Topics in Geometric Mechanics: Week 10

Week 10

We left off with:

the symplectic form;
the Hamiltonian;
invariants of the Hamiltonian vector field.

Volume Preservation

If $X$ is a Hamiltonian vector field with Hamiltonian function $H$, then $H$ is constant along flow lines, the volume form $\Omega^n$ is preserved and the restriction of $V = \alpha \wedge \Omega^{n-1}$ to $H^{-1}(c)$ is preserved, where $\alpha$ is a smooth $1$-form on a nhd of $H^{-1}(c)$ such that $\interiorproduc{X}{\alpha}=1$.

$H$ is constant along flow lines: $$ \lieder{X}{H} = \ip{\D{H}}{X} = \Omega(X, X) = 0, $$ since $\Omega$ is skew symmetric. $\Omega^k$ is preserved since $$\lieder{X}{\Omega^k} = (\lieder{X}{\Omega^{k-1}}) \wedge \Omega + \Omega^{k-1} \wedge \lieder{X}{\Omega}.$$ Finally, since $\interiorproduc{X}{\alpha} = 1$ near $H^{-1}(c)$, $\lieder{X}{\alpha}$ annihilates $X$ so \begin{align*} \lieder{X}{V} &= (\lieder{X}{\alpha}) \wedge \Omega^{n-1} && \text{since $\Omega^{n-1}$ is preserved}\\ &= f \D{H} \wedge \Omega^{n-1}. \end{align*} Note that $V$ is independent of the choice of $\alpha$.

Poisson Algebras

A Poisson algebra $(A, \cdot, \pb{}{})$ where $(A, \cdot)$ is an abelian algebra, and $(A, \pb{}{})$ is a Lie algebra s.t. \begin{align*} \pb{f}{g \cdot h} &= \pb{f}{g} \cdot h + g \cdot \pb{f}{h} && \forall f, g, h \in A. \end{align*}

Examples

Let $(A, \cdot)$ be an abelian algebra, define $\pb{f}{g} = 0$ for all $f, g \in A$.
Let $\liealg{g}$ be a Lie algebra. Let $A = \symalg{*}{\liealg{g}}$ with the symmetric product. For $f, g \in \symalg{1}{\liealg{g}} = \liealg{g}$, define \[ \pb{f}{g} = \liebracket{f}{g}. \] Extend this by Leibnitz' rule.
Let $\liealg{g}$ be a reflexive Lie algebra. Let $A = \cinfty{\liealg{g}^* ; \R}$ with pointwise multiplication and \begin{align*} \pb{f}{g}(x) &= \ip{x}{\liebracket{df(x)}{dg(x)}} && \forall f, g \in A, x \in \liealg{g}^*. \end{align*}
Let $(M, \Omega)$ be a symplectic manifold. Let $A = \cinfty{M;\R}$ with pointwise multiplication and \begin{align*} \pb{f}{g} &= -\Omega(X_f, X_g) && \forall f, g \in A. \end{align*}

First, \begin{align*} X_{\pb{f}{g}} &= \liebracket{X_f}{X_g}. \end{align*} Second, \begin{align*} & \pb{f}{\pb{g}{h}} + \pb{g}{\pb{h}{f}} + \pb{h}{\pb{f}{g}} \\ &= \lieder{X_f}{\lieder{X_g}{h}} - \lieder{X_g}{\lieder{X_f}{h}} - \lieder{X_{\pb{f}{g}}}{h} \\ &= \lieder{\liebracket{X_f}{X_g}}{h} - \lieder{X_{\pb{f}{g}}}{h} \\ &= 0. \end{align*} Third, \begin{align*} \pb{f}{g \cdot h} &= \lieder{X_f}{g \cdot h} = (\lieder{X_f}{g}) \cdot h + g \cdot (\lieder{X_f}{h})\\ &= \pb{f}{g} \cdot h + g \cdot \pb{f}{h}. \end{align*}

Subalgebras

Let $(M, \Omega)$ be $T^* \R^n$ with its canonical Poisson structure: \begin{align*} \pb{f}{g} &= \ip{\didi{f}{p}}{\didi{g}{x}} - \ip{\didi{f}{x}}{\didi{g}{p}}. \end{align*}

The vector space of quadratic hamiltonians is a Poisson subalgebra.

Let $z=(x, p)$. Relative to this, we can write the Poisson bracket as \begin{align*} \pb{f}{g} &= \dotp{J \nabla f}{\nabla g}, && J = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}. \end{align*} Let $f(z) = \frac{1}{2} \dotp{Az}{z}$, $g(z) = \frac{1}{2} \dotp{Bz}{z}$ for some $2n \times 2n$ symmetric matrices $A, B$. Then \begin{align*} \pb{f}{g} &= \dotp{J Az}{Bz} = \frac{1}{2} \dotp{Cz}{z}, && C=BJA-AJB. \end{align*}

The Symplectic Group Action

Let $\Symp{\R^{2n}}$ be the group of linear transformations that preserve $\Omega$. We have the actions \begin{align*} (g, z) &\mapsto gz \in \R^{2n} && g \in \Symp{\R^{2n}}, z \in \R^{2n}\\ (\xi, z) &\mapsto \xi z \in T_z\R^{2n} && \xi \in \symp{\R^{2n}}, z \in \R^{2n}. \end{align*}

Let $\symalg{2}{\R^{2n}}$ be the Poisson algebra of quadratic hamiltonians on $\R^{2n}$. Define \begin{align*} \psi : \symp{\R^{2n}} \to \symalg{2}{\R^{2n}} &&& \xi \mapsto f_{\xi}(z) = \frac{1}{2} \dotp{J\xi z}{z}. \end{align*} Then $\psi$ is a Lie algebra isomorphism.

First, $\xi \in \symp{\R^{2n}}$ implies that $J \xi = (J \xi)'$. Second, if $\xi, \eta \in \symp{\R^{2n}}$, then \begin{align*} \pb{f_{\xi}}{f_{\eta}}(z) &= \frac{1}{2} \dotp{Cz}{z} \\ &\text{where} && C = (J \eta)J(J \xi) - (J \xi)J(J \eta) = J\liebracket{\xi}{\eta}\\ &= f_{\liebracket{\xi}{\eta}}(z). \end{align*} This shows that $\psi : \xi \mapsto f_{\xi}$ is a Lie algebra homomorphism; it is clearly one-to-one. To see it is onto, let $A=A'$ and define $\xi = JA$. Then $\xi' J + J \xi = 0$, so $\xi \in \symp{\R^{2n}}$.

Let's note that $X_{f_{\xi}} = J(J\xi z) = - \xi z$.

Momentum map

Let $\liealg{g}$ be a Lie algebra, and $\liealg{h}$ a Lie algebra of smooth hamiltonians on $(M, \Omega)$. 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*}

Examples

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*}