[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*}[/step]

[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: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]