![[FSF Associate Member 10686]](http://static.fsf.org/nosvn/associate/fsf-10686.png)
|
Last Updated: Fri 06 Apr 2012 01:11:20 PM EDT |
| Navigation |
|---|
| MTH-696a |
| Course outline |
| Assignments & Solutions |
| Tutorials & Solutions |
| Schedule |
| Lecture Notes |
| Sources |
| Home Page of Leo Butler |
We left off with:
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.
Let $\psi : M \to \liealg{g}^*$ be a momentum map. Then $\psi$ is a Poisson map.
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.
Let $(T^* M, \Omega)$. Let $f : M \to M$ be a diffeomorphism of $M$. Define the canonical lift of $f$ to $T^*M$ by \begin{align*} F(x, p) &= (f(x), (d_x f^{-1})^* p). \end{align*}
The canonical lift is a symplectomorphism.
First, assume that $M$ is an open subset of $\R^n$. Let $x=(x_i)$ be a coordinate system in a neighbourhood of $x_0 \in M$ and $y_i=f_i(x)$ a coordinate system in a neighbourhood of $y_0=f(x_0)$. Let $p_i$ (resp. $q_i$) be linear coordinates on $T_x^* M$ (resp. $T_y^* M$) so that the Liouville $1$-form is $\theta_x = \sum_{i} p_i \D{x_i}$, $\theta_y = \sum_{i} q_i \D{y_i}$. Observe that $q = (d_x f^{-1})^* p$, i.e. $p = (d_x f)^* q$ and that $\D{y_i} = \sum_{\alpha} \didi{f_i}{x_\alpha} \D{x_{\alpha}}$. \begin{align*} F^* \theta &= \sum_i F^*(q_i \D{y_i}) = \sum_{i, \alpha} q_i \didi{f_i}{x_{\alpha}} \D{x_{\alpha}} = \sum_{\alpha} \left[ (df)^* q \right]_{\alpha} \D{x_{\alpha}} \\ &= \sum_{\alpha} p_{\alpha} \D{x_{\alpha}} = \theta. \end{align*} Since $F^*(\D{\theta}) = \D{(F^* \theta)} = \D{\theta}$, we are done.
Let $X$ be a vector field on $M$.
The hamiltonian vector field $Y$ of \begin{align*} H_X(x, p) &= \ip{p}{X(x)} \end{align*} satisfies \begin{align*} \D{\pi}(Y) &= X, && \text{where } \pi : T^*M \to M : (x, p) \mapsto x. \end{align*}
We call $Y$ the canonical lift of $X$.
We have that $\D{\pi}(\dot{x}, \dot{p}) = \dot{x}$. Now, use Hamilton's equations \begin{align*} \dot{x} &= \didi{H_X}{p} = X(x). \end{align*}
Let $\Liegp{G}$ be a Lie group that acts smoothly on $M$. Then, the canonical lift of this action is hamiltonian, with momentum map \begin{align*} \psi : T^* M \to \liealg{g}^* && \ip{\psi(x, p)}{\xi} = \ip{p}{\xi_M(x)} \end{align*} where $\xi_M(x) = \ddtat{t}{}{0} \exp(t \xi) \cdot x$ for all $x \in M$, $\xi \in \liealg{g}$.
It suffices to verify that if $X, Y$ are smooth vector fields on $M$, then \begin{align*} \pb{H_X}{H_Y} &= H_{\liebracket{X}{Y}}. \end{align*}
|
|
Last Updated: Fri 06 Apr 2012 01:11:20 PM EDT |