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I have the following problem in my differential topology class: Let $M$ be an $n$-dimensional manifold and let $\omega$ denote the standard symplectic form on the cotangent space $T^*M$. Let $f \in C^{\infty}(M)$ and put $L_f = \{(p,\mathrm{d}f_p) \in T^*M: p \in M\}$. Prove that $L_f$ is a Lagrangian submanifold of $T^*M$ and that $L_f$ is diffeomorphic to $M$.

I can prove that $L_f$ is a submanifold of $T^*M$ without any difficulty and the proof that $L_f$ is diffeomorphic to $M$ is similarly easy. I am having difficulty with the computation that $L_f$ is Lagrangian however, so I was hoping that someone could at least get me started in the right direction here.

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up vote 2 down vote accepted

Here's a proof of a more general result. Here I let $\alpha \in \Omega^1(M)$, I define $$s_\alpha: M \longrightarrow T^\ast M,$$ $$p \mapsto (p, \alpha_p),$$ and I write $M_\alpha$ for the image of $s_\alpha$. Recall that the canonical symplectic form $\omega_{\text{can}}$ on $T^\ast M$ is exact, $$ \omega_{\text{can}} = - d\lambda,$$ where $\lambda$ is the Liouville form given locally by $$\lambda = \sum_{i = 1}^n \xi_i ~dx_i$$ and globally by $$\lambda_{(x,\xi)} = (d\pi_{(x,\xi)})^\ast \xi,$$ where $\pi: T^\ast M \longrightarrow M$ is the bundle projection.

With the above setup, I claim that $$s_\alpha^\ast \lambda = \alpha.$$ To see this, recall that by definition, $$\lambda_p = (d\pi_p)^\ast \xi$$ where $p = (x, \xi) \in T^\ast X$ and $\pi: T^\ast X \longrightarrow X$ is the bundle projection. Then we have $$\lambda_{s_\alpha(x)} = (d\pi_{s_\alpha(x)})^\ast \alpha_x,$$ and hence \begin{align*} (s^\ast_\alpha \lambda)_x & = (ds_\alpha)^\ast_x \lambda_p \\ & = (ds_\alpha)^\ast_x (d\pi_{s_\alpha(x)})^\ast \alpha_x \\ & = (d\pi_{s_\alpha(x)} \circ (ds_\alpha)_x)^\ast \alpha_x \\ & = d(\pi \circ s_\alpha)_x^\ast \alpha_x \\ & = \alpha_x, \end{align*} since $s_\alpha$ is a section and thus $\pi \circ s_\alpha = \mathrm{Id}_X$. Therefore $$s_\alpha^\ast \lambda = \alpha.$$

Theorem. Let $\alpha \in \Omega^1(M)$. Then $M_\alpha$ is a Lagrangian submanifold of $(T^\ast M, \omega_\mathrm{can})$ if and only if $\alpha$ is closed.

Proof. Since $M_\alpha$ is the image of the closed embedding $s_\alpha$, we have that \begin{align*} M_\alpha \text{ is Lagrangian} & \iff s_\alpha^\ast \omega_\text{can} = 0 & & \\ & \iff s_\alpha^\ast d\lambda = 0 & & (\omega_\text{can} = -d\lambda) \\ & \iff ds_\alpha^\ast \lambda = 0 & & (d \text{ commutes with pullbacks}) \\ & \iff d\alpha = 0, & & (s_\alpha^\ast \lambda = \alpha) \end{align*} from which the result follows.

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Recall that $\omega$ is defined as $d\theta$ for a certain canonical 1-form $\theta$ on $T^*M$. You have a map (diffeomorphism) $s:M\to L_f$, given by $s(p)=(p,df_p)$. Prove that $s^*\theta=df$ and conclude what is $s^*\omega$.

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