Homework 4 Problem 3

#### Problem 3

The Lie group $(\mathbb{S}^1, +)$ acts on $\mathbb{T}^2$ such that $\left(\alpha, (\theta, \phi)\right) \mapsto (\theta + \alpha, \phi)$. Show that the lifted action on $T \mathbb{T}^2$ leaves the Lagrangian from the preceding problem invariant, and compute the corresponding momentum map. Show that this momentum map is $\text{Ad}^*$ equivariant.

#### Shu's solution.

The Lagrangian for this system is

$L=KE-PE=\frac{1}{2}M(\frac{1}{2}R\dot{\theta})^{2}+\frac{1}{2}J\dot{\theta}^{2}+\frac{1}{2}m\left[ \left( \frac{1}{2}R\dot{\theta}\right) ^{2}+\left( l\dot{\phi}\right) ^{2}\right] +mgl\cos \phi =\frac{1}{8}(M+m)R^{2}\dot{\theta}^{2}+\frac{1}{2}J\dot{\theta}^{2}+\frac{1}{2}ml^{2}\dot{\phi}^{2}+mgl\cos\phi$

The fiber derivative is:

$\mathbb{F}L:TQ{\large \ \rightarrow }T^{\ast }Q:(\theta ,\phi ,\dot{\theta},\dot{\phi})\longmapsto (\theta ,\phi ,\frac{\partial L}{\partial \dot{\theta}},\frac{\partial L}{\partial \dot{\phi}})=(\theta ,\phi ,\frac{1}{4}(M+m)R^{2}\dot{\theta}+J\dot{\theta},ml^{2}\dot{\phi})$

The tangent lifted action is give by Jacobian $\left[ \begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$, i.e.

$\left( \alpha ,\left( \theta ,\phi ,\dot{\theta},\dot{\phi}\right) \right)\longmapsto \left( \theta +\alpha ,\phi ,\dot{\theta},\dot{\phi}\right)$

Since the Langrangian of the system doesn't include $\theta$ item, so the lifted action leaves the Langrangian invariant.

The momentum map is $\mathbb{J}:TQ\longrightarrow \mathfrak{g}^{\ast }:\left( \theta ,\phi ,\dot{\theta},\dot{\phi}\right) \longmapsto \left[ \mathbb{J}_{1},\mathbb{J}_{2}\right]$.

Defined by

$\left\langle \mathbb{J},\xi \right\rangle =\left\langle \mathbb{F}L,\mathbb{\xi }_{Q}\right\rangle \Longrightarrow \left[ \mathbb{J}_{1},\mathbb{J}_{2}\right] \left[ \begin{array}{c}\mathbb{\xi }_{1} \\ \mathbb{\xi }_{2}\end{array}\right] =\left\langle \frac{\mathbb{\partial }L}{\partial \dot{\theta}}d\theta +\frac{\mathbb{\partial }L}{\partial \dot{\phi}}d\phi ,\mathbb{\xi }_{1}\frac{\partial }{\partial \theta }+\mathbb{\xi }_{2}\frac{\partial }{\partial \phi }\right\rangle$, i.e.

$\mathbb{J}_{1}\mathbb{\xi }_{1}+\mathbb{J}_{2}\mathbb{\xi }_{2}=\frac{\mathbb{\partial }L}{\partial \dot{\theta}}\mathbb{\xi }_{1}+\frac{\mathbb{\partial }L}{\partial \dot{\phi}}\mathbb{\xi }_{2}$, i.e.

$\mathbb{J}_{1}=\frac{\mathbb{\partial }L}{\partial \dot{\theta}}=\frac{1}{4}(M+m)R^{2}\dot{\theta}+J\dot{\theta}$

$\mathbb{J}_{2}=\frac{\mathbb{\partial }L}{\partial \dot{\phi}}=ml^{2}\dot{\phi}$

The adjoint map $Ad_{(\alpha )}$ is given by $\left[ \begin{array}{c}\eta _{\theta } \\ \eta _{\phi }\end{array}\right] \longmapsto \left[ \begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \left[ \begin{array}{c}\eta _{\theta } \\ \eta _{\phi }\end{array}\right]$

$Ad_{(\alpha )}^{\ast }$ is given by $\left[ \begin{array}{c}\mathbb{J}_{\theta } \\ \mathbb{J}\phi \end{array}\right] \longmapsto \left[ \begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \left[ \begin{array}{c}\mathbb{J}_{\theta } \\ \mathbb{J}_{\phi }\end{array}\right]$

Therefore

$\left\langle \mathbb{J},Ad_{(\alpha )}\eta \right\rangle =\mathbb{J}_{\theta }\mathbb{\eta }_{\theta }+\mathbb{J}_{\phi }\mathbb{\eta }_{\phi}=\left\langle Ad_{(\alpha )}^{\ast }\mathbb{J},\eta \right\rangle$

So the momentum map is $\text{Ad}^*$ equivariant.

page revision: 19, last edited: 16 May 2007 21:56