Topics in Geometric Mechanics: Week 13

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Recall

We left off with:

• momentum maps;
• Poisson manifolds.

Recall - 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.

If $\psi$ is a momentum map, then it is a Poisson map.

The Momentum Map & Lie Groups

Recall, we showed that if $\Liegp{G}$ is a Lie group acting smoothly on $M$, then the canonical lift of this action to $T^* M$ has momentum map.

There are two (generally distinct) actions of $\Liegp{G}$ on itself: \begin{align*} \textrm{Left translation}: &&& (h, g) &&\mapsto&& L_h(g) = h \cdot g \\ \textrm{Right translation}: &&& (h, g) &&\mapsto&& R_{h^{-1}}(g) = g \cdot h^{-1}. \end{align*} For each $\xi \in T_1 \Liegp{G}$, there are two extensions: \begin{align*} \textrm{Left translation}: &&& \xi_-(g) = d R_g \xi,\\ \textrm{Right translation}: &&& \xi_+(g) = -d L_g \xi.&& \textrm{Note transposition.} \end{align*}

Let $\Liegp{G}$ act on $\Liegp{G}$ by left (resp. right) translations. The momentum map $\psi_-$ (resp. $\psi_+$) of the left (resp. right) translation action is \begin{align*} \psi_-(g, \mu_g) &= \coAd{g^{-1}}{\mu}, && \psi_+(g, \mu_g) = -\mu, \end{align*} where $\mu = (d L_{g^{-1}})^* \mu_g \in T_1^* \Liegp{G}$.

Let $g \in \Liegp{G}$, and $\mu_g \in T_g^* \Liegp{G}$. Then $\mu_g = (d L_{g^{-1}})^* \mu$ for some $\mu \in T_1^* \Liegp{G} \equiv \liealg{g}^*$. Let $\xi \in T_1 \liegp{G}$, so that $\xi_-(g) = d R_g \xi$. \begin{align*} \ip{\psi_-(g, \mu_g)}{\xi} &= h_{\xi_-}(g, \mu_g) = \ip{\mu_g}{\xi_-(g)} = \ip{(d L_{g^{-1}})^* \mu}{(d R_g) \xi} \\ &= \ip{(d R_g)^* (d L_{g^{-1}})^* \mu}{\xi} = \ip{\coAd{g^{-1}}{\mu}}{\xi}.\\ \ip{\psi_+(g, \mu_g)}{\xi} &= h_{\xi_+}(g, \mu_g) = \ip{\mu_g}{\xi_+(g)} = -\ip{(d L_{g^{-1}})^* \mu}{(d L_g) \xi} \\ &= -\ip{\mu}{\xi}. \end{align*}

Equivariance

Let $\Liegp{G}$ act by hamiltonian diffeomorphisms on $(M, \pb{}{})$ with momentum map $\psi : M \to \liealg{g}^*$. The momentum map is equivariant, that is, \begin{align*} \psi(g \cdot m) &= \coAd{g^{-1}} \psi(m) && \forall g \in \Liegp{G}, m \in M. \end{align*}

A Change of Variables Formula

Let $f : (M, \pb{}{}_M) \to (N, \pb{}{}_N)$ be a Poisson diffeomorphism, $h \in \cinfty{M}$. The hamiltonian vector $X_h$ satisfies \begin{align*} X_{h \compose f^{-1}} = d f \cdot X_h \compose f^{-1}. \end{align*}

Let $k \in \cinfty{N}$. Then \begin{align*} \ip{X_{h \compose f^{-1}}}{dk} &= \pb{h \compose f^{-1}}{k \compose f \compose f^{-1}}_N \\ &= \pb{h}{k \compose f}_M \compose f^{-1} = \ip{X_h \compose f^{-1}}{df^*(dk)} \\ &= \ip{df(X_h \compose f^{-1})}{dk}. \end{align*}

Let $g \in \Liegp{G}$ and $m \in M$. Let $\sigma_g(m) = g \cdot m$ be the diffeomorphism induced by $g$. For $\xi \in \liealg{g}$, we have \begin{align*} X_{h_{\xi} \compose \sigma_g}(m) &= d \sigma_{g^{-1}} \cdot X_{h_{\xi}} \compose \sigma_g(m) \\ &= \ddtat{t}{}{0} \sigma_{g^{-1}} \compose \sigma_{\exp(t \xi)} \compose \sigma_g(m) \\ &= \ddtat{t}{}{0} \sigma_{\exp(t \Ad{g^{-1}}{\xi})}(m) \\ &= X_{h_{(\Ad{g^{-1}}{\xi})}} (m). && \textrm{Thus,} \\ h_{\xi} \compose \sigma_g &= h_{(\Ad{g^{-1}}{\xi})}. \end{align*} This implies equivariance.

Coadjoint orbits

Let $\mu \in \liealg{g}^*$. \begin{align*} \coorbit{\mu} &= \set{ x=\coAd{g}{\mu} \mid g \in \Liegp{G}} \end{align*} is the coadjoint orbit of $\mu$.

Each coadjoint orbit is a symplectic manifold.