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Every cotangent bundle $T^{\ast }M$ is an orientable smooth manifold even when its base is not. The intrinsic canonical one-form on a cotangent bundle ${\lambda }_{(p,\alpha )}(W)=\alpha (d\pi W)$ has $d\lambda ={\sum }_{i}d{a}_i\wedge d{x}^i$. Its $n$th wedge power is the nowhere-zero form $n!d{a}_1\wedge d{x}^1\wedge \cdots \wedge d{a}_n\wedge d{x}^n$, which defines an orientation of the $2n$-dimensional total space. Using $-d\lambda$ gives another standard sign convention; either is a symplectic form.

For coordinates $x,y$ on a smooth manifold, fibre coefficients of a cotangent space element obey $b_j={\sum }_i{a}_i\partial x^i/\partial y^j$. Thus induced cotangent bundle charts change by $(x,a)\mapsto (y(x),b)$ with smooth, invertible, fibre-linear changes. The canonical one-form on a cotangent bundle satisfies ${\sum }_i{a}_{i}d{x}^i={\sum }_j{b}_{j}d{y}^j$, and differentiating proves invariance of its exterior derivative.

Translation by a differential one-form $\beta$ is $(x,\xi )\mapsto (x,\xi +{\beta }_x)$. It changes the canonical one-form on a cotangent bundle by ${\pi }^{\ast }\beta$. A closed differential form gives a symplectomorphism; an exact differential form gives a Hamiltonian diffeomorphism when compact support is not required.

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.

The cotangent fiber translation $f(q,p)=(q,p+{\alpha }_q)$ is a smooth diffeomorphism with inverse $(q,p)\mapsto (q,p-{\alpha }_q)$, and $\pi \circ f=\pi$. For a tangent vector $v$ at $(q,p)$, the intrinsic definition of the canonical one-form on a cotangent bundle givesThus $f^{\ast }\theta =\theta +{\pi }^{\ast }\alpha$. Taking its exterior derivative with the fixed sign convention yieldsIf $\alpha$ is a closed differential one-form, the last term vanishes, proving that $f$ is a symplectomorphism. In fact the converse holds too: the pullback by the zero section shows that ${\pi }^{\ast }(d\alpha )=0$ implies $d\alpha =0$.

A symplectic manifold is a smooth even-dimensional manifold $X^{2n}$ with a closed, nondegenerate differential two-form $\omega$. A Lagrangian submanifold $L\subset X$ has dimension $n$ and $\omega {|}_{TL}=0$; equivalently $T_{x}L=(T_{x}L{)}^{\omega }$ at every point.

On $T^{\ast }M$, let $\pi :T^{\ast }M\to M$ and define the canonical one-form on a cotangent bundle by ${\theta }_{(q,p)}(v)=p(d\pi (v))$. In local coordinates $\theta ={\sum }_i{p}_{i}d{q}_i$, so our sign convention givesThe intrinsic definition of $\theta$ makes this form globally defined. It is closed because $d^2=0$, and its displayed coordinate matrix is invertible. The zero section pulls $\theta$, and therefore ${\omega }_0$, back to zero and has half the total dimension. It is a Lagrangian submanifold.

We prove the Weinstein neighborhood theorem, rather than merely invoking it. Along a Lagrangian submanifold $L\subset (X,\omega )$, choose a compatible almost complex structure $J$; its existence is proved by the metric construction in Question 5. Then $JL$ at the tangent-bundle level means $J(TL)$, andIndeed, $Jv\in TL$ with $v\in TL$ would imply $0=\omega (v,Jv)$, forcing $v=0$. Also $\omega (Ju,Jv)=\omega (u,v)=0$ for $u,v\in TL$, so $J(TL)$ is a Lagrangian complement. Identify a normal vector $n\in J(TL)$ with the covector $\eta (v)=\omega (v,n)$. This is a vector-bundle isomorphism with $T^{\ast }L$, and under itwhich is exactly the form ${\omega }_0$ along the zero section. The tubular neighborhood theorem therefore supplies a smooth map $F$ from a neighborhood of that section to a neighborhood of $L$, restricting to the identity and with this derivative along $L$.

Write ${\omega }_1=F^{\ast }\omega$ and $\delta ={\omega }_1-{\omega }_0$. This closed differential form vanishes as an entire bilinear form along the zero section. The relative Poincaré lemma gives $\delta =d\beta$ with $\beta$ zero along the section. Explicitly, if $r_s(q,p)=(q,sp)$ and $R=\sum p_i{\partial }_{p_i}$, useThe homotopy identity gives $d\beta =\delta -r_0^{\ast }\delta =\delta$; the vanishing at the section ensures smoothness of this integral. The forms ${\omega }_t={\omega }_0+t\delta$ are nondegenerate on a sufficiently small neighborhood of the section for all $0\le t\le 1$. Solve uniquelyThe vector field $Y_t$ vanishes on the section. After another neighborhood shrink its flow ${\varphi }_t$ exists up to time one and fixes the section. For noncompact $L$, the sizes of these neighborhoods may vary with the base point; no compactness of $L$ is required for the local conclusion. By Cartan's magic formula,Thus $(F\circ {\varphi }_1{)}^{\ast }\omega ={\omega }_0$. This establishes the universal neighborhood model, with the Lagrangian submanifold identified with the zero section.

For the section $s_{\alpha }(q)=(q,{\alpha }_q)$, the canonical differential one-form satisfies $s_{\alpha }^{\ast }\theta =\alpha$. HenceIts dimension is already half that of $T^{\ast }M$, so vanishing of this pullback is sufficient as well as necessary.

Suppose now that a smooth open six-dimensional region were a fiber bundle with Lagrangian submanifolds diffeomorphic to $S^3$ as its fibres. Fix one compact fibre $L$. Smooth local triviality and compactness imply that the other sufficiently nearby fibres are $C^1$-close embeddings of $L$. In the Weinstein neighborhood, projection to $L$ restricts on each such fibre to a local diffeomorphism close to the identity. It has degree one, and compactness makes it a covering map, so it is a diffeomorphism. Each nearby fibre is therefore a graph of a closed differential one-form $\alpha$ on $S^3$. Since $H^1(S^3;\mathbb{R})=0$, this is $df$ for a smooth real function $f$. A maximum of $f$ exists by compactness, and there $df=0$. The graph meets the original zero section, contradicting disjointness of different fibres. No such Lagrangian three-sphere fibration exists. More generally this is the compact first-cohomology obstruction to a Lagrangian fibration.

A Lagrangian submanifold of a $2n$-dimensional symplectic manifold $(M,\omega )$ is an $n$-dimensional submanifold $L$ such that the pullback of $\omega$ to $L$ is zero. Equivalently, each tangent space $T_{x}L$ is its own symplectic orthogonal. The dimension condition makes it maximally isotropic: in a nondegenerate alternating vector space no isotropic subspace has dimension greater than $n$.

In geometric quantization, a line bundle with a connection on a vector bundle whose curvature is proportional to $\omega /\hslash$ provides the initial quantum-state space. A real polarization foliates phase space locally by Lagrangian leaves; requiring sections to be covariantly constant along these leaves reduces their independent variables by half. The curvature restricts to zero on a Lagrangian leaf, allowing local flat sections. Globally its holonomy can impose the familiar Bohr-Sommerfeld quantization condition. Thus Lagrangian submanifolds describe choices of quantum configuration data and the support of semiclassical states, rather than requiring simultaneous sharp values of all position and momentum coordinates.

For the thermodynamic application, regard $(S,V,N)$ as base coordinates $q^i$ and $(T,-P,\mu )$ as fiber coordinates $p_i$ on $T^{\ast }{\mathbb{R}}^3$. The canonical one-form on a cotangent bundle and the position-first symplectic form areThe negative sign in the pressure coordinate is essential. A smooth potential $M(S,V,N)$ determines the embeddingwhere the last three coordinates are ordered as $(T,-P,\mu )$: explicitly $T=M_S$, $P=-M_V$, and $\mu =M_N$, so the second fiber coordinate is $-P=M_V$. Accordingly, in the actual $p_i$ coordinates the graph isProjection to the base is the identity, so this is an embedded three-dimensional submanifold, even if the Hessian of $M$ is degenerate. The first law of thermodynamics gives ${\iota }_M^{\ast }\theta =dM$. HenceIt has half the dimension of the ambient cotangent bundle and is therefore Lagrangian. This proves the thermodynamic potential as a Lagrangian graph construction for every smooth choice of $M$, locally on any domain on which the potential is defined. No invertibility, extensivity or stability assumption is needed for this geometric conclusion.

In components the same vanishing follows from equality of mixed derivatives. The thermodynamic conjugate variables satisfyThese cancel the three coefficients of the pulled-back two-form. They are the integrability relations behind the local thermodynamic potential, and demonstrate directly why the first-law graph is Lagrangian.

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.

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.

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 $1$.

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 cotangent fiber translation $f$ is a diffeomorphism, with inverse $(x,\xi )\mapsto (x,\xi -{\beta }_x)$. Because $\pi \circ f=\pi$, the definition of the canonical one-form on a cotangent bundle givesTaking the exterior derivative yieldsIf $\beta$ is a closed differential form, this is $f^{\ast }\omega =\omega$, proving that the cotangent fiber translation is a symplectomorphism.

For a smooth function $M(q)$, its differential defines a Lagrangian submanifold of the cotangent bundle: pullback of the canonical one-form on a cotangent bundle is $dM$, so pullback of the symplectic form $-d\theta$ is zero. The graph has half the ambient dimension. With $q=(S,V,N)$ and momentum coordinates $(T,-P,\mu )$, the first law of thermodynamics identifies the equilibrium graph with $dM$. This geometric statement does not require extensivity or thermodynamic stability.