All Questions
7 questions
7
votes
1
answer
279
views
Question about an example in symplectic geometry
Let M be a coadjoint orbit of dimension 6 of $SU(3)$, and let T be the maximal torus in $SU(3)$. If we denote $\mu : M \longrightarrow \mathbb{R}^2$ the moment map associated to the action of T on M, ...
- 133
3
votes
1
answer
105
views
Define a symplectic structure on $G \times_{G_\beta} V$, where $V$ is symplectic
Let $G$ be a compact Lie group with algebra $\mathfrak{g}$. Let $\beta $ be an element in the dual of the Lie algebra $\mathfrak{g}$. We denote by $G_\beta$ the stabilizer subgroup of $\beta$ by ...
- 139
3
votes
0
answers
74
views
Coordinates for quasiperiodic motion after reconstruction
Consider a free action of $SO(3)$ on a manifold $M$ and some (reducible) dynamics vector field $X$ on $M$. Suposse that the reduced dynamics $X_{red}$ on $M/SO(3)$ has only fixed points and periodic ...
- 141
3
votes
0
answers
135
views
Moment map of $\mathrm{O}(n)$-action on $\mathbb{C}^n$
Let $(\mathbb{C}^n, \omega_0)$ be the complex Euclidean space of dimension $n$ with the standard Kähler structure $\omega_0$. I am looking for a Hamiltonian $\mathrm{O}(n)$-action on $(\mathbb{C}^n, \...
- 369
2
votes
0
answers
100
views
Effective actions by non-commutative groups have non-commuting fundamental vector fields?
I have a bit of a contradiction in my brain and I was hoping once again that excellent Mathoverflow community could help me out :)
Let $\rho_g$ be the action associated to a non-abelian Lie Group $G$ ...
- 979
2
votes
1
answer
123
views
Hamiltonian Group action with infinitely many stabiliser types
What is an example of a connected symplectic manifold $(M,\omega)$, with a Hamiltonian action of $G = U(1) =S^{1}$ with infinitely many stabiliser types?
Infinitely many stabiliser types means that ...
- 6,995
1
vote
0
answers
48
views
Relation between weight spaces of fixed loci of Hamiltonian $S^1$-actions
Consider an almost Kähler manifold $(M,\omega,I)$ with a $I$-(pseudo)holomorphic $\mathbb{C}^*$-action, whose $S^1$-part is Hamiltonian and the fixed locus
$F=M^{S^1}$ is compact. Then, it breaks $F=\...
- 1,677