Master Algèbre géométrie et théorie des nombres Final exam of differential geometry Lecture notes allowed

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1 Université de Bordeaux U.F. Mathématiques et Interactions Master Algèbre géométrie et théorie des nombres Final exam of differential geometry Lecture notes allowed Exercise 1 We call H (like Heisenberg) the manifold R 3 endowed with the following group structure: (x, y, z).(x, y, z ) = (x + x, y + y, z + z + xy ). We denote by e the neutral element of H. ) } 1) Show that H can also be seen as, (x, y, z) R 3 deduce that H is a Lie group. Compute its centre Z. The map {( 1 x z 0 1 y F : (x, y, z) ( 1 x ) z 0 1 y is injective, smooth and satisfies F ((x, y, z).(x, y, z ) = F (x, y, z)f (x, y, z ). It implies that H can be seen as a closed subgroup of GL(3, R) and is therefore a Lie group. { 1 0 z } The center of H is clearly 0 1 0, z R ) Show that X = x, Y = y + x z and Z = z are left invariant vector fields on H. Compute their brackets. What is the flow of Y? What is exp H (Y )? For any (a, b, c) R 3, there exists a unique vector field T such that T e = a b + c. For any g = (x, y, z) H we compute the Jacobian of L y z g at e: JL g e = x 1 It follows that T g = dl g.t e = a x + b y + (c + xb) z. x + Taking (a, b, c) = (1, 0, 0), (0, 1, 0) and (0, 0, 1) respectively answers the first question. Let f C (H), we have [X, Y ].f = X(Y.f) Y (X.f) = x ( f y + x f z ) ( y + x z )( f x ) = f z Therefore [X, Y ] = Z. Likewise, [X, Z] = 0 and [Y, Z] = 0. The flow of Y is the solution of the Cauchy problem c (t) = Y (c(t)) and c(0) = 1

2 (x, y, z). Therefore φ Y t (x, y, z) = (x, y + t, z + xt). We also have φ Y 1 = R exph (Y ) and therefore exp H (Y ) = (0, 1, 0). 3) Give a basis of the space of left invariant 1-forms of H. Deduce the sets of left invariant 2-forms and 3-forms. The space of left invariant 1-forms of H is the dual of the space of left-invariant vector fields on H. Therefore it is spanned by dy, dx and dz xdy, the elements of the dual basis of (X, Y, Z). The space of left invariant 2-forms is also 3 dimensional, it is spanned by dx dy, dy dz = dy (dz xdy) and dx (dz xdy). The space of left invariant 3-forms is 1 dimensional, it is spanned by dx dy (dz xdy) = dx dy dz. 4) Let Γ be a discrete subgroup of a Lie group G. Show that the action of Γ on G by left translations is free and properly discontinuous. The action is obviously free. Let us prove that it is properly discontinuous. Let A be a compact subset of H. Let g Γ such that g.a A. Clearly g AA 1 = {g.h 1 g A, h A}. By continuity of the map (g, h) gh 1, the set AA 1 is also compact. The subgroup Γ being discrete, it follows that Γ AA 1 is finite and therefore {g Γ g.a A } is also finite. 5) Let Γ = {(a, b, c) H (a, b, c) Z 3 } acting on H by left translation. Show that every Γ-orbit contains a point of [0, 1] [0, 1] [0, 1]. Prove that Γ \H is (in a natural way) a compact manifold. Let (x, y, z) be a point of H. We denote by x the integer part of x ie the integer such that 0 x x < 1. The element ( x, y, 0) belongs to Γ therefore (x x, y y, z x y) is in the Γ-orbit of (x, y, z). Likewise, (x x, y y, z x y (z x y) ) which is clearly in [0, 1] 3 is also in the Γ-orbit of (x, y, z). By the former question, the action of Γ on H is free and properly discontinuous, therefore Γ \H has a unique manifold structure such that the canonical projection p : H Γ \H is a smooth covering. We just proved that Γ \H = p([0, 1] 3 ), it is therefore compact. 6) Show that the map π 0 : H R 2, (x, y, z) (x, y) induces a smooth submersion π between Γ \H and the torus R 2 / Z 2. We denote by q the canonical projection from R 2 to R 2 /Z 2. The map q π 0 is a smooth submersion. For any g Γ and any (x, y, z) H, we have q π 0 (g.(x, y, z)) = q π 0 (x, y, z). Therefore there exists a map π : Γ \H R 2 / Z 2 such that q π 0 = π p. Using local sections of p we see that π is also a submersion (it reads locally as a composition of submersions). 7) Prove that the action of Γ on H induces an action of Z 2 Γ/ (Γ Z) on (Γ Z) \H R 2 R/ Z. Find the expression of the action. Deduce that Γ \H is diffeomorphic to Z 2 \(R 2 R/ Z ). 2

3 The map Γ Z 2 defined by (a, b, c) (a, b) is a surjective group homomorphism. Its kernel is Γ Z = {(0, 0, c) c Z}. For any (x, y, z) H and c Z, (0, 0, c).(x, y, z) = (x, y, z+c) therefore (Γ Z) \H is diffeomorphic to R 2 R/ Z. Let g Γ and [h] an element of (Γ Z) \H (we denote by [.] the class modulo Γ Z). In order to define an action of Γ on (Γ Z) \H we set g.[h] = [g.h]. Is it well defined? Yes, if [h] = [h ] then h = h.d with d H Z; it follows that g.h = g.h.d = d.g.h (as d Z) ie [g.h] = [g.h ] and the action is well defined. Clearly Γ Z is in the kernel of this action therefore this action is in fact an action of Γ/ (Γ Z). Denote by z the class of z in R/Z. If we identify (Γ Z) \H with R 2 R/Z and Γ/ (Γ Z) with Z 2 then for any (x, y, z) H, [x, y, z] = (x, y, z) and for any (a, b, c) Γ, [a, b, c] = (a, b). [a, b, c].[x, y, z] = [x + a, y + b, z + c + ay] = (x + a, y + b, z + c + ay) = (x + a, y + b, z + ay) = (a, b).(x, y, z). This action is clearly free and properly discontinuous therefore the quotient space is smooth. As Γ Z Γ, we have a natural map Φ 0 from (Γ Z) \H to Γ \H induced by p. It is clearly a local diffeomorphism. By definition of the action of Z 2, Φ 0 ((a.b).(x, y, z)) = Φ 0 (x, y, z). It means that it induces a map Φ from Z 2\(R 2 R/ Z ) to Γ \H. The map Φ is a local diffeomorphism and it is not difficult to see that it is bijective. Remark that π Φ x, y, t = (x, y). 8) Use the diffeomorphism above to find local sections of π and prove that π : Γ \H R 2 / Z 2 is a circle bundle (ie a fibre bundle with fibre R/Z). Is this bundle trivial? The map q (see question 6) is a smooth covering. Hence, for any (x 0, y 0 ) in R 2 /Z 2 there exists a local section s 0 of q defined on a neighborhood U 0 of it. Denote by r the natural projection from R 2 R/Z to Z 2\(R 2 R/ Z ). The map r is a smooth covering. The map σ 0 defined by (x, y) r(s 0 (x, y), 0) is then a local section of π Φ (and Φ 1 σ 0 is a section of π). Indeed, for any (x, y) U 0, we have π Φ σ 0 (x, y) = (x, y). The map Ψ between U 0 R/Z and (π Φ) 1 (U 0 ) given by Ψ((x, y), t) = r(σ 0 (x, y), t) is a diffeomorphism (the restriction of r to σ 0 (U 0 ) R/Z is injective). It satisfies p 1 = (π Φ) Ψ, where p 1 is the first projection from U 0 R/Z to U 0. Thus π Φ (and π) is a circle bundle. The group Γ is the fundamental group of Γ \H and is not Abelian. Therefore Γ \H is not homeomorphic to R 2 /Z 2 R/Z whose fundamental group is Abelian. 9) First do exercise 2 Prove that (x, y, z) x induces a map π : Γ \H R/ Z which is a fibre bundle with fibre R 2 /Z 2. 3

4 Exercise 2 Let SU(2) = {A GL(2, C) t AA = Id and det(a) = 1}, SO(3) = {A GL(3, R) t AA = Id and det(a) = 1} and S n be the the unit sphere of R n+1 endowed with its canonical scalar product.,.. 1) a) Show that SU(2) is a closed Lie subgroup of GL(2, C). Compute its tangent space at Id (recall that the differential of the map determinant at Id is the trace). What is its dimension? Give a basis of its Lie algebra su(2). SU(2) is stable by product and inverse so it is a subgroup of GL(2, C). It is closed as the maps A t AA and det are continuous. By Cartan s theorem, SU(2) is therefore a closed Lie subgroup. If X is an element of its tangent space at Id then there exists a path c in SU(2) such that c (0) = X (and c(0) = Id). By differentiation of the equality t c(t)c(t) = Id at t = 0, we find t X + X = 0 and by derivation of det(c(t)) = 1 we find tr(x) = 0. } i i 0 Therefore X span R {A =, B =, C =. Moreover these matrices are tangent respectively to t, to t 1 0 i 0 0 i cos t sin t sin t cos t cos t i sin t e and to t it 0 i sin t cos t 0 e it which are path in SU(2). { } i i 0 Therefore T Id SU(2) = span R,,. 1 0 i 0 0 i It follows that dim(su(2)) = 3. b) Show that the map SU(2) S 3 C 2 given by A A(1, 0) is a diffeomorphism. We prove that SU(2) acts simply transitively on S 3. For any (a, b) S 3 a b, the matrix SU(2). Therefore SU(2) acts b ā transitively {( on ) S 3 (any point } of S 3 is the orbit of (1, 0)). The stabilizer of 1 c (1, 0) is SU(2) = {Id}. The action is simply transitive. 0 d 2) a) Show that SO(3) is a closed Lie subgroup of GL(3, R). Compute its tangent space at Id. What is its dimension? Give a basis of its Lie algebra so(3). As before, the group SO(3) is clearly a subgroup of GL(3, R) that is closed and is therefore a closed Lie subgroup of GL(3, R) by Cartan s theorem. As above, we find that the element of T Id SO(3) are skew symmetric and trace free ie T Id SO(3) span { X = ( 0 1 ) , Y = ( 0 0 ) 1 ( 0 0 )} , Z = Again we can find paths showing that we have in fact equality. Therefore dim(so(3)) = 3. b) Let T 1 S 2 = {(x, v) S 2 S 2 x, v = 0}. Show that SO(3) is diffeomorphic to T 1 S 2. 4

5 Again we prove that the group acts simply transitively. For any (x, v) T 1 S 2, the matrix A whose columns are x, v and X v is in SO(3) therefore SO(3) acts transitively. The matrix A is the only one in SO(3) sending ((1, 0, 0), (0, 1, 0)) on (x, v). c) Find a closed subgroup H of SO(3) such that S 2 is diffeomorphic to SO(3)/H. Deduce that SO(3) is connected (hint: take a continuous function f : SO(3) {±1} and prove it is constant). The group SO(3) acts transitively on S 2. Denoting by H, the stablilizer of (1, 0, 0) we therefore have S 2 SO(3)/H. The group H is clearly connected. If f is a continuous map from SO(3) to {±1} then it is constant on H (and every gh) therefore there exists a continous map f : S 2 SO(3)/H {±1} such that f p = f, where p is the canonical projection from SO(3) to S 2. The sphere being connected f is constant and so f is constant. 3) a) Find a Lie algebra isomorphism Ψ between su(2) and so(3) (recall that the brackets on gl(2, C) and gl(3, R) are given by [A, B] = AB BA). We first compute the brackets: [A, B] = 2C; [A, C] = 2B; [B, C] = 2A [X, Y ] = Z; [X, Z] = Y ; [Y, Z] = X Let Ψ be the linear from su(2) to so(3) defined by Ψ(A) = 2X, Ψ(B) = 2Y and Ψ(C) = 2Z. It is clearly a linear isomorphism. It is also a homomorphism of Lie algebra as Ψ([A, B]) = Ψ(2C) = 4Z = [Ψ(A), Ψ(B)] Ψ([A, C]) = Ψ( 2B) = 4Y = [Ψ(A), Ψ(C)] Ψ([B, C]) = Ψ(2A) = 4X = [Ψ(B), Ψ(C)] Prove that there exists a Lie group covering p : SU(2) SO(3) (ie a group homomorphism which is also a covering). As SU(2) is simply connected (being diffeomorphic to S 3 ) there exists a Lie group homorphism p : SU(2) SO(3) such that dp Id = Ψ. The map Ψ being an isomomorphism and SO(3) being connected the map p is also a covering. b) Prove that ker(p) is contained in the centre of SU(2) and deduce a upper bound on its cardinal. ker p is discrete and normal and so contained in the center. Indeed, for any h ker p, the map φ : SU(2) ker p defined by φ(g) = ghg 1 is continuous. So φ(su(2)) is connected and then equal to {h}. i 0 The center of SU(2) is {± Id} (the matrices commuting with are 0 i ( diagonal and the only diagonal matrices of SU(2) commuting with are ±Id). Hence ker p ) 5

6 c) For your favorite X 0 su(2), compute exp su(2) (tx) and exp so(3) (t Ψ(X)), for any t R. Deduce from it the cardinal of ker(p) and the fundamental group of SO(3) e exp su(2) (tc) = it 0 0 e it, whereas exp so(3) (t Ψ(C)) = 0 cos(2t) sin(2t). 0 sin(2t) cos(2t) p(exp su(2) (πc)) = exp so(3) (π Ψ(C)) = Id and exp su(2) (πc)) = Id. Therefore ker p = 2 and the fundamental group of SO(3) is Z/2Z {±Id}. 4) a) Show that if there exists a nowhere vanishing vector field X on S 2 then there exist a smooth map X 0 : S 2 S 2 such that x, X 0 (x) = 0, for any x S 2. We see T S 2 as {(x, v) R 3 R 3 x S 2, x, v = 0}. The vector field X is then a map from S 2 to R 3 such that x, X(x) = 0 for any x S 2 and X 0 is just X. X b) Show that that in such a case, the map Φ : S 2 R/Z T 1 S 2 defined by Φ(x, t) = (x, cos(2πt)x 0 (x) + sin(2πt)(x X 0 (x)) is a diffeomorphism (here a b is the vector (or cross) product 1 on R 3 ). The map Φ is smooth and clearly bijective as (X 0 (x), x X 0 (x)) is an orthonormal basis of x. The map R/Z S 1 R 2, t (cos(2πt), sin(2πt)) is a smooth diffeomorphism. We call θ its inverse. Then map Φ 1 (x, v) = (x, θ( X 0 (x), v, x X 0 (x), v )). So Φ 1 is smooth. c) Conclusion? It is absurd because the fundamental group of S 2 R/Z is Z and the one of SO(3) is Z/2Z. Hence every vector field on S 2 vanishes somewhere. 1 more precisely by definition a b is the vector of R 3 such that, for any v R 3, v, a b = det(a, b, v) in an orthonormal basis 6