MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.