symplectic

Definition (Symplectic Manifold): Recall that \(M^{2n}\) is a symplectic manifold iff \(W\) is smooth of even dimension and admits a ^395eb1

closed: \(d\omega = 0\) Motivation: the Lie derivative of \(\omega\) along \(V_H\) is 0, i.e. \(\mathcal{L}_{V_H}(\omega) = 0\).
nondegenerate \(\omega_p: T_p M \times T_p M \to {\mathbf{R}}\); \(\omega_p(\mathbf{v},\mathbf{w}) = 0~~\forall \mathbf{w} \implies \mathbf{v}= 0\).
- Motivation: for every \(dH\) there exists a vector field \(V_H\) such that \(dH = \omega(V_H, {-})\).
skew-symmetric: \(\omega_p(\mathbf{v}, \mathbf{w}) = \omega_p(\mathbf{w}, \mathbf{v})\).
- Motivation: \(H\) should be constant along flow lines, i.e. \(dH(V_H) = \omega(V_H, V_H) = 0\)
bilinear: Lifts to a map \(T_pM\otimes T_P M \to {\mathbf{R}}\)
- Motivation: send any form \({\left\langle {{-}},~{{-}} \right\rangle}: V\times V \to k\) to the linear map \(f: V \to V {}^{ \vee }\) where \(v\mapsto f(v) \coloneqq{\left\langle {v},~{{-}} \right\rangle}\). If the pairing is nondegenerate, \(\ker f = 0\), and we get an identification \(V\cong V {}^{ \vee }\).
- Yields \(TM \cong T {}^{ \vee }M\), which can be combined with \(\iota\) to obtain an isomorphism \(\mathfrak{X}(M) \cong \Omega^1(M)\) between vector fields and 1-forms.
2-form \begin{align*} \omega \in \Omega^2(M) = {{\Gamma}\qty{ { {\bigwedge}^{\scriptscriptstyle \bullet}} _{\mathbf{R}}^2 {\mathbf{T}} {}^{ \vee }M} } .\end{align*} An important consequence: to any \(f\in C^\infty(M \to {\mathbf{R}})\), we can associate to it a vector field \(X_f\). So there is a map \(C^\infty(M\to {\mathbf{R}}) \to {{\Gamma}\qty{{\mathbf{T}}M} }\)?

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Motivations

Relation to string theory: attachments/Pasted%20image%2020220424165246.png

See open string, closed string, Floer homology, quantum cohomology, symplectic cohomology, Lagrangian Floer cohomology, wrapped Floer cohomology, Fukaya category, homological mirror symmetry.

Results

Proposition: \((M, \omega \in \Omega^2(M))\) is symplectic iff \(\omega^n \neq 0\) everywhere (c.f. Mike Hutchings).

Corollary: Every symplectic manifold is orientable (since it has a nonvanishing volume form).

Important Remark: Symplectic structures on smooth manifolds give us a way to generate flows on a manifold (by defining a Hamiltonian or a symplectic vector field).

Definition (Exact Symplectic Manifold): \(W\) is an exact symplectic manifold iff there exists a 1-form \(\lambda \in \Omega^1(W)\) such that \(d\lambda \in \Omega^2(W)\) is non-degenerate. ^9a87d0

Remark: If \((W, \lambda)\) is exact symplectic then \((W, d\lambda)\) is symplectic. \(\lambda\) is sometimes referred to as a Liouville form.

Important Remark: If \((W, \lambda)\) is exact and \(H: {\mathbf{R}}\times M \to {\mathbf{R}}\) is smooth, then the Hamiltonian flow \(\phi_H^t: M \to M\) is defined for all time and is an exact symplectomorphism.

Theorem: There are no closed (compact and boundaryless) exact symplectic manifolds.

Proof: \begin{align*} \int_{{{\partial}}M} \lambda \wedge \omega^{n-1} &= \int_M d(\lambda \wedge \omega^{n-1}) \\ &= \int_M d\lambda \wedge \omega^{n-1} + (-1)^{{\left\lvert {\lambda} \right\rvert}}\lambda\wedge d\omega^{n-1} \\ &= \int_M \omega \wedge \omega^{n-1} + (-1)^{{\left\lvert {\lambda} \right\rvert}} \lambda \wedge 0 \\ &= \int_M \omega^n \\ &= \mathrm{Vol}_{\text{Sp}}(M) \\ &> 0 ,\end{align*}

so \({{\partial}}M \neq 0\), and thus \(M\) can not be closed.

Definition (isotropic): Let \(\Lambda\) be the image of an embedded sphere \(S^k \to W\). Then \(\Lambda\) is isotropic iff \({\left.{{\lambda}} \right|_{{\Lambda}} } = 0\).

Given a almost complex structure,…?

Reference: p 68-70 of Cannas da Silva

Symplectic vector spaces

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