Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose I have a Hamiltonian $H$ with first integrals $L_1, \ldots, L_k$ on a symplectic manifold $M$ such that $\{H,L_i\}=0$, and suppose that there are constants $c_{pq}^r$ such that $\{L_p,L_q\}=\sum_r c_{pq}^r L_r$. In this case $c_{pq}^r$ are structure constants of a Lie algebra $\mathfrak{g}$ (the Lie algebra spanned by $L_i$'s).

I have trouble in how to the construct the momentum map $\mu: M \rightarrow \mathfrak{g}$ in this general setting. Is $\mu$ just given by $(L_1,L_2, \ldots , L_k)$? Could somebody explain this to me?

share|improve this question
    
The moment map has values in the dual of $\mathfrak{g}$, i.e. $\mu:M\to\mathfrak{g}^*$. It is, indeed, just $(L_1,\dots,L_n)$. A value of $\mu$ is a value for each of $L_i$'s, that's why it's in the dual. –  user8268 Jul 8 '12 at 18:12
    
@user8268 So what does an element $P \in \mathfrak{g}^*$ correspond to with respect to $\mu$? Does it correspond to fixing the $L_i$ to a constant? –  Novo Jul 8 '12 at 19:05
    
Yes, that's it. –  user8268 Jul 8 '12 at 19:16
    
try mathoverflow? –  john mangual Jul 9 '12 at 3:47
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.