All Questions
Tagged with dg.differential-geometry moment-map
8 questions
5
votes
1
answer
564
views
Geometric invariants of a Riemannian manifold encoded in certain moment map
Let $(M,g)$ be a Riemannian manifold with isometric group $G=Iso(M,G)$. The metric gives an isomorphism between tangent and cotangent bundle of $M$. So $g$ induce a natural symplectic structure on $...
1
vote
1
answer
345
views
What is general expression for the moment map of a Kaehler Hamiltonian G-manifold
A Kaehler Hamiltonian G-manifold is a Kaehler manifold with a Hamiltonian G-action, i.e., a G-action generated by a moment map. In particular, the Killing vector fields which generate the G-action are ...
1
vote
0
answers
121
views
Moment map of isometries on Kähler mainfolds
Let us assume we are given a Kähler manifold $M$, equipped with its metric $g_{\imath\bar\jmath}$ and with the associated symplectic form
$$
\Omega = i\, g_{\imath \bar \jmath}dz^\imath \wedge d\bar z^...
22
votes
2
answers
3k
views
Origin of the name ''momentum map''
Why is the momentum map in the differential geometry of symmetries called the ''momentum'' (or ''moment'') map?
3
votes
1
answer
386
views
Set of singular points for momentum map (with coisotropic action)
Let $G$ be a Lie-group acting on a connected symplectic manifold $M'$ in a hamiltonian way, with an $\operatorname{Ad}^*_G$-equivariant momentum map. Assuming $G$ acts properly on $M'$, we can ...
2
votes
1
answer
131
views
Polynomials pulled back by momentum maps
Let $G$ be a Lie group acting Hamiltonian on some real analytic symplectic manifold $(M, \omega)$, with an $G$-equivariant momentum map $\Phi \colon M \to \mathfrak{g}^*$.
Assuming I can find ...
2
votes
1
answer
121
views
coisotropic action on $TS^{2n+1}$
Let $S^{2n+1}$ be the $m$-dimensional sphere in $\mathbb{C}^{n+1}$. Endow $S^{2n+1}$ with the standard metric. Let $S^1$ act by multiplication on $S^{2n+1}$. Then $S^1$ and the canonical action of $SU(...
1
vote
0
answers
80
views
Momentum Map on cotangentbundle as submersion
Let $N$ be a homogeneous space. Therefore we find a Liegroup $G$ and a isotropy-subgroup $K$ of $G$, such that we can identify $N = G/K$. Then we have a canonical action $l\colon G \times G/K \to G/K$ ...