Homework 4 Problem 6

#### Problem

The Lie group $SE(2)$ acts on $\mathbb{R}^2$ such that $\left((x, y, \theta), (q, p)\right) \mapsto (q \cos \theta - p \sin \theta + x, q \sin \theta + p \cos \theta + y)$. Endowed with the symplectic form $\Omega = dq \wedge dp$, $\mathbb{R}^2$ is a symplectic manifold and this action is canonical. Show that this action has a nonequivariant momentum map given by $\mathbb{J}: (q, p) \mapsto \left(-\frac{1}{2} \left(q^2 + p^2\right), p, -q\right)$.

#### Solution

I have been working on this, and I think the momentum map should be :

$\mathbb{J}: (q, p) \mapsto \left(p, -q , -\frac{1}{2} \left(q^2 + p^2\right) \right)$ corresponding to $\mathbb{J}_x, \mathbb{J}_y, \mathbb{J}_\theta$

-Here is what I think is the solution:

Lets call $\mathbb{R}^2$ as $P$, and the group action as $\Phi$, and the triple $(z,y,\theta)$ as $g$

(1)
\begin{align} \Phi(g,(q,p) ) \mapsto (q\cos\theta - p\sin\theta + x, q\sin\theta + p\cos\theta + y) \end{align}

First we try and find the infinitesimal generator $\xi_P$ due to $SE(2)$.

The exponential map of $SE(2)$ is given by:

(2)
\begin{align} (\xi_x,\xi_y,\xi_\theta) \mapsto \left( \frac{1}{\xi_\theta}(-\xi_x + \xi_y \cos \xi_\theta + \xi_x \sin \xi_\theta) , \frac{1}{\xi_\theta}( \xi_x - \xi_x \cos \xi_\theta + \xi_y \sin \xi_\theta) , \xi_\theta \right) \end{align}

Calculating $exp (t \xi)$

(3)
\begin{align} exp ( t \xi) \mapsto \left( \frac{1}{\xi_\theta}(-\xi_x + \xi_y \cos t\xi_\theta + \xi_x \sin t\xi_\theta) , \frac{1}{\xi_\theta}( \xi_x - \xi_x \cos t\xi_\theta + \xi_y \sin t\xi_\theta) , t\xi_\theta \right) \end{align}
(4)
\begin{align} \Phi(exp (t\xi), (q,p)) = \left( q\cos t\xi_\theta - p \sin t\xi_\theta + \frac{1}{\xi_\theta} (- \xi_y + \xi_y \cos t\xi_\theta + \xi_x \sin t \xi_\theta),q \sin t\xi_\theta + p \cos t\xi_\theta + \frac{1}{\xi_\theta} (- \xi_x + \xi_x \cos t\xi_\theta + \xi_y \sin t \xi_\theta) \right) \end{align}

Differentiating the above equation to get the infinitesimal generator,

(5)
\begin{align} \left. \xi_P &=\frac{d}{dt} \right |_{t = 0}(\Phi(exp(t\xi),(q,p))) \\ &= (-p \xi_\theta + \xi_x)\frac{\partial}{\partial q} + (q \xi_\theta + \xi_y)\frac{\partial}{\partial p} \end{align}

The momentum map satisfies:

(6)
\begin{align} < \mathbb{J}(q,p), \xi > = J(\xi)(q,p) \end{align}

where $J(\xi)$ satisfies:

(7)
\begin{align} X_{J(\xi)} = \xi_P \end{align}
(8)
\begin{align} X_{J(\xi)} (f) &= \{f , J(\xi) \} = \frac{\partial f}{\partial q} \frac{\partial {J(\xi)} }{\partial p} - \frac{\partial {J(\xi)} }{\partial q} \frac{\partial f}{\partial p} \end{align}

This gives,

(9)
\begin{align} X_{J(\xi)} = \frac{\partial {J(\xi)} }{\partial p}\frac{\partial }{\partial q} - \frac{\partial {J(\xi)} }{\partial q} \frac{\partial }{\partial p} \end{align}

Comparing Equation (5) and Equation (9), we get two PDE's

(10)
\begin{align} \frac{\partial J}{\partial p} &= -p \xi_\theta + \xi_x \\ J &= \frac{-p^2}{2}\xi_\theta + \xi_x p + c1 \end{align}
(11)
\begin{align} \frac{\partial J}{\partial q} &= -q \xi_\theta - \xi_y \\ J &= \frac{-q^2}{2}\xi_\theta - \xi_y q + c2 \end{align}

Choosing $c1$ and $c2$ so that $J$ is linear in $\xi$ , we get the following momentum map:

(12)
\begin{align} \mathbb{J}: (q, p) \mapsto \left(p, -q , -\frac{1}{2} \left(q^2 + p^2\right) \right) \end{align}

which is different from what is specified in the problem in that the order is permuted.

Note: This gives $J(\xi)(q,p)$ as

(13)
\begin{align} J(\xi)(q,p) = p \xi_x - q \xi_y - \frac{1}{2}\left( q^2 + p^2 \right) \xi_\theta \end{align}

To verify that the momentum map is indeed not equivariant, we will use equation (11.5.8) on page 379 from the text book, which says an equivariant momentum map satisfies:

(14)
\begin{align} J(Ad_g \xi)(\Phi(g,( q,p)) = J(\xi)(q,p) \end{align}

The right hand side is Equation (13). Lets compute the left hand side.

From lecture notes, the Ad map on $SE(2)$ is given by:

$Ad_g \xi \mapsto (\xi_x \cos \theta - \xi_y\sin\theta + x\xi_\theta, \xi_x \sin \theta + \xi_y\cos\theta + y\xi_\theta,\xi_\theta)$

Substituting this in the left of Equation (14), which after many algebraic manipulations, gives:

(15)
\begin{align} J(Ad_g \xi)(\Phi(g,( q,p)) &= (p + y \cos \theta - x \sin \theta)\xi_x + (-q - y \sin\theta - x \cos\theta)\xi_y \\ - \frac{1}{2} (q^2 + p^2 + 2 (qx\cos\theta - px \sin\theta + qy\sin\theta + py\cos\theta) \\ + xq\sin\theta + xp\cos\theta -yq\cos\theta + yp\sin\theta )\xi_\theta \end{align}

This is certainly different from Equation (13), and hence the momentum map is not equivariant.

—Done by Jehanzeb

page revision: 15, last edited: 19 May 2007 02:11