Lagrangian Submanifolds under symplectomorphisms Let $(X_1, \omega_1)$ and $(X_2, \omega_2)$ be two sympletic manifolds, and let $\varphi:(X_1, \omega_1) \to (X_2,\omega_2)$ be a symplectomorphism, i.e., $\varphi^*\omega_2 = \omega_1$.
Is it true that $L\subset(X_1, \omega_1)$ is a lagrangian submanifold iff $\phi(L)\subset(X_2,\omega_2)$ is a lagrangian submanifold?
I tried to prove this using the pullback of the respective inclusion mappings, but wasn't able to, even though it seems reasonable for it to be true.
 A: Remark that if $f:M\rightarrow N$ is a diffeomorphism and $L$ a submanifold of dimension $p$, $f(L)$ is a submanifold of dimension $p$ of $N$.
$L$ is Lagrangian is equivalent to $dim(L)=dim(X_1)/2$, and for every $x\in L, u,v\in T_xL, \omega^1_x(u,v)=0$ where $\omega^1$ is the symplectic form of $X_1$.
$dim(\phi(L))=dim(L)=dim(X_1)/2=dim(X_2)/2$. Let $u',v'\in T_{\phi(x)}\phi(L)$, there exists $u,v\in T_xL$ such that $d\phi_x(u)=u', d\phi_x(v)=v'$. We have $\omega^2_{\phi(x)}(u',v')=\omega_x(u,v)=0$
A: *

*$\dim L = \dim \varphi[L]$ since $\varphi$ is in particular a diffeomorphism.


*Let $\iota_L:L\to X_1$ and $\iota_{\varphi[L]}:\varphi[L]\to X_2$ be the inclusions. Compute: $$\iota_{\varphi[L]}^*\omega_2 = (\varphi\circ \iota_L\circ\varphi^{-1})^*\omega_2=(\varphi^{-1})^*\iota_L^*\varphi^*\omega_2=(\varphi^{-1})^*\iota_L^*\omega_1.$$So $\iota_L^*\omega_1=0$ (i.e., $L$ is isotropic) if and only if $\iota_{\varphi[L]}^*\omega_2=0$ (i.e., $\varphi[L]$ is isotropic), since $(\varphi^{-1})^*$ is a linear isomorphism.
Combine (1) and (2) to conclude $L$ is Lagrangian if and only if $\varphi[L]$ is Lagrangian.
