Timeline for Question about the definition of hamiltonian group action.
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2011 at 23:26 | comment | added | José Figueroa-O'Farrill | The question on whether it is a homomorphism or anti-homomorphism is purely a matter of convention. It has to do with how you define $X^*$. Make a choice and stick with it. Nothin of relevance changes. | |
| Sep 26, 2011 at 21:53 | comment | added | Gigou | Lie algebra anti-homomorphism, I find indeed that $[X,Y]^*=-[X^*,Y^*]$. So I ask you: did you meant to say "anti-homomorphism" in your post, and if so, what is wrong with my proof of $[X,Y]^*=[X^*,Y^*]$? | |
| Sep 26, 2011 at 21:50 | comment | added | Gigou | Hey. I thought I was done with this but it came back to haunt me! You said the map X-->X* is a Lie algebra homomorphism, and I thought I could prove this using Paul Skerritt's insight that $X^* =\Phi^p_*(X)$ by using the fact that if $V_i$, $W_i$ (i=1,2) are F-related at p, then so are $[V_1,V_2]$ and $[W_1,W_2]$. Hence, $[X,Y]^*_p=\Phi^p_*([X,Y]_e)=[\Phi^p_*(X),\Phi^p_*(Y)]_p=[X^*,Y^*]_p$. BUT, Silva says that X-->X* is supposed to be a Lie-algebra anti-homomorphism. And if I assume the existence of a moment map (as per your definition), then using that $f\rightarrow X_f$ is a | |
| Sep 21, 2011 at 22:32 | comment | added | José Figueroa-O'Farrill | "locally function" of course means "locally constant function". (I wish I could edit comments...) | |
| Sep 21, 2011 at 22:32 | comment | added | José Figueroa-O'Farrill | The locally function in 1. cannot always be chosen so that 2. is satisfied. The obstruction is the class of the cocycle $c$ in the Lie algebra cohomology and that need not be zero. In a way, of course, this is just a tautology, but phrasing it in terms of Lie algebra cohomology allows you to conclude that in many cases (e.g., semisimple Lie algebras) there is no obstruction. $$ $$ The map $X \to X^*$ is a Lie algebra homomorphism independently of whether the action is hamiltonian or even symplectic. It's simply the fact that you have an action of a Lie group on a manifold. | |
| Sep 21, 2011 at 21:27 | comment | added | Gigou | By the way, just so we're clear, I agree that X-->X* is a Lie algebra homomorphism PROVIDED the action is hamiltonian (i.e., provided there exists a moment map for it, and in particular, provided X* is hamiltonian for each X in Lie(G)). But of course this is irrelevant for my question because I work with the hypothesis that I have only a basis of X's whose X*'s are hamiltonian. And using only this hypothesis, I want to show that every Y in Lie(G) induces a hamiltonian Y*. One way to do this is if we had (X+Y)*=X*+Y*, but without a moment map, how do we prove this? | |
| Sep 21, 2011 at 20:45 | comment | added | Gigou | Hello José, thanks for answering my question. I am not saying that 1. implies 2., but I am saying that the locally constant function in 1. can be chosen so that 2. is satisfied. Is this false? (Sorry, I don't understand the Lie algebra cohomology argument.) Now concerning my actual question, which you answer by the affirmative (i.e. (X+Y)*=X*+Y*), can you provide an explanation or a reference? Can you read this property just from properties of the exponential map? | |
| Sep 21, 2011 at 19:56 | history | answered | José Figueroa-O'Farrill | CC BY-SA 3.0 |