Phd past qualifying exams, Exams of Mathematics

Download Phd past qualifying exams and more Exams Mathematics in PDF only on Docsity!

Harvard University PHd

MatHeMatics

QUalifying exaMinations

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 2-sphere, 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) =

p∈A

B(p).

(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 )}.

Show that dH defines a metric on the set X˜ := {A ⊂ X

∣ (^) A is closed and bounded}.

3. (AT) Let T n^ = Rn/Zn, the n-torus. Prove that any path-connected 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 ⊂ P^2 be two plane cubics (that is, curves of degree 3), intersecting transversely in 9 points {p 1 , p 2 ,... , p 9 }. Show that p 1 ,... , p 6 lie on a conic (that is, a curve of degree 2) if and only if p 7 , p 8 and p 9 are colinear.

6. (AG)

Let C be the smooth projective curve over C with affine equation y^2 = f (x), where f ∈ C[x] is a square-free 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^2 n−^1 as the unit sphere in Cn, and consider the inclusions

· · · → S^2 n−^1 → S^2 n+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^2 n−^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

(k + 1)(n − k)

then X does not contain any k-plane Λ ⊂ Pn.

3. (DG) Let H^2 := {(x, y) ∈ R^2 : y > 0 }. Equip H^2 with a metric

gα := dx^2 + dy^2 yα where α ∈ R.

(a) Show that (H^2 , gα) is incomplete if α 6 = 2.

(b) Identify z = x + iy. For

a b c d

∈ SL(2, R), consider the map z 7 → azcz++db. Show that this defines an isometry of (H^2 , g 2 ).

(c) Show that (H^2 , g 2 ) is complete. (Hint: Show that the isometry group acts transitively on the tangent space at each point.)

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 2-sphere, 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 → Rk, v 7 → fi(p(v))Ti(v) which extend (by zero) to all of V and which we denote by fiTi. Together, the fiTi 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 non-zero. 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 R^3 in the usual way, then

T S^2 ⊕ NS 2 = T R^3 |S 2

where NS 2 is the normal bundle to S^2 in R^3. Dualizing we get

T ∗S^2 ⊕ (NS 2 )∗^ = T ∗R^3 |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) =

p∈A

B(p).

(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 )}. Show that dH defines a metric on the set X˜ := {A ⊂ X

A is closed and bounded}.

Solution: We need to show that ( X, d˜ H ) is a metric space. First, since com- 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 ∩ Zc. Since Z is com- pact, it is closed, so there exists r > 0 such that Br(p) ⊂ Zc. 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 ⊂ B 1 (Z), Z ⊂ B 1 (Y ), Z ⊂ B 2 (W ), W ⊂ B 2 (Z)

Then dH (Y, Z) <  1 , dH (Z, W ) <  2. Let us prove that Y ⊂ B 1 + 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 ∈ N 1 (z). By our choice of  2 there exists a point w ∈ W such that z ∈ B 2 (w). Then

d(y, w) ≤ d(y, z) + d(z, w) ≤  1 +  2 so y ∈ B 1 + 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 n-torus. Prove that any path-connected 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 ⊆ π 1 (T n). We have π 1 (T n) = π 1 (S^1 ) × · · · × π 1 (S^1 ) = Zn, and Zn^ is acting on Rn^ by translation. Thus, G ⊆ Zn^ is free. Choose a Z-basis (v 1 ,... , vm) of G, and consider the (real!) change of basis taking (v 1 ,... , vm) to the first m standard basis vectors (e 1 ,... , 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.

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⊇I p prime

p.

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 /∈

1. 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 ) ∈/ 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 (

(^0) S ) = π−^1

p⊆S prime

p

p⊆S prime

π−^1 (p) =

q⊇I q prime

q.

2. (DG) Let c(s) = (r(s), z(s)) be a curve in the (x, z)-plane which is parame- terized 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 (s, θ) = (r′(s) cos θ, r′(s) sin θ, z′(s)) ∂ϕ ∂θ

(s, θ) = (−r(s) sin θ, r(s) cos θ, 0)

The coefficients of the first fundamental form are:

E = r′(s)^2 + z′(s)^2 = 1, F = 0, G = r(s)^2

Curvature: K = −

G

∂^2

∂s^2

G = −

r′′(s) r(s) To get K = −1 we need to find r(s), z(s) such that

r′′(s) = r(s), r′(s)^2 + z′(s)^2 = 1.

A possible solution is r(s) = e−s^ with

z(s) =

1 − e−^2 tdt = Arcosh(r−^1 ) −

1 − r^2.

3. (RA) Let f ∈ L^2 (0, ∞) and consider

F (z) =

0

f (t)e^2 πiztdt

for z in the upper half-plane.

(a) Check that the above integral converges absolutely and uniformly in any region Im(z) ≥ C > 0. (b) Show that sup y> 0

0

|F (x + iy)|^2 dx = ‖f ‖^2 L (^2) (0,∞).

Solution: For Im(z) ≥ C > 0 we have

|f (t)e^2 πizt| ≤ |f (t)|e−^2 Cπt

thus with the Cauchy–Schwarz inequality ∫ (^) ∞

0

|f (t)e^2 πizt|dt ≤

0

|f (t)|^2 dt

0

e−^4 Cπtdt

Solution: The universal covers of RP 2 and RP 3 are S^2 and S^3 , respectively. Moreover, these covers are both 2-sheeted. Hence, we have

π 1 (X) = π 1 (RP 3 ) × π 1 (S^2 ) = π 1 (RP 3 ) = Z/ 2 Z

π 1 (Y ) = π 1 (RP 2 ) × π 1 (S^3 ) = π 1 (RP 2 ) = Z/ 2 Z.

Also, πk(RP j^ ) = πk(Sj^ ), for k > 1, j = 2, 3 so that

πk(X) = πk(S^2 ) × πk(S^3 ) = πk(Y ), k > 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 well-known singular homology groups (with integral coefficients)

H 0 (Sn) = Hn(Sn) = Z, Hi(Sk) = 0, i 6 = 0, n,

H 0 (RP 2 ) = H 2 (RP 2 ) = Z, H 1 (RP 2 ) = Z/ 2 Z, Hi(RP 2 ) = 0, i 6 = 0, 1 , 2 H 0 (RP 3 ) = Z, H 1 (RP 3 ) = Z/ 2 Z, Hi(RP 3 ) = 0, i 6 = 0, 1

Now, the Kunneth theorem in singular homology (with Z-coefficients) gives an exact sequence

i+j=

Hi(RP 3 )⊗Z Hj (S^2 ) → H 2 (X) →

i+j=

T or 1 (Hi(RP 3 ), Hj (S^2 )) → 0

Since Hk(S^2 ) is free, for every k, we have

H 2 (X) '

i+j=

Hi(RP 3 ) ⊗Z Hj (S^2 ) = Z

Similarly, we compute

H 2 (Y ) '

i+j=

Hi(RP 2 ) ⊗Z Hj (S^3 ) = Z/ 2 Z.

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 H 2 i(B) = Z, for i = 0,... , 3, and Hk(B) = 0 otherwise.

The Kunneth theorem for singular cohomology (with Z-coefficients), combined with the fact that Hk(Sn) is free, for any k, gives

Hk(A) '

i+j=k

Hi(S^2 ) ⊗ Hj (S^4 ).

Hence, H 2 i(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 C^4 − { 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 π 3 (S 1 ) = π 4 (S^7 ), shows that π 4 (CP 3 ) = 0. How- ever, π 4 (A) = π 4 (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 square-free 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 Riemann-Hurwitz: the 2 : 1 map from C to the x-line is ramified above the roots of f and nowhere else (not even at infinity because deg f is even), so

2 − 2 g(C) = χ(C) = 2χ(P^1 ) − deg P = 4 − 2 n,

whence g(C) = n − 1. For the second, let ω 0 = 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 ′(x) dx takes care of the Weierstrass points.) Hence the space of holomorphic differentials contains

Ω := {P (x) ω 0 | deg P < g},

which has dimension g. Thus Ω is the full space of differentials, with basis {ωk = xkω 0 , k = 0,... , g − 1 }.

The fiber of Γ over the point [Λ] ∈ G(k, n) is just the subspace of PN^ corre- sponding to the vector space of polynomials vanishing on Λ; since the space of polynomials on Pn^ surjects onto the space of polynomials on Λ ∼= Pk, this is a subspace of codimension

(k+d k

in PN^. We deduce that

dim Γ = (k + 1)(n − k) + N −

k + d k

in particular, if the inequality of the problem holds, then dim Γ < N , so that Γ cannot dominate PN^.

3. (DG) Let H^2 := {(x, y) ∈ R^2 : y > 0 }. Equip H^2 with a metric

gα := dx^2 + dy^2 yα where α ∈ R.

(a) Show that (H^2 , gα) is incomplete if α 6 = 2.

(b) Identify z = x + iy. For

a b c d

∈ SL(2, R), consider the map

z 7 → azcz++db. Show that this defines an isometry of (H^2 , g 2 ).

(c) Show that (H^2 , g 2 ) 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) = (^) ∂y∂. In order for (H^2 , 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

1. Thus ( ˙y)

2 yα^ = 1, or in other words,

y˙ = yα/^2.

If α 6 = 2, then the solution to this ODE is

y(t) =

α 2

)t + 1

) 1 /(1− α 2 )

thus, this geodesics persists only as long as (1 − α 2 )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 g 2 = 4 dz ⊗ dz¯ |z − z¯|^2 Let A ∈ SL(2, R), we compute

A∗dz =

adz cz + d − c

(az + b)dz (cz + d)^2 = (ad − bc)

dz (cz + d)^2

dz (cz + d)^2

and so A∗dz¯ = (^) (c¯zd+z¯d) 2. It remains to compute

A∗z − A∗^ ¯z = az + b cz + d

a¯z + b c¯z + d

z − z¯ |cz + d|^2

where we have used that A ∈ SL(2, R). Putting everything together we get

A∗g 2 = 4 dz ⊗ d¯z |cz + d|^4

|cz + d|^4 |z − ¯z|^2

= g 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 γ 0 (t)– through the point (0, 1) in the direction (0, 1) exists for all time. Let z = x + iy be any point in H^2. 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 TiH^2 , for then the geodesic through i with tangent vector V will be nothing but A−^1 (γ 0 (t)), and hence will exist for all time. First, observe that A(i) = i, if and only if A =

a b −b a

Consider the rotation matrix

A =

cos θ − sin θ sin θ cos θ

A straightforward computation shows that, in complex coordinates,

A∗V =

(cos θ + i sin θ)^2

V = e−^2

√ − 1 θV,

that is, A∗ : TiH^2 → TiH^2 acts as a rotation. Since θ is arbitrary, and the rotations act transitively on S^2 , we’re done.

4. (RA)

49. In particular, we have g 3 g 7 g 3 − 1 = gj 7 for some nonzero j ∈ Z/7. Now we use the order of g 3 :

g 7 = g 3 g 3 g 3 · g 7 · g− 3 1 g 3 − 1 g 3 −^1 = g 3 g 3 (g 7 j)g 3 − 1 g 3 −^1 = g 3 (gj

2 7 )g

− 1 3 = gj

3 7 , 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 g 3 g 7 g− 3 1 = g^27 , then by replacing the generator g 3 with g^23 we instead see

g^23 g 7 (g 32 )−^1 = g 3 g^27 g 3 − 1 = g^47.

We have the following conjugacy classes of elements:

• {e} forms a class of its own.
• {g 7 , g^47 , g^27 } and {g^37 , g^57 , g 76 } 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 xg 3 x−^1 = g 32 , since G has only elements of odd order. Hence, there are two final conjugacy classes, each of size 7: those elements conjugate to g 3 and those conjugate to g^23.

These five conjugacy sets give rise to five irreducible representations, which must be of dimensions 1, 1, 1, 3, and 3 (since these square-sum 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(10|z|) for all z ∈ C.

Solution: The functions satisfying these conditions are precisely the C-linear combinations of e−^2 πiz^ , 1, and e^2 πiz^. Indeed such f is readily seen to satisfy the two conditions. Conversely (1) means that f descends to a function of q := e^2 πiz^ ∈ C∗, say f (z) = F (q), and then by (2) there is some M ′^ such that |F (q)| ≤ M ′^ max(|q|−^5 /π, |q|^5 /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 R^3 where the coordinates (x, y, z) obey x^2 + y^2 = 1 + z^2. 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 Levi-Civita

connection of the metric that you computed in Part a).

PROBLEM 2 (T)

Let X be path-connected and locally path-connected, and let Y be a finite Cartesian

product of circles. Show that if π 1 (X) is finite, then every continuous map from X to Y is

null-homotopic. (Hint: recall that there is a fiber bundle Z → R → S^1 .)

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 = (z^6 + 1)/z^3.

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.