Symplectic manifolds: -Kahler manifolds -Surfaces with volume forms Also: -The cotangent bundle of anything. There is a canonical 1-form $\theta$ on any cotangent bundle; $d\theta=\sum dp_i \wedge dq_i=\omega$ is a symplectic form. -coadjoint orbits of semisimple real Lie algebras Complex vs. symplectic geometry Given a complex manifold $(M,J)$, $J$ integrable. Then $\omega$ is a {\em Kahler form} if $d\omega=0$, $\omega(J -,-)$ is a Riemannian metric. Let $K=\{\kappa \in H^2(M;\R):there exists a Kahler form \omega representing \kappa\}$. Then $K$ is a convex subset of $H^2(M;\R)$; this is a convex "cone". If $M$ is closed, then, for every $\kappa \in K$, we obtain a symplectic manifold $(M,\omega)$; this is unique up to isomorphism by Moser's theorem. Conversely, if we have a symplectic manifold $(M,\omega)$, then we can choose a compatible almost-complex structure $J$ (and this is as good as it gets...) Automorphism groups: If $(M,\omega)$ is a symplectic manifold, its group of symplectic automorphisms (usually carrying a $C^\infty$-topology) is denoted $\Sym(M,\omega)$ or $\Aut(M,\omega)$. Recall: $\LSymp(M,\omega)$ is not simple (because it consists of closed 1-forms...) Hamiltonian vector fields give a notion of {\em Hamiltonian isotopy} in $\Symp(M,\omega)$ (namely, an isotopy $(\phi_t)$ such that $X_t=\phi^*_t(\partial \phi_t/\partial t)$ is Hamiltonian). Submanifolds: Suppose $(M,\omega)$ symplectic, $N \subset M$ is a closed symplectic submanifold. Two theorems we want to translate over from the topological setting. Tubular n'hood thm: The structure of $(M,\omega)$ locally near $N$ depends only on $(N,\omega|N)$ and $(\nu_N,\omega|\nu_N) \to N$ as a symplectic vector bundle. Note that $\nu_N$ as a manifold carries no preferred symplectic structure (hence the somewhat odd phrasing). Also note that, in order to make $\nu_N$ a symplectic vector bundle, we must use the fact that $N$ is a symplectic submanifold. Eg: $M=M^4$, $N=N^2$. The local structure depends only on the genus of $N$, $\int_N \omega$, and the intersection number of $N$ with itself. Up to isomorphism, the last determines its normal bundle; the first determines its topology; and the second determines the symplectic structure. Isotopy thm: Suppose $N_t \subset M$ is a smooth family of symplectic submanifolds. Then there exists some family of smooth symplectic automorphisms $(\phi_t)$ such that $\phi_t(N_0)=N_t$; this can be chosen to be a Hamiltonian isotopy. Pf: Let $N=N_0$ be the initial submanifold. We have the orthogonal tangent directions $TN^{\perp}$. Infinitesimally, we consider $"\frac{d}{d\epsilon} N_\epsilon"$; we take this to be lie in $TN^{\perp}$. Behave similarly to the proof in the 0-dimensional case, picking your vector field in tangential directions in such a way as to make it Hamiltonian. Note that we can't influence precisely how $N_0$ is identified with $N_t$... Example: Take $\C P^n$ with the Fubini-Study form, $X \subset \C P^n$ a smooth complex hypersurface of degree $d$. Then the pair $(\C P^n,X)$ depends only on $d$ up to symplectic automorphism (i.e., not only can we symplectomorph two distinct degree-$d$ varieties to each other, we can do so in a globally-extendible way). Isotropic, coisotropic submanifolds: Isotropic manifolds $N \subset M$ are such that $TN^{\perp_\omega} \supset TN$ (and so $\dim N \leq \dim M/2$); equivalently, $\omega$ vanishes on $N$. Coisotropic manifolds are such that $TN^{\perp} \subset TN$ (and so $\dim N \geq \dim M/2$). Lagrangian submanifolds are isotropic and coisotropic (and so $\dim N=\dim M/2$); equivalently, $\omega$ vanishes on $N$ and $N$ is of max'l dimension for this. Examples: $M=T^*N$ with the standard symplectic structure. Let $\alpha \in \Omega^1(N)=C^\infty(T^* N)$. Then $\alpha:N \to T^*N$. Recall the universal 1-form $\theta$. Then $\alpha^* \theta=\alpha$. Hence $\Gamma_\alpha=\im(\alpha:N \to T^*N)$ is Lagrangian iff $\alpha$ is closed. (Since it's Lagrangian iff $\omega$ vanishes on it; thus it's Lagrangian iff the pullback of $\theta$ to $M$ is closed; but this pullback is precisely $\alpha$). Suppose $\phi:(M_1,\omega_1) \to (M_2,\omega_2)$ is a symplectic isomorphism. Consider $(M_1 \times M_2,\omega_2-\omega_1)$. Then $\Gamma_\phi=\{x_2=\phi(x_1)\}$ is Lagrangian, and conversely. Ex: Suppose $(M,J,\omega)$ is a complex manifold; let $\iota$ be an involution such that $\iota^* J=-J$ (so $D\iota(JX)=-JD\iota(X)$), $\iota^*\omega=-\omega$. Then the fixed-point set of $\iota$ is Lagrangian. e.g. $\R P^n$ is Lagrangian in $\C P^n$. We want: 1) Tubular n'hood thms 2) Isotopy thms For these kinds of submanifolds. Lemma (Tubular n'hood): Suppose $L \subset M$ Lagrangian. Then there exist n'hoods $U \subset M$ and $V \subset $T^* L$ of the zero section which are symplectically isomorphic. (Locally, $L \subset M$ looks like $L \subset T^* L$). This claims that the normal bundle of $L$ in $M$ is always $T^* L$; i.e., $\nu_L=TM/TL$ has a nondegenerate pairing $\nu_L \otimes TL \to \R$. We get this pairing via $(\overline{X},Y)=\omega(X,Y)$; this is well-defined as $\omega$ vanishes, and nondegenerate as $L$ is max'l wrt $\omega$ vanishing. For instance, if $M=M^4$ orientable, $L=L^2$, then $L \cdot L=-\chi(L)$. Ex: On $\C P^2$ every self-intersection number is non-negative; thus there are no Lagrangian 2-spheres in $\C P^2$. Now, what about the isotopy theorem? Let $\{L_t\}$ be a smooth family of Lagrangian submanifolds. Then we have "$dL_t/dt$", a section of $\nu_{L_t}$. Canonically, this is a 1-form on $L_t$. Moreover, this 1-form is closed (because $L_t$ is Lagrangian, applying the tubular n'hood theorem). So, if $dL_t/dt$ is exact, we call $L_t$ an {\em exact Lagrangian isotopy}. Why is this relevant? Consider the 2-dimensional case. Every curve is a Lagrangian submanifold. Take a simple closed curve and contract it. There can't possibly be a symplectic (i.e., area-preserving) automorphism that extends this! Lemma: If $(L_t)$ be an exact Lagrangian isotopy. Then there exists a Hamiltonian isotopy $(\phi_t)$ with $\phi_0=\id_0$, $\phi_t(L_0)=L_t$. Pf: The isotopy gives a $t$-dependent exact 1-form, and hence a $t$-dependent function. Extend to a function on all of $M$, take the corresponding Hamiltonian vector field, and integrate! There are also tubular n'hood thms for isotropic and coisotropic submanifolds. For an isotropic $N \subset M$, the local structure depends on $N$ and the symplectic vector bundle $TN^{\perp}/TN \to N$. (In the top-dimensional case, this bundle is 0-dimensional, so carries no information; in the 0-dimensional case, it's the whole tangent bundle and so also carries no information). For coisotropic $N \subset M$, the local structure depends on $(N, \omega|N)$. How are these theorems proved: The same kind of Moser-type method used to prove Darboux's theorem; build the intended local model, and show that you can deform the original space... All this holds equivariantly under action by compact Lie groups $G$ (since it's proved by constructing average-able things). References for the next part: Guillemin, Moment Maps and Combinatorial Invariants Atiyah, Convexity and commuting Hamiltonians Guillemin and Sternberg, Birational equivalences Let $(M,\omega)$ be a symplectic manifold, $K \equiv T^r$ acting on $(M,\omega)$. So if $k=Lie(K)$ we have the Killing field $X_\gamma$, which is symplectic, and $[X_{\gamma_1},X_{\gamma_2}]=0$. Def'n: The action is {\em Hamiltonian} if all the $X_\gamma$ are Hamiltonian. Then there is $\mu:M \to k^*$ such that $X_\gamma$ is the Hamiltonian vector field of $\<\mu,\gamma\>:M \to \R$. Such a $\mu$ is called a {\em moment map}. Note that $\mu=(\mu_1,\dots,\mu_r):M \to \R^r$, then $\{\mu_i,\mu_j\}=d\mu_i(X_{\mu_j})=0$ (clearly it's constant as the corresponding vector fields commute; $d\mu_i(X_{\mu_j})$ gives the change of $\mu_i$ in the $\mu_j$-direction; since, if you go in the $\mu_j$-direction you eventually get back to where you started, it must average to 0 over such an orbit and hence be zero) Ex: $M=\R^2$, $S^1$ acts by rotations, $\mu=(|x|^2+|y|^2)/2$. Moreover, $\mu$ is $K$-invariant (Noether's theorem). Lemma: $x \in M$ is a regular point of $\mu$ iff $K_x$ (stabilizer) is a finite group.