Tautological and Canonical Forms

Tautological Form

Let $(\mathcal{U},x_1,\dots ,x_n)$ be a coordinate chart for $X$, with associated cotangent coordinates. Let $X$ be any $n$-dimensional manifold and $M=T^{\ast }X$ its cotangent bundle. If the manifold structure on $X$ is described by coordinate charts $(\mathcal{U},x_1,\dots ,x_n)$ with $x_i:\mathcal{U}\to \mathbb{R}$, then at any $x\in \mathcal{U}$, the differentials $(d{x}_1{)}_x,\dots ,(d{x}_n{)}_x$ form a basis of $T_x^{\ast }X$. Namely, if $\xi \in T_x^{\ast }X$, then $\xi ={\sum }_{i=1}^n{\xi }_i(d{x}_i{)}_x$ for some real coefficients ${\xi }_1,\dots ,{\xi }_n$. This induces a map. $T^{\ast }\mathcal{U}⟶{\mathbb{R}}^{2n};(x,\xi )⟼(x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n).$ The chart $(T^{\ast }\mathcal{U},x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n)$ is a coordinate chart for $T^{\ast }X$. Define the 1-form on $T^{\ast }\mathcal{U}$ by $\alpha ={\sum }_{i=1}^n{\xi }_{i}d{x}_i$

Canonical Symplectic 2-Form

By the same setting for Tautological Form, define a 2-form $\omega$ on $T^{\ast }\mathcal{U}$ by $\omega ={\sum }_{i=1}^{n}d{x}_i\wedge d{\xi }_i$ We have $\omega =-d\alpha$

Remark

The above $\alpha$ and $\omega$ are all intrinsically defined.

Proposition

Let $X_1$ and $X_2$ be $n$-dimensional manifolds with cotangent bundles $M_1=T^{\ast }{X}_1$ and $M_2=T^{\ast }{X}_2$, and tautological 1-forms ${\alpha }_1$ and ${\alpha }_2$. Suppose that $f:X_1\to X_2$ is a diffeomorphism. Then there is a natural diffeomorphism $f_{♯}:M_1\to M_2$ which lifts $f$; namely, if $p_1=(x_1,{\xi }_1)\in M_1$ for $x_1\in X_1$ and ${\xi }_1\in T_{x_1}^{\ast }{X}_1$, then we define $f_{♯}(p_1)=p_2=(x_2,{\xi }_2)$ with $$\begin{gathered} x_2=f(x_1)\in X_2\text{ and} \\ {\xi }_1=(d{f}_{x_1}{)}^{\ast }{\xi }_2\text{ ,} \end{gathered}$$ where $(d{f}_{x_1}{)}^{\ast }:T_{x_2}^{\ast }{X}_2\stackrel{\cong }{\to }{T}_{x_1}^{\ast }{X}_1$, so $f_{♯}{|}_{T_{x_1}^{\ast }}$ is the inverse map of $(d{f}_{x_1}{)}^{\ast }$. The lift $f_{♯}$ of a diffeomorphism $f:X_1\to X_2$ pulls the tautological form on $T^{\ast }{X}_2$ back to the tautological form on $T^{\ast }{X}_1$, i.e., $(f_{♯}{)}^{\ast }{\alpha }_2={\alpha }_1.$ The lift $f_{♯}$ of a diffeomorphism $f:X_1\to X_2$ is a symplectomorphism, i.e., $(f_{♯}{)}^{\ast }{\omega }_2={\omega }_1$, where ${\omega }_1,{\omega }_2$ are the canonical symplectic forms. In summary, a diffeomorphism of manifolds induces a canonical symplectomorphism of cotangent bundles: $\begin{array}{rl}{f}_{♯}:& T^{\ast }{X}_1⟶T^{\ast }{X}_2\\ & \uparrow \\ f:& X_1⟶X_2\end{array}$

Symplectomorphism of $T^{\ast }X$

Let $(M,\omega )$ be a symplectic manifold, and let $\alpha$ be a 1-form such that $\omega =-d\alpha .$ There exists a unique vector field $v$ such that its interior product with $\omega$ is $\alpha$, i.e., ${\iota }_v\omega =-\alpha$. If $g$ is a symplectomorphism which preserves $\alpha$ (that is, $g^{\ast }\alpha =\alpha$), then $g$ commutes with the one-parameter group of diffeomorphisms generated by $v$, i.e., $(\exp tv)\circ g=g\circ (\exp tv).$

Corollary

Combine the previous two proposition, we actually get $\text{Diff}(X)\cong \{g\in \text{Symp}(T^{\ast }X)\mid g^{\ast }\alpha =\alpha \}$

Remark

If we consider $X$ as configuration space, $TX$ would corresponds to velocity space, $T^{\ast }X$ as the phase space, which means ${\xi }_i$ corresponds to momentum. Then $\alpha$ measures the projection of momentum along a displacement in configuration space and $\omega$ measures the oriented area in phase space.