Let $(M,\omega)$ be a symplectic manifold. To keep it simple, let us take $M = \mathbb{R}^{2n}$ with linear coordinates $(x^1,\ldots,x^n,y^1,\ldots,y^n)$ and the standard symplectic form $\omega = \sum_{i=1}^n dx^i \wedge dy^i$.

A submanifold $\Lambda$ of $M$ is said to be Lagrangian if, for each $p \in \Lambda$, $T_p\Lambda$ is a lagrangian subspace of $T_pM$, that is, $\omega_p \vert_{T_p\Lambda} \equiv 0$ and $\dim T_p\Lambda = \frac{1}{2}\dim T_pM$. Equivalently, if $\iota : \Lambda \hookrightarrow M$ is the inclusion map, then $\Lambda$ is Lagrangian if and only if $\iota^*\omega = 0$ and $\dim \Lambda = \frac{1}{2} \dim M$.

This seems like a reasonably easy definition, but in practice, how does one really construct a lagrangian submanifold of $(M,\omega)$? For instance, given a smooth function $f : U \subset \mathbb{R}^2 \to \mathbb{R}^4$, can one explicitely construct a Lagrangian immersion out of $f$?

I read a little bit about generating functions of Lagrangian submanifolds, but am unsure if this relates here.

Thank you for your help.

  • $\begingroup$ What do you mean by «construct a Lagrangian immersion out of f?»? $\endgroup$ Aug 10, 2014 at 0:23
  • $\begingroup$ Sorry for being unclear. What I meant is, can I find a map involving $f$ (or partials of $f$) such that its image forms a Lagrangian submanifold of $(M,\omega)$? $\endgroup$
    – Hubble
    Aug 10, 2014 at 0:27
  • $\begingroup$ Why would you expect such a thing? $\endgroup$ Aug 10, 2014 at 0:29
  • $\begingroup$ I find that graphs of functions are easier to work with, at least to some extent. By definition, it seems like it should be possible, since we only require the pullback of $\omega$ to vanish and the dimension condition, although this might be a hard problem. If you drop my example from my question, would you be able to answer it? $\endgroup$
    – Hubble
    Aug 10, 2014 at 0:34

1 Answer 1


Here are some facts that might help you, but I'm not sure if it is what you are looking for.

  • Consider a $1$-form $\mu$ as a map $\mu:M\to T^\ast M$. We know that the cotangent bundle always has a canonical symplectic structure. Let $\omega=-d\alpha$ be the canonical $2$-form on $T^\ast M$. It is not too hard to show that the graph of $\mu$, that is $\Gamma_\mu=\{(p,\mu_p) \ , \ p\in M\}$, is a Lagrangian submanifold of $T^\ast M$ if and only if $\mu$ is closed.
  • Let $(M_1,\omega_1)$ and $(M_2,\omega_2)$ be two symplectic manifolds. It's not too hard to check that $M_1\times M_2$ is symplectic. In fact, $a\pi_1^\ast\omega_1+b\pi_2^\ast\omega_2$ is a symplectic form for all non-zero $a$ and $b$. Taking $a=1$ and $b=-1$, we obtain the twisted form $\widetilde\omega=\pi_1^\ast\omega_1-\pi_2^\ast\omega_2$. Given a diffeomorphism $\varphi:M_1\to M_2$ it is a proposition that $\varphi$ is a symplectomorphism if and only if $\Gamma_\varphi$ is a Lagrangian submanifold of $(M_1\times M_2,\widetilde\omega)$.

We can combine these 2 bullets to find something related to what your question is about. Let $(M_1,\omega_1)$ and $(M_2,\omega_2)$ be symplectic manifolds. Let $f\in C^\infty(M_1\times M_2)$. It follows that $df$ is a closed $1$-form on $T^\ast M_1\times T^\ast M_2$. Hence by bullet 1, we have that $\Gamma_{df}$ is a Lagrangian submanifold of $T^\ast M_1\times T^\ast M_2$, with the natural symplectic form. In fact, its not hard to show that its a Lagrangian submanifold of $(T^\ast M_1\times T^\ast M_2,\widetilde\omega)$, where $\widetilde\omega$ is the twisted product form described above.

Hence, using bullet 2, the goal is to find a diffeomorphism $\varphi:T^\ast M_1\times T^\ast M_2$ such that $\Gamma_\varphi=\Gamma_{df}$. If you can find such a $\varphi$ then you know it is a symplectomorphism and so its graph is a Lagrangian submanifold.

It's often possible, locally, to find such a $\varphi$. In fact, finding such a $\varphi$ is governed by the implicit function theorem.

  • 1
    $\begingroup$ Not exactly what I was looking for, but very enlightening. I'll accept. $\endgroup$
    – Hubble
    Aug 10, 2014 at 19:26

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