# Momentum maps and matrix poisson brackets.

I'm trying to understand how this solution works. The question was to deduce momentum maps for right and left actions of $SO(3)$ on $GL(3,R)$, and got them as $J_R = \frac{1}{2}(Q^TP-P^TQ)$ and $J_L=\frac{1}{2}(PQ^T-QP^T)$ respectively, for $Q\in GL(3)$, $P\in GL(3)^*$. However, then I need to find if they Poisson commute. The solution is:

$$\{ J_L,J_R \} = \frac{1}{4}\{(PQ^T-QP^T),(Q^TP-P^TQ)\} = \frac{1}{4}((P-P^T)(Q-Q^T)-(Q^T-Q)(P-P^T))$$

But I don't understand how this works (specifically, how to compute a poisson bracket with matrices). There is no explicit definition of the poisson bracket in the solution. Elsewhere in the text there was a Poisson bracket defined as

$$\{F,H\}=\operatorname{ tr}\left(\frac{\partial F}{\partial Q}\frac{\partial H}{\partial P}-\frac{\partial H}{\partial Q}\frac{\partial F}{\partial P}\right)$$

but that is for scalar functions $F,H$. I guess you could apply that formula to every entry of the $P,Q$'s but there must be a shortcut. Sorry if i've not given enough context, please ask. Thanks for any help.

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Perhaps there is a problem with the notation to begin with: I guess your Poisson brackets should include a tensor product somewhere. Could you write down the Poisson brackets of the entries of your matrices? –  mathphysicist Jun 30 '11 at 12:32

Below I write as I would proceed, hoping it could be useful to you, but, by the way, could I know the motivation of this problem?

I would prefer to work on the real associative unitary algebra $g\equiv\mathfrak{gl}(n)$, for arbitrary $n$, instead of its unit group $GL(n)$. Let $\Phi_{R(L)}$ be the natural right (resp. left) action of $SO(n)$ on $g$, and $Psi_{R(L)}$ its lift to $T^\ast g$.

The canonical symplectic $2$-form $\omega$ on $T^\ast g=g\times g^\ast$ is constant and given by the bilinear product $\omega((A_1,f_1),(A_2,f_2))=f_1(A_2)-f_2(A_1)$, for all $(+(A_1,f_1),(A_2,f_2)\in g\times g^\ast$.

Let us identify $g$ with $g^\ast$ through the linear isomorphism $A\mapsto \langle A,\cdot\rangle$, where $\langle,\rangle$ is the scalar product on $g$ defined by $\langle A,B\rangle=\textrm{tr}(A^T B)$ for all $A,B\in g$. Consequently let us identify $T^\ast g$ with $g\times g$.

With this idenifications we find the following expression for the symplectic form and the actions:

$\omega((A_1,B_1),(A_2,B_2))=\mathrm{Tr}(B_1^T A_2-B_2^T A_1)$,

$\Psi_R(O,(A,B))=(AO,BO)$, $\Psi_L(O,(A,B))=(OA,OB)$.

For any $X\in\mathfrak{so}(n)$, let $\zeta_{R(L)}^X$ dentote the fundamental vector field on $g\times g$ corresponding to $X$ w.r.t. the action $\Psi_{R(L)}$. We find the following expressions:

$\zeta_R^X(A,B)=(AX,BX)$, $\zeta_L^X(A,B)=(XA,XB)$.

Now it is easy to find that:

$(\zeta_R^X\omega)(A,B):(P,Q)\to\mathrm{Tr}((BX)^T Q-P^T (AX))=\mathrm{Tr}(X^T(B^T Q+P^T A)$ is equal to the differential of $J_R^X(A,B)=\mathrm{Tr}(X^T(B^T A))$, and

$(\zeta_L^X\omega)(A,B):(P,Q)\to\mathrm{Tr}((XB)^T Q-P^T (XA))=\mathrm{Tr}(X^T(QB^T+AP^T))$ is equal to the differential of $J_L^X(A,B)=\mathrm{Tr}(X^T(AB^T))$.

So the actions $\Psi_{R(L)}$ are hamiltonian with momentum maps $J_{R(L)}:T^\ast g\cong g\times g \to\mathfrak{so}(n)^\ast$ defined by $J_{R(L)}(A,B):X\in\mathfrak{so}(n)\to J_{R(L)}^X(A,B)$.

Finally, for arbitrary $X,Y\in\mathfrak{so}(n)$, we find that $\{J_R^X,J_L^Y\}\equiv\omega(\zeta_R^X,\zeta_L^Y)=0$.
Infact its value at an arbitrary $(A,B)\in T^\ast g\cong g\times g$ is equal to $\omega((AX,BX),(YA,YB))=\mathrm{Tr}((BX)^T YA-(YB)^T AX)=\mathrm{Tr}(-XB^T YA+B^TYAX)=0$.

This completes the proof.

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