 + y 2 < 1}.
For e2 = (0, 1) ∈ R2 , map H to B by the following diffeomorphism. v 7−→ x = −e2 +
2(v + e2 ) . kv + e2 k2
(i) Verify that the image of the above map is indeed B. (Hint: Think of the standard inversion in the circle.) (ii) Consider the following metric on B: g=
dx2 + dy 2 . (1 − kxk2 )2
Put a metric on H such that the above map is an isometry. (iii) Show that H is complete. 2c. Let C ⊂ CP 2 be a smooth projective curve of degree 4. (i) Find the genus of C and give the RiemannRoch formula for the dimension of the space of sections of a line bundle M of degree d on the curve C. (ii) If l ∈ CP 2 is a line meeting C at four distinct points p1 , . . . , p4 , prove that there exists a nonzero holomorphic differential form on C vanishing at the four points pi . (Hint: Note that OCP 2 (1) restricted to C is a line bundle of degree 4. Use the RiemannRoch formula to prove that this restriction is the canonical line bundle KC .) 3c. Let A be the ring of realvalued continuous functions on the unit interval [0, 1]. Construct (with proof) an ideal in A which is not finitely generated.
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4c. Construct a holomorphic function f (z) on C satisfying the following two conditions: (i) For every algebraic number z, the image f (z) is algebraic. (ii) f (z) is not a polynomial. (Hint: The algebraic numbers are countable.) 5c. Let q < p be two prime numbers and N (q, p) the number of distinct isomorphy types of groups of order pq. What can you say, more concretely, about the number N (q, p)? 6c. Let i : S 1 ,→ S 3 be a smooth embedding of S 1 in S 3 . Let X denote the complement of the image of i. Compute the homology groups H ∗ (X; Z).
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday October 2, 2001 (Day 1)
Each question is worth 10 points, and parts of questions are of equal weight. 1a. Let X be a measure space with measure µ. Let f ∈ L1 (X, µ). Prove that for each > 0 there exists δ > 0 such that if A is a measurable set with µ(A) < δ, then Z A
f dµ < .
2a. Let P be a point of an algebraic curve C of genus g. Prove that any divisor D with deg D = 0 is equivalent to a divisor of the form E − gP , where E > 0. 3a. Let f be a function that is analytic on the annulus 1 ≤ z ≤ 2 and assume that f (z) is constant on each circle of the boundary of the annulus. Show that f can be meromorphically continued to C − {0}. 4a. Prove that the rings C[x, y]/(x2 − y m ), m = 1, 2, 3, 4, are all nonisomorphic. 5a. Show that the ellipsoid x2 +2y 2 +3z 2 = 1 is not isometric to any sphere x2 + y 2 + z 2 = r. 6a. For each of the properties P1 through P4 listed below either show the existence of a CW complex X with those properties or else show that there doesn’t exist such a CW complex. P1. The fundamental group of X is isomorphic to SL(2, Z). P2. The cohomology ring H ∗ (X, Z) is isomorphic to the graded ring freely generated by one element in degree 2. P3. The CW complex X is “finite” (i.e., is built out of a finite number of cells) and the cohomology ring of its universal covering space is not finitely generated. P4. The cohomology ring H ∗ (X, Z) is generated by its elements of degree 1 and has nontrivial elements of degree 100.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday October 3, 2001 (Day 2)
Each question is worth 10 points, and parts of questions are of equal weight. 1b. Prove that a general surface of degree 4 in P3C contains no lines. 2b. Let R be a ring. We say that Fermat’s last theorem is false in R if there exists x, y, z ∈ R and n ∈ Z with n ≥ 3 such that xn + y n = z n and xyz 6= 0. For which prime numbers p is Fermat’s last theorem false in the residue class ring Z/pZ? 3b. Compute the integral
Z∞
cos(x) dx. 1 + x2
0
4b. Let R = Z[x]/(f ), where f = x4 − x3 + x2 − 2x + 4. Let I = 3R be the principal ideal of R generated by 3. Find all prime ideals ℘ of R that contain I. (Give generators for each ℘.) 5b. Let S4 be the symmetric group on four letters. Give the character table of S4 , and explain how you computed it. 6b. Let X ⊂ R2 and let f : X → R2 be distance nonincreasing. Show that f extends to a distance nonincreasing map fˆ : R2 → R2 such that fˆX = f . Does your construction of fˆ necessarily use the Axiom of Choice? (Hint: Imagine that X consists of 3 points. How would you extend f to X ∪ {p} for any 4th point p?)
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday October 4, 2001 (Day 3) Each question is worth 10 points, and parts of questions are of equal weight. 1c. Let S ⊂ P3C be the surface defined by the equation XY − ZW = 0. Find two skew lines on S. Prove that S is nonsingular, birationally equivalent to P2C , but not isomorphic to P2C . 2c. Let f ∈ C[z] be a degree n polynomial and for any positive real number R, let M (R) = maxz=R f (z). Show that if R2 > R1 > 0, then M (R2 ) M (R1 ) ≤ , n R2 R1n with equality being possible only if f (z) = Cz n , for some constant C. 3c. Describe, as a direct sum of cyclic groups, the cokernel of ϕ : Z 3 → Z3 given by left multiplication by the matrix 3 5 21 3 10 14 . −24 −65 −126
4c. Let X and Y be compact orientable 2manifolds of genus g and h, respectively, and let f : X → Y be any continuous map. Assuming that the degree of f is nonzero (that is, the induced map f ∗ : H 2 (Y, Z) → H 2 (X, Z) is nonzero), show that g ≥ h. 5c. Use the Rouch´e’s theorem to show that the equation zeλ−z = 1, where λ is a given real number greater than 1, has exactly one root in the disk z < 1. Show that this root is real.
6c. Let f : R → R be a bounded function such that for all x and y 6= 0, f (x + y) + f (x − y) − 2f (x) ≤ B, y
for some finite constant B. Prove that for all x 6= y, 1 + , f (x) − f (y) ≤ M · x − y · 1 + log x − y
where M depends on B and kf k∞ , and log+ (x) = max(0, log x). 3
QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday, February 25, 1997 (Day 1)
1. Factor the polynomial x3 − x + 1 and find the Galois group of its splitting field if the ground field is: a) R,
b) Q,
c) Z/2Z .
2. Let A be the n × n (real or complex) matrix
0 0 . .. 0 1/n
1 0 .. .
0 1 .. .
0 1/n
0 1/n
0 0 .. . . ... 1 . . . 1/n ... ... .. .
Prove that as k → ∞, Ak tends to a projection operator P onto a onedimensional subspace. Find ker P and Image P .
3. a. Show that there are infinitely many primes p congruent to 3 mod 4. b. Show that there are infinitely many primes p congruent to 1 mod 4. 4. a. Let L1 , L2 and L3 ⊂ P3C be three pairwise skew lines. Describe the locus of lines L ⊂ P3C meeting all three. b. Now let L1 , L2 , L3 and L4 ⊂ P3C be four pairwise skew lines. Show that if there are three or more lines L ⊂ P3C meeting all four, then there are infinitely many. 5. a. State the Poincar´e duality and Kunneth theorems for homology with coefficients in Z (partial credit for coefficients in Q). b. Find an example of a compact 4manifold M whose first and third Betti numbers are not equal, that is, such that H 1 (M, Q) and H 3 (M, Q) do not have the same dimension.
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6. Compute
Z 0
∞
log x dx x2 + b2
for b a positive real number.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday, February 26, 1997 (Day 2)
1. Define a metric on the unit disc {(x, y) ∈ R2 : x2 + y 2 < 1} by the line element ds
2
dr2 + r2 dθ2 = . (1 − r2 )p
Here (r, θ) are polar coordinates and p is any real number. a. For which p is the circle r = 1/2 a geodesic? b. Compute the Gaussian curvature of this metric. 2. Let C be the space C[0, 1] of continuous realvalued functions on the closed interval [0, 1], with the sup norm f ∞ = max f (t) . t∈[0,1]
Let C 1 be the space C 1 [0, 1] of C 1 functions on [0, 1] with the norm f  = f ∞ + f 0 ∞ . Prove that the natural inclusion C 1 ⊂ C is a compact operator.
3. Let M be a compact Riemann surface, and let f and g be two meromorphic functions on M . Show that there exists a polynomial P ∈ C[X, Y ] such that P (f (z), g(z)) ≡ 0. 4. Let S 3 = {(z, w) ∈ C2 : z2 + w2 = 1}. Let p be a prime and m an integer relatively prime to p. Let ζ be a primitive pth root of unity, and let the group G of pth roots of unity act on S 3 by letting ζ ∈ G send (z, w) to (ζz, ζ m w). Let M = S 3 /G. a. compute πi (M ) for i = 1, 2 and 3. b. compute Hi (M, Z) for i = 1, 2 and 3. c. compute H i (M, Z) for i = 1, 2 and 3.
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√ 5. Let d be a squarefree integer. Compute the integral closure of Z in Q( d). Give an example where this ring is not a principal ideal domain, and give an example of a nonprincipal ideal.
6. Prove that
π2 = sin2 πz
∞ X
1 . (z − n)2 n=−∞
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday, October 17, 1996 (Day 3)
1. Let α : (0, 1) → R3 be any regular arc (that is, α is differentiable and α0 is nowhere zero). Let t(u), n(u) and b(u) be the unit tangent, normal and binormal vectors to α at α(u). Consider the normal tube of radius ² around α, that is, the surface given parametrically by φ(u, v) = α(u) + ² cos(v)n(u) + ² sin(v)b(u) . a. For what values of ² is this an immersion? b. Assuming that α itself has finite length, find the surface area of the normal tube of radius ² around α. The answers to both questions should be expressed in terms of the curvature κ(u) and torsion τ (u) of α.
2. Recall that a fundamental solution of a linear partial differential operator P on Rn is a distribution E on Rn such that P E = δ in the distribution sense, where δ is the unit Dirac measure at the origin. Find a fundamental solution E of the Laplacian on R3 3 X ∂2 ∆ = ∂x2i i=1
that is a function of r = x alone. Prove that your fundamental solution indeed satisfies ∆E = δ. Hint: Use the appropriate form of Green’s theorem.
3. The group of rotations of the cube in R3 is the symmetric group S4 on four letters. Consider the action of this group on the set of 8 vertices of the cube, and the corresponding permutation representation of S4 on C8 . Describe the decomposition of this representation into irreducible representations.
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4. Suppose ai , i = i, . . . , n are positive real numbers with a1 + . . . + an = 1. Prove that for any nonnegative real numbers λ1 , . . . , λn , n X
ai λ2i ≥
Ã n X
i=1
!2 ai λi
i=1
with equality holding only if λ1 = . . . = λn . 5. a. For which natural numbers n is it the case that every continuous map from PnC to itself has a fixed point? b. For which n is it the case that every continuous map from PnR to itself has a fixed point? 6. Fermat proved that the number 237 − 1 = 137438953471 was composite by finding a small prime factor p. Suppose you know that 200 < p < 300. What is p?
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday, March 12 (Day 1)
1. Let X be a compact ndimensional differentiable manifold, and Y ⊂ X a closed submanifold of dimension m. Show that the Euler characteristic χ(X \ Y ) of the complement of Y in X is given by χ(X \ Y ) = χ(X) + (−1)n−m−1 χ(Y ). Does the same result hold if we do not assume that X is compact, but only that the Euler characteristics of X and Y are finite?
2. Prove that the infinite sum X 1 1 1 1 = + + + ... p 2 3 5
p prime
diverges. 3. Let h(x) be a C ∞ function on the real line subset of R2 containing the xaxis such that
R.
Find a C ∞ function u(x, y) on an open
∂u ∂u +2 = u2 ∂x ∂y and u(x, 0) = h(x).
4. a) Let K be a field, and let L = K(α) be a finite Galois extension of K. Assume that the Galois group of L over K is cyclic, generated by an automorphism sending α to α + 1. Prove that K has characteristic p > 0 and that αp − α ∈ K. b) Conversely, prove that if K is of characteristic p, then every Galois extension L/K of degree p arises in this way. (Hint: show that there exists β ∈ L with trace 1, and construct α out of the various conjugates of β.)
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5. For small positive α, compute Z 0
∞
xα dx . x2 + x + 1
For what values of α ∈ R does the integral actually converge? 6. Let M ∈ Mn (C) be a complex n × n matrix such that M is similar to its complex conjugate M ; i.e., there exists g ∈ GLn (C) such that M = gM g −1 . Prove that M is similar to a real matrix M0 ∈ Mn (R).
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday, March 13 (Day 2)
1. Prove the Brouwer fixed point theorem: that any continuous map from the closed ndisc Dn ⊂ Rn to itself has a fixed point. 2. Find a harmonic function f on the right halfplane {z ∈ C  Re z > 0} satisfying ½ lim f (x + iy) =
x→0+
1 −1
if y > 0 . if y < 0
3. Let n be any integer. Show that any odd prime p dividing n2 + 1 is congruent to 1 (mod 4).
4. Let V be a vector space of dimension n over a finite field with q elements. a) Find the number of onedimensional subspaces of V . b) For any k : 1 ≤ k ≤ n − 1, find the number of kdimensional subspaces of V . 5. Let K be a field of characteristic 0. Let PN be the projective space of homogeneous polynomials F (X, Y, Z) of degree d modulo scalars (N = d(d + 3)/2). Let W ⊂ PN be the subset of polynomials F of the form F (X, Y, Z) =
d Y
Li (X, Y, Z)
i=1
for some collection of linear forms L1 , . . . , Ld . a. Show that W is a closed subvariety of PN . b. What is the dimension of W ? c. Find the degree of W in case d = 2 and in case d = 3.
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6. a. Suppose that M → Rn+1 is an embedding of an ndimensional Riemannian manifold (i.e., M is a hypersurface). Define the second fundamental form of M . b. Show that if M ⊂ Rn+1 is a compact hypersurface, its second fundamental form is positive definite (or negative definite, depending on your choice of normal vector) at at least one point of M .
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday, March 14 (Day 3)
1. In
R3 , let S, L and M
be the circle and lines S = {(x, y, z) : x2 + y 2 = 1; z = 0} L = {(x, y, z) : x = y = 0} 1 M = {(x, y, z) : x = ; y = 0} 2
respectively.
R3 \ (S ∪ L). Compute the homology groups of the complement R3 \ (S ∪ L ∪ M ).
a. Compute the homology groups of the complement b.
2. Let L, M, N ⊂ P3C be any three pairwise disjoint lines in complex projective threespace. Show that there is a unique quadric surface Q ⊂ P3C containing all three. 3. Let G be a compact Lie group, and let ρ : G → GL(V ) be a representation of G on a finitedimensional Rvector space V . a) Define the dual representation ρ∗ : G → GL(V ∗ ) of V . b) Show that the two representations V and V ∗ of G are isomorphic. c) Consider the action of SO(n) on the unit sphere S n−1 ⊂ Rn , and the corresponding representation of SO(n) on the vector space V of C ∞ Rvalued functions on S n−1 . Show that each nonzero irreducible SO(n)subrepresentation W ⊂ V of V has a nonzero vector fixed by SO(n − 1), where we view SO(n − 1) as the subgroup of SO(n) fixing the vector (0, . . . , 0, 1). 4. Show that if K is a finite extension field of Q, and A is the integral closure of Z in K, then A is a free Zmodule of rank [K : Q] (the degree of the field extension). (Hint: sandwich A between two free Zmodules of the same rank.)
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5. Let n be a nonnegative integer. Show that µ ¶ 1 X l = −1 (−1)l k 0≤k≤l 0 k+l=n
if n ≡ 0 (mod 3) if n ≡ 1 (mod 3) . if n ≡ 2 (mod 3)
(Hint: Use a generating function.)
6. Suppose K is integrable on
Rn and for ² > 0 define x K² (x) = ²−n K( ). ²
R Suppose that Rn K R R = 1. a. Show that Rn K² = 1 and that x>δ K²  → 0 as ² → 0. b. Suppose f ∈ Lp (Rn ) and for ² > 0 let f² ∈ Lp (Rn ) be the convolution Z f² (x) =
y∈Rn
f (y)K² (x − y)dy .
Show that for 1 ≤ p < ∞ we have kf² − f kp → 0 as ² → 0. c. Conclude that for 1 ≤ p < ∞ the space of smooth compactly supported functions on Rn is dense in Lp (Rn ).
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Extra problems: Let me know if you think these should replace any of the ones above, either for balance or just by preference. 1. Suppose that M → RN is an embedding of an ndimensional manifold into N dimensional Euclidean space. Endow M with the induced Riemannian metric. Let γ : (−1, 1) → M be a curve in M and γ : (−1, 1) → RN be given by composition with the embedding. Assume that k dγ dt k ≡ 1. Prove that γ is a geodesic iff d2 γ dt2 is normal to M at γ(t) for all t. 2. Let A be a commutative Noetherian ring. Prove the following statements and explain their geometric meaning (even if you do not prove all the statements below, you may use any statement in proving a subsequent one): a) A has only finitely many minimal prime ideals {pk  k = 1, . . . , n}, and every prime ideal of T A contains one of the pk . n b) k=1 pk is the set of nilpotent elements of A. Sn c) If A is reduced (i.e., its only nilpotent element is 0), then k=1 pk is the set of zerodivisors of A. 4. Let A be the n × n matrix
0 0 . .. 0 1/n
1 0 .. .
0 1 .. .
0 1/n
0 1/n
0 0 .. . . ... 1 . . . 1/n ... ... .. .
Prove that as k → ∞, Ak tends to a projection operator P onto a onedimensional subspace. Find the kernel and image of P .
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Tuesday, October 24, 1995 (Day 1)
1. Let K be a field of characteristic 0. a. Find three nonconstant polynomials x(t), y(t), z(t) ∈ K[t] such that x2 + y 2 = z 2 b. Now let n be any integer, n ≥ 3. Show that there do not exist three nonconstant polynomials x(t), y(t), z(t) ∈ K[t] such that xn + y n = z n .
2. For any integers k and n with 1 ≤ k ≤ n, let S n = {(x1 , . . . , xn+1 ) : x21 + . . . + x2n+1 = 1} ⊂ be the nsphere, and let Dk ⊂
n+1
n+1
be the closed disc n+1
Dk = {(x1 , . . . , xn+1 ) : x21 + . . . + x2k ≤ 1; xk+1 = . . . = xn+1 = 0} ⊂
.
Let Xk,n = S n ∪ Dk be their union. Calculate the cohomology ring H ∗ (Xk,n , ).
3. Let f :
2
→
be any C ∞ map such that ∂2f ∂2f + ≡ 0. ∂x2 ∂y 2
Show that if f is not surjective then it is constant.
4. Let G be a finite group, and let σ, τ ∈ G be two elements selected at random from G (with the uniform distribution). In terms of the order of G and the number of conjugacy classes of G, what is the probability that σ and τ commute? What is the probability if G is the symmetric group S5 on 5 letters? 1
5. Let Ω ⊂
be the region given by Ω = {z : z − 1 < 1
and
z − i < 1}.
Find a conformal map f : Ω → ∆ of Ω onto the unit disc ∆ = {z : z < 1}.
6. Find the degree and the Galois group of the splitting fields over polynomials: a. x6 − 2 b. x6 + 3
2
of the following
QUALIFYING EXAMINATION Harvard University Department of Mathematics Wednesday, October 25, 1995 (Day 2)
√ 1. Find the ring A of integers in the real quadratic number field K = ( 5). What is the structure of the group of units in A? For which prime numbers p ∈ is the ideal pA ⊂ A prime?
2. Let U ⊂ 2 be an open set. a. Define a Riemannian metric on U . b. In terms of your definition, define the distance between two points p, q ∈ U . c. Let ∆ = {(x, y) : x2 + y 2 < 1} be the open unit disc in 2 , and consider the metric on ∆ given by dx2 + dy 2 . ds2 = (1 − x2 − y 2 )2 Show that ∆ is complete with respect to this metric. 3. Let K be a field of characteristic 0. Let N be the projective space of homogeneous polynomials F (X, Y, Z) of degree d modulo scalars (N = d(d + 3)/2). Let U be the subset of N of polynomials F whose zero loci are smooth plane curves C ⊂ 2 of degree d, and let V ⊂ N be the complement of U in N . a. Show that V is a closed subvariety of N . b. Show that V ⊂ N is a hypersurface. c. Find the degree of V in case d = 2. d. Find the degree of V for general d. 4. Let n be real projective nspace. a. Calculate the cohomology ring H ∗ ( n , /2 ). b. Show that for m > n there does not exist an antipodal map f : S m → S n , that is, a continuous map carrying antipodal points to antipodal points.
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5. Let V be any continuous nonnegative function on , and let H : L2 ( ) → L2 ( ) be defined by −1 d2 f + V · f. H(f ) = 2 dx2 a. Show that the eigenvalues of H are all nonnegative. b. Suppose now that V (x) = 12 x2 and f is an eigenfunction for H. Show that the Fourier transform Z ∞ e−ixy f (x)dx fˆ(y) = −∞
is also an eigenfunction for H.
6. Find the Laurent expansion of the function f (z) =
1 z(z + 1)
valid in the annulus 1 < z − 1 < 2.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics Thursday, October 26, 1995 (Day 3)
1. Evaluate the integral
Z
∞ 0
sin x dx. x
2. Let p be an odd prime, and let V be a vector space of dimension n over the field with p elements. a. Give the definition of a nondegenerate quadratic form Q : V → p b. Show that for any such form Q there is an ² ∈ p and a linear isomorphism
p
n p
φ : V −→
v 7−→ (x1 , . . . , xn )
such that Q is given by the formula Q(x1 , x2 , . . . , xn ) = x21 + x22 + . . . + x2n−1 + ²x2n c. In what sense is ² determined by Q?
3. Let G be a finite group. Define the group ring R = [G] of G. What is the center of R? How does this relate to the number of irreducible representations of G? Explain. 4. Let φ : n → n be any isometry, that is, a map such that the euclidean distance between any two points x, y ∈ n is equal to the distance between their images φ(x), φ(y). Show that φ is affine linear, that is, there exists a vector b ∈ n and an orthogonal matrix A ∈ O(n) such that for all x ∈ n , φ(x) = Ax + b.
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5. Let G be a finite group, H ⊂ G a proper subgroup. Show that the union of the conjugates of H in G is not all of G, that is, G 6=
[
gHg −1 .
g∈G
Give a counterexample to this assertion with G a compact Lie group. 6. Show that the sphere S 2n is not the underlying topological space of any Lie group.
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QUALIFYING EXAMINATION Harvard University Department of Mathematics . October 8, 1991 (Day 1) 1. Think of 5 1 x 5 1 as R2 /Z 2 •
a) What is
11' 1 (5
1
x 5 1 )?
b) Define X to be the space obtained by taking
[O, 1] x (5 1 x S 1 ) and identifying (O, (Yi, Y2 )) with ( l, (y2, Y1 )).
a) Compute Hi(X ). b) Compute r.i(X). c) Is X homeomorphic to RIP 2 x 5 1 ? Justify your answer. 2. Fermat proved that if pis prime, then aP =a (mod p) for all a. Generalize this to prove the following:
.
..
Let f n( a)= Lµ(d)an/d with µ(d) defined as zero if dis not square fr.ee and µ(d)
=1
din
'o·.··; '~~
or 1 as d has an even or odd number of prime divisors otherwise. Thus, f5(a) = a6  a 3  a 2 +a. Prove that for all positive integers n and all integers a, .fn(a) =: 0 (mod n).
3. Let e : ( 0, 1)   R 2 be an arc parametrized by arc length, i.e., such that the derivative e' (t) has norm 1 for all t. \Ve call t•( t)
= e' (t)
the unit tangent vector to the arc e at
time t, and define both the unit vector n( t), called the unit normal vector, and the scalar 1'(t)
~
0, called the cur'Vature. by the equation i·'(t) = n(t) · n(t).
a) Assume 1'(t) > 0 and show that n (t) is orthogonal to v( t) for all t, and that the derivative of n is given by n'(t)
= K(t) · v(t). 1
;
Day 1, 10/8/91 b. Again asswning 1e(t)
> 0,
we define the o.tcula.ting circle to the arc e at time t
to be the circle of radius l/K.(t) with center e(t) + n(t)/K.(t)  that is, a circle of radius l/K.(t) tangent to the arc at e(t). Show that if the curva.ture function K(t) i.s monotone, then the osculating circle& to the a.re a.re ne.sted.
4. Let
f, g : C 
C be analytic on a simply connected region containing a simple closed
curve "Y· Suppose that y misses all the zeros off and g, and that on "Y If(=) g(=)I
Show that
f
< lf(z)I.
and g have the same number of zeros, counting multiplicity, inside y, i.e.,
prove Rouche 's theorem. 5. Let ( X, d) be a compact metric spa.ce and suppose
f
is a map X  X such that for
every x ::/= y, d(f(;z:),f(y))
Show that there exists a unique x 0 EX satisfying f(xo)
6. a) Let V
(
< d(x,y).
= xo.
= Hom( cm, en) be the vector space of m x n matrices, and let A c V be any
linear subspace such that for all .4 E A, the rank of A is at most 1. Show that either i) there exists a hyperplane U ~ cmI C cm such that U C ker( A) for all A E A; or ii) there exists a line W 9! C C C" such that Im( A) b) Lets: PC: 1 x
Pc
1
+
cW
for all A E A.
PC:" 1 be the Segre map given by
Show that the image of s is an algebraic subvariety X of PenI, and describe all linear subspaces of PC'" 1 lying on .Y.
2
QUALIFYING EXAMINATION Harvard University Department of Mathematics October 9, 1991 (Day 2) 1. a) State the MeyerVietoris, Excision and Kunneth properties of the singular homology
theory. b) Consider the trefoil knot
I\ in R3 U oo = S3
Compute H.(S 3

!.
E
c•
acts on C 2
\
{O}
as the transformation
Let
1r :
C2
\
{O}
+
CP 1 be the projection map.
a) Write down a nontrivial vector field inker
Tr ••
b) Show that
is the pull back by r. of a_real valued 2form on CP 1 •
1
., ' '7
Day 2 (10/9/91) 4. Let
q EC obey 0 < lql < 1.
For,\ EC, consider
a) Exhibit a Laurent expansion for c,:?(..\) which converges for jql < j..\j < 1. b) Exhibit one which converges for 1 < j..\j < jqj 1 •. ~
c) Show that c,:?( ,\) defines a meromorphic function on C \ {0} with poles at {qn} nEZ. d) Prove ~,
..J'
1+
x + a, for a E F, and let N be the normalizer of P in S,
i.e., the largest subgroup of Sin which Pis normal. (i) Show that N is the group of permutations of the form x
i+
cz + a, for a E F,
c E F, c ;l: 0.
(ii) Show how Sylow's theorems for the group Simply Wilson's theorem: (p 1)!
=
l(mod p).
(iii) Suppose /(X) E O[X] is an irreducible polynomial of degree
p which has two
roots o:,{J E C such that the field K = O(a,{J) is a Galois extension of(). Show that the Galois group G = Gal ( K / 0) is isomorphic to a subgroup of N containing P, where N and
P are as above.
3
I\
QUALIFYING EXAMINATION
Harvard University Department of Mathematics December 6, 1988 (Day 1)
1. (a)
For which values of the real constants (a,b,c,d) does the function f(:r:, y)
obey /(:r:,y)
= a:r: 2 + b:r:y + cy2 + d
> 0 for all (:r:,y) E R 2 ?
(b) Prove that an n x n symmetric matrix has real eigenvalues. (c)
Derive necessary and sufficient conditions on an n x n symmetric matrix for its eigenvalues to be positive.
2. Determine the Galois group over k of :r: 5  3 = 0 when
=Q 21ri k = Q(e 7 ) k = the finite field of 11 elements.
(a) k (b)
(c)
3. Compute: 00
1 0
(cos a:r: ...... cos b:r:) d 2 :r;. x
1
4. Let M be an ndimensional manifold, with nonzero Euler characteristic.
Let X be a vector field on M, and suppose Wi, • • • ,wn are closed 1forms on M which are invariant under the flow generated by X. Show that w1 A···/\ Wn
5. (a)
=0.
Write down the homotopy exact sequence for a fi.bration.
(b) Let Gl(n, C) denote the group of n x n, invertible matrices with complex entries. Show that the inclusion of Gl(n, C) in Gl(n + 1, C) (it sends g to (
g ~)) induces
an isomorphism 7r1i:(Gl(n, C))
.:+
j['11:(Gl(n + 1,C)) for n large, and k fixed.
(c) Find, as a function of k, a number N(k) so that (b) is true for n > N(k).
6. Let V1 , V2 and V3
c C" be pairwise complementary linear subspaces of dimension 2.
Show that there is a nondegenerate symmetric bilinear form Q : c" x
c" + c
such that the subspaces Vi are isotropic for Q, i.e., Q(Vi, Vi) is unique up to multiplication by a scalar.
2
= 0 for each i.
Show that Q
QUALIFYING EXAMINATION Harvard University Department of Mathematics December 7, 1988 (Day 2)
1. Wha.t is the order of the group G of invertible linear transformations of an ndimensional
vector space Vn over the field Z/pZ? Fix a complete flag 0 C V1 C V2 C • • • C Vn1 C V of linear subspaces (with dim
v, =
i) and let B be the subgroup of G stabilizing this flag. What is the order of B?
Describe a pSylow subgroup P of G which is contained in B.
2. Recall that a fundamental solution of a linear partial differential operator on Rn, P, is a distribution E on Rn such that PE
= 6 in the distribution sense, where o is the unit
Dirac measure at the origin. Find a fundamental solution of the Laplacian A= on R 3 , which is a function of r =
lzl only.
L:a,a, i
Prove that your fundamental solution, P, obeys
AP=6.
3. When n is an integer, let n° a)
= ~8~·
Explain why sin (n°) and cos (n°), where n is an ordinary integer, are algebraic numbers.
b) Of the following two statements, one is true, the other fa.lse. State which is which and prove your answer. sin 36° E Q(cos 36°) cos 36° E Q(sin 36°).
1
4. The Zariski closure of a subset
locii of all polynomials
r c en is defined to be the intersection of the zero
I (Zi, ... 'Zn) vanishing on
r.
Find the Zariski closure of the
image of the map
c c 3
tp:
given by
and prove your answer is correct.
5. Show that there is no convex polyhedron in R 3 with 5 sides, each having the same number of edges. You will have demonstrated that no "fair" 5sided die exists.
6. Let E
,...,,_ ~;;:
c.......+
R 3 be an immersed surface.
a} Define the mean curvature, H, of E. b} Define the Gauss curvature, K. c} H H = O, show that K ~ 0. d} Suppose H = 0 (E is called a minimal surface). Consider a closed, geodesic polygon (it bounds a cell}, and show that there at least 2 vertices.
Msedeslr. "":!men~ ,verte~
2 ·\
\ l ....
QUALIFYING EXAMlNATION
Harvard University Department of Mathematics December 8, 1988 (Day 3)
1. Let A be a commutative integral domain with identity. Recall that an Amodule M is
said to be torsion free if for any f EA and m EM, fm
= 0 implies that f = 0 or m = 0.
a) Let A be a principal ideal domain and M, N two Amodules. If M and N are torsion free, show that M ®A N is torsion free. b) Give an example of an integral domain A and two Amodules M, N such that M and N are torsion free but M ®AN is not torsion free.
2. a) Compute the cohomology ring of RP" with Z/2Z coefficients. b) Show that there is no continuous map f : R & + R 3 which obeys
/(z) = /(z) and which has
3. Take 8 2
1 1 (0)
0 such that for any rational number ~' Ii  QI~ e~>.
Hint: Consider
f(i) where f is the polynomial over
/(Q) = 0.
2
Q of smallest degree such that
QUALIFYING EXAMINATION Harvard University Department of Mathematics April 26, 1988 (Day 1)
.'
Let f : 8 3
1.
+
8 2 be a smooth map. Let w be a 2form on 8 2 •
(a) Prove that 3 a E 0 1 (8 3 ) such that da = (b) Show that
11(a,w)::
1 sa
rw. QA
rw
is independent of the choice of a E 0 1 obeying da
= r w.
.,._
(c)· Show that 11 is unchanged if w+ w +du for any c E 0 1 (8 2).
.,
(d) Show that 11 depends only on the homotopy class of the map f. )
,..
~
2.
Evaluate
1
lzl=20000
T H
3.
zlOOldz
z 1002 + 708z 97 + 59z + 10 ·
Let X be the curve in P 2 defined by the homogeneous equation f = zy 2 +x3 8z 3 • Let
Ii, /2, fa be the restrictions of the meromorphic functions x / y, y / z, z / x to
x. a) Find the following divisors on X.
(Ii)
b) Three points on X with coordinates (x, y, z) are given by Pl = (0, VS, 1)
P2 = (0, v'B, 1)
p3=(0,1,0).
Is the divisor 2[p 1] linearly equivalent to (P2]
+ [Pa]'l
1 \' '\ i
Let 8 3
c\ 4.
Let
= {(z1, z2} E C 2 : lz11 2 + jz:ll2 =1}.
z,, act on 8 3 by (r, (z1,z2}}
. {rzi,rmz2} where mis relatively prime to
p and r is a primitive p'th root of unity. Let p(m) denote this action.
Let M = 8 3 /p(m}. (a} compute w1{M} for i = 1,2,3 (b) compute H,(M; Z) for i
"L
= 1,2,3
"\
(c} compute H 1(M; Z} for i = 1,2,3 "\ (d) How do {b), {c} change when Z is replaced by Q?
("'\
Let {~
5.
= {z11,z12,···)}:1 and~
J..
= {z00 i,z 00 2,·· ·} be sequences with
nonnegative entries such that 00
00
I:xi; = LZoo; < oo. i=l
If
{zi;}:, 1
converges to
z00;
i=l
for ea.ch j, show that
{:L;: 1 lxi;  Xoo;l}:, 1
converges to zero. Put another way: If a sequence of probability measures on {1, 2, · · ·} converges pointwise, then they converge in total variation.
\'\4
6.
a) Define a Noetherian ring. b)
Prove Hilbert's theorem: Let A be a Noetherian ring. Then A[x] is Noethetian.
2
•
QUALIFYING EXAMINATION Harvard University Department of Ma.thematics April 27, 1988 (Day 2)
1.
Let
f, g be two meromorphic functions on a compact, complex Riemann
surface. Show that there exists a polynomial F(X,Y) such that F(f,g)
2.
= 0.
Calculate the Galois group of the splitting field of the polynomial /(:c) = :z:4 2 over Q, R and Z/SZ.
3.
Let
n· = {z E Cj O Y 2 :> Y3
I
5.
~ •••
is a
sequence of closed sets then 3r s.t. Yr= Yr+l = Yr+2 = · · ·. a) Show that A~ is a noetherial topological space in the Zariski topology. (You I
n I
may assume the Hilbert basis theorem.)
?::
~
b)
Let X be the closed subset of
Ab defined by the zero set of the equation
x 3 + x 2 y + x 2 yz + x 2 y 2 z in C[x, y, z]. Show that X
= X 1 u X 2 • • • u Xr where Xi is closed, Xi~ X; for i =ft j
and Xi
is irreducible (that is, if Xi= Z1 UZ2 where Z1 and Z2 are closed then Z1 =Xi or Z2 =Xi)
3.
Let X be a compact 3dimensional manifold and p E X a given point. Let D
cX
be an open neighborhood of p with a diffeomorphism
tp:
D+ {(xi,x2,xa) E R 3
:
lx1l 2 + lx21 2 + lxal 2 < 1}.
Is there a vector field v, on X which is nonvanishing on
X\lntD.
1
.·
,>
}e, /} '
4. Let D = {(:i:,y) E R 2 : l:i:l 2 + IYl 2 < l}.
(a) Solve for a continuous function, u, on D which obeys
8 2u 8:i:2
8 2u
+ 8712
= 16%2.
{b) Find a function u on D which solves (•) and which vanishes on the boundary ~ of D. (c) Is there a solution to ( *) on D whose normal derivative vanishes on the bound ) ary of D?
rDp ,
5.
Justify your answer.
Let G be a connected Lie group.
 ,...
a) Let G+ G be the universal
cov~
of G.
Show that there is a group law on 't
G such that 1r is a homomorphism. b) Show that ker1r lies in the center of G. ~ c) Show that 1r 1{G) is abelian.
·~
T
6.
a) Find the integral closure, A, of Zin the quadratic field Q(w) where w 3 = 1 and w
:I= 1.
b) Describe the behavior of rational primes p in the ring A.
2
i
..
I :•
,; •
.
QUALIFYING EXAMINATION Ha.rvaia University Department of Mathematics December 8, 1987 (Da.y 1)
1. (a) Define the notion of "tra.nsversality" of two manifolds N 1 and N2 in a third one, M. (b) Let f: M+ M be a smooth ma.p. A fixed point p off is called transversal if the graph of f meets the diagonal in M
x M transversally. Give a criterion for the transversality of p in terms of the differential of f at p. (c) Define the degree of the map at such a fixed point and explain how to compute it.
2. Compute
fo
00
;~,;;
ck
for
b > 0.
3. Let p be a prime. (a.)
Show tha.t the ring k = Z/pZ is a. field.
{b) Let k be an algebraic closure of k. Show that for each integer n
> 1 there
is a unique subfield k,,, of k with (k,,, : k) = n.
(c)
4.
Fix
Show that k,,, is a Galois extension of k and describe its Galois group.
eas a. primitive
nth
root of unity and let T : C(:z:, y)
+
C(:z:, y)
be defined by
Tp(:z:,y)
= p(ex, e 1y).
(a) Give polynomials P1, ••• , Pm tha.t generate the ring of invariants R 1
= {p I Tp = p}.
,
...
•
,
(b) Compute the kernel of the map
\
sending %i to Pi· (c) Let X be the variety with R as coordinate ring. Describe the singular points of this variety and compute its dimension. (d) For n = 2 blow up X at each singularity to construct a nonsingular variety.

5. (a) Describe the cohomology of the n sphere, S". Indicate how you would
compute it. (b) Describe the cohomology and homotopy groups of S" x S"' in terms of the corresponding groups of S". (c) Let t:t: S"' x S"... S" x S" be the involution which exchanges the factors. Compute the Lefschetz number of a and deform a to a map with the mjnimum number of fixed points.
6.
Let C'°((O, l]) be the Banach space of continuous functions on (0, 1) with norm
II/  911 = Let K : (o, 1) x (0, 1)
+
R obey
sup (IKl(z,y)
(s,sr)
Define H:
sup l/(:r:)  g(:r:)I·
se(o.1)
a
+la Kl(:r:,y)) < 10. :r;
c 0 ([0, l]) + C 0 ([0, 1)) by
fo dy K(:r:,y)/(y). 1
(H /){z) =
(a) Prove that His a bounded operator. (b) Let {/i}:!1
c c 0 ([0,l))
be a bounded sequence. Pro'Ve that {H/in'! 1 has a
convergent subsequence. 2
.
~/
~/
QUALIFYING EXAM'.I.NATION Harvard University Department of Mathematics December 9, 1987 (D~y 2)
1.
Let R be a commutative ring which is a unique factorization domain. Show that
R(:z:] is a unique factorization domain.
2. Let :z:1, ••• ,:Z:m be distinct points of S 2' and set
x = S 2 \ {:z:i, ••• ,:z:m}·
.. (a) Compute "'i(X) . (Give generators with relations.) (b) How many 2fold coverings of X are there? (c) Say that a covering space of X is ez:tendable if it is the restriction of a covering space of X U { :z:&} for some i. How many covering spaces of degree 2 of X are not extendable?
I~ j
3.
fa.) If :z:, y, z
are the homogeneous coordinates in CP 2 give criteria on the complex
number c:r for which the surface S[c:r) given by: y
2
z = (:z:  z)(:z: + z)(:z:  c:rz)
is nonsingular.
(b) What is the genus of S[et] when et= 2? when c:r = 3? Justify your answer. (c) Let Uz be the subset where z :f; 0 and let X = :z:/z, Y = y/z be coordinates there.
Is the differential form dX
Jex  i)(x + i)(x 
2)
holomorphic on S(2] fl U.? Does it extend to a holomorphic form on all of S(2)? Justify your answers. 1
4. Let A{t), t E R, be an n x n matrix valued, continuous function oft. Suppose that the entries of A(t) are bounded for all t. Consider the ODE for a vector valued function oft:
:(t) = A(t) • :(t) :(0) = %o e R". (a) Solve ( •) for A a. constant matrix.
(b) Generate a series expansion for :(t) when A varies with t. (c) Prove that this expansion converges uniformly for t bounded.
5. (a.).
Let R: C+ C be a complex: polynomial of degree n
> 2 with distinct zeroes
,1, ... , '"" Show that
L"
i=l
(b).
A fixed point,
equation P(z)
= z.
zo,
1 R'(,,·) =0. .
of a complex: polynomial P : C _. C is a solution to th~
It is called a.n attractor if IP'(zo)I < 1, a repdlor if IP'(zo)I > 1 a..•.:.
an indifferent point if IP'{Zo)I
= 1.
Show that if P is a complex: polynomial, then P must either have an indifferent or a repelling fixed point.
6. Define a metric on {(:z:,y) E R 2
J
:
f:z:I + 1Yf2 < 1} by the line element
2 d8 d.s2 _dr _2 _ _2 +r_  (1 r2)P •
Here (r, 8) are polar coordinates a.nd p ER. (a) Compute the connection form relative to the orthonormal frame ,. e
= (1 
dr r2)p/2 '
I
e
{b) Compute the 2nd fundamental form of the circle r (c) For which pis t~ circle a geodesic? (d) Compute the Gaussian curvature of this metric. 2
r d8
= (1 
r2)p/2 •
= 1/2.
QUALIFYING EXAMINATION Harvard University Department of Mathematics December 10, 1987 (Day 3)
1. Let
I :R 2

R be a. smooth, harmoni~ function, i.e.,
Suppose f is not surjective. Prove that f is constant.
.
2. Let A be the ring of continuous functions f: R+ R such that /(z+ 1) = /(z) for all z E R. Let M be the Amodule consisting of the continuous functions u : R  R such that u(z + 1)
= u(z) for all :z: ER, the action of A
on M being simply multiplication
of functions. Let c and a denote the elements of M defined by c(:z:) s(:z:)
= cos
w:i:
and
= sin wz. (a) Show u EM==> u(:z:) = 0 for some z ER. (b) Show that M cannot be generated by one element, i.e., M '::/:: Au for all u E M.
(c) Show M is not isomorphic to A as an Amodule. However, Mis generated by c ands, i.e., M =Ac+ As, because u = (uc)c + (us)s for all u EM, and uv EA for u,v EM.
(d) Show M E9 M
~A E9
(e) Show M®AM
~A
3. Let T 2
A as an Amodule. (Here, E9 denotes "direct sum".)
as an Amodule.
= {(z,w) e C 2 : lzl = lwl = 1}. Define a map, I: T~+ T 2 by f(z,w)
~
1
(zw" ,w).
:/
..
/
r• .L. (a) Prove that

'
c
f is a diffeomorphism.
(b) Choose a basis for H 1 (T2 ; Z) and compute
the matrix which represents ·
(c) Define the mapping torus of/ to be the following 3dimensiona.1 manifold:
Take (0, 1) x T 2 and identify (0, (z,w))  (1, (zw" ,w)). This defines a compact manifold without boundary, X. Compute H 1 (X; Z).
4. Let E
'+
CP" x cpm be a: closed subvariety.
Prove that E is a projective variety.
5. Let D = Let
{z e C : fzJ < l}.
I :D
+
D be a conformal map, onto. (That is,
M = O,
~ :/: 0
and f is
onto.)
(a) (b)
=
Show that lim lf(z)I 1. l•t1 If /(0) 0 and /'(O) is real and positive, prove that /(z)
=
(Hint: consider g(z)
= z.
= 11fl : D + C).
6.
(a) Prove that every nonsingular real matrix .A can be written in the form:
A=O·S where 0 is orthogonal and S is symmetric.
(b) What is the infinitesimal analogue of this decomposition? (Think of the tangent space to the identity in GL(n,R)). (c) What is the rank and signature of the quadratic form on the space cf n x n matrices which sends
•
QUALIFYING EXAMINATION Hana.rd University Department of Mathematics April 28, 1987 (Day 1)
1. Suppose Mis a smooth oriented surface in R 3 , and z EM.
a) Define the second fundamental form Ss: TsM x TsM+ R, the Gaussian curvature K z and the mean curvature H
=·
b) An asymptotic direction at z EM is a line in the tangent apace TsM spanned by a vector v such that Ss(v,v)
= 0.
Show that if Ks> 0 there are no asymptotic directions,
while if Ks< 0 there are exactly two. c) Compute the asymptotic directions for the hyperboloid z 2 . . (. !
)"
mt 1, 2 ,1.
..
.
.
.
+ 4y 2 
z2
= 1 at the
2. Let L be a linear differential operator acting on functions on the real line. What is a Green function (fun~amental solution) of L? Compute the Green function for
L
= £s + 1.
3. Consider the ring R =C(xJ,z]. a) Describe all maximal ideals of R. b) Describe all minimal prime ideals of R. c) Give an example of a prime ideal in R which is neither minimal nor maximal. 1
4. Let F be a field of characteristic rf: 2. PGL(2,F) form Q
Give an isomorphism between the group
= GL(2,F)/F• and the special orthogonal group SO(Q) of the quadratic
=:e2 
y2 
z2.
5. Let C be the space C[O, 1] with the sup norm with the norm
11/11
00 •
Let C 1 be the space C 1 [0, 1]
II/II = 11/lloo + 11/'lloo· Prove that the natural embedding C 1 c.. c is a
compact operator.
6...
.
a) Defiiie ·the real projective space P 3 of dimension 3. Compute its homology and homotopy groups up to dimension 3 and descn"be explicitly the Hurewitz map in this range. b) Is it possible to find a retraction of P 3 on the projective plane P 2 your assertion.
2
c P 3?
Prove
;
QUALIFYING EXAMINATION Harvard University Department of Mathematics April 29, 1987 Day 2
1. Let II/II
= (i; J; .. l/(ei1)1 2d8) 112 be a norm on the space V = C[z) of polynomials.
a) Show that the completion B of V with respect to this norm is a Hilbert space. Find an orthonormal basis of B. b) Compute the norm of the functional tp : V
+
C, 'P(/)
= /(j), with respect to
II II· c) For any r (0 < r S 1) define the norm
II llr on V by
~
J'
Let V,. denote the space V equipped with the norm morphism V1
+
II llr·
V,. is a compact operator for every r
Prove that the identity
< 1.
2. Compute
l
oo
COS:&
oo 1 +:&+:&
,.1_
2 Chi;.
3. Let F be a field and P(:i:) E F[z) a monic irreducible polynomial of degree 17. Let E be a normal extension of F. Prove that either the polynomial P(z) stays irreducible in the algebra E[z], or it splits completely, i.e., there are .\i in E such that P(:i:)
= Il(z .\i)·
1
\z
..
4. Let
I:
C2
xP3
..... pT
be a holomorphic map (here pA: is a kdimensional complex
projective space). Suppose that /(0 x P 1) is a constant map. Prove that there exists a map A: C 2
_. pT
such that /(z,JI) 5 h.(z).
5. Let G be the group of all isometries of the Euclidean plane R 2 • Introduce a coordinate system on G, and in these coordinates describe the invariant measure on G.
6. Find a formula for the general solution u( z, 11) of the differential equation :c Uz + y u~ = A (here l is a parameter). For which l does this equation have nonzero solutions which are continuous on the entire
(%, 11)plane?"
2 J...
QUALIFYING EXAMINATION Harvard University Department of Mathematics April 30, 1987 Day 3
1. Let 6 be a complex root of the equation
z 8 3z+3 = 0.
a) Prove that the additive group of the ring Z[I] is finitely generated. Is it a free Zmodule? b) Prove that the additive group of the ring Z[s 1) is not finitely generated. Is it a free Zmodule? c) Wha.t if the equation had been
2. Let F 9 be a finite field with qelements, q odd. a) Define the finite projective plane P 2 (F 9 ), and count the number of points on it. b) Count the number of points on the conic za + %~
+ %~
= O, where Zo,
% 1t %2
a.re
homogeneous coordinates on P 2 (F 9). c) Count the number of points on the conic zoz1+z~=0.
3. Denote by D the open unit disc in C and by D its closure in C. Suppose that f is a holomorphic function, defined in some neighborhood of D, and suppose that it maps D strictly inside D. Bow ma.ny fixed points can / have on D (a fixed point is a point
z such that /(z)
= z)?. 1
4. Let
f be a holomorphic function, defined on D*
= D{O}, where DCC is the unit
disc.
a) Suppose that
JD· l/(z)l 2dzd71 < oo, where z =z + iy.
Prove that
I extends to a
holomorphic function on D. b) Suppose that
JD• l/'(z) lcizdy < oo. Prove that I extends to a holomorphic function
onD.
5. Let G be the group of all isometries of the Euclidean plane R 2 • a) Let H CG be a. finite subgroup. Prove that all elements of H hav~ a common fixed point :
e R 2.
b) List all finite groups H, which can occur as subgroups of G.
6. Let X be a path connected topological space, covered by two open subsets U and V. Suppose that U n V is path connected. Prove from first principles that for any point
zo in Un V, the group "'1(X,:o) is generated by the images of the groups 11"1(U, zo) and "'1 (V, zo).
2. ..,
!>. ,... ...·.1
"'.:',..
t/j(::l'..,
..
...

_, Qualifying Examination Harvard University Department of Mathematics November 18, 1986
Day 1
1 a.
Let X •
a a v ii  x iY
. 2 be a vector field on R •
2 1) At what points in R is X tangent to the curve
2
2
2x  4xy + 3y • 0 • 3 2) What is the Lie derivative of (3x dx v2dy) in the direction of the vector field Xl 3) Cive the general formula of the Lie derivative of an n·form
bv
a vector field on a smooth manifold. 4) Explain the relation of 3) to horTOtopy invariance of De Rham theory.
I ~ ~·
a•.
a) Define the Euler •function •Cn).
,
i~

2•i
Let n be an
integer~ 1
b) Pruve that (K(z):K)
and let
~ •(n)
for every subfield K of C.
c) Prove that (Clz):Q1J • t(n). (Partial credit if you C!an do this for n prime.)
Ja.
Compute the Fourier transform of the function on It
which sends to
1
(1+t 1 ) 1 •
Please turn page
I
/\/..,. V. I
2 4a.
a)
Define the projective space
.
,t' ,f' /:. /
C Pn , of
dim n over C
and explain why it is a
differentiable orientable manifold. b) State the Poincare duality theorem for such manifolds.
c) Explain how you would prove that . .2q1 q • 1,. •• ,n, and tr (C P ) •
n
~q(C Pn)
• Z for
o.
d) What. does P0 teach you concemina the ring structure of I
"+ n1
e) Prove that the inclusion C P
induced by the usual inclusion of
admits no retraction, i.e., there is no map C P
n1
continuous on (0,1 ] •
.'
Prove: Clven t c (0,1 ), there exists x c (O,t) such that k•1
f(t) 
~
j0
4b.
Let C be a finite aroup and o a representation of C by nxn c°"1)lex matrixes. a) Show ttiat the trace f111ction X, aiven by:· (X)I • Trace 0 determines
o up to conjuaacy.
b) CQf1'1JUte the sum
where
X, and x2 are the traces of two irreducible representations o1 and o 2 of
dimension
'i
and
"2 •
c) How many distinct (non Isomorphic) representations does a finite aroup havel
Please turn paae
3
r
Sb.
(a).
Show that a compact orientable surface with 4 handles cannot be a cowrina space of a coq>act orientable surface with 3 handles.
(b).
Let X be a con.,act si"l>ly connected 3 manifold with boundary. Show that the boundary
6b.
aX
is a union of l•spheres.
Consider the ODE
2
2
x w"(x)  (2+x )w •
for a f111ction w
~
on (O,•)
0
c2cco,•)).
a) ·Express two linearly Independent solutions as power series expansion . '~
In the variable x.
1..,••)
b) Describe the behavior of solutions as
Jt •
•·
c) Can a solution to • be bounded from below and negative at some point. Explain.
, ...
\......... ·.i
, Quallfylna Examination Harvard University Department of Mathematics
·~
November 20, 1986
·· Oay3
1 c:.
Let U(n) be the unitary aroup of COf11>lex nxn matrices A, satisfyina
°'i!'•A
•I •
a) What Is the real dimension of U(n) 1 2
b) If X Is the set of points In U(n) satisfyina the equation A • 1, find:
1 ) the numb4!r of compunents in X.
2) the dimension of each component. 3) What Is the dearee of the map
U(n) + U(n) 2 which sends A to A •
.......
2c.
An open domain U in
c1
is equipped with the Hilbert space
'\
' .
H(U) • square Int ear able holOtROrphic fwiction h: U • C •· a)
Show that
function on H.
bv
evaluation at some point • c U
Hence every
• c
u
one. obtains a contl,.,.ous linear
defines a holomorphlc function K(r;•) of· z such
that
h(.. , •
b)
It
dzAdz
h(z) K(z,..)
=ii .
Show how to construct an explicit orthonormal base for
H(U)
when U is a unit
disc:
'· u • n>1

,, ~) +c}
..
_.. \
'J,.
b)
t
C)
Show K •
d)
Describe the aroup of autcmorphisms of
What is the degree (F(a):F].
F~~
~~pMt
m  x n)
~
,. ~.I\
lsi 4 c F(a)l
far\.
and (K:F(y)J • S.
detennine the order of this aroup.
f(l)
• •
~~
1
which are identity on
~
~ A.4
F; in particular,
~
11(11 /ts
3
. '. :... C."·· ,"'.,.l ··.• • .... ,. Show that for any elliptic curve over C, '/End E is ccinmutative. (~~(...; . 2 3 3 3 Over 1'5 , let E bi curve in r aiven bv {(x:y:z): x • y • z • O} with r 1.
s.
(a)
(b)
~
.*'
•·.•
(1 : •1 : 0) as its zero element. Show that End E is not commutative.
•.
6.
Let
I: ak
be a converaent series of real runbers with ak ~ O Vic. Prove that
.
r·
1
n
~
lim ·  L kak • ~· n k1
o.
..
C1 i
••
~· .,
.,
 1 
Qualifying Examination
Harvard University Department of Mathematics Novenber 20, 19tSS
Let
1.
f:
X • Y
be a nonconstant morphism between projective, nonsingular algebraic
curves, X and Y, over O.
,....,.. 6.
let f:
B"
+
In be a smooth map, where l:Sn is the closed unit ball. Suppose that for
some open ray
K in
In
based at the origin, there is a unique x c Snn1
f(x) c R; suppose moreover that if T is the plane tangent to S is a hyperplane transverse to R. Prove that f has a zero.
,,
1
such that
at x, then df (T) x
.· ~·...::
.
~.
, 
.,
Qualifying Examination Harvard University Department of Mathematics November 21 )985
1
Consider the nonlinear ODE, for u c
1.
J~ • u3 =c,
c:2((0,1J)
with boundary conditions u(O)
=u(1) =o.
dt
Here, c c ct Is a constant. Show that a
c2
2 solution, u[cJ(•) c (x dy  y dx).
Explain your reasoning carefully.
Let·· A be a bounded operator on the complex Banach space Show that
A ha~ a no~empty ·spectriim, i.e., there exi.sts
a complex number A such that the operator
AA·1v is not
invertible. Hint.
~ 2UP}.
Remember your complex analysis.
Determine the Galois group (with a justification) of the splitting field of the following polynomials· f (x) over t: (1)
f (x) • x6 + XS + x4 + XJ
(2)
f (x)
(3)
f (x)
Hint.
= x3 = x3
+·x~ + X + l
+ 2x + 2  Jx + l
For a cubic polynomial XJ + ax + b, if
are its roots in
a
then
al,a2,a3
v.
Qual. Exam. 4/5/84
.: DD/14J,!aY 4.
2•
True or False. The complement of the ·diagonal in 2 2 5 x 5 · is homeomorphic to 5 2 x m2 •
Find the integer x 13
x
such that
~ 2 19821 45917 30833 04870 13369.
Be sure to explain your method.
~o 71· 6:
Let· T
be. the toros
Prove that if
T
·_six s 1 •
is smoothly imbedded in
m3
then
there is at least one point where the curvature is negative.
{
,
HARVARD UNIVERSITY DEPARTMENT OF MATHEMATICS QUALIFYING EXAMINATION 4 April 1984 ' \
1.
:.. r
.>· ~~2.
a)
Construct a field
F
b)
Find the roots of
x6 • 1
Let
with 25 elements. in
F.
K be the integral operator given by the kernel K(x,y) •
cos(xy)
operating on the smooth functions on the circle S
) 3.
= IR/21rZ. a)
Find its spectrum  with multiplicity.
b)
Relate this example to the Peter Weyl theorem.
Let
Mn
denote the vector space of
For each ad X:
X c Mn
Mn +Mn
nxn matrices.
define the linear operator by ad X(Y)
= XY
 YX.
Explain the relation between the eigenvalues of and
ad
X
x.
'
.
,,.(.
Qual. Exam. 4/4/84
"B~7"t·
4.
2.
Let
X be a (compact oriented) Riemann surface of genus
Let
w: Y >X be an unramified connected covering of
degree 5.
Compu.te the genus of
s.
Y.
_....L

~A.,.:: / ... r.N1 5:
Compute
f
xl/3
~ dx •
0 l+x
CL111..=!1n 6 •
Suppose
~
a plane and a)
and
y
are cartesian coordinate functions in v • xy.
u =
Find explicitly a maximal connected open set containing
on which
u
and
v
D
can serve
as coordinate functions. b)
Express the Laplacian operator on
D in terms of u,v,
and partial derivatives with respect to c)
u
and
v.
Find a basis for the linear space of those harmonic functions on
D that can be expressed as a homogeneous
polynomial of degree two in
u
and
v.
BARVAIU> UNIVERSITY
._ . _
DEPARTMENT OF MATHEMATICS QUALIFYING EXAMINATION 28 October 1983 l.
Consider the upper half plane B • { z I Im ( z) > O} , endowed with the Riemannian metric ds • .1 c;,z I , i.e. ,
~.2
w
•
(dx)
2
+(dy) Y2
2
,
(z • x + iy).
(i) Prove that f9r real a,b,c,d with adbe • 1, the map az+b z ....._. cz+d is an isometry of H onto itself. (ii) Prove that the vertical lines x • const. are geodesics. (iii) Prove that any two points in H can be joined by a unique geodesic. 2.
Compute the homology groups of the space x • p 3 (a:)  P 1 (a:) obtained by removing a complex projective line from complex projective 3space. Suppose G is an open subset of a:. Let A 1 (G) denote the set of ·holomorphic functions f on G for which the L1 norm
II f '111 
If I
f ( z) I dx dy
G
is finite.
(i) Show' that if the open disk of radius rand center a is contained in G, then for f € A1 (G),
(ii)
Show A1 (G) is complete in the L1 norm.
Continued on ne.'\.t page .;,
'
,~
Qual1fyin9 Ex~ina~ion October 28, 198~
4.
Let ·P(X) ·be an irreducible polynomial with coefficients in ~, and let a be a complex root of P(X). Let n be the deqree of P{X) and r the number of. roots of P(X) in the field t(a). Prove that r divides n, and that if the group of P(X) ' (i.e., the Galois qroup of a splitting field of P(X) over ~) ·is abelian, then r • n.
s.
Consider the compact real Lie qroup the matrices zl
z2
z2
zl
•I
.zl,z2 €
In terms of the coordinates give:
G • SU(2) consisting of
c,
2 2 lz 1 1 +lz 2 1 = 1.
x 1 ,y1 ,x 2 ,y 2 ,
where
z a  x a +iya ,
a) a basis for the leftinvariant difterentials on G, and the Baar measure on G (i.e., the invariant volume element), normalized so that the total volume of G isl. b)
6.
Let f (t,x) be a continuous realvalued function.defined in a neighborhood O of the origin in the (t,x)plane, which satisfies · the Lipschitz condition:
Prove that in some interval dx dt •
f(~,
x(t)),
I ti
< t , the problem
x(O) • 0
has a unique differentiable solution x(t).
(End of Exam. 1
.. HARVARD UNIVE:RSITY DEPARTMENT OF MATHEMATICS QUALIFYING EXAMINATION 27 October 1983
Describe the structure of the group of automorphisms
1.
of a cyclic group of order 34000.
Let
2.
~nalytic
.cs l
~.
f
in lzl < l and such that
If Cz) I = l
Prove that if f has no zero in lzl
·a)
~
be a function continuous on 1zl
1,
on lzl • ·l •
< l
then
f
is constant. b)
Iz I
in
3.
Describe the f's· which have exactly one zero . < i.
For 9 ::. 0, let
Sg
be a compact oriented surf ace
Prove that for
9 < 9' there is no continuous
of genus 9. a)
map
s9
~s,
g
b) 5
2
___. 5
with degree
~ o~
For which integers n is there a continuous map 2
with degree n?
1~Q·~ 
ro9e two 
4.
a)
curve
198~
The field of rational functions
Kf
on a plane
f (X,Y) • 0 is, by definition, the fraction field
of the ring then
Uctooer
~I
Kf b)
C[X,YJ/(f).
Prove that if
f •
x2
is a pure transcendental extension of
+ Y2  1, ~.
Give a general definition of the notion of genus
of a plane curve, and show that the curve
x2
+ Y2
=1
has genus O according to your definition.
Suppose
5.
f: It
is infinitely differentiable
~ lR
and satisfies lfcx>I < n1
and

for all
x
and all n > 1.
€ It
f (n) (0) •
O
Prove f is constant •
.. 6.
Let
v
be a finite dimensional Euclidean space.
For a subset
Sc V, define {x € V
S• •
where
xs
I
xs >
o for all s
€
s} ,
denotes the scalar product of x and s.
s
Prove that if
is closed, nonempty, and satisfies
the two conditions
s
a)
s,t
€
b)
s €
s, c
:::m::m:>
s+t
€ S,
and ~
0 .....
cs € s,
then S •
(S')'.
The End!
ft.V iJP'"
HARVARD UNIVERSITY DEPARTMENT OF MATHEMATICS QUALIFYING EXAMINATION 26 October 1983
l>
a)
Give (without proof) the structure of the
x
fundamental group of the compact orientable surface of genus 2 pictured below. b) s
1
to
Prove or disprove that the continuous map from X
indicated by the loop in the middle of the
picture is homotopic to a constant map.
2>
Let
R
an ideal in a) of
be a commutative ring with unity, and
..
R.
Show that the radical of
of the prime ideals of
·R
(the set of elements
the radical of ideals of
R
I.
State the Hilbert Nullstellensatz and explain
polynomials over the field
c)
is the intersection
which contain
how that theorem implies that if
in
I
R, some power of which is in I)
b)
I
R
I
E
R
is a ring of
of complex numbers, then
is the intersection of all maximal
which contain
I.
Give.an example of a ring
R
and an ideal
I
whose radical is not the .intersection of the maximal
ideals which contain it.
particular solution of y••  Jy• + 3y •  y • e
2x
•
What is the most general solution? Discuss the solution of y• + y • 0 for the
(b)
•two endpoint• conditions y(a) •
a, y(b) • s.
s 4,
the group of
Give the character table of
4)
permutations of 4 objects, and explain how you compu~ed
it.
Evaluate
5)
(a>O)
• 11,
6)
Consider the unbounded operator
given by a) and that b)
I"' I
s
11',
A
• on
L2(Sl)
(Af) (0)  f  f• • Prove that
A
has a welldefined inverse
is a compact operator.
B
Give a continuous function
Ie I
$
1f I
K('&l,8)
on
such that I
( Bg ) ( 14>) •
J''
K('&l,0)g(0)d6
•
11'
The End!
B
·~ril
(1)
QUALIFYING EXAM
27, 1983
The curve
(x2 + y2)2 + 3x2y. ~ •·O has the origin as an ordinary triple point. Find a rational parameterization of this curve, i.e., rational functions x • x(t), y • y(t) satisfying the equation.
I
l'ff""' .;
I~
( 2)
Consider the equation
Describe the asymptotic behavior of the solution with y (0)
=  1•
Aprf 1
( 3)
27. 1983
a)
Define the degree of a smooth map of an n ·sphere into itself in terms of de R
theory and lncllcate a proof that this degree ls an Integer. b)
Prove that the rational function f (z) • (z • 1) (z; 2) (z • 3)
extends to a smooth map of
( 4)
s2 = a: U •
, and compute its degree.
Construct two noncommutattve groups of order 8.
April 27. 1983 · r;)
a)
.
Prove that If n • 561, lhen nery integer a an·l • 1 (mod
b)
Prave that If
D •
p, q ,
11)
relatively prime to
p, q diatinct primes, lhen there are integers a
11 1
a •
;
1 (mod n)
Let S be lhe square with sides of length 1 bl the complex plane. Let f
be holomorphic on S
such that
( I t 12 m dy
ls. Prove that
If (center of
S)
I
0 , 0 < r < 1 , such that
lan I < C rlnl
Z.
[Hint: ccmalderholomorphlc extensions of f
( 2)
1 A (5 ) if and·
.
only 1f there exists constaDta
for all a
€
to a neighborhood of S' in Il
Describe the geodesics on the upper half plane relative to the metric ds
2
=
mc2 + y
2
dy
2
. . 26. 1983 I
Por each of the following equations decide whether the inclusion of the set of real solutions Into the set of complex solutions ls a homotopy equivalence. (Real solutions are in
2
2
Jl ; complex solutions in '1. • )
+
a)
:r.2
b)
xy (1  x  y)
c)
x2
d)
x
2
,2 •  1
+ 12  y
2
s
0
• 1 • 1
n
I~
( 4)
Prove that there ls no field with 6 elements.
. ;I!
~~
. . . 26. 1983 (5)
Compute
o
( 6)
F ' :> F , as follows: E
if p > 0 • Prove that E
F • Let
E
=F •
if p
is generated by a single element over
=O F
F••
· 3.
· Does there exist a function with at least 10 zeros in
holomorphic in { Iz I ;: 300} , with
f (z)
I I ;:
{ z
sup
lzl ~ 300
f (0)
=1 ,
100} , and such that
If(z) I
< 1024 ?
Either produce such a function, or prove that no such function exists.
/
'f ':
Qualifying Exam 17 November 19 
4.
_..._.
 ·"·
.. ..:...==a. ... ~~ ..    ...
=..... 
!::"""..::.._.
......
 
Ptnd twO linearly independent series solutions for the differential equation
· 2xy" + y• + xy = 0 near the origin.
5.
Determine the Galois group (with a justification) of the splitting field of the following polynomials f (x) (l)
4 f(:x) = x
(2)
f (x) = x
(3)
.!:!!!!:
~
over
3
.3
f (x)
= x
:
+ x3 + x2 +
1
+ :x  1 ..
 3
x+
. 1
•
3 Por a cubic polynomial x + ax + b , 1f cr , 1
=  4a
6.
Let D
={z : lz I
< R}
and assume that
o1
==
3
~, ~
 27b
{z :
a:
th
2
lz  z 0 )
of S ls
= Jnf {J,&(E): E € e. , E :a· X}
O(X)
and the Inner measure is l(X) • sup {J,&(E): E Let
A
O'·field
e.'
E ;: x}
be a non ·measurable subset of S • Show that generated by
between I (A) and
6.
f:
e.
and
A
u
can be extended to the
with u (A) taken to be any prescribed number
0 (A) •
Set h(z)·
where
= Jor
D is the unit disk in the
~
d ~ dn
_z_·_(..._~_+....1,,...) •
'l"1'
plane and z c CC •
1) Show that the integral exists as a Lebesgue integral for all
ii) Evaluate the integraL
(Suggestion: Consider an annulus and a smaller concentric
disk. Use power series. ) W) At what points of CC
z .
ls h
holomorphic7
HARVARD UNIVERSITY DEPARTMENT OF MAniEMATICS QUALIFYING EXAMINATION 5 Aprll 1982
1.
2.
3.
4.
State and prove the Chinese Remainder Theorem.
Compute [• .
cos x dx + x.4 1
Suppose
f : JR  JR
point of
JR  {O} and that
ls continuous. Suppose also that
Um
xo
u(x, 0) ~u ~t
S.
Let a)
(x,O)
is differentiable at 0.
of the wave equation
satisfying the initial conditions: For all x

is differentiable at each
f' (x) exists. Prove that f
= u(x,t)
Find explicitly a solution u
f
€
JR,
= sin x =
1 1 +
2
x
V be a finite dimensional vector space over a field of characteristic not 2 • Define a guadratic form cm
V.
Continued on next
pa~.
.•
Qualifying Exam 5 Aprll 1982
..JS caidnued. b)
Given a quadratic form q on V , show that, with suitably chosen linear coordinate functions
x , ••• , 1 2
al xl
6.
+
~
11
2
a2 x2
on V ,
+ • •• +
q can be expressed 2
an xn
•
Let Y be a metric space. a)
Show that every open covering of Y has a flnlte subcoverillg ff and only ff every . lnfln!te sequence ln Y has a convergent subsequence.
b)
Is thls true ill general (Le., 11011metrlc) topological spaces? If not give a coUD.terexample.
... QUAUFYING EXAM November 12, 1981  Day 3 3l)
Prove the Ba.nach If
is a cCQplete metric space,
(M,d)
exists a
fixed point theorem: f: M > .M ,
and there
such that
Ol,
is
not
ll
whenever
R.
f: En :> m. 1 ,
11.
Show that a continuously differentiable function is not
R•
n>m •
f: En '> JR m
21)
f(t ,t > be a he111ogenous polynomial of degree 4. 0 1 f(t,l) • O has four distinct roots. Let
can
22)
As;sume that
contain S/ 7 ~ ?
Aut(f)
2 2 Let i5' • { (x,y): x + y ~ l}
be the closed disk in
m 2 and
p. a(x,y)o/ax + b(x,y)o/oy
where
~.
a)
a,~
are
c!° in O •
Assuming the existence theory for O.O.E.'s show that the
~~ation
Pu + u • 0
(*}
has a solution in the neighborhood of any point (x0 ,y0 )
b)
If
E
O •interior of
x a(x,y) + y b(x,y) > 0
O where
an
laI+ Ibex0 ,y.o >I~ o
2 2 oD • {(x,y):x + y • l} ,
show that there is no solution of
(•) in i5'
(Hint: try the maximum principle).
c)
Establish the topological result that, if x a(x,y) + y b(x,y) > O on ao, then P must have a zero in
o.
1 I
23)
a)
Let
c~
on
lRn •
(l't
n)
be the space of compactly supported
Define the Fourier transform of
f
f(t) •
eitx f
(.x)dx ,
f
c•
functions
C~ tin)
(x) t:
by
t t: IRn* •
IRn
Show that
.
f(t)
has rapid decrease; i.e.
1£ Ct> I for any
b)
N > O.
Is the converse true?
Denote points. ·in :R n+l •
+
JR n+l
{ (x,t) :t~O} ,
(x,t) • (x ,x ,t) n 1,
by
~
. x
.  Ii· a21 axi2
and set
•
Consider the wave equation, for an unknown function
u
£
c•cm:+l),
(**)
*
2 ct,t> • I ltl 1 zct,t> •
o
,.. zct,o> • f
I c)
Osing
a)
solution to
az
,.
rt • gCt>
and inverse Fourier transforms, justify the formal given in
(**)
b)
.. I
j~
I ...
~;
24)
a)
Define the local intersection number of two algebraic curves in the complex projective plane.
b)
Coni~te
x
3
0
the intersection number of
 x x x • 0 , 0 2 1
[x ,x ,x2 ] 0 1
c)
at the point
2
3
x x  x "" 0 , 2 0 1
with
[x ,x ,x ] • [0,1,0] , 0 1 2
where
are homo9enous coordinates.
State Bezout's theorem.
Why is this an interesting result
(if it is)?
11
25)
In
a)
n
this problem '2 is a nice open set in IR
Define what is meant by a differential form (linear differential fo:nn) in '2 • d& (you may do this by a formula)
b)
•
tat
V
be a vector space with dual
hyperplane { t • O} •
Suppose that
e
Define the exterior derivative
v*, o ~ ~ 2 4> € A v*
bilinear foz:m on v such that 4>lw • O (i.e., all w,w' € W). Show that
c)
Suppose that· for each x M
x
c:.n
€
0
~(w,w')
• 0
there is a smooth hypersurface
M
x
• 0
(i.e., the 1form 0 restricts to zero on M ). x
e ,. . d& • 0 .
c v*, and w the is an alternating
such that
eI
on
of degree one
o
Is the converse true?
Show that
for
26)
a)
Define what is meant by a of
b)
o
2
CW canplex.
Give a
CW deccmposition
(• canple.x projective plane).
Let x be a cw canplex with cells e0 ,e ,e (dim ei • i). 2 4 It is a true fact that X is determined up to homotopy equivalence by the class [a] £ n . Let n £ &2 or, z) ; II 3 2 (why?) be a generator. Show that, if [a] • O , then the c:upproduct
nu n
0 •
Qualify;ng Exam
11)
Let
 November 9, 1981
R be an integral danain,
F
its field of fractions.
Call
R
neat if it has the following property: If q(X)
and
h(X)
are monic (i.e. highest coefficient l)
polyncmials with coefficients in
F and their product
has coefficients in
and
R,
then
g
h
q(X) h(X)
each have coefficients in
a)
Prove that if
R has unique factorization then
R is neat.
b)
Is neatness equivalent to a more standard property of integral danains?
c)
Give a counterexample with q not every R is neat
and
h
linear which shows that
R.
· , .l2)
a)
IAt f: [a,b] > JR 2 be an imnersed curve (i.e., df " 0). Define the arc lenqth sa(t) and curvature ic(a).
b)
IAt
f:S
1
> :it
2
be an immersed closed curve.
L2'1T •(
K(s)ds • k
Show that
Z
£
sl
What is the geanetric interpretation of
c)
Show that .
1 2'IT
Is l
lie Ids
with equality if, and only if, ~
""'""''"
k?
~1
,
fCs 1 > is a convex plane curve.
.·.
~
l3)
Let a)
s
c:
m3
be a smooth surface
In (at least) two •ys define the Gaussia:: :·.::"'1ature State
S > :R •
Gauss' "theorem eqreqium"
Now assume that
b)
1t
What is
2
I
;
S
is ccmpact
K c!A ,
where
(• closecU.
c!A • is the
r.:~:ace
area of
S?
s wha:
c)
Show that
l 2'1T
r J
litlc!A
s
is a convex surface.
> 2
•
with
equa!.~ ~::·
any) significance
!.!, and only if,
s
f.
l4)
For l 1 do there exist n·polygons on S , Le. homeomorphic
=
images of a closed disc whose boundary breaks up into n geodesics?
2.
Prove that if A is a noetherian ring, then so is the formal power series ring A[[X , ••• , X D. . 1 D
3.
Let H be a separable HUhert space, a)
Prove: if
T : H • H a bounded linear operator.
~ ITvD1 H. S.
depends only on T • 'Ibis is the ~ of T •
Continued on next page.
2
Quallfying Exam 21 March 1980 ~
. DtlmJecL
Bxbfhlt a HilbertSchmidt opemtor
e)
ofdieuaboundedopontor
f)
Let
JC(8 , cp) he a
defined
~
T!(8) •
lhat trace T •
•
C
(
2 1 2 1 R: L (S ) • L (S ) , which ls the inverse
·d~ + 1).
1 1 21 21 function on S x S , T : L (S ) • L (S ) the operator
AJ
ft JJC(9 ,
IC(9 , 'Ill ftp) d'IJ • l'ro9e lhat T
8) d 8 • (Hint:
T • (
d~
+
1)
I• of ttace
class,
and
11 a Hilbert •Schmidt
operator.)
Prove that th~e ls only one "palindromic" prime number P with an even number of Jigits (palindromic means that in decimal notation, die number ls aa , or
abba , or
abc cba , etc. ) • Generalize this to other bases.
~~~.,,
I
'....#
5.
Prove that there is a real number which ls not algebraic over a
padic
number and do the same for a
p·adlc
~
: construct one. Define
number not algebraic over
~.
Qualifying Exam 21 March 1980
6.
fzl
a) Let f(z) be holomorpllle for
< R and sattsfy f(O) • 0, f'(O) ; 0 , f(z) ; O for
0 < lzf < r • R. Let C be the circle
fzf • p wbere p < 0. Show that
1
g(w) •
2"../'=i c
w
define a holomorphlc function of
for
tf '(t) dt f(t>  w
fwl
18
< m • min jf(pe ) I , and that 8
z • g(w) ls the unique solution of f(z) •
w
that tends io zero with w •
b) Find the Taylor's expansion of g(w) , and apply this to find the explicit series expansion of the root of the equation
z that tends
i
J
to zero with
w•
3·
+ 3z
... w • 0
'\
....
:
QUALIFYING EXAMINA noN
1.
31 October 1979
2 Jn the plane 1t , let x2 (resp.. >Ca> be the union of two (resp. three) dlstlnct ~
late.rrals with the same middle point:
Prove that
x2
and
x3
arc not homeomorphic. Do the aame with closed intervals
replaced by open Intervals.
2.
(a)
State and Prove Taylor's theorem (with an explicit formula for the remainder ·term).
(b)
Using only elementary properties of the exponeD!'al function, give a rough upper ·bound for
(c) I~
3.
(l. e., with roundoff error
< 10

3
) • Justify your method.
Let · CQ be the algebr.iic closure of CQ • Show that If R ls a

type contained in CQ ,
4.
e < 4).
Agatri by elementary me:ins, compute the first four terms of the decimal expansion of e
~)'
c. (e.g.
% ·algebra
of finite
R cannot be a field.
Define Hilbert space, and state the fundamental facts about orthonormal bases and expansions in terms of them. Relate this abstract theory to expansions in Fourier aeries and Legendre polynomials.
.'

,
Qualifying Exam
c
31October1979
Dl D2 Dr Prove that the product of spheres S x S x • • • x S la smoothly embeddable in •1 +n2+ ••• +•r +l
• 6.
(l)
•
•
Let f(z) be a bolomorphlc function defined ln a ~elghhorbood of the disc
and let M • . max jf(z) fzj • R dlac
fz I < r (r
0,
I:•
" • i
\
•
II
Suppose dlat f(z) ls an .+£
"
< •
and
R
.. QUALIFYING EXAMINAnoN
1.
1 November 1979
Let ABC he a triangle 1n the plane. Prove (using rc:il or complex linear algebra ndler tban Euclid) that
2.
(a)
the Unes jotning vertices to the midpoints of the opposite sides meet in a poiat
(b)
the llnes from vertices pcrpendtcular to the opposite sides meet in a point.
Let D he • region ID Oil
D,
a: •
Let A be the additive group of functions f
B tbe subset of A of functions nowhere zero on D •
multlpllcatlon. 'Ibe map exp : A .. B defined by (exp f)
analytic
B ls a group U11der
Cz) • e' > 0 •
3.
Plnd all analytic automorphisms of the (open) first qu:idrant of the complex plane.
4.
Let JC be a field of characteristic zero 1n which every cubic polynomial hll s a root.
Let f(x>. be an irreducible quartic polynomial with coefficients In IC whose dl8crlmlnant la a aqua.re ln
S.
K • What ts the Galois group of f(x) 1
If u ls a real number, denote by (u)
'< u)
~ Inf lu • n I • D. £
Let ex •
1
z
+45 2
• Show that
convergence of the power series
(an )  O as n  • • What is the radius of
I:• (o.n) zn 1 Hint: D.•0
1n
the distance of u to the nearest integer:
QNS>.
ex is an algebraic integer
..
.·
Qual1fy1Dg Ezam 2November1979
Piaf the followtq game. · St&1't with any 3 numbers a, b, c oa the vertices of tria.Dgle.
a
c
b After one move, replace each number by the arithmetic mean of tbe 2 adjacent numbers:
b+c ,
a +b , _ _ _ _ _ _
~
!.±.E. 2
Repeat. Show that ID tbe l1m1t a • b • c • Find the rate of convergence of the game to tbla limit. What happens In the same game with 5 numbers
\
011
the vertices of a penttt ·
.·
21March197"1
QUALIFYING EXAM
Let f
a)
be analytic and bounded by
1 on the open unJ.t disc
D • {z 1
I .,z I
I ~ n la1 I
•
1•1
2.
Let V be a normed vector space over R , let I c:: R be an open interval, and let
f : I ... V be a differentiable function. Show that if is a constant functioo.
3.
(H~t:
First do the case
Define the degree of a map from one
r
ls identically zero, then
f
V • R •)
n·maldfold to anotber n·manifold. What is the
degree of the Gaussmap of the following surface S in
m3 ?
s (Recall that the Gauss map sends a point in S to the translate of its outward unft normal to the origin. )
·2
'·..
Quallfying Bxam 21 March 1979
4.
Let IC be the spJJtting field of the polynomial
x4  1or + 1 over the rational number field
Q • Give explicitly a linear basis for
K over Q •
Describe the Galois group both as an abstract group, and in terms of lts action on K. Describe all subfields of IC •
5.
Let F
be a field of characterlstlc
e , • • • , e n , and T : V • V 2
1
0,
V a vector space over F
with basis
a linear map such that the associated map
2 2 2 A T: A V  > A V
leaves the element e A en+l + • • • +en A e n invariant. Shpw that det T • 1. 1 2
6.
a)
Let G be a finite group,
N a normal subgroup, and cp
the natural map of
G onto G/N • Give examples to show that 1)
There need be no subgroup of G which is mapped isomorphically onto
2)
G/N by cp.
There may be such a subgroup as ln ( 1) and yet G need not be isomorphic to
b)
N x (G/N) •
Give an example of a group (containing more than 6 elements) exactly half of whose elements are of (exact) order 2 • Show, however, that no such group ls of order 8 •
''.
QUALIPYING EXAM
1.
a)
22 March 1979
Define the notion ot curvuure and torsion for a smooth curve ill Euclidean three space and deriVe the Prenet•Serret formulas (clifferentlal equations cODllecdng the curvature and torsion with certalll vectors associated with the curve). State apllcitly the
appropriate meaning of smooth ill this context. b)
Show that, lf the torsion of a curve is everywhere 0 , it lies ln a plane.
c)
Give the prlmitlve (1. e. as the limit of a geometrical constrUCtion) deflnition of the circle of curvature of a curve at a point and ~how it ls related to the notions above.
2.
Let F
~ a finite field with q = p
11
q
elements, where p is a prime. Let D : F • F q
q
be the Frobenius automorphism D(x) • ~
n
Prove that
•
considered as a linear map over F p is cUagonaH:r.able if and only if n
n
..•.,,,.
divides p  1 •
3.
The Bessel differential equation ·of order zero is u" a)
+ !u• +u x
=0.
lhis differential equation has a power series solution of the form u =1 Obtain an explicit formula for the coefficients
+
• 2k 2: · c x • k= 1 k
ck •
b)
Show that this power series converges for all x •
c)
From the differential equation, deduce the oscillatory behavior of the general (nonzero) solution on (O, •) • (Hint: you may find lt helpful to consider v
=u ./i . )
Qualifying Exam
22 March 1979
4.
Calculate the fundamem:at groups for the complements of the following links Jn JR
3
c )
a)
h)
Justify your answers.
s.
Show that
la
\
•
0
x0 22
(1 +x)
dx •
tr(l  a)
, for
 l
Consider a geodesic
1>
2
obtained by revolving the circle (x  R)
+ y2 •
l
·X
y · on
T , and a point
one of the circles . y • const.
Wha~
P , at which
'Y ts tangent to
can you say about the behavior of ,,
near P ? (i.e.. does· .,, crosll the circle at . P , and lf not; on which side of the drcle does 'Y . lie near P 7) . b)
Let 'Y be a geodesic which ls tangent to the top circle (y • 1) at some
point. Show that ,, a.lways remains
QD
the outer half of T ,
osc~ting
back and forth between the top circle and the bottom circle.
This problem .c:L:t be done without e::tensive calculations!
6.
State carefully all the theorems you know about passing to the limit under the integral
sign 1n the context of the Lebesgue integral.
"' .
.rQUALIPYlNG BXAM 21 April 1978
L
If f(z) ls analytic tn the complex plane, real on the real axis, and purely lmagina.?y on the Imaginary axis, show that f(z) ls odd.
2.
Ptnd the solud.on to the differential equation (with lnltlal condltlon)
· X'(t) • AX(t)
,
X(O) •
(i)
where
X(t) •
3.
a)
(~~~) z(t) •
and
~
1)
•
Give aa example of two slmply connected cellcomplexes wtth identical Integral
cohomology rtnp but which are not homotopy equivalent. b)
Suppose X ls a stmply ·connected cell ·complex with the same
rtn1 as complex projective n • space ,
ln~gral
cohomolo:
G:P(n) • Show that X ls homotopy·
equivalent to G:P(n) •
4.
Let z , • • • , zD be distinct complex numbers. Let f 1
I
of degree
a
and g be polynomials,
n • 2 and g(z) • (z • z ) • • • (z • z ) 1 n
Continued on next page
·./ /.
2
QuaUfytDg !um 21Apdl1978
14 continued. •)
Sbowtbat
•
b)
Show that there
atst. a polynomial of degree
•
D •
2 with
If and only If
•
c)
Given a· aequence of complex number• z , z , • • • 2 1 does there exllt an entire function
h, ,~·
such that
with ft.z ) •• a ?
f
1
1
Iz n I . • • ,
cAn JOU Write thls
tunctlon down?
s.
Let X be a linearly ordered set. Assume
t) Por all x < y tn X there exists a z tn X IUCh that x < z < y. U) Por any two aubseu x
1
~
s1 , 52 c
x ) , there ls an x 2
£
X with
X with
Prove that any orderreversing bijection
f
s1 !: s2
s1 !:
x
~
(Le. lf x
1
£
s1
5 • 2
of X baa a fixed polnt.
Continued on next
page
then
'·
3
Qualifying Bu
21April1978
6.
Coalllder tbe following trlanguladon of the disk.
1..~~...,..."'
i
O~+E~~t
Can you ftnd an Integer valued cocycle with value l on ( 0 • 1 ) that vanishes on all .other edgea of the bOundary of tbe dtak?
Explain your coacluatona.
. '.
.1· QUAlJFYING EXAM
20April1978
a
Let G be a group with p
a)
. elementt, p a prime, a aa lateger
?
1•
Prove G bu It least one subgroup of Index p , and that each such
subgroup ls normal. b)
2.
P1"0ft that If G bas exactly one IUbgroup of Index
p , then G la
Let A be a commutative ring. Suppoee that c:ver:y ideal of A la f!nltely generated. Show that nery Ideal of A [ X] la finitely gcncr.ited.
3.
1
·.
.. CP(3).
a)
Define complex projective 3space,
b)
Prove that U a quadric eurface Q contains 3 points of a line J. then lt
~;
contains the entire line. c)
Given 3 lines
1. , t
1
2
, 1.
3
in Cl1(3) prove thero exists a qundric surface
Q contalnlng all 3 lines.
4.
Evaluate
·1·
2 0 (x
x •ln x
+ l)(x2 + 4)
dx
Continued on next page
Qua11fyiD.g Exam 20 April 1978
Por a aet S let PS denote the free group generated by S • Let · I' : P { x,
J} .. P { z} be the homomorphism defined by tr (x) • z , I' (y) • 1 •
Give generators and %elatlou for ker (tr) •
6.
rtna
Z( ..(:'2] la Euclidean.
a)
Prove that the
b)
uam, a) ' ftDd all integer eolutlons to the equation .,2
+ 2 • x3
•
.:....;..,i • t·
;, ..•
QU.AUFYJNG EX.AM 19April1978
L
2.
Factor the polynomlal
Let T
Into lrn:ducil>lc polynomlals with coemclent1 1n
a)
'Ibe field
b)
The tleld of rational numbers.
c)
1be field wlth 2 elements.
d)
'Ibe field wlth 5 elements.
e)
1be field With 73 elements.
of.
real numbers.
be a camp:i.ct self adjoint operator from a separable HUbe..""t sp:lcc
Prove that
3.
r' •1
H tc itsc!!.
H bas .•n ortb~rmal basis consisting ~f eigenvectors of T.
Let :x and y be the usual coordlMte funcdons ori
2 m • For each real number
find the critical points of the function
x
2
+
2xy
2
1 4
+ ay • y 2
and describe: their nature (i. c. maxima, minima, a.addle points).
Continued on next page
et
......
Quallfytng Exa 19 April 1978
4.
Let s
3
be Cbe symmetric group of { 1, 2 , 3 }· and
the eet of group homomotphlsms of
s3
to OL2(CQ). Two elements a and JI
of X are said to be equivalent 1f cbere is an element g of GL (
.&2
0
!> Ba ,&1

·o
t
l
a2
As
0
l>:A2
"2
l
.. E2
l
l>
li1
!>Cs
J
0
!> c2
Al
.t2
"1
l0
l>
0
l
!>Bl
!> B2 ,Ell
"2
~
~
!> 0
t1l >Cl
!>
0
l0
i ~ i 't
,"''.
If all 3 columns and the first two rows are short exact, then the last row ls also
short exact.
6.
Show that the center of a flnJ.te
p  group
interseCts normal subgroups nontrivially.
. . .....• .. ',...
·'· .
~
(
HARVARD UNIVERSITY DEPARTMENT OF MAniEMATICS QUALIFYING EXAMINAnoN
October 24, 1975
1.
Define the Gaussian curiratUre of a smooth surface in 3apace. Construct examples of surfaces with constant positive and constant negative curvature •
.. 2.
Let I be a nilpotent ideal in a ring R. , let M and N be R·modules, and let f : M  N be an R·homomorphism. Show that 1f the induced map
f:
M/IM  N/IN is surjective, then
1• surjective.
3.
Let A(t) be a nonsingular matrix whose elements are differentiable functions of a real variable
t • Let
A'(t)
denote the matrix formed by the derivatives of the elements.
Show that the derivative of the determinant d dt
( det A)
= det A
det A
• trace (A ' A1 )
•
satisfies
f
.. ..  ...
Harvard UDlverslty Qua.Ufylug Examlnatlon, Page 2
October 24, 1975
4.
State the ma.in theorem and definitions comiecting the Lebesgue integral cm a product apace. of finitely many •pace• with the Lebe•gue tmegrals over the factors. Include a discussion of how the product n:ieaaure ia constructed but omit proofs.
. n2 2
s. i
1
12
Hint: For
i.
1~·
f(X) •
i~~
! .1
JC
Show that
I
c [ 0 ' 1 l , consider the function
Jen
Also, consider f(l  x} •
n=l n2
6.
. 2 log 2
t
5
t
+l
is irreducible over
Z
·;
 .. .• '
~
HARVARD UNIVERSITY DEPARTMENT OF MA'niEMATICS
QUALIFYING EXAMINATION October 23, 1975
1.
Let V be a fintte dimensional vector space over C With basis
{v1,···, vn} • Let a beapermutatlonon {v , ••• , vn} and 1 ·thus lnduce a linear transformation A on V • Show that A ~
2.
dtagaaalizable.
Let m be an integer > 0 • Dlscuss Sn detail the suucture of tbe group of automorphisms of a cyclic group of order m 4 2 mustrate your answer by the case m • 2 • 5 • 17 •
I l
•
21
3.
Let G be a relatively compact set in lR. X IR • Let f : G ... m.n be continuous and satisfy a Lipschitz condition
in the sec:ond variable: there exists some constant K such that tlf{t,x)  f (t,y)ll
Show that for any about the point t
cp : I  m.n
~
$
K tlx 
(t ,x )
0 0 , and
YU
when
(t,x) E! G and (t,y) eG •
e G , there exists an interval I
a unique solution 0 for the initial value problem:
= f (t,cp(t))
and
l
.. _,. Iarvard University Qualifying Examination, Page 2 October 23, 1975
4.
i
denote the set of all sequences of complex numbers 2 2 that 1 + 2 + · • • and let 1, 2 , • • •
Let
c c
such
Uc1,c2,... ff
lc 1 lc l
Jfcl 12 + le2 1·2 + ...
i
and
• .
2
are 1n l it may be shown 1hat x 1  y1, x 2  y2 ·~· 2 2 . is in l and that l becomes a metric space if one defines
y • y1,y2 , •••
llx1  Y1,
···II •
x  y , Tal::ingtbls.for granted 2 2 prove tbat the result::lDg metric .space ls complete. Give a direct p(x,y) •
proof. Do not deduce the result from the corresponding (and more difficult t!17orem about square
s.
summable
functions.
Map the· disk { Iz I < 1} with slits along the segments
[a, 1], [1, b] (0 = 1 (b)
•
Cii for
all
torsion free. with
i= 0,1,Z,···
Formulate and solve an analogous problem using instead of
H (X) 1
of
'll'i(X)
I.
.,
..
~,.
HARVARD UNIVER~TY DEPARTMENT OF MATHEMAncs
QUALIFYING EXAMINAnON October 22, 1975
1.
Let A be a symmetrical real· n
xn
matrix.
.
Show that there
exists an n x n. real orthogonal matrix U such that
lJ . '\. U
a diagonal matrix.
2.
3.
Solve
x
5
 1
=0
by radicals •
Prove that an infinite product of topological spaces ls connected if and only if each factor is connected.
is
. ...
.
~
_.,.
,,.
,,. HanUd University Qua\lfylDg Examtp.a.tion, Page 2 October 22, 1975
4.
5.
Sta.te and prove the i!Older inequality for integrals.
Use contour integration and careful estimates to compute
•
.J 0
x•·l cos .
x dx.
0
5. P.mve that the
~
product of any mual>er of connected Hausdorff spaces
t. c:ozmected. •
:
!J•·
.
6. Suppose tbat·tbe power series f(z) •
Iz I < 1l •
8n·~ a
2
.
z +.a z + ••• converges for 1 2
Derive expressiom for
2•
·'.
;,
I
...
I f(re1i >1 2 dlJ
0
· ·tor
0~·r n M')"
for all lmegers n 2 k •
"
"""""
3. Let X • s1uete obtained by attachJDg 2cell e (• {w cCt I lw\ ~ l}) to the 2 2 1 1 . 1 5 circle S (~ {zectl lzl •l}) with a map !Z: S (•?>e )__.S by cx(w)=z • 2 PiDd the integral homology and cohcxnology groups of X •
. .4. Prove thr.
i.e. if f
~
rcapping theorem for an analytic fm1ction of a complex variable,
is an analytl.c function defined on an open subset U of ct then f(U)
is open in ct ,. provided that U is cmmected ad f ls nae constant.
5. Prove that a closed smooth surface in R 3 bas a point of positive curvature. CD
(Closed smooth surface means C
.
2manifold without boundary Cid compact.)
over+
6. a) If I.,/X 18 a flD1te eep~~
Uteuion,_
shaw that • TrL/K(u • v)
glwa a DODdegenerata qaadntlc fOJm on L , nprded as f1D.lted:lmeuicmal ftetor
space b9er the field E •
..
.
...
b) Prove that the ring of. a1gelsra1c Integers ID a flDlte algebraic number field . .. ·' (a 1lDite ezteuion. of 0 , if both solutions exist on [O, x].
2. a) Oefln.e the Riemann integral of a tuned.on f : [a, b)i> JR • b) Prove that the Riemann integral exists for any continuous function f :(a, b};>JR.
c) Suppose that the Riemaml integral of a tuned.cm f exists. Prove that the set of points of dtscontlnul.ty of f
A~
ls a set of Lebesgue measure zero.
3. Deflne algebraic integer and show that the compla numbers which are algebraic in~gers
....
form a ring whose intersection wlth the rational field ls the ring of rational
~tegers.
.
4. Compute the integral
m
log x dx
0J x2 +a2 , ...
(a> 0)
5. State and prove the Cayley·Hamilton identity between an endomorphism and itS characteristic polynomlaL
6. State and prove Brouwer'a fixed point theorem, using the fundamental facts of homology theory lf you wish.
October 15, 1971 1. Bessel's equation of order '' 11 X2 d2!
dx
+ X ~ + (X2 • y2)y • 0 . ; dx
and lf '' is not an integer then J , the Be11el function of order v, is defined to be II
.
.
.
that solution y • J (x) of BeHel '1 equation which satisfies II
as x .. 0 • Prove the identity x
Ldx JII (x) + v JII(x) • x J11• 1(x) .
either ·by finding the power series expansion for

J; + x • 1., J . eatisfies~Bessel'··· equatiOn of or~r II
II
near x • O
~
J
II
(x) or by verifying that 
11 l · and
by any other method you can 1IUnk of.
bas the correct• behavior
· ·. ·
.}~;'
i...i>
2. Let A denote the ring ~[X]/(X3·1). What are the possible ring homomorphisms f. : A  ~
such that f(l) • 1? By using these homomorphisms to determine a map
A  CU , or otherwise, show that lf each of two rational numbers can be written in the form
3 3 3 a + b + c  3 abc with a, b, c rational, then so can their product.
3. Let T
II T II l>1 >
be a bounded linear operator on a Banach space
of T • Prove that
X.• Define the norm
T • >. has an everywhere defined bounded inverse whenever
llTO. continued
i~iathematics
Qualifying Bxamins.tion . ..~~tober 14, 1971 · ·· '
(
. ''
1. Let
u
be
a real valued,
c2 function cm the disc
lzl < 1
satisfying the
Laplace equation
Prove that there ts a function v(x, y) on the same domain such that f(z) • u(x, y) + 1 v (x, y) ls an analytic function of z • x + .iY • To what extent la
v unique?
2. What are the Galola groups over the field of rational numbers of the following four cubic polynomials?
I
\
a) x3  3x+3
c) x3 3x + 1
b) x3sx+2
d) x 3 3x .•
Prove all your Statements, espedally ones about trreduclbillty.
3. Let ;.2 denote the set of all infinite sequences a , a , a ,... of real numbers 1 2 3
t;
a! one and conformally onto B •
6. Carefully define the fundamental group
with base point
Xo.
U Yo
11
1
(X,x ) of a topological.space X 0
is another point of X , what can you say about the
relationship between " 1(X, Xo) and
11
(X,y0)7 1
6. a) Let A and
B be two commuting semisimple (1. e. diagomlizable)
linear transformations of a complex vector apace V • Let V • V 1+. ;. +Vr (resp. V • W +•.. + W ) be the decomposition of V into the direct sum of 1 8 elgenspac~
WJ of B) • Show that
V of A (resp. 1
wj •
r
!'
l•l
wJnv1
b) Show tbl!t there ls a neighborhood V
of the identit"f ln the unitary
group U(D) With the follrwtng property: If h€ V and g € U(D) and g commutes With h • lgh, then g commutes with h •
'.....
MATHEMATICS QUALIFYING EXAM May 14,1971
2
1. Compute the homology of
2.
tr. .. X ,
when
X•
{s 1 ,z2 ls~ +z:
• 0} •
Prove that every finite 1roup haa a unique l&rgeat normal aolvable aubgroup. Show by ezample that thia aaaertion would be falae if the word 11 aolvable" were replaced by the word "abelian".
3. a.) If
f{z)
ia analytic for
lr(f)
=JJ
I• I < R
what ia the 1eometric meaning of
2 lf'{z) 1 dxdy
(z • s+iy, 0 < r < R)
lzl 0 • d) Give an example of a compact B.iemannian surface which is diffeomorph to neither a sphere nor a torus.
;
..
MATHEMATICS QUALIFYING EXAM May 13,1971
Which le ha :rder to compute 1
1. Define the homotopy group• of a apace.
3 homotopy or homology? Why? What are the groups ,,. (S ) and 2
11'
3
2 (S )?
How woulcl you ao about proving your answer•?
2.
Let
be a metric 1pace with metric d • A fu.nction f : M> JR is
M
called lower semi continuous 1l and only if (Yx E M)(Yc>O) (3:6> O)(Yyi l.!) d(x,.y) < 6 • £(y)> f{x) e
Prove that every
l~wer
aemi continuous, everywhere positive function
.
.
f: M!> JR is th• !Unit pointwise of an increasing sequence at .positiv~ contuu,1ou1 function• g : M
n
~
JR, and conversely,
Hint: Consider, for integral values of n inf { f(y} +nd(x,y>} y
3. Assume that
f(z,t} depends holomorphica.lly on z and continuously
on both variables in
a:
X
JR • Assume that there ii no value of
that f{z,t) is identically zero in z.
of
a:.
For each fixed value of
zeroe1 of function of
f(z,t) lying in t
E
m..
Q.
t
1
Let let
such
be a bounded open subset
Q
n(t)
Prove that
t
denote the number of
n(t) ii a lower sem.icontinuou1:
Mathematic• Qualifying
4. Let
Es&m/ May 13, 1971/ pa1e two
CD denote the field of rational number• md Call a complex number a
complex number•.
E
C the field of
ct con1tructible
if there exi1t1 a aeque:a.ce of field•
auch that each
a
£
Fn
and auch that
F 1+ /Fi
ii a quadratic ex:ten1ion for
i.
a) Let a£ G:. Show that if [EPARTMENT OF MATHEMATICS QUAUFYINO EXAMINATION MAY 1970
m 1. a.
Ciive the ba•ic definition• and theorem• of Oalob Theory (without
proofs). b. Prove that for every finite 1roup aroup bomorpht.c to
0
, there exi•t• a Oalois
Ci.
2. State and prove the Hahn Banach theorem about the extension of continuous linear functional.a.
3. Let
V
be an nclimensional vector apace over the reals. Let
be a linear transformation from (I
= identity) • Prove : n
· 4. Let
V
to
V
•uch that
T
2
T
= 1
ia even.
a. : JR __. JR be continuous and increasing. Show that there
ia a Borel measure
closed intervals
u
such that
µ([a, b]) = a.(b)  a. (a)
for au·
[a, b] •
./
\~
5. Let
A
be a tmiqu• factorization domain •uch that
invertible in
A •
Let a E A
i• an intearally closed
the psoJ'• ge&Leral caee in which the
domain. (You may prove
exponent 2 is replaced by an arbitrary positive integer
6. a. For the initial condition• f(t)
satisfying · If(t) I
x(t)
of the
1
x(O) • 1
x(TJ
Find the least such
and
x"(t)
= 1 , x'(T)
T •
c. Prove your answer correct••
n.)
x'(O) • 0 , find an
such that, for some
differential •Clu&tion
conditions aatisfie• b.
i
is
be a noninvertible equare free
B • A[X]/QC 2 a)
element. Show that
Z • 1A
T > 0, the solution
= f( t) for the stated initial
=o.