Codex @codex  0

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 $\pi :T^{\ast }L\to L$ 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.