Let $(M,\omega)$ be a symplectic manifold, $(T^*M, \omega_{\rm can})$ its cotangent bundle, and consider the twisted product $M\times \overline{M} = (M\times M, \omega \oplus (-\omega))$.
Consider a symplectomorphism $\varphi \in {\rm Sp}(M,\omega)$ which is $C^1$-close to the identity ${\rm Id}_M: M \to M$.
We know that:
- the diagonal $\Delta \subseteq M\times \overline{M}$ is Lagrangian;
- the graph $\Gamma_\varphi\subseteq M \times \overline{M}$ is Lagrangian;
- $M \cong 0^{T^*M} \subseteq T^*M$ is Lagrangian.
Ok. Assume that $U$ is a neighborhood of $\Delta$ in $M\times M$ which contains $\Gamma_\varphi$, $U_0$ is a neighborhood of $M \cong \Delta$ in $T^*M$, and that $F: U \to U_0$ is a symplectomorphism such that $F(x,x) = x \cong 0_x$, for all $x \in M$.
Let $L = F(\Gamma_\varphi)$.
I want to check that there is a closed form $\sigma \in \Omega^1(M)$ $C^1$-close to the zero section such that $L = {\rm Im}\;\sigma$.
Since $F$ is a symplectomorphism, we have that $L$ is also Lagrangian, so once we find $\sigma$, it will be closed (I already have the result "${\rm Im}\,\sigma$ is Lagrangian iff ${\rm d}\sigma = 0$"). So I'd guess that it suffices to show that $L$ intersect each cotangent space to $M$ exactly once. But it is not clear to me that if $x \in M$, then $F(x,\varphi(x)) \in T_x^*M$.
The $C^1$-close to the zero section I guess it is automatic, in some sense, since $L\subseteq U_0$ (it was never made clear what exactly is meant by "$C^1$-close", although one can guess).
I'd like some help filling up these details, and to be sure nothing I said so far is wrong. Thanks!