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

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Symplectic Differential Geometry Definition 1.1. a Two Form Ω on A 1. Lecture 2: Symplectic differential geometry Definition 1.1. A two form ! on a manifold M is symplectic if and only if 1) 8x 2 M, !(x) is symplectic on TxM; 2) d! = 0 (! is closed). Examples: 1) (R2n; σ) is symplectic manifold. 2) If N is a manifold, then ∗ T N = f(q; p)jp linear form on TxMg 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 ! = dp ^ dqi: i=1 One can check that ! 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 = !. 3) Projective algebraic manifolds(See also K¨ahlermanifolds) 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 2 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 sym- plectic structure. Indeed, consider σjM . It's obviously skew-symmetric and closed. We must prove that σjM is non-degenerate. This is true because if ξ 2 TxMf0g and Jξ 2 TxM, so !(x)(ξ; Jξ) 6= 0 Definition 1.2. A submanifold in symplectic manifold (M; !) is La- 1 grangian if and only if !jTxL = 0 for all x 2 M and dim L = 2 dim M. Lemma 1.3. (Moser) Let N be a compact submanifold in M. Let !t be a family of symplectic forms such that !tjTN M is constant. Then ∗ there is a diffeomorphism ' defined near N such that ' !1 = !0 and 'jN = idjN . 1 2 Proof. We will construct a vector field X(t; x) whose flow 't satisfies 0 t ∗ ' = id and (' ) !t = !0. Differentiate the last equality d d ( ('t)∗)! + ('t)∗( ! ) = 0: dt t dt t Then d ('t)∗L ! + ('t)∗( ! ) = 0: X t dt t Since 't is diffeomorphism, this is equivalent to d L ! + ! = 0: X t dt t Using Cartan's formula LX = d ◦ iX + iX ◦ d; we get d d(i ! ) + ! = 0: X t dt t ∗ Since !t is nondegenerate, the map TxM ! (TxM) which maps X to !(X; ·) is an isomorphism. Therefore, for any one form β, the equation iX ! = β has a unique solution Xβ. It suffices to solve for βt, d dβ = − ! : t dt t We require βt = 0 on N for all t, because we want 'jN = idjN . On the d other hand, the assumption that !t = !0 on TN M implies ( dt !t) ≡ 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 Poincar´e'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 = f(x; ξ) 2 TN M j ξ ? TNg). Lemma 1.4. (Poincar´e)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 1 for the C topology, i.e. if α = dβ" + γ" and lim"!0 γ" = 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 2 TN and ξ 2 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; !) be a symplectic manifold. Then for each z 2 M, there is a local diffeomorphism ' from a neigh- borhood of z to a neighborhood of o in R2n such that '∗σ = !. Proof. According to Lecture 1, there exists a linear map L : TzM ! 2n ∗ R such that L σ = !(z). Hence, using a local diffeomorphism '0 : U ! W such that d'0(z) = L, where U and W are neighborhoods of z 2 M and o 2 R2n respectively, we are reduced to considering the case ∗ ∗ where '0σ is a symplectic form defined in U and !(z) = ('0)σ. ∗ Define !t = (1 − t)'0σ + t! in U. It's easy to check !t satisfies the assumptions of Moser's Lemma, therefore, there exists such that ∗ !1 = !0, i.e. ∗ ∗ ! = '0σ: −1 Then ' = '0 ◦ is the required diffeomorphism. 4 Exercices: (1) Show the analogue of Moser's Lemma for volume forms. (2) Let !1, !2 be symplectic forms on a compact surface with- out boundary. Then there exists a diffeomorphism ' such that ∗ R R ' !1 = !2 if and only if !1 = !2. Proposition 1.6. (Weinstein) Let L be a closed Lagrangian subman- ifold in (M; !). Then L has a neighborhood symplectomorphic to a ∗ neighborhood of OL ⊂ T L. (Here, OL = f(q; 0)jq 2 Lg is the zero section.) Proof. The idea of the proof is the same as that of Darboux Lemma. First, for any x 2 L, find a subspace V (x) in TxM such that 1) V (x) ⊂ TxM is Lagrangian subspace; 2) V (x) \ TxL = f0g; 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 2 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 2 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 !(L0u; L0v) = σ(u; v); ∗ ∗ where u 2 Tx(Tx L) = Tx L and v 2 TxL. This defines L0 uniquely. ∗ Again, we can find '0 from a neighborhood of L in T L to a neighbor- ∗ hood of L in M such that d'0jTL(T L) = L0. By the construction of L0, one may check that ∗ ∗ '0! = σ on TL(T L): Define ∗ !t = (1 − t)'0! + tσ; t 2 [0; 1]: !t is a family of symplectic forms in a neighborhood of OL. Moreover, ∗ !t ≡ !0 on TL(T L). By Moser's Lemma, there exists Ψ defined near 5 ∗ ∗ ∗ −1 OL such that Ψ !1 = !0, i.e. Ψ σ = '0!. Then '0 ◦ Ψ is the diffeomorphism we need. Exercice: Let I1, I2 be two diffeomorphic isotropic submanifold in !1 !2 (M1;!1), (M2;!2). 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 Diff! Since Klein's Erlangen's programme, Geometry has meant the study of symmetry groups. The group playing the first role here is Diff!(M). Let (M; !) be a symplectic manifold. Define ∗ Diff!(M) = f' 2 Diff(M)j' ! = !g: This is a very large group since it contains Ham(M; !), which we will now define. Let H(t; x) be any smooth function and XH the unique vector field such that !(XH (t; x); ξ) = dxH(t; x)ξ; 8ξ 2 TxM: Here dx means exterior derivative with respect to x only. Claim: The flow of XH is in Diff!(M). To see this, d ('t)∗! = ('t)∗(L !) dt XH t ∗ = (' ) (d ◦ iXH ! + iXH ◦ d!) = ('t)∗(d(dH)) = 0: Definition 2.1.
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