Practice Qualifying Exam Questions, Differentiable Manifolds, Fall, 2009.

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1 Practice Qualifying Exam Questions, Differentiable Manifolds, Fall, Solutions (1) Let Γ be a discrete group acting on a manifold M. (a) Define what it means for Γ to act freely. Solution: Γ acts freely on M if and only if for any x M, γ x = x implies γ = e the identity map. (b) Define what it means for Γ to act properly discontinuously. Solution: Γ acts properly discontinuously on M if and only if the two following conditions are met: Each x M has a neighborhood U so that {γ Γ : γu U = } is finite. If x, y M and y / Γx, there are neighborhoods U x and V y which satisfy U ΓV =. (c) Let V be a finite-dimensional real vector space equipped with the usual topology. Let v 1, v 2 V. Let Γ be the group Z 2 with addition as the group law. If γ = (n 1, n 2 ) Γ, define the action of γ on V by γ x = n 1 v 1 + n 2 v 2 + x. Find necessary and sufficient conditions on {v 1, v 2 } for the action of Γ on V to be free and properly discontinuous. Solution: Γ acts freely and properly discontinuously on V if and only if {v 1, v 2 } are linearly independent. To prove =, we prove the contrapositive. If {v 1, v 2 } are linearly dependent, we may assume v 1 = λv 2 for λ R. Then there are two cases: λ = p/q is rational. Then qv 1 pv 2 = 0 and Γ cannot act freely. λ is irrational. The Γ-orbit of 0 is contained in the line spanned by v 2. Consider the quotient map π : R R/Z. Then the map φ : Z R/Z given by φ(n) = π(nλ) is injective. Since R/Z is compact, there is an accumulation point lim i φ(n i ) = x R/Z. In other words, if we consider x [0, 1), there are integers n i, m i so that lim i n iλ + m i = x. 1

2 2 Then for p = xv 2 V, we have n i v 1 + m i v 2 = n i λv 2 + m i v 2 p. There are two cases now: If p Γ0, then infinitely many (n i, m i ) p are contained in any neighborhood of p, which violates the first condition of properly discontinuous. On the other hand, if p / Γ0, then 0 and p cannot have neighborhoods U and V so that U ΓV =. To prove =, if v 1, v 2 are linearly independent, they can be extended to a basis {v 1, v 2,..., v n } of V, where n is the dimension. In terms of this basis, introduce the standard Euclidean metric on V (so that v 1 = v 2 = 1 in particular). It is easy to see that the action of Γ is free. To see it is properly discontinuous, note that Γ acts by isometries. For the first condition, take x V and let U be the open ball of radius 1/2 around x. Then {γ Γ : ΓU U = } consists only of the identity element. Secondly, if x, y V and x / Γy, then it is not hard to see that d = min dist(x, γ y) > 0. γ Γ Now choose open balls U, V around x, y respectively, each of radius d/2. Then U ΓV =. So Γ acts freely and properly discontinuously when v 1, v 2 are linearly independent. (2) (a) If U is a manifold and F = U R r is the trivial vector bundle, write down an explicit connection on F. Solution: A section of F is equivalent to an r-tuple of functions e 1,..., e r on U. Define, for s = s α e α and X a vector field on U, X s = (Xs α )e α. It is straightforward to check this is a connection (b) Let M be a smooth manifold of dimension n, and let E be a vector bundle of rank r over M. Use a partition of unity argument to construct a connection over E. Solution: Choose an open cover {U α } of M so E can be locally trivialized over each U α. Choose a local trivialization over each U α and part (a) shows that there is a connection α of E restricted to U α. Now let ρ α be a partition of

3 3 unity subordinate to {U α }. Define on all of M via X s = α ρ α α Xs. Now check is a connection. is C -linear in X, since each α is. is R-linear in s, since each α is. For the Leibniz rule, compute X (fs) = α = α ρ α α X(fs) ρ α (f α Xs + (Xf)s) = f α ρ α α Xs + (Xf)s α ρ α = f X s + (Xf)s since α ρ α = 1. (3) (a) Let M n (C) denote the set of n n complex-valued matrices. Show M n (C) is a manifold of (real) dimension 2n 2. Solution: M n (C) is a complex vector space of dimension n 2, and so by taking real and imaginary parts, is a real vector space of dimension 2n 2. (b) Let f : M n (C) V be defined by f(a) = AĀ, and V = {B M n (C) : B = B }. Show f is a submersion at each A U(n), where U(n) = {A M n (C) : AĀ = I}. Solution: We first check this at A = I: In this case, compute f (C) = d dɛ f(i + ɛc) = C + C. ɛ=0 Therefore, if B V, f ( 1B) = B. Therefore f 2 is onto, and so f is an immersion at A = I. In the more general case of A U(n), compute f (C) = d dɛ f(a + ɛc) = CĀ + A C. ɛ=0 If B V, f ( 1 2 BA) = 1 2 BAĀ A(BA) = 1 2 B B = B since AĀ = I and B = B. Therefore, f is onto at any A U(n) and f is a submersion.

4 (c) Show U(n) is a Lie group. Find its dimension. Solution: U(n) = f 1 (I) is a submanifold of M n (C), since f is a submersion for all A U(n). The dimension is equal to dim M n (C) dim V. To compute dim V, recall it consists of all Hermitian-symmetric matrices in M n (C). As such, we can write a general element of V as a 1 b 12 + ic 12 b 13 + ic b 1n + ic 1n b 12 ic 12 a 2 b 23 + ic b 2n + ic 2n b 13 ic 13 b 23 ic 23 a 3... a 3n + ic 3n b 1n ic 1n b 2n ic 2n b 3n ic 3n... a n This shows the dimension of V is n a j s plus n(n 1)/2 b jk s plus n(n 1)/2 c jk s to make the dimension n 2. Thus the dimension of U(n) = 2n 2 n 2 = n 2. To show U(n) is a group, let A, B U(n). Then compute (AB)(AB) = AB B Ā = AIĀ = AĀ = I. Therefore, AB U(n). Also, let A U(n). Compute (A 1 ) = ((Ā) 1 ) = (Ā ) 1 (note (C 1 ) = (C ) 1 ). This implies A 1 (A 1 ) = I and so A 1 U(n). It is standard that the group law and inverse operation are smooth on nonsingular matrices. (d) Calculate the tangent space T I U(n). Solution: Since U(n) = f 1 (I) for f a submersion, T I U(n) = ker f (I) = {C : C + C = 0} the set of skew-hermitian complex n n matrices. (4) Let (M, g) be a compact Riemannian manifold. If ω = ω i dx i is a one-form, define ω 2 g = g ij ω i ω j for g ij the inverse matrix of g ij. Let f be a function on M. Define E g (f) = df 2 g dv g M for dv g the volume density of the metric g. Let u be a positive smooth function. Assume the dimension of M is 2. Show that E ug (f) = E g (f). Solution: Recall that in local coordinates x 1, x 2, dv g = det g ij dv Eucl, where dv Eucl is the standard volume form on Euclidean space. 4

5 5 Therefore, dv ug = det(ug ij ) dv Eucl = u 2 det g ij dv Eucl = u dv g, since the dimension is 2. On the other hand, the inverse matrix of ug ij is u 1 g ij, and df 2 ug = u 1 g ij i f j f = u 1 df 2 g. Therefore, in any local coordinate chart, the integrands are df 2 g dv g = u 1 df 2 g u dv g = df 2 ug dv ug. So given a partition of unity ρ α subordinate to an atlas of M, we may compute E ug (f) = ρ α df 2 ug dv ug = ρ α df 2 g dv g = E g (f). α α (5) Recall a differential form η on a manifold M is called closed if dη = 0. (a) Let M be a simply connected manifold, let η be a closed one-form on M, and let p M. Show that f(q) = γ η, 1 0 [a,b] is well-defined independently of the interval [a, b] and the smooth path γ : [a, b] M from p to q. (Feel free to use any result proved in class.) Solution: We saw in class that if γ 1 and γ 2 are two homotopic paths from [0, 1] M from p to q, then f(q) is independent of γ 1 or γ 2. Therefore, we need only show f(q) is independent of the choice of interval [a, b]. It suffices to show that if s = s(t) is a diffeomorphism from [0, 1] to [a, b], γ η = f(t) dt, then γ η = 1 0 f(t) dt = b a f(s) ds dt dt = b a (γ s) η. (b) Under the assumptions of part (a), show that η = df. (Hint: In local coordinates, consider paths near q in each coordinate direction.) Solution: To show η = df, then we work in local coordinates {x i } near q. Then if we have η = η i dx i, we need only show that η i = f. Pick an i, and by part (a), we x i may choose a path γ with domain [ 1, 0] from p to q so

6 that near q, the path γ(t) = (c 1,..., t + c i,..., c n ) so that in our coordinates q = (c 1,..., c i,..., c n ). So, near q, γ η = η i dt. Then compute for t small and q t = q + (0,..., t,..., 0), f(q t ) = t 1 γ η, and at q t x f = d i dt f(q t) = d γ η = d η i dt = η i (0) dt 1 dt c for c a small negative constant (corresponding to the part of γ which is constant in the x i direction). (6) (a) Give, without proof, an example of a connected Riemannian manifold (M, g) and points p, q in M so that there is no path from p to q of length equal to the distance from p to q. Solution: Consider M = R 2 \ {(0, 0)}, equipped with the standard metric dx 2 + dy 2 on R 2. Then if p = (1, 0) and q = ( 1, 0), the distance from p to q is 2, but any path in M from p to q has length strictly greater than 2. (b) Give, without proof, an example of a complete Riemannian manifold (M, g) and points p, q M so that there is more than one geodesic from p to q. Solution: If x S n with the standard metric induced from R n+1, then there is an infinite family of geodesics from x to x corresponding to each great circle through x. (c) Give, without proof, an example of a Riemannian manifold (M, g) and a complete geodesic curve which is dense in M. Solution: Recall that geodesics on (R 2, δ) are straight lines. If Z 2 acts on R 2 by translation t (n 1, n 2 ) (x 1, x 2 ) = (n 1 + x 1, n 2 + x 2 ). All theses translations are isometries, and so the flat metric δ descends to the quotient T = R 2 /Z 2 (the standard flat torus). More specifically, let π : R 2 T be the quotient map. Then there is a metric g on T so that δ = π g. The geodesics of g are then all of the form π γ for γ a geodesic of δ on R 2. Recall that any line of irrational slope in R 2 projected to a curve in T with dense image. Any such curve is a dense geodesic in (T, g). 6