Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let's assume $M$ is a symplectic manifold with the group action $G$. If $Lie(G)$ is semi simple then why the Hamiltonian condition, which requires the existence of linear map $Lie(G)\to C^{\infty}(M,R)$ is always satisfied?

share|improve this question
add comment

1 Answer

up vote 7 down vote accepted

Let $\mathfrak{g}=Lie(G)$. The action of $G$ on $M$ gives a morphism of Lie algebras $a:\mathfrak{g}\rightarrow Vect_{symp}(M)$.

Since $\mathfrak{g}$ has trivial abelianization, $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, i.e. any element can be decomposed into commutators. An easy computation shows that a commutator of two symplectic vector fields is a Hamiltonian vector field, so you can define a linear map $b:\mathfrak{g}\rightarrow C^\infty(M)$, which is not a morphism of Lie algebras in general.

Pick any two elements $x,y\in\mathfrak{g}$ and observe, that $b([x,y])-\{b(x),b(y)\}$ is a constant, since $\{b(x),b(y)\}$ is a Hamiltonian function for $[x,y]$. Call it $c(x,y)$: it defines a two-cocycle on $\mathfrak{g}$ which is furthermore trivial (by semisimplicity $H^2(\mathfrak{g})=0$). Therefore, there is an element $f\in\mathfrak{g}^*$, such that $b([x,y])-\{b(x),b(y)\}=c(x,y)=f([x,y])$.

Finally, define the map $\mathfrak{g}\rightarrow C^\infty(M)$ by $x\mapsto b(x)-f(x)$, you can easily check that it is a morphism of Lie algebras.

share|improve this answer
1  
A repackaging of the above: the map $C^\infty(M) \to Vect_{symp}(M)$ can be regarded as a central extension of Lie algebras (with kernel $H^0$ and cokernel $H^1$, both trivial Lie algebras). Pull it back along $a$, to get a central extension $\mathfrak g'$ of $\mathfrak g$, and a commuting square. Semisimple algebras have only trivial central extensions (the $H^2=0$ condition), i.e. there is a splitting ${\mathfrak g} \to {\mathfrak g}'$, which one follows with the map to $C^\infty(M)$. –  Allen Knutson Nov 30 '11 at 3:43
add comment

Your Answer

 
discard

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.