[proofplan]
We identify the graph $\Gamma_\alpha$ with $Q$ by the section $\alpha:Q\to T^*Q$. Under this identification, the restriction of the canonical symplectic form to the graph is detected by the pullback $\alpha^*\omega$. The tautological one-form identity gives $\alpha^*\lambda=\alpha$, so $\alpha^*\omega=-d\alpha$. Therefore the graph is isotropic exactly when $\alpha$ is closed, and since the graph has dimension $n$ while the ambient cotangent bundle has dimension
\begin{align*}
\dim T^*Q = 2n
\end{align*}
isotropicity is equivalent to being Lagrangian.
[/proofplan]
[step:Identify the graph as an embedded copy of $Q$]
Regard the one-form $\alpha\in\Omega^1(Q)$ as the smooth section
\begin{align*}
\alpha:Q\to T^*Q
\end{align*}
of the cotangent bundle projection $\tau:T^*Q\to Q$, so that $\tau\circ\alpha=\operatorname{id}_Q$. Its image is exactly $\Gamma_\alpha$.
The map $\alpha:Q\to T^*Q$ is an embedding onto $\Gamma_\alpha$. Indeed, it is injective because $\tau(\alpha_q)=q$, and its inverse on the image is the restriction
\begin{align*}
\tau|_{\Gamma_\alpha}:\Gamma_\alpha\to Q.
\end{align*}
Thus $\Gamma_\alpha$ is an embedded submanifold of $T^*Q$ diffeomorphic to $Q$, and
\begin{align*}
\dim \Gamma_\alpha=n.
\end{align*}
Since $\dim T^*Q=2n$, the graph is half-dimensional.
[/step]
[step:Compute the pullback of the canonical symplectic form to the graph]
Let
\begin{align*}
j:\Gamma_\alpha\hookrightarrow T^*Q
\end{align*}
denote the inclusion, and let
\begin{align*}
\beta:Q\to \Gamma_\alpha
\end{align*}
denote the diffeomorphism $q\mapsto \alpha_q$. Then $\alpha=j\circ\beta$. Since $\omega=-d\lambda$, naturality of [exterior derivative](/theorems/1525) under pullback gives
\begin{align*}
\alpha^*\omega=\alpha^*(-d\lambda)=-d(\alpha^*\lambda).
\end{align*}
By [citetheorem:10044], applied to the one-form $\alpha:Q\to T^*Q$ viewed as a section of $\tau$, we have
\begin{align*}
\alpha^*\lambda=\alpha.
\end{align*}
Therefore
\begin{align*}
\alpha^*\omega=-d\alpha.
\end{align*}
[guided]
The goal is to translate a geometric condition on the graph into an equation on the original one-form $\alpha$. The graph itself sits inside $T^*Q$, so the restricted two-form is $j^*\omega$, where
\begin{align*}
j:\Gamma_\alpha\hookrightarrow T^*Q
\end{align*}
is the inclusion. To compare this with a form on $Q$, use the diffeomorphism
\begin{align*}
\beta:Q\to \Gamma_\alpha
\end{align*}
defined by $\beta(q)=\alpha_q$. By construction, the original section factors as
\begin{align*}
\alpha=j\circ\beta.
\end{align*}
Hence
\begin{align*}
\alpha^*\omega=(j\circ\beta)^*\omega=\beta^*(j^*\omega).
\end{align*}
Now we compute $\alpha^*\omega$ using the convention in the statement:
\begin{align*}
\omega=-d\lambda.
\end{align*}
Pullback commutes with exterior derivative, so
\begin{align*}
\alpha^*\omega=\alpha^*(-d\lambda)=-d(\alpha^*\lambda).
\end{align*}
The needed identity for the tautological one-form is [citetheorem:10044]: for a one-form $\alpha:Q\to T^*Q$ viewed as a section of the cotangent bundle, its pullback of $\lambda$ is exactly the original one-form:
\begin{align*}
\alpha^*\lambda=\alpha.
\end{align*}
Substituting this identity into the previous display gives
\begin{align*}
\alpha^*\omega=-d\alpha.
\end{align*}
This is the central computation: the symplectic form restricted to the graph vanishes precisely when the exterior derivative of $\alpha$ vanishes.
[/guided]
[/step]
[step:Translate vanishing of the pullback into isotropicity of the graph]
Because $\beta:Q\to\Gamma_\alpha$ is a diffeomorphism, the pullback map
\begin{align*}
\beta^*:\Omega^2(\Gamma_\alpha)\to\Omega^2(Q)
\end{align*}
is injective. Since $\alpha^*\omega=\beta^*(j^*\omega)$, the equality
\begin{align*}
j^*\omega=0
\end{align*}
holds if and only if
\begin{align*}
\alpha^*\omega=0.
\end{align*}
Using the computation from the previous step, this is equivalent to
\begin{align*}
-d\alpha=0,
\end{align*}
and hence to
\begin{align*}
d\alpha=0.
\end{align*}
Thus $\Gamma_\alpha$ is isotropic in $(T^*Q,\omega)$ if and only if $\alpha$ is closed.
[/step]
[step:Use half-dimensionality to conclude the Lagrangian criterion]
For each point $q\in Q$, the tangent space $T_{\alpha_q}\Gamma_\alpha$ is an $n$-dimensional subspace of the symplectic [vector space](/page/Vector%20Space)
\begin{align*}
T_{\alpha_q}(T^*Q).
\end{align*}
This ambient symplectic vector space has dimension $2n$. By [citetheorem:10040], an isotropic subspace of dimension $n$ in a $2n$-dimensional symplectic vector space is Lagrangian.
Therefore $\Gamma_\alpha$ is Lagrangian if and only if it is isotropic. From the previous step, this holds if and only if
\begin{align*}
d\alpha=0.
\end{align*}
This proves the claimed equivalence.
[/step]
