Topics in Geometric Mechanics: Week 11

Recall

We left off with:

Poisson algebras;
Poisson structures;

Poisson manifolds

Let $M$ be a smooth manifold, $\pb{}{}$ a Poisson bracket on $\cinfty{M}$. We call $(M, \pb{}{})$ a Poisson manifold. We call $\bivector{P}$, \begin{align*} \bivectorwargs{P}{df}{dg} &:= \pb{f}{g}, && \forall f, g \in \cinfty{M} \end{align*} a Poisson structure.

Example

Let $M=\R^3$ with its dot and cross products. Define, at $x \in \R^3$, \begin{align*} \bivectorwargs{P}{df}{dg}_x &= x \cdot (\nabla f \crossproduct \nabla g) && \forall f, g \in \cinfty{\R^3}. \end{align*}

The Poisson structure is a $(2, 0)$ skew-symmetric tensor field on $M$. Conversely, given a skew-symmetric $(2, 0)$ tensor field $\bivector{P}$, the equation \begin{align*} \bivectorwargs{P}{df}{dg} &:= \pb{f}{g}, && \forall f, g \in \cinfty{M} \end{align*} defines a Poisson bracket iff \begin{align*} J^{ijk} = \sum_{\alpha} \bivector{P}^{i \alpha} \didi{\bivector{P}^{jk}}{x_{\alpha}} + \bivector{P}^{j \alpha} \didi{\bivector{P}^{ki}}{x_{\alpha}} + \bivector{P}^{k \alpha} \didi{\bivector{P}^{ij}}{x_{\alpha}} \end{align*} where $(x_i)$ is a coordinate system and $i, j, k = 1, \ldots, \dim M$.

Properties

If $\bivector{P}$ is a Poisson structure and $X$ is a hamiltonian vector field with respect to $\bivector{P}$, then $\lieder{X}{\bivector{P}} = 0$.

Let $f \in \cinfty{M}$ and $X = \bivector{P} \cdot df$. Then $\pb{f}{g} = \bivectorwargs{P}{df}{dg} = \lieder{X}{g}$ for all $g \in \cinfty{M}$. Then \begin{align*} (\lieder{X}{\bivector{P}})(dg, dh) &= \lieder{X}{(\bivectorwargs{P}{dg, dh})} - \bivectorwargs{P}{\lieder{X}{dg}}{dh} - \bivectorwargs{P}{dg}{\lieder{X}{dh}}\\ &= \pb{f}{\pb{g}{h}} - \pb{\pb{f}{g}}{h} - \pb{g}{\pb{f}{h}} \\ &= \pb{f}{\pb{g}{h}} + \pb{h}{\pb{f}{g}} + \pb{g}{\pb{h}{f}} = 0 \\ \forall g, h \in \cinfty{M}. \end{align*} Since the cotangent space at any point is spanned by differentials of functions, this shows that $\lieder{X}{\bivector{P}} = 0$.

Symplectic manifolds

A Poisson structure $\bivector{P}$ is nowhere degenerate iff the $(0,2)$ tensor field $\Omega = \bivector{P}^{-1}$ is a symplectic form.

First, note that if $X = \bivector{P} \cdot df$, then $\iota_X \Omega = \Omega \cdot X = df$ since $\Omega \cdot \bivector{P} = 1$. % We need $\Omega$ to be non-degenerate and closed. The former is clear. For the latter, let $p \in M$ and $x, y, z \in T_p M$. There are smooth functions $f, g, h$ such that $x=\bivector{P}_{p} \cdot df_p, y=\bivector{P}_{p} \cdot dg_p, z=\bivector{P}_{p} \cdot dh_p$. Let $X = \bivector{P} \cdot df$, etc. so that $x=X(p)$, etc. \begin{align*} \cmag{X}{Y}{Z} \\ \cmag{Y}{Z}{X} \\ \cmag{Z}{X}{Y} \\ \implies 3 d\Omega(X, Y, Z) &= 0. \end{align*}

Poisson maps

A map $\phi : (M,\pb{}{}_M) \to (N,\pb{}{}_N)$ is a Poisson map iff it preserves Poisson brackets \begin{align*} \pb{f \circ \phi}{g \circ \phi}_M &= \pb{f}{g}_N \circ \phi && \forall f, g \in \cinfty{N}. \end{align*}

Examples

Let $\varphi_t : M \to M$ be the time-$t$ map of the hamiltonian flow of $f \in \cinfty{M}$. Then $\phi = \varphi_t$ is a Poisson map.
Let $\iota : \liealg{h} \hookrightarrow \liealg{g}$ be a subalgebra, let $\pi = \iota^* : \liealg{g}^* \to \liealg{h}^*$. Then $\pi$ is a Poisson map.
Let $f : M \to N$ be a diffeomorphism. Then $\phi = \D{f}^* : T^*M \to T^*N$ is a Poisson map.

Momentum map

Let $\liealg{g}$ be a Lie algebra, and $\liealg{h}$ a Lie algebra of smooth hamiltonians on $(M, \pb{}{})$. An homomorphism \begin{align*} \Psi &: \liealg{g} \to \liealg{h} && \xi \mapsto h_{\xi} \\ \text{induces}\\ \psi &: M \to \liealg{g}^* && \ip{\psi(x)}{\xi} = h_{\xi}(x). \end{align*} We call $\psi$ a momentum map.

Examples

Let $h \in \cinfty{M}$ and $\liealg{h} = \R \cdot h$, $\liealg{g} = \R$, and $\Psi(1) = h$. Then $\psi(x) = h(x)$. This is the tautological momentum map.
For $\liealg{g} = \symp{\R^{2n}}$, $\liealg{h} = \symalg{2}{\R^{2n}}$ from above, \begin{align*} \psi(z) &= \frac{1}{2} zz'J && \psi : \R^{2n} \to \symp{\R^{2n}}^* \equiv \symp{\R^{2n}}. \end{align*}
For $\liealg{g} = \sorth{\R^n}$ acting on $\R^{2n} = T^*\R^n$, \begin{align*} \psi(z) &= \frac{1}{2} (px' - xp') && \psi : \R^{2n} \to \sorth{\R^n}^* \equiv \sorth{\R^n}. \end{align*}

Let $\psi : M \to \liealg{g}^*$ be a momentum map. Then $\psi$ is a Poisson map.

Momentum maps and symplectic manifolds

Let $(M, \Omega)$ be a symplectic manifold, $\liegp{G}$ a Lie group that acts by Poisson diffeomorphisms. For $\xi \in \liealg{g}$, define the vector field on $M$ \begin{align*} \xi_M(x) &= \ddtat{t}{}{0} \exp(t \xi) \cdot x && \forall x \in M. \end{align*}

If $H^1(M)$ and $H^2(\liealg{g})$ vanishes, then there is a momentum map $\psi : M \to \liealg{g}^*$ such that the hamiltonian of $\xi_M$ is $\psi^* \xi$ for all $\xi \in \liealg{g}$.

If $H^1(M)$ vanishes, then all closed $1$-forms are exact. The locally hamiltonian vector field $\xi_M$ therefore has a hamiltonian $h=h_{\xi}$. We get a linear map $\Psi : \liealg{g} \to \cinfty{M}$, $\xi \mapsto h_{\xi}$. % We know that $h_{\liebracket{\xi}{\eta}}$ is a hamiltonian of $\liebracket{\xi}{\eta}_M$ and that $\pb{h_{\xi}}{h_{\eta}}$ is a hamiltonian, too. Therefore, \begin{align*} c(\xi, \eta) &= h_{\liebracket{\xi}{\eta}}(x) - \pb{h_{\xi}}{h_{\eta}}(x) \end{align*} is independent of $x \in M$ and so it defines a skew-symmetric $2$-form on $\liealg{g}$. It is closed, and therefore exact, so there is a $\mu \in \liealg{g}^*$ such that \begin{align*} c(\xi, \eta) &= \ip{\mu}{\liebracket{\xi}{\eta}}. \end{align*} If we define \begin{align*} H_{\xi} &= h_{\xi} - \ip{\mu}{\xi}, \end{align*} then $\xi \mapsto H_{\xi}$ is a Lie algebra homomorphism.