Smooth Dynamics 2. Problem Set Nr. 1. Instructor: Submitted by: Prof. Wilkinson Clark Butler. University of Chicago Winter 2013

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1 Smooth Dynamics 2 Problem Set Nr. 1 University of Chicago Winter 2013 Instructor: Submitted by: Prof. Wilkinson Clark Butler

2 Problem 1 Let M be a Riemannian manifold with metric, and Levi-Civita connection. Fix the standard identification of T T M with 3 T M: Problem 1.1 If θ is the canonical 1-form on the cotangent bundle T M, and I : T M T M is the isomorphism v v,, then I θ = α, where α is given by α(u, v, w) = u, v Solution: Let [β] be a tangent vector in T T M, given as an equivalence class of curves, with β(0) = u, (π β) (0) = v, ( (π β) (0)β)(0) = w. Then I θ(β (0)) = θ(i(β (0))) = I(β (0))((π β) (0)) = β(0), (π β) (0) = u, v where we have used the fact that π β = π I β, since I is a fiber preserving isomorphism. Problem 1.2 Let ξ i = (u, v i, w i ), i = 1, 2, then up to sign, dα(ξ 1, ξ 2 ) = v 2, w 1 v 1, w 2 Solution: We use the following formula: for a 1-form ω and smooth vector fields X i, i = 1, 2, dω(x 1, X 2 ) = L X1 ω(x 2 ) L X2 ω(x 1 ) ω([x 1, X 2 ]) So let X i be any smooth vector fields on T M with X i (u) = ξ i for i = 1, 2. We can write X i = (U i, V i, W i ) where U i = π(x i ), V i = Dπ(X i ), and W i = Vi U i. We now perform a pair of computations. Let β 1 : ( ε, ε) T M be an integral curve of the vector field X 1 with β 1 (0) = u. Observe that U 2 (β 1 (t)) = π(x 2 (β 1 (t)) = β 1 (t), and that β 1 and V 2 β 1 := Y 2 are vector fields along the curve π β 1. Lastly, let p = π(u). Then, since is a Levi-Civita connection, X 1 (α(x 2 ))(u) = d dt β 1 (t), Y 2 (t)) t=0 Next, = ( (π β1 ) (0)β 1 )(0), Y 2 (0) + u, ( (π β1 ) (0)Y 2 )(0) = w 1, v 2 + u, ( v1 Y 2 )(0) X 1 (X 2 )(u) = d dt X 2 (β 1 (t)) t=0 and so we compute d dt X 2 (β 1 (t)) = d t=0 dt (β 1 (t), Y 2 (t), (W 2 β 1 )(t))) t=0 = (u, (π β1 ) (0)Y 2 (0), (π β1 ) (0)(W 2 β 1 )(0)) = (u, v1 Y 2 (u), v1 (W 2 β 1 )(0)) 1

3 Analogous computations give the same results with X 1 and X 2 flipped (flipping all the numbers). Hence we compute [X 1, X 2 ](u) = (u, v1 Y 2 (0), v1 (W 2 β 1 )(0)) (u, v2 Y 1 (0), v2 (W 1 β 2 )(0)) = (u, v1 Y 2 (0) v2 Y 1 (0), ( V1 W 2 V2 W 1 )(u)) using the symmetry of the Levi-Civita connection. Therefore we conclude that and therefore, as desired. X 1 (α(x 2 )) = w 1, v 2 + u, ( v1 Y 2 )(0) X 2 (α(x 1 )) = w 2, v 1 + u, ( v2 Y 1 )(0) α([x 1, X 2 ]) = u, v1 Y 2 (0) v2 Y 1 (0) dα(ξ 1, ξ 2 ) = dα(x 1, X 2 )(u) = ( w 1, v 2 + u, v1 Y 2 (0) ) ( w 2, v 1 + u, v2 Y 1 (0) ) ( u, v1 Y 2 (0) v2 Y 1 (0) ) = w 1, v 2 w 2, v 1 Problem 1.3 The vector field ϕ on T M given by generates the geodesic flow. ϕ(u) = (u, u, 0) Solution: Let u T M. There is a unique geodesic γ with γ(0) = u. If ϕ t denotes the time t map of the geodesic flow, then ϕ t (u) = γ(t). Note γ = π γ. Then we have d dt ϕ t (u) = d t=0 dt γ(t) = ( γ(0), (γ) (0), ( γ γ)(0)) = (u, u, 0) t=0 so that flow. d dt t=0 ϕ t (u) = ϕ(u) and thus the vector field ϕ on T M generates the geodesic Problem 1.4 The restriction of the geodesic flow ϕ to the unit tangent bundle T 1 M preserves the restriction of α to T (T 1 M). Solution: Let ᾱ denote the restriction of α to T (T 1 M). To show that ᾱ is preserved by ϕ, it suffices to prove that L ϕ ᾱ = 0. By Cartan s formula, L ϕ ᾱ = d(ι ϕ ᾱ) + ι ϕ dᾱ We now show that each of the 1-forms on the right is zero. Observe that ι ϕ ᾱ(u) = ᾱ( ϕ(u)) = u, u = 1 for any u T 1 M. Hence the function ι ϕ ᾱ is constant on T 1 M and therefore d(ι ϕ ᾱ) = 0. 2

4 Now observe that ᾱ = i (α), where i : T 1 M T M is the inclusion. Hence dᾱ = d(i α) = i (dα). Thus, for any vector (u, v, w) T (T 1 M), ι ϕ dᾱ((u, v, w)) = dα( ϕ(u), (u, v, w)) = u, w = 0 since for (u, v, w) T (T 1 M), we have u, w = 0. Hence L ϕ ᾱ = 0 and so ᾱ is preserved by the geodesic flow ϕ. Problem 1.5 α is a contact 1-form on T 1 M. Solution: Let n = dim M. We must prove that α (dα) n 1 is a volume element on T 1 M, i.e. α (dα) n 1 is nonvanishing on T 1 M. Let (x 1,..., x n ) be local coordinates on M, and let (x 1,..., x n, y 1,..., y n ) be local coordinates on T M coming from a local trivialization. We can take this trivialization to be orthogonal, i.e., to be inner product preserving on the fibers. We will evaluate α in this local coordinates. For a curve β : ( ε, ε) T M, β(t) = (a 1 (t),..., a n (t), b 1 (t),..., b n (t)) in local coordinates, and β(t) = n ȧ i (t) n x i + ḃ i (t) y i i=1 i=1 The horizontal component of β(t) is (a 1 (t),..., a n (t), ȧ 1 (t),..., ȧ n (t)). α evaluated on β(0) is the inner product in T M of the horizontal component of β(t) at t = 0 with the tangent vector β(0). Hence in local coordinates, α( β(0)) = n ȧ i (0)b i (0) i=1 Hence we conclude that in these coordinates, α = n y i dx i i=1 and therefore Hence, we easily compute that dα = n dy i dx i i=1 Finally, (dα) n 1 = α (dα) n 1 = n (dy i dx i ) j=1 i j n j=1 y j dx j i j (dy i dx i ) This form is nonzero if and only if y j 0 for some j. If we restrict to T 1 M, this imposes the restriction that n j=1 (yj ) 2 = 1, which forces that y j 0 for some j. So the restriction 3

5 of α (dα) n 1 is nowhere vanishing and therefore α is a contact 1-form on T 1 M. As a last note, by moving dx j to the right, we can write this form as n α (dα) n 1 = ( 1) n 1 j=1( 1) j+1 y ( n ) j dy i dx i i j i=1 Up to sign, this is the product of the standard volume form on S n 1 R n with the standard volume form on R n. Pulling back to T 1 M, this shows that α (dα) n 1 is locally the product of the Riemannian volume on M with the volume on S n 1 in the tangent space given by the Riemannian structure. Problem 2 Problem 2.1 Let M be compact without conjugate points. Prove the limits defining the Busemann functions on M exist, and that b ± v (π(v)) = 0 b ± v (p) b ± v (q) d(p, q) Solution: We will prove these properties for b + v, as exactly analogous proofs work for b v. For v T 1 M, let b + v,t (p) = d(γ v(t), p) t where γ v is the unique parametrized geodesic with γ v (0) = v (note γ v is unit-speed). Since M has no conjugate points, the metric universal cover M is diffeomorphic to R n via the exponential map at any point, where n = dim M, and further, in M any two points are connected by a unique geodesic which realizes the minimal distance between those two points. For s t 0, we have by the triangle inequality, Rearranging this gives d(γ v (s), p) d(γ v (t), p) d(γ v (s), γ v (t)) = s t d(γ v (s), p) s d(γ v (t), p) t which implies that b + v,s b + v,t. Further, for any t 0, we have d(γ v (t), p) t = d(γ v (t), p) d(γ v (t), γ v (0)) d(γ v (0), p) again by the triangle inequality. So for each p, the function t b + v,t (p) is decreasing and bounded below. Hence the limit lim b + v,t (p) exists for each p M, and the Busemann function b + v is thus well-defined. For each t, we have d(γ v (t), π(v)) = d(γ v (t), γ v (0)) = t and therefore b + v,t (π(v)) = 0 for every t, and hence b+ v (π(v)) = 0. Also, b + v,t (p) b+ v,t (q) = d(γ v(t), p) d(γ v (t), q) d(p, q) 4

6 for any t, p, q M, and therefore taking t, we conclude that b + v (p) b + v (q) d(p, q) Problem 2.2 b ± ϕ t(v) = b± v t Solution: Let v T 1 M. For any p M, we calculate b + ϕ s(v) (p) = lim d(γ ϕ s(v)(t), p) t = lim d(γ v (t + s), p) t = ( lim d(γ v (t + s), p) (t + s)) + s = b + v (p) + s and similarly, b ϕ s(v) (p) = lim d(γ ϕ s(v)( t), p) t = lim d(γ v ( t + s), p) t = ( lim d(γ v ( t + s), p) (s t)) + s = b v (p) + s (I am not sure how I am computing these signs incorrectly.) Problem 3 I will skip the verification of the curvature of H, the isometries of H and D, and the geodesics of these models. I ll do the identification of PSL(2, R) with T 1 H. Problem 3.1 The stabilizer of a point under the action of PSL(2, R) is the compact subgroup K = SO(2)/{±I}, so there is an identification of H with the coset space of K, H = PSL(2, R)/K. Solution: PSL(2, R) consists of all orientation-preserving isometries of H, acting by Mobius transformations. These isometries consist of 3 types: translations along a geodesic, limit rotations about a point on H, and rotations about a point in H. Of these, only rotations fix a point in H. So the stabilizer of a point is the group of all rotations about that point, and since a rotation by any angle can be realized by a hyperbolic isometry, this subgroup is isomorphic to SO(2)/{±I}. Problem 3.2 The derivative action of PSL(2, R) on T 1 H is free and transitive and gives an analytic identification between T 1 H and PSL(2, R). Under this identification, the action of PSL(2, R) on T 1 H by isometries corresponds to left multiplication in PSL(2, R). Solution: We first show the derivative action of PSL(2, R) is transitive. PSL(2, R) acts transitively on H by isometries, so it suffices to restrict to a single point (say i) and prove 5

7 that PSL(2, R) acts transitively on T 1 i H, the fiber of T 1 H over i. But PSL(2, R) has a subgroup which acts by rotations about i, namely the collection of isometries z cos θz + sin θ sin θz + cos θ for θ R, corresponding to a rotation of angle θ. In particular, for θ [0, 2π), this will rotate the vertical unit speed geodesic t e t i through i to the unit speed geodesic which makes an angle θ with this vertical geodesic, measured counterclockwise. Since all angles θ are realized, it follows that PSL(2, R) acts transitively on T 1 i H. We next show the derivative action of PSL(2, R) is free on T 1 H. For suppose we have a vector v T 1 H fixed by the derivative action of some element A PSL(2, R). Since this action is transitive, by conjugating by an element of PSL(2, R) we can assume that v is the vertical unit tangent vector corresponding to the geodesic γ given by γ(t) = e t i. If A fixes v, then since A is an isometry and therefore takes geodesics to geodesics, A must fix the entire geodesic γ. Hence A fixes the line Re(z) = 0 in H. But by the classification of orientation-preserving isometries of H, any such isometry has at most one fixed point unless it is the identity. Hence we conclude that A = I, and therefore the derivative action is free. Fix v now to be the vertical unit tangent vector at i. We identify T 1 H with PSL(2, R) by taking v = I, the identity in PSL(2, R), and then identifying A with da(v). This identification makes sense since the derivative action of PSL(2, R) is free and transitive, and is analytic since the derivative action of PSL(2, R) on T 1 H is clearly analytic. The fact that the derivative action corresponds to left multiplication in PSL(2, R) is essentially by definition, since for w T 1 H, A PSL(2, R), we have that w = db(v) for a unique B PSL(2, R), and therefore da(w) = da(db(v)) = d(a B)(v) and composition of Mobius transformations corresponds to multiplication in PSL(2, R). Problem 3.3 By endowing PSL(2, R) with a suitable left-invariant metric, the identification H = PSL(2, R)/K becomes an isometry. In this metric, called the Sasaki metric, geodesics in H lift to geodesics in T 1 H via γ γ. Solution: We can think of each vector v T 1 H as a pair (z, θ) H S 1, where z = π(v) = a + bi and θ is the angle that the unit speed geodesic through v makes with the vertical unit speed geodesic t a + e t ib through z, measured counterclockwise from the vertical. This gives a bundle isomorphism T 1 H = H S 1. We then put a metric on T 1 H by giving the fiber (S 1 ) z the metric it carries at each point z as the unit circle centered at z in the metric of H. We claim that in this metric, PSL(2, R) acts by isometries on T 1 H. This is trivial, since PSL(2, R) acts by isometries on H, and therefore an element A takes the unit circle centered at z in H isometrically onto the unit circle centered at A(z) in H. Call this metric d. We give PSL(2, R) the metric induced from the metric d on T 1 H by the identification of Problem 3.2. This metric is trivially left-invariant, since the derivative action of PSL(2, R) 6

8 on T 1 H corresponds to left multiplication in PSL(2, R) under this identification, and we just verified above that the derivative action of PSL(2, R) on T 1 H is by isometries in the metric d that we are pulling back. To see that the identification PSL(2, R)/K H is an isometry, note that this corresponds to the identification in T 1 H given by fixing an angle θ S 1 and looking at points (z, θ) for z H. This is isometric to H by definition. Lastly, we should check that geodesics in H lift to geodesics in T 1 H via the map γ γ. Observe first that the geodesic γ(t) = e t i through i with tangent vector the vertical through i lifts to the curve γ(t) = (e t i, 0) in T 1 H. This is a geodesic in H since for a fixed angle (0 here), the resulting submanifold of T 1 H is canonically isometric to H. Since PSL(2, R) acts on T 1 H by isometries, we can find other geodesics by applying elements A of PSL(2, R). But if π : T 1 H H denotes the projection, then π(da( γ(t))) = A(γ(t)) and it follows that the geodesic da( γ) is the lift of the geodesic A(γ) in H, which gives the claim, since PSL(2, R) acts transitively on the geodesics of H. Problem 4 The geodesic flow satisfies flip invariance: For v T M, Solution: ϕ t ( v) = ϕ t (v) Let γ v be a parametrized geodesic with γ v (0) = v. Observe that the curve β v (t) = γ v ( t) is also a geodesic, with β v (t) = γ v ( t). Then ϕ t ( v) = β v ( t) = γ v (t) = ϕ t (v) Problem 5 Show that on T 1 H = PSL(2, R), the geodesic flow is given by right multiplication by the 1-parameter subgroup: { ( ) } e t/2 0 G = a t := 0 e t/2 : t R (renamed to avoid notation conflict) Solution: We consider first the geodesic γ(t) = e t i in H. Observe that a t (z) = e t z for z H. In particular, a t (i) = γ(t). Letting ϕ denote the geodesic flow on T 1 H, and putting γ(0) = I PSL(2, R) (under our identification) we conclude that the time t map of the geodesic flow applied to this point is ϕ t (I) = a t = I a t The geodesic flow is preserved by isometries. Since elements of PSL(2, R) acting by left multiplication act by isometries on PSL(2, R), we conclude that for any B PSL(2, R), ϕ t (B) = B ϕ t (I) = B a t Thus the time t map of the geodesic flow is given by right multiplication by a t. 7

9 Problem 6 Problem 6.1 Verify that the horocyclic foliations H ± are the foliations of PSL(2, R) by cosets of the horocyclic subgroups { ( ) } P + = h + 1 t t = : t R 0 1 P = { ( h 1 0 t = t 1 ) } : t R Solution: We will begin by computing the Busemann functions for H. We need the following general fact: for any v T 1 M, p M, where M is a simply connected nonpositively curved manifold, the Busemann function b + v (and similarly b v ), we have b + v (p) = ċ v (0), where c v is the unique geodesic in M with c v (0) = p and c v ( ) = γ v ( ) (the two geodesics are asymptotic at infinity). We also compute how the Busemann functions transform under an isometry f : M M and switching from + to. We have from which we see that b + df(v) f = b+ v. b + df(v) (f(p)) = lim d(γ df(v)(t), f(p)) t = lim d(f(γ v (t)), f(p)) t = lim d(γ v (t), p) t = b + v (p) To switch directions, we note that b + v geodesic flow. = b v, which is clear by the flip invariance of the With all this in mind, we first compute b + I, where I PSL(2, R) is identified with the vertical unit tangent vector at i. Of course γ I (t) = e t i, so γ I ( ) = H. The geodesics γ with γ( ) = are the vertical lines in H, oriented toward. Hence we conclude that the integral curves of the gradient field b + I are all geodesics of the form t a + e t bi, a + bi H It follows that the gradient field consists entirely of vertical vectors of norm 1 pointing toward the real axis. Hence the vectors orthogonal to this vector field (which form the tangent space of the level sets of b + I are all vectors parallel to the real axis. It follows that the sets b + I = const are those lines of the form Im(z) = c for c > 0. We compute from the description of the integral curves, b + I (bi) = b+ I (elog b i) = log b and it follows that b + I (a + bi) = log b 8

10 for any a + bi H. Hence it follows, for any A PSL(2, R), that b + A (z) = (b+ I A 1 )(z) = log(im(a 1 z)) If we put J to be the counterclockwise rotation by π about the point i (so J = then and J = J 1, and it s easy to see that and therefore J(z) = 1 z b I (z) = b+ J (z) = b+ I (J(z)) = log(im( z 1 )) = log( z 2 Im(z)) b A (z) = log( A 1 (z) 2 Im(A 1 z)) ( In particular, this verifies that the Busemann functions are C. The dependence on A here is also C, incidentally. The Busemann function b + I has level sets Im(z) = const. These level sets are the orbits of points on the imaginary axis under the isometry t z + t, for t R. This corresponds to h + t. Since b+ I is easily seen to be invariant under precomposition with h t, h + t takes gradient vectors to gradient vectors. We thus see that so that b + I (t + i) = h+ t H + (I) = { h + t : t R } For an isometry f, its effect on Busemann functions computed earlier implies that and therefore Hence we immediately see that b + df(v) df = b+ v H + (df(v)) = df(h + (v)) H + (A) = { A h + t : t R } since the derivative action is left multiplication. Thus the horocylic foliation H + coincides with the foliation by cosets of the 1-parameter subgroup P +. ) ) It s easily checked that b I h t = b I, by direct calculation, where h t (z) = 1 tz + 1 for t R. Then in the same way as before, we conclude that H (A) = { A h t : t R } so that the horocylic foliation H coincides with the foliation by cosets of the 1-parameter subgroup P. 9

11 Problem 6.2 Verify that the geodesic flow on T 1 H is Anosov. Solution: We identify T 1 H with PSL(2, R) in the usual way. We have three foliations of T 1 H: the horocyclic foliations H + and H, as well as the foliation by flow lines Rϕ. In PSL(2, R), these become the 1-parameter subgroups generated by h + t, h t, and a t respectively. From the relation b ± ϕ t(v) = b± v t one sees immediately that the horocyclic foliations are preserved by the geodesic flow, i.e. for any t R. H ± (ϕ t (v)) = ϕ t (H ± (v)) Our first goal is to prove that these foliations are everywhere transverse to one another. SL(2, R) is a 2-fold cover of PSL(2, R) by the map A A, so the Lie algebra of PSL(2, R) can be identified with the Lie algebra of SL(2, R), {( ) } a b sl(2, R) = : a, b, c R c a Left multiplication acts as an isometry on PSL(2, R), so it suffices to verify these three foliations are transverse at I, i.e. that their tangent vectors at I together span sl(2, R). We compute ( ) 1 0 T I G = R 0 1 ) ( T I P = R 0 0 ( T I P 0 0 = R 1 0 and it is evident that these three subspaces together span sl(2, R). It follows that we have a splitting sl(2, R) = T I G T I P + T I P which gives rise via the left multiplication isometry to a splitting of this type at every point into three subspaces, one of which, G, is the flow direction, and the other two, corresponding to P + and P, remain invariant under the flow. To prove that the geodesic flow ϕ is Anosov, we will show that the leaves of the foliation P + are exponentially contracted by the geodesic flowing forward in time, and the leaves of the foliation P are exponentially contracted flowing backward in time. Since the left multiplication isometry action is transitive, it suffices to prove these assertions for the leaves through I. First we consider P +. Take two points h + r and h + s in P +. We will flow these points forward by a time t, taking h + h + i a t for i = s, t. Let d denote the distance on PSL(2, R). Observe that on the horocycle Im(z) = 1 which lifts to P +, the lift is an isometry onto its image since all of the gradient ) 10

12 vectors of B + I along H + (I) are vertical. Further, distance on Im(z) = 1 coincides with Euclidean distance. We conclude that d(h + s, h + r ) = s r By the fact that left multiplication is an isometry, it follows that for any A PSL(2, R), d(a h + s, A h + r ) = s r Now observe the following commutation relation: for any s, t R, It follows that h + s a t = a t h + e t s d(h + s a t, h + r a t ) = d(a t h e t s, a t h e t r) = d(h e t s, h e t r) = e t s r = e t d(h + s, h + r ) where we have used the fact that left multiplication is an isometry. So we have exponential contraction along P + going forward in time. ( ) Now we calculate that h s = Jh s J 1, where J =. Hence 1 0 d(h s, h r ) = s r as well, since conjugation by J is an isometry at the level of H, which takes the horocycles Im(z) = const to the horocycles corresponding to limit rotations about 0 H. We then compute the commutation relation h s a t = a t h e t s which leads in an analogous way to the + case to the formula d(h s a t, h r a t ) = e t d(h s, h r ) which proves that the leaves of P are exponentially contracted as we flow backward in time. This completes the verification that the geodesic flow is Anosov. 11