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Let $\pi :T^{\ast }N\to N$ be the projection and $s_{\alpha }(q)=(q,{\alpha }_q)$ the section with image ${\Gamma }_{\alpha }$. For the canonical one-form on a cotangent bundle,Since $\pi \circ s_{\alpha }=\mathrm{id}$, evaluation on a tangent vector $v\in T_{q}N$ givesUse the same position-first standard symplectic form as before, $\omega =-d\theta$. Pullback commutes with the exterior derivative, soThe graph of a differential one-form is an embedded submanifold, because its projection is the inverse of the section on the graph. Its dimension is $\dim N$, half that of $T^{\ast }N$. Therefore the vanishing of the pulled-back symplectic form is exactly the Lagrangian submanifold condition, andThis proves that the graph of a closed one-form is Lagrangian using the intrinsic one-form rather than a coordinate guess.

A Lagrangian submanifold of a $2n$-dimensional symplectic manifold is an embedded $n$-dimensional submanifold on which the symplectic form restricts to zero. The dimension requirement distinguishes it from a lower-dimensional isotropic submanifold.

A differential one-form $\sigma$ on $L$ gives an embedded section of its cotangent bundle, since $\pi \circ \sigma =\mathrm{id}$. Directly from the canonical one-form on a cotangent bundle,Its graph already has half the ambient dimension. ThereforeThis proves the graph of a closed one-form is Lagrangian criterion, in both directions.

Use the Weinstein neighborhood theorem: a neighborhood of a compact Lagrangian submanifold is symplectomorphic to a neighborhood of the zero section of its cotangent bundle, with the identification equal to the identity on that submanifold. Take the canonical sign $-d\lambda$ in this identification. We also use C1 openness of diffeomorphisms: on a compact manifold, all smooth self-maps sufficiently close in the $C^1$ topology to a fixed diffeomorphism are themselves diffeomorphisms.

In $M\times M$ put $\Omega =-{\mathrm{pr}}_1^{\ast }\omega +{\mathrm{pr}}_2^{\ast }\omega$. The diagonal $\Delta$ is Lagrangian. For a symplectomorphism $\varphi$, its graph ${\Gamma }_{\varphi }$ is also Lagrangian, because its pullback of $\Omega$ is $-\omega +{\varphi }^{\ast }\omega =0$.

If $\varphi$ is sufficiently $C^1$-close to the identity, ${\Gamma }_{\varphi }$ lies in the fixed Weinstein neighborhood of $\Delta$. Its image in $T^{\ast }\Delta$ is transverse to the cotangent fibers and is a section: the projection of that image to $\Delta \cong M$ is $C^1$-close to the identity, hence is a diffeomorphism on compact $M$. Reparametrizing by this projection identifies the image with $\mathrm{graph}(\sigma )$ for a small differential one-form $\sigma$.

The preceding graph of a closed one-form is Lagrangian criterion gives $d\sigma =0$. Since $H_{\mathrm{dR}}^1(M)=0$, the de Rham cohomology definition gives $\sigma =df$. Intersections with the zero section are precisely the critical points of $f$; under the neighborhood identification these are the intersections ${\Gamma }_{\varphi }\cap \Delta$, hence the fixed points of $\varphi$.

On a nonempty compact manifold without boundary, $f$ has a maximum and a minimum. If it is nonconstant, these occur at distinct critical points. If it is constant, $df=0$ everywhere, so the whole graph is the diagonal and every point is fixed. For positive-dimensional $M$, there are at least two distinct fixed points. This is the nearby exact Lagrangian intersection lemma applied to the diagonal. No connectedness assumption is needed.

The usual positive-dimensional convention is necessary for the assertion: if zero-dimensional symplectic manifolds are allowed, a single-point $M$ has $H^1=0$ and only one fixed point. That is a literal exception to the printed statement.

Let $\pi :T^{\ast }X\to X$ be the cotangent bundle projection and $\lambda$ its canonical one-form on a cotangent bundle, defined by ${\lambda }_{(x,\xi )}(v)=\xi (d\pi (v))$. Use the symplectic form $\omega =-d\lambda$, so in local coordinates $\omega ={\sum }_{j}d{x}_j\wedge d{\xi }_j$.

The graph map $s_{\beta }(x)=(x,{\beta }_x)$ is a smooth embedding, with $s_{\beta }^{\ast }\lambda =\beta$, and henceIts image has dimension $\dim X$, half the dimension of $T^{\ast }X$. By the definition of a Lagrangian submanifold,This is the graph of a closed one-form is Lagrangian criterion. The opposite conventional sign for the symplectic form gives the same criterion.

The Weinstein neighborhood theorem identifies a neighborhood of $X$ in $M$ with a neighborhood of the zero section in $T^{\ast }X$. If $j$ is sufficiently $C^1$-close to the inclusion, projection of its image to $X$ is a diffeomorphism. After reparametrization, $j(X)$ is therefore the graph of a small one-form $\beta$.

The graph of a closed one-form is Lagrangian criterion says that $j(X)$ is Lagrangian exactly when $d\beta =0$. Since $H^1(X;\mathbb{R})=0$, write $\beta =df$. Intersections of $j(X)$ with $X$ are the critical points of $f$. A smooth function on a compact manifold has a maximum and a minimum; if they coincide as points because $f$ is constant, every point is an intersection. Thus there are at least two intersection points, proving the nearby exact Lagrangian intersection lemma.

The cohomology hypothesis is necessary. Take the zero section in $T^{\ast }{S}^1$ and the graph of the arbitrarily small nowhere-zero closed one-form $\epsilon ,d\theta$. Both are Lagrangian and disjoint.

With $\omega =-d\alpha$ and ${\omega }_{\sigma }=\omega +{\pi }^{\ast }\sigma$, the one-form graph of $\theta$ is a Lagrangian submanifold exactly when $d\theta =\sigma$. Indeed its pulled-back symplectic form is $\sigma -d\theta$ and its dimension is half that of the total space. This extends graph of a closed one-form is Lagrangian.