3
$\begingroup$

I am trying to prove the following: Let $(M,\omega)$ be a symplectic manifold and suppose that $X$ is a compact lagrangian submanifold of $M$ with $H^{1}_{\text{dR}}(X)=0$. Then, every lagrangian submanifold $Y$ of $M$ which is $C^1$-close to $X$ intersects $X$ in at least two points.

(A submanifold $Y$ of $M$ is $C^1$-close to $X$ if there is a diffeomorphism $h\colon X\longrightarrow Y$ such that $\iota \circ h$ is $C^1$ close to the inclusion $X\longrightarrow M$.

In order to prove it, due to the similarity, I think that it can be used the following theorem which I have already proven: If $(M,\omega)$ is a compact symplectic manifold such that $H^{1}_{\text{dR}}(M)=0$, then any symplectomorphism $\varphi \colon M \longrightarrow M$ $C^1$-close to $\text{id}$ has at least two fixed points.

However, I have no idea how to define $\varphi$ in order to apply the theorem because I need a symplectomorphism from $X$ to $X$.

Can anyone give me a clue? Or due you believe it is not possible to use the theorem? Is there any better idea to prove it? Thanks in advance.

(Both of the theorems are applications of Lagrange Neighborhood Theorem)

$\endgroup$

1 Answer 1

4
$\begingroup$

As $X$ is Lagrangian, the normal bundle of $X\subset M$ is naturally isomorphic to $T^*X$, the cotangent bundle of $X$. This means that every submanifold $X'\subset M$ which is $C^1$-close to $X$ can be thought of as the graph of a $1$-form on $X$.

Now, since the submanifold $Y$ in question is also Lagrangian, it corresponds to the graph of a closed $1$-form. By assumption, every closed $1$-form on $X$ is exact and thus vanishes at least at two points.

$\endgroup$
1
  • $\begingroup$ Why can X' be thought of as a the graph of a 1-form in X? Could you, please, say something more about this? $\endgroup$ Oct 10, 2018 at 20:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .