Timeline for Understanding moment maps and Lie brackets
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 6, 2010 at 23:44 | comment | added | Sam Lewallen | Sorry for yet another comment. I think I see what you're saying now: if our manifold M is compact, then a canonical splitting can be chosen, where we take the set of all functions which integrate to 0. Is that right? Therefore on a compact symplectic manifold, an "exact" action by G is always a "Hamiltonian" one. | |
| Feb 6, 2010 at 23:37 | comment | added | Sam Lewallen | Sorry, I never referred to the "exactness requirement" in my post, but I just mean precisely that the Lie algebra g gets sent to Hamiltonian vector fields. Then the "Hamiltonian" condition (see eg McDuff and Salamon) is precisely what I said above, I think. | |
| Feb 6, 2010 at 23:34 | comment | added | Sam Lewallen | In particular, as Ben said, there is a "S.E.S." of lie algebras 0 --> R ---> C^inf(M) --p-> Lie(Ham(M)) ---> 0 Using our rho : g ---> Lie(Ham(M)), which exists because of the exactness condition, we get a sequence 0 ---> R ---> p^-1(rho(g)) ---> rho(g) ---> 0, and the Hamiltonian condition corresponds to being able to find a lie-algebra splitting of this sequence. is that right? | |
| Feb 6, 2010 at 23:29 | comment | added | Sam Lewallen | Hmm I thought I was starting to understand things but now this comment confuses me again. Lie(Ham(M)) is a sub-lie-algebra of Lie(Symp(M)), right, because Ham(M) is just a subgroup of Symp(M)? If this is not right then I am misunderstanding your definitions. If it is right, then by "factor through the the Lie group homomorphism..." you just mean that the image lies in the subgroup Ham(M). In which case this is the "exactness" requirement that I referred to, and the "Hamiltonian" requirement is something more. | |
| Feb 6, 2010 at 23:00 | comment | added | Ilya Grigoriev | Right. I guess 1) wouldn't even make sense - composition of two symplectomorphisms from time-independent Hamiltonian might not come from a time-independent Hamiltonian. 3) should be equivalent to 2), I hope. And an element of Lie(Ham(M)) is just a Hamiltonian (which determines an infinitesmall Hamiltonian symplectomorphism, as they always do). So it's in $C^\infty (M)$. Since adding constants to a Hamiltonian doesn't change anything, we can assume it's normalized. | |
| Feb 6, 2010 at 22:40 | comment | added | user1835 | The definition of Ham(M) is your #2, namely symplectomorphisms that come from time dependent Hamiltonians. | |
| Feb 6, 2010 at 22:31 | comment | added | Ilya Grigoriev | Thank you, this seems to clear things up quite a bit. However, what is your exact definition of Ham(M)? Three possible definitions come to my mind, and I'm not sure which of them are the same. 1) Symplectomorphisms that come from time-independent Hamiltonians. 2) Symplectomorphisms that come from dependent Hamiltonians. 3) Symplectomorphisms that come from moment maps on some Lie group. (The last would make talking about it quite tautological, but I'm pretty sure it's equivalent to number 2). The reason this feels important is that I'm not sure how exactly you calculate Lie(Ham(M)). | |
| Feb 6, 2010 at 22:20 | history | answered | user1835 | CC BY-SA 2.5 |