Codex @codex  0

A symplectic manifold is a smooth manifold $P$ with a closed, nondegenerate two-form $\omega$. Nondegeneracy forces $\dim P=2n$. Examples are ${\mathbb{R}}^{2n}$ with $\omega ={\sum }_{i}d{q}^i\wedge d{p}_i$, an oriented surface with a nowhere-zero area form, and every cotangent bundle with its canonical form.

A Poisson manifold has a bivector $\Pi$ such that $\{f,g\}=\Pi (df,dg)$ is an antisymmetric bilinear Poisson bracket, is a derivation in each argument, and satisfies the Jacobi identity. Its rank need not be full or constant. The zero bivector gives a Poisson manifold on any smooth manifold, including an odd-dimensional or nonorientable one. A nonzero example on ${\mathbb{R}}^3$ is the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$: it has rank two off the origin and rank zero at the origin, so it does not arise from a symplectic form on the whole three-dimensional space. Even on ${\mathbb{R}}^2$, $\Pi =x{\partial }_x\wedge {\partial }_y$ is Poisson, since its Jacobi condition is a three-vector and hence vanishes, but its degeneracy on $x=0$ prevents it from being the inverse of a global symplectic form. A manifold carrying a degenerate Poisson structure might separately admit some other symplectic structure; these are distinct claims.

At each point of a symplectic manifold, linear nondegeneracy gives a basis with $\omega ={\sum }_i{e}^i\wedge f^i$. Its top wedge product of differential forms isThus ${\omega }^n/n!$ is a global nowhere-zero volume form and selects a consistent symplectic orientation. This proves every symplectic manifold is orientable without choosing a global frame.

Let $\pi :T^{\ast }M\to M$ and let a point of the cotangent bundle be $(x,p)$ with $p\in T_x^{\ast }M$. Define the canonical Liouville one-form intrinsically by ${\theta }_{(x,p)}(V)=p(d\pi (V))$. In local coordinates it is $\theta =p_{i}d{q}^i$, soThis definition is global, $d\omega =0$ by $d^2=0$, and its coordinate matrix has invertible blocks. Hence $T^{\ast }M$ is symplectic for every smooth $M$, even if $M$ itself is not orientable.

The coordinate Jacobi condition for a Poisson bivector isNecessity follows by applying the Jacobi identity to three coordinate functions. For arbitrary functions, the second-derivative terms in the Jacobiator cancel by antisymmetry of $\Pi$; the terms left are exactly this coefficient multiplying $f_i{g}_j{h}_k$, proving sufficiency.

For a symplectic form, take ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. With these signs the matrix of $\Pi$ is $-{\omega }^{-1}$. Differentiation gives ${\partial }_{\ell }\Pi =\Pi ({\partial }_{\ell }\omega )\Pi$. The cyclic expression above consequently equalswhich vanishes because $d\omega =0$. This directly proves that a symplectic manifold defines a Poisson manifold.

Define the cotangent space $T_p^{\ast }M={\mathrm{Hom}}_{\mathbb{R}}(T_{p}M,\mathbb{R})$ and the cotangent bundle $T^{\ast }M={\coprod }_p{T}_p^{\ast }M$, with projection $(p,\alpha )\mapsto p$. A coordinate chart $x$ gives the dual basis $d{x}^1{|}_p,\dots ,d{x}^n{|}_p$. Write $\alpha ={\sum }_i{\xi }_{i}d{x}^i{|}_p$ and define the induced local trivialization byIf $y=f(x)$ and $\alpha ={\sum }_j{\eta }_{j}d{y}^j$, the chain rule gives $\xi =Df(x{)}^T\eta$. Therefore the cotangent coordinate transition isThe inverse Jacobian matrix is smooth wherever its determinant is nonzero, so this transition and its inverse are smooth. The cocycle follows either from the chain rule or by dualizing the tangent transitions.

Use these product charts to define the topology and smooth atlas. The Hausdorff and second-countability arguments are the same as for the tangent bundle: separate distinct fibres using the base, separate distinct covectors in one product chart, and use a countable coordinate cover. The fibre transitions are smooth linear isomorphisms, so the projection is a smooth vector bundle of rank $n$. Using every compatible base coordinate chart shows that this structure is natural and independent of chart choices.

If a vector bundle $E\to M$ has rank $k$, form the exterior power of a vector bundle fibrewise: its fibre at $p$ is ${\Lambda }^r{E}_p$. A vector bundle trivialization over $U$ induces one with fibre ${\Lambda }^r{\mathbb{R}}^k$. Transition matrices $A_{\alpha \beta }(p)$ induce ${\Lambda }^r{A}_{\alpha \beta }(p)$, which are smooth and obey the cocycle law because exterior powers preserve composition. They glue to a smooth bundle of rank $\left(\genfrac{}{}{0ex}{}{k}{r}\right)$; for $r>k$ it is the zero bundle. At $r=k$ it is the determinant line bundle.

A real line bundle is trivial precisely when it has a nowhere-zero smooth section: a section $s$ gives the bundle isomorphism $(p,t)\mapsto ts(p)$ from $M\times \mathbb{R}$. Apply this to ${\Lambda }^n{T}^{\ast }M$. Suppose first that an oriented atlas is given. Its local coordinate formsare positive multiples of one another on every overlap. Choose a subordinate smooth partition of unity ${\rho }_{\alpha }\ge 0$. Extend ${\rho }_{\alpha }{\nu }_{\alpha }$ by zero outside its chart and set $\nu ={\sum }_{\alpha }{\rho }_{\alpha }{\nu }_{\alpha }$. Local finiteness makes this smooth. At any point at least one weight is positive, and all summands are nonnegative multiples of the same positive top form. Therefore $\nu$ never vanishes, proving triviality of the line bundle.

Conversely let $\nu$ be a nowhere-zero section of ${\Lambda }^n{T}^{\ast }M$. In each sufficiently small connected chart write $\nu =a_{\alpha }d{x}_{\alpha }^1\wedge \cdots \wedge d{x}_{\alpha }^n$. Its coefficient has constant nonzero sign. If necessary reverse the first coordinate, making $a_{\alpha }>0$. On overlaps the equality of the two expressions givesso each transition determinant is positive. These charts cover $M$, provingFor dimension zero both assertions hold with the empty exterior product equal to one.

For the orientability of a vector bundle total space, write $\pi :E\to M$, with rank $r$ and orientable base. The vertical tangent vectors at $v\in E_p$ identify with $E_p$ by differentiating translation in that fibre. The differential of $\pi$ therefore gives the exact sequence of vector bundles on $E$Taking determinant line bundles givesTo see the determinant identification directly, wedge a vertical frame with any lifts of a base frame; altering a lift by a vertical vector does not change that top wedge. Since $M$ is orientable, $\det TM$ is trivial. Consequently $\det TE\cong {\pi }^{\ast }\det E$. If $\det E$ is trivial, so is its pullback, making the total space orientable. Conversely, if $E$ is orientable as a manifold, its tangent determinant is trivial; restricting this isomorphism to the zero section gives triviality of $\det E$. The tangent and cotangent determinant lines are dual, so either tests orientability. We obtain

For a cotangent bundle the base need not be orientable. If base coordinates change from $x$ to $y=y(x)$ with Jacobian $J$, fibre coordinates of the same covector change from $\xi$ to $\eta =J^{-T}\xi$. The full coordinate change on $T^{\ast }M$ has block JacobianThus every induced cotangent coordinate change preserves orientation, even when the base change reverses it. These charts form an oriented atlas on the total space, proving every cotangent bundle is orientable. Equivalently the canonical symplectic form of the cotangent bundle has a nowhere-zero top exterior power, as in cotangent bundle orientation.

The Riemannian volume form of an oriented $n$-dimensional Riemannian manifold is the unique positive $n$-form taking value one on every positively oriented orthonormal frame. In positive coordinates,The metric induces an inner product on exterior powers of the cotangent bundle, with wedges of distinct orthonormal covectors forming an orthonormal basis. The Hodge star is the pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ specified byIt is an isometry and satisfies $\ast \ast \alpha =(-1{)}^{p(n-p)}\alpha$. With the positive-sign convention, the Hodge Laplacian on differential forms isHere $d$ is the exterior derivative and $\delta$ its formal adjoint; $\delta$ on functions is zero. On functions this is the Laplace-Beltrami operatorThe convention $\mathrm{div}\mathrm{grad}$ gives the negative of this operator. All subsequent Laplacians use the nonnegative convention above.

On a compact oriented manifold without boundary, the Hodge decomposition theorem gives an orthogonal decomposition of smooth real forms,The three component forms are uniquely determined, although the potentials in the last two summands need not be unique. The space ${\mathcal{H}}^p$ is finite-dimensional, and each de Rham cohomology class has a unique harmonic representative. Integration by parts gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic forms are exactly the closed and coclosed forms. Set ${\Omega }^k=0$ outside $0\le k\le n$. Compactness without boundary is a hypothesis of this version of the theorem; orientation alone does not supply it.

A connection on $TM$ induces a dual connection on $T^{\ast }M$ and then connections on its tensor and exterior powers by the Leibniz rule. Explicitly, for a $p$-form $\alpha$,To prove the parallelism of the Riemannian volume form, use a local positively oriented orthonormal frame $E_i$, its dual coframe ${\theta }^i$, and write ${\nabla }_X{E}_j={\sum }_i{a}_{ij}(X)E_i$. Metric compatibility makes the matrix $a(X)$ skew-symmetric. Since ${\omega }_g={\theta }^1\wedge \cdots \wedge {\theta }^n$, differentiating the wedge yieldsOff-diagonal terms vanish because they repeat a coframe factor. Thus $\nabla {\omega }_g=0$; equivalently parallel transport preserves both the metric and the continuously chosen orientation.

Now let $M$ be compact, oriented and without boundary. For representing functionals on harmonic forms by wedge pairing, choose an $L^2$-orthonormal basis $h_1,\dots ,h_b$ of ${\mathcal{H}}^p$ and defineIf $\phi ={\sum }_a{c}_a{h}_a$, linearity givesThis also works when $b=0$, using the empty sum. Since the Hodge star commutes with the Hodge Laplacian, ${\psi }_0$ is harmonic of degree $n-p$.

For the full ambiguity of harmonic wedge-pairing representatives, write another representative as ${\psi }_0+\chi$. Because the star is an isometry,The forms $\ast \phi$ range over all of ${\mathcal{H}}^{n-p}$. Hence $\chi$ gives zero functional exactly when its harmonic projection is zero. Hodge decomposition proves that the complete answer isBoth summands in this ambiguity are permitted; it is not only an exact-form ambiguity. The exact and coexact component forms are orthogonal to every harmonic form, so every displayed choice works. Conversely a nonzero harmonic component would pair nontrivially with some $\ast \phi$, and therefore cannot occur. The harmonic representative $\ast h_f$ is unique; the arbitrary smooth representative generally is not. The formula also covers the endpoint degrees and any exceptional case where the ambiguity space is zero.

For a concrete coexact ambiguity, take the torus $(\mathbb{R}/2\pi \mathbb{Z}{)}^2$ with metric $d{x}^2+d{y}^2$. The form $\chi =\delta (\sin ydx\wedge dy)=\cos ydx$ annihilates every harmonic one-form by orthogonality, yet $d\chi =\sin ydx\wedge dy$ is nonzero. Thus an annihilating representative need not even be closed, let alone exact.

In cotangent bundle coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$, the Liouville one-form and the canonical symplectic form areThe twisted cotangent symplectic form is closed: pullback of a differential form commutes with the exterior derivative, soTo prove the form is nondegenerate, write a tangent vector in these coordinates as $(u,v)$ and test its pairing with an arbitrary vertical vector $(0,w)$. The base differential two-form has zero pairing with vertical vectors, henceA vector in the kernel of a linear map therefore has $u=0$. Pairing it next with every $(z,0)$ gives $-z^i{v}_i=0$, so $v=0$. Thus the kernel is zero at every point, independently of the size or rank of $\sigma$. Together with closedness, this proves ${\omega }_{\sigma }$ is a symplectic form. The argument is local only for convenience; the nondegenerate property is intrinsic and the coordinate charts cover the whole cotangent bundle.

The differential one-form $\theta$ defines a smooth section of a vector bundle, and its one-form graph is an embedded copy of $X$ because $\pi \circ \theta ={\mathrm{id}}_X$. The defining property of the Liouville one-form gives ${\theta }^{\ast }\alpha =\theta$. ConsequentlyThe one-form graph has dimension $n$, half the dimension of the cotangent bundle. It is therefore a Lagrangian submanifold precisely when this pullback of a differential form vanishes:This is the twisted Lagrangian graph criterion.

Now suppose the projection restricts to a diffeomorphism on a Lagrangian submanifold $L$. Its inverse followed by the inclusion defines a smooth section $s:X\to T^{\ast }X$, hence a global differential one-form $\theta$ with image $L$. The criterion forces $\sigma =d\theta$, so its class in de Rham cohomology is zero. Thus $[\sigma ]\ne 0$ rules out every such Lagrangian section. This cohomological obstruction to a Lagrangian section concerns graphs over the entire base; it does not claim that all Lagrangian submanifolds are absent.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

First observe that the Liouville one-form is zero as an ambient covector exactly on the zero section. Indeed, ${\alpha }_{(x,p)}=p\circ d\pi$, and $d\pi$ is surjective. Since $dg$ is invertible, $g^{\ast }\alpha =\alpha$ therefore implies that $g$ maps the zero section onto itself. It induces a diffeomorphism $f:X\to X$ defined by $g(x,0)=(f(x),0)$.

The map $g$ is a symplectomorphism, since it preserves $-d\alpha$. Let $E={\sum }_i{p}_i{\partial }_{p_i}$ be the vertical Liouville vector field. With the chosen sign convention,Preservation of both $\omega$ and $\alpha$ forces $g_{\ast }E=E$ because $\omega$ is nondegenerate. Its complete cotangent fiber-dilation flow isUniqueness of integral curves gives $g\circ D_s=D_s\circ g$. If $g(x,p)=(y,P)$, let $s\to -\infty$. Continuity and this commutation yieldThus $y=f(x)$ for every $p$: the whole cotangent fiber over $x$ maps into the fiber over $f(x)$. This step proves that $g$ covers $f$; it was not assumed.

Now write $g(x,p)=(f(x),P(x,p))$. For any $u\in T_{x}X$, choose a tangent vector to $T^{\ast }X$ projecting to $u$. Evaluating $g^{\ast }\alpha =\alpha$ on it givesSince $d{f}_x$ is invertible, $P(x,p)=p\circ (d{f}_x{)}^{-1}$. Thereforeand $f$ is unique. This rigidity of cotangent Liouville-form preservation uses smoothness at the zero section; it follows from the canonical form and its dilation dynamics on the entire cotangent bundle.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.

The passage to Hamiltonian mechanics uses the Legendre transform in mechanics. Assume a regular Lagrangian: the velocity Hessian matrix $({\partial }^{2}L/\partial v^i\partial v^j)$ is invertible. Thendefines a locally invertible map from the tangent bundle to the cotangent bundle. Write its local inverse as $v=v(t,q,p)$ and define the HamiltonianDifferentiating this expression, all terms containing $dv$ cancel because $p_i=L_{v^i}$. HenceThe Euler-Lagrange equations become Hamilton's equations:Conversely, a solution of Hamilton's equations satisfies $\dot{q}=v(t,q,p)$, hence $p=L_v(t,q,\dot{q})$, and $\dot{p}=L_q$ recovers the Euler-Lagrange equations. This proves local equivalence of the two descriptions.

On the cotangent bundle, choose the canonical symplectic form ${\omega }_{\mathrm{can}}={\sum }_{i}d{q}^i\wedge d{p}_i=-d\lambda$ and the convention ${\iota }_{X_H}{\omega }_{\mathrm{can}}=dH$. Then $X_H={\sum }_i(H_{p_i}{\partial }_{q^i}-H_{q^i}{\partial }_{p_i})$, so its integral curves are exactly the phase-space equations above. This sign convention is used throughout these solutions. For a hyperregular Lagrangian, the Legendre transform in mechanics is globally invertible and gives global equivalence; regularity alone only gives local equivalence. A singular velocity Hessian matrix may instead produce constraints, so the ordinary unconstrained argument does not apply to every Lagrangian.

Let $\pi :T^{\ast }L\to L$ be the cotangent bundle projection. The canonical one-form on a cotangent bundle is intrinsically defined byNo metric or coordinate choice is needed. In local coordinates $\lambda ={\sum }_i{p}_{i}d{q}^i$. ChooseIt is closed because $d^2=0$, and its coordinate matrix is $\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$, which is invertible. The intrinsic definition of $\lambda$ makes the forms agree under all cotangent coordinate changes. This is the canonical symplectic form; choosing $d\lambda$ instead is the opposite common sign convention.

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.

A tangent vector at $p$ is a derivation at a point: a real-linear map $v:C_p^{\infty }(M)\to \mathbb{R}$ on germs of smooth functions satisfyingAddition and scalar multiplication preserve this identity, defining the tangent space $T_{p}M$ as a vector space. A derivation annihilates constants. In a manifold chart $x=(x^1,\dots ,x^n)$ centered near $p$, the local factorization $f-f(p)={\sum }_i(x^i-x^i(p))f_i$, with $f_i(p)={\partial }_{i}f(p)$, gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$. The factorization follows by integrating the first derivatives along a straight segment in a sufficiently small coordinate ball. ThereforeConversely these coordinate derivatives satisfy the derivation identity, so they really give a basis rather than merely a spanning family.

The cotangent space is the dual vector space $T_p^{\ast }M=(T_{p}M{)}^{\ast }$. Its basis $d{x}^i{|}_p$ is characterized by $d{x}^i({\partial }_j)={\delta }_j^i$. If $\alpha =a_{i}d{x}^i=b_{j}d{y}^j$, the chain rule gives the cotangent coordinate transitionThe cotangent bundle is the disjoint union $T^{\ast }M={\coprod }_{p\in M}{T}_p^{\ast }M$, with projection $\pi (p,\alpha )=p$. For each base manifold chart $(U,x)$, defineIts inverse sends $(z,a)$ to $(x^{-1}(z),a_{i}d{x}^i{|}_{x^{-1}(z)})$. On overlaps the transition is $(x,a)\mapsto (y(x),b)$, with the displayed fibre-linear transformation. Its coefficients and inverse are smooth, so these maps give a smooth atlas in dimension $2n$. Give the total space the topology obtained by transporting the product topology through these charts. Different base points are separated by disjoint base neighborhoods; distinct points over the same base point are separated within one bundle chart. A countable base smooth atlas and countable product bases give second countability. Thus the total space is a Hausdorff, second-countable smooth manifold. The same charts are vector bundle trivializations, since $\pi$ is product projection and their fibre changes are invertible linear maps.

The intrinsic canonical one-form on a cotangent bundle isIn the bundle chart it is $\lambda ={\sum }_i{a}_{i}d{x}^i$. This definition is coordinate independent, hence its exterior derivative is the globally defined smooth differential two-formThis sign follows the fibre-first order in this problem; the equally common position-first symplectic form is $-d\lambda$. The volume formnever vanishes. A smooth manifold is orientable exactly when it admits a nowhere-zero top-degree form; its positive ordered bases determine a consistent orientation. Consequently $T^{\ast }M$ is an orientable smooth manifold, even when $M$ is not. This is cotangent bundle orientation. In dimension zero the same conclusion uses the nowhere-zero zero-form $1$.

A smooth vector field is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.