Q:

Proving that a particular submanifold of the cotangent space is Lagrangian

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. 

A:

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.