Lagrangian Submanifold

Lagrangian Submanifold

A Lagrangian submanifold of a symplectic manifold $(M,\omega )$ is a submanifold $L\subset M$ such that $\dim L=\frac{1}{2}\dim M$ and $\omega {|}_L=0$. In other words, the restriction of the symplectic form to $L$ vanishes. Equivalently, if $i:L\to M$ is the inclusion map, then $L$ is lagrangian if and only if $i^{\ast }\omega =0$ and $\dim L=\frac{1}{2}\dim M$.

Proposition

For $X$ an n-dimensional manifold, $T^{\ast }X$ the cotangent bundle, then $X_0:=\{(x,\xi )\in T^{\ast }X|\xi =0\text{in}{T}_x^{\ast }X\}$, the 0-section is a lagrangian.

Theorem

Let $X_{\mu }$ be (the image of) another section, that is, an n-dimensional submanifold of $T^{\ast }X$ of the form $X_{\mu }=\{(x,{\mu }_x)|x\in X,{\mu }_x\in T_x^{\ast }X\}$, with associated de Rham 1-form $\mu$, then $X_{\mu }$ is Lagrangian if and only if $\mu$ is closed, that is $d\mu =0$. That is to say there’s one to one correspondence between the set of lagrangian submanifolds of $T^{\ast }X$ of the form $\{(x,{\mu }_x)|x\in X,{\mu }_x\in T_x^{\ast }X\}$ and the set of closed 1-forms on X.

Conormal Bundle

Let $S$ be any $k$-dimensional submanifold of an $n$-dimensional manifold $X$. The conormal space at $x\in S$ is $N_x^{\ast }S=\{\xi \in T_x^{\ast }X\mid \xi (v)=0,\text{ for all }v\in T_{x}S\}.$ The conormal bundle of $S$ is $N^{\ast }S=\{(x,\xi )\in T^{\ast }X\mid x\in S,\xi \in N_x^{\ast }S\}.$

Proposition

Let $i:N^{\ast }S\hookrightarrow T^{\ast }X$ be the inclusion, and let $\alpha$ be the tautological 1-form on $T^{\ast }X$. Then $i^{\ast }\alpha =0$ .

Corollary

For any submanifold $S\subset X$, the conormal bundle $N^{\ast }S$ is a lagrangian submanifold of $T^{\ast }X$. Actually the 0-section $X_0$ is a special case of conormal bundle by taking $S=X$.

Weinstein Lagrangian Neighborhood Theorem

Let $M$ be a $2n$-dimensional manifold, $X$ a compact $n$-dimensional submanifold, $i:X\hookrightarrow M$ the inclusion map, and ${\omega }_0$ and ${\omega }_1$ symplectic forms on $M$ such that $i^{\ast }{\omega }_0=i^{\ast }{\omega }_1=0$, i.e., $X$ is a lagrangian submanifold of both $(M,{\omega }_0)$ and $(M,{\omega }_1)$. Then there exist neighborhoods ${\mathcal{U}}_0$ and ${\mathcal{U}}_1$ of $X$ in $M$ and a diffeomorphism $\phi :{\mathcal{U}}_0\to {\mathcal{U}}_1$ such that $\begin{array}{ccc}{\mathcal{U}}_0& \stackrel{\phi }{\to }& {\mathcal{U}}_1\\ & \nwarrow i\nearrow i& \\ & X& \end{array}$ $\text{commutes and }{\phi }^{\ast }{\omega }_1={\omega }_0$

Coisotropic Embedding Theorem

Let $M$ be a manifold of dimension $2n$, $X$ a submanifold of dimension $k\ge n$, $i:X\hookrightarrow M$ the inclusion map, and ${\omega }_0$ and ${\omega }_1$ symplectic forms on $M$, such that $i^{\ast }{\omega }_0=i^{\ast }{\omega }_1$ and $X$ is coisotropic for both $(M,{\omega }_0)$ and $(M,{\omega }_1)$. Then there exist neighborhoods ${\mathcal{U}}_0$ and ${\mathcal{U}}_1$ of $X$ in $M$ and a diffeomorphism $\phi :{\mathcal{U}}_0\to {\mathcal{U}}_1$ such that $\begin{array}{ccc}{\mathcal{U}}_0& \stackrel{\phi }{\to }& {\mathcal{U}}_1\\ & \nwarrow i\nearrow i& \\ & X& \end{array}$ $\text{commutes and }{\phi }^{\ast }{\omega }_1={\omega }_0$

Weinstein Tubular Neighborhood Theorem

Let $(M,\omega )$ be a symplectic manifold, $X$ a compact lagrangian submanifold, ${\omega }_0$ the canonical symplectic form on $T^{\ast }X$, $i_0:X\hookrightarrow T^{\ast }X$ the lagrangian embedding as the zero section, and $i:X\hookrightarrow M$ the lagrangian embedding given by inclusion. Then there are neighborhoods ${\mathcal{U}}_0$ of $X$ in $T^{\ast }X$, $\mathcal{U}$ of $X$ in $M$, and a diffeomorphism $\phi :{\mathcal{U}}_0\to \mathcal{U}$ such that$\begin{array}{ccc}{\mathcal{U}}_0& \stackrel{\phi }{\to }& {\mathcal{U}}_1\\ & \nwarrow i_0\nearrow i& \\ & X& \end{array}$ $\text{commutes and }{\phi }^{\ast }\omega ={\omega }_0$