Symplectic Differential Geometry Definition 1.1. a Two Form Ω on A

Home , Symplectic manifold

1. Lecture 2: Symplectic differential geometry Definition 1.1. A two form ω on a manifold M is symplectic if and only if 1) ∀x ∈ M, ω(x) is symplectic on TxM; 2) dω = 0 (ω is closed). Examples: 1) (R2n, σ) is symplectic manifold. 2) If N is a manifold, then ∗ T N = {(q, p)|p linear form on TxM} is a symplectic manifold. Let q1, ··· , qn be local coordinates on N and let p1, ··· , pn be the dual coordinates. Then the symplectic form is defined by n X i ω = dp ∧ dqi. i=1 One can check that ω does not depend on the choice of coordinates and is a symplectic form. We can also define a one form n X i λ = pdq = p dqi. i=1 It is well defined and dλ = ω. 3) Projective algebraic manifolds(See also Kählermanifolds) CP n has a canonical symplectic structure σ and is also a complex manifold. The complex structure J and the symplectic form σ are compatible. CP n has a hermitian metric h. For any z ∈ CP n, h(z) n is a hermitian inner product on TzCP . h = g + iσ, where g is a Riemannian metric and σ(Jξ, ξ) = g(ξ, ξ). Claim: A complex submanifold M of CP n carries a natural symplectic structure. Indeed, consider σ|M . It’s obviously skew-symmetric and closed. We must prove that σ|M is non-degenerate. This is true because if ξ ∈ TxM{0} and Jξ ∈ TxM, so ω(x)(ξ, Jξ) 6= 0 Definition 1.2. A submanifold in symplectic manifold (M, ω) is La- 1 grangian if and only if ω|TxL = 0 for all x ∈ M and dim L = 2 dim M.

Lemma 1.3. (Moser) Let N be a compact submanifold in M. Let ωt be a family of symplectic forms such that ωt|TN M is constant. Then ∗ there is a diﬀeomorphism ϕ deﬁned near N such that ϕ ω1 = ω0 and ϕ|N = id|N . 1 2

Proof. We will construct a vector field X(t, x) whose flow ϕt satisfies 0 t ∗ ϕ = id and (ϕ ) ωt = ω0. Differentiate the last equality d d ( (ϕt)∗)ω + (ϕt)∗( ω ) = 0. dt t dt t Then d (ϕt)∗L ω + (ϕt)∗( ω ) = 0. X t dt t Since ϕt is diffeomorphism, this is equivalent to d L ω + ω = 0. X t dt t Using Cartan’s formula

LX = d ◦ iX + iX ◦ d, we get d d(i ω ) + ω = 0. X t dt t ∗ Since ωt is nondegenerate, the map TxM → (TxM) which maps X to ω(X, ·) is an isomorphism. Therefore, for any one form β, the equation iX ω = β has a unique solution Xβ. It suﬃces to solve for βt, d dβ = − ω . t dt t

We require βt = 0 on N for all t, because we want ϕ|N = id|N . On the d other hand, the assumption that ωt = ω0 on TN M implies ( dt ωt) ≡ 0 on TN M. Denote the right hand side of the above equation by α, then α is defined in a neighborhood U of N. The solution of βt is given by Poincaré’sLemma on the tubular neighborhood of N. Here by a tubular neighbourhood we mean a neighbourhood of N in M diffeomorphic to the unit disc bundle DνN M of νM N the normal bundle of N in M (i.e. νM N = {(x, ξ) ∈ TN M | ξ ⊥ TN}). Lemma 1.4. (Poincaré)If α is a p− form on U, closed and vanishing on TN M, then there exists β such that α = dβ and β vanishes on N. Proof. 1 This means that for a tubular neighbourhood H∗(U, N) = 0. Indeed, let rt be the map on νN M defined by rt(x, ξ) = (x, tξ). Since ∗ r0 sends νN M to its zero section, N, we have r0α = 0.

1The proof is easier if one is willing to admit that the set of exact forms is closed ∞ for the C topology, i.e. if α = dβε + γε and limε→0 γε = 0 then α is exact. This follows for example from the fact that exactness of a closed form is checked by veryﬁying that its integral over a ﬁnite number of cycles vanishes. 3

Then d (r )∗(α) = r∗(L α) = d (r∗(i α)) dt t t V t Vt ∗ Note that rt (iVt α) is well deﬁned, continuous and bounded as t goes to zero, since writing (locally) (u, η) for a tangent vector to T(x,ξ)νN M were u ∈ TN and ξ ∈ TN M, we have

∗ (rt (iVt α))(x, ξ)((u2, η2)....(up, ηp)) = α(x, tξ)((0, ξ), (u2, tη2)...(up, tηp) is a well-deﬁned smooth form including for t = 0. let us denote by βt the above form. Thus

∗ lim(rt) (iV α) = 0 t→0 t and we can write

Z 1 Z 1 ∗ ∗ d ∗ ∗ α = r1(α) − r0(α) = (rt) (α)dt = d (rt) (iVt α)dt = dβ 0 dt 0 where Z 1 ∗ 1 β = (rt) (iVt α)dt = int0βtdt 0 but βt vanishes on N, since

βt(x, 0)((u2, η2)...(up, ηp) = α(x, 0)((0, 0), (u2, tη2)...(up, tηp) = 0 This proves our lemma. As an application, we have Proposition 1.5. (Darboux) Let (M, ω) be a symplectic manifold. Then for each z ∈ M, there is a local diﬀeomorphism ϕ from a neighborhood of z to a neighborhood of o in R2n such that ϕ∗σ = ω.

Proof. According to Lecture 1, there exists a linear map L : TzM → 2n ∗ R such that L σ = ω(z). Hence, using a local diffeomorphism ϕ0 : U → W such that dϕ0(z) = L, where U and W are neighborhoods of z ∈ M and o ∈ R2n respectively, we are reduced to considering the case ∗ ∗ where ϕ0σ is a symplectic form defined in U and ω(z) = (ϕ0)σ. ∗ Define ωt = (1 − t)ϕ0σ + tω in U. It’s easy to check ωt satisfies the assumptions of Moser’s Lemma, therefore, there exists ψ such that ∗ ψ ω1 = ω0, i.e. ∗ ∗ ψ ω = ϕ0σ. −1 Then ϕ = ϕ0 ◦ ψ is the required diffeomorphism. 4

Exercices: (1) Show the analogue of Moser’s Lemma for volume forms. (2) Let ω1, ω2 be symplectic forms on a compact surface with- out boundary. Then there exists a diffeomorphism ϕ such that ∗ R R ϕ ω1 = ω2 if and only if ω1 = ω2. Proposition 1.6. (Weinstein) Let L be a closed Lagrangian submanifold in (M, ω). Then L has a neighborhood symplectomorphic to a ∗ neighborhood of OL ⊂ T L. (Here, OL = {(q, 0)|q ∈ L} is the zero section.) Proof. The idea of the proof is the same as that of Darboux Lemma. First, for any x ∈ L, find a subspace V (x) in TxM such that 1) V (x) ⊂ TxM is Lagrangian subspace; 2) V (x) ∩ TxL = {0}; 3) x → V (x) is smooth. According to our discussion in linear symplectic space, we can find such V (x) at least pointwise. To see 3), note that at each point x ∈ L the set of all Lagrangian subspaces in TxM transverse to T xL may be ∗ identified with quadratic forms on (TxL) . It’s then possible to find a smooth section of such an “affine bundle”. Abusing notations a little, we write L for the zero section in T ∗L. ∗ ∗ Denote by TL(T L) the restriction of the tangent bundle T L to L. Denote by TLM the restriction of bundle TM to L. Both bundles are over L. For x ∈ L, their fibers are ∗ ∗ Tx(T L) = TxL ⊕ Tx(Tx L) and

TxM = TxL ⊕ V (x). ∗ Construct a bundle map L0 : TL(T L) → TLM which restricts to iden- ∗ tity on factor TxL and sends Tx(Tx L) to V (x). Moreover, we require

ω(L0u, L0v) = σ(u, v), ∗ ∗ where u ∈ Tx(Tx L) = Tx L and v ∈ TxL. This defines L0 uniquely. ∗ Again, we can find ϕ0 from a neighborhood of L in T L to a neighbor- ∗ hood of L in M such that dϕ0|TL(T L) = L0. By the construction of L0, one may check that ∗ ∗ ϕ0ω = σ on TL(T L). Define ∗ ωt = (1 − t)ϕ0ω + tσ, t ∈ [0, 1].

ωt is a family of symplectic forms in a neighborhood of OL. Moreover, ∗ ωt ≡ ω0 on TL(T L). By Moser’s Lemma, there exists Ψ deﬁned near 5

∗ ∗ ∗ −1 OL such that Ψ ω1 = ω0, i.e. Ψ σ = ϕ0ω. Then ϕ0 ◦ Ψ is the diﬀeomorphism we need.

Exercice: Let I1, I2 be two diﬀeomorphic isotropic submanifold in ω1 ω2 (M1, ω1), (M2, ω2). Let E1 = (TI1) /(TI1) and E2 = (TI2) /(TI2). E1,E2 are symplectic vector bundlesover I1 and I2. Show that I1 and ∼ I2 have symplectomorphic neighborhoods if and only if E1 = E2 as symplectic vector bundles.

2. The groups Ham and Diffω Since Klein’s Erlangen’s programme, Geometry has meant the study of symmetry groups. The group playing the first role here is Diffω(M). Let (M, ω) be a symplectic manifold. Define ∗ Diffω(M) = {ϕ ∈ Diff(M)|ϕ ω = ω}. This is a very large group since it contains Ham(M, ω), which we will now define. Let H(t, x) be any smooth function and XH the unique vector field such that ω(XH (t, x), ξ) = dxH(t, x)ξ, ∀ξ ∈ TxM.

Here dx means exterior derivative with respect to x only. Claim: The flow of XH is in Diffω(M). To see this, d (ϕt)∗ω = (ϕt)∗(L ω) dt XH t ∗ = (ϕ ) (d ◦ iXH ω + iXH ◦ dω) = (ϕt)∗(d(dH)) = 0. Definition 2.1. The set of all diffeomorphism ϕ that can be obtained as the flow of some H is a subgroup Diff(M, ω)) called Ham(M, ω). To prove that Ham(M, ω) is a subgroup, we proceed as follows: first notice that the Hamiltonian isotopy can be reparametrized, and still yields a Hamiltonian isotopy ϕs(t) satisfies d d ( ϕ ) = s0(t)( ϕ ) = s0(t)X (s(t ), ϕ ) dt s(t) t=t0 ds s s=s(t0) H 0 s(t0 which is the Hamiltonian flow of s0(t)H(s(t), z) Therefore we may use a function s(t) on [0, 1] such that s(0) = 0, s(1/2) = 1, s0(t) = 0 for t close to 1/2 and we find a Hamiltonian flow ending at ϕ1 in time 1/2 and such that H vanishes near t = 1/2. Similarly if 6

ψt is the flow associated to K(t, z) we may modify it in a similar way using r(t) so that K ≡ 0 for t in a neighbourhood of [0, 1/2]. We can then consider the flow associated to H(t, z) + K(t, z) = L(t, z) it will be ϕs(t) ◦ ψr(t) and for t = 1 we get ϕ1 ◦ ψ1. −1 That ϕ1 is also Hamiltonian follows from the fact that −H(t, ϕt(z)) −1 has flow ϕt . Exercice: Show that (ϕt)−1ψt is the Hamiltonian flow of

L(t, z) = K(t, ϕt(z)) − H(t, ϕt(z)) This immediately proves that Ham(M, ω) is a group.

Remark 2.2. Denote by Diffω,0 the component of Diffω(M) in which the identity lies. It’s obvious that Ham(M, ω) ⊂ (Diffω,0(M). Remark 2.3. If H(t, x) = H(x), then H ◦ ϕt = H. This is what physi- cists call conservation of energy. Indeed H is the energy of the system, and for time-independent conservative systems, enregy is preserved. This is not the case in time-dependent siutations.

Remark 2.4. If we choose local coordinates q1, ..., qn and their dual p1, ..., pn in the cotangent space, pi, qi, the ﬂow is given by the ODE ∂H q˙i = ∂pi ∂H p˙i = − ∂qi

Question: How big is the quotient Diffω0 /Ham(M, ω)?

Given ϕ ∈ Diffω0 , there is an obvious obstruction for ϕ to belong to ∗ Ham(M, ω). Assume ω = dλ. Then ϕ λ−λ is closed for all ϕ ∈ Diffω, since d(ϕ∗λ − λ) = ϕ∗ω − ω = 0. t If ϕ is the ﬂow of XH , d ((ϕt)∗λ) = (ϕt)∗(L λ) dt HX t ∗ = (ϕ ) (d(iXH λ) + iXH dλ) t ∗ = (ϕ ) d(iXH λ + H) t ∗ = d((ϕ ) (iXH λ + H)). This implies that ϕ∗λ − λ is exact. In summary, we can deﬁne map 1 Flux : (Diffω)0(M) → H (M, R) ϕ 7→ [ϕ∗λ − λ] We know Hamω(M) = ker(F lux). Examples: 7

∗ 1 (1) On T T the translation ϕ :(x, p) −→ (x, p + p0) is symplectic, but Flux(ϕ) = p0. (2) Similarly if M = T 2 and σ = dx ∧ dy, the map (x, y) −→ 2 (x, y + y0) is not in Ham(T , σ) for y0 6≡ 0 mod 1. Indeed, since the projection π : T ∗T 1 −→ T 2 is a symplectic covering, any Hamiltonian isotopy on T 2 ending in ϕ would lift to a Hamiltonian isotopy on T ∗T 1 (if H(t, z) is the Hamiltonian on T 2, H(t, π(z)) is the Hamiltonian on T ∗T 1) ending to some lift of ϕ. But the lifts of ϕ are given by (x, y) −→ (x + m, y + 2 y0 + n) for (m, n) ∈ Z , with Flux given by y0 + n 6= 0. Exercices:

(1) Prove the Darboux-Weinstein-Givental theorem: Let S1,S2 be two submanifolds in (M1, ω1), (M2, ω2). Assume there is a map ϕ : S1 −→ S2 which lifts to bundle map

Φ: TS1 M1 −→ TS2 M2

coinciding with dϕ on the subbundle TS1, and preserving the ∗ symplectic structures, i.e. Φ (ω2) = ω1. Then there is a symplectic diﬀeomorphism between a neighbourhood U1 of S1 and a neighbourhood U2 of S2. (2) Use the Darboux-Weinstein-Givental theorem to prove that all closed curves have symplectomorphic neighbourhoods. Hint: Show that all symplectic vector bundle on the circle are trivial. (3) (a) Prove that the Flux homomorphism can be deﬁned on (M, ω) as follows. Let ϕt be a symplectic isotopy. Then d dt ϕt(z) = X(t, ϕt(z)) and ω(X(t, z)) = αt is a closed form. Then Z 1 1 Flux(] ϕ) = αtdt ∈ H (M, R) 0

depends only on the homotopy class of the path ϕt. If Γ is the image by Flux of the set of closed loops, we get a well defined map 1 Flux : Diff(M, ω)0 −→ H (M, R)/Γ (b) Prove that when ω is exact, Γ vanishes and the new defini- ton coincides with the old one.