5
$\begingroup$

Let $G$ be a compact Lie group and $\frak g$ be its Lie algebra. Then by Marsden-Weinstein reduction theory we know that if we take $M=T^*G$ and $J \colon M\to \frak g^*$ be its moment map then the reduced space $$S=J^{-1}(\mu)/G_\mu$$ is exactly $G/G_\mu$ where $\mu\in \frak g^*$ and $G_\mu$ is the isotropy subgroup of $G$ at the point $\mu$

What we must choose for the space $M$ such that the reduced space $S$ is exactly $G^\mathbb C/(G_\mu)^\mathbb C$ where $G^\mathbb C$ is the complexification of the Lie group $G$ and we have $G^\mathbb C\cong G\times\frak {g}^*$ and so $$G^\mathbb C\cong {T^*G}?$$

$\endgroup$
  • $\begingroup$ This is true for every Lie group, compact or non-compact, real or complex. $\endgroup$ – Ben Webster Apr 27 '14 at 16:05
  • $\begingroup$ No, we need $J$ be equivariant and then it will have coadjoint orbit and for non-equivariant case we have Soriau 1-cocycle and if $G$ be compact or semisimple then $J$ is equivariant and $G\mu$ and in other case $S=J^{-1}(\mu)/H$ which $H$ in general is not isotropy group. But $G^\mathbb C$ IS NEVER COMPACT $\endgroup$ – user21574 Apr 27 '14 at 16:26
  • $\begingroup$ Now that I read your question again, I'm just confused about what you want. For every group $G$, the reduced space $S$ is $G/G_\mu$ if you choose the obvious moment map on $T^*G$. You seem to know this, so what are you asking? Are you asking if there are other spaces that also realize this? $\endgroup$ – Ben Webster Apr 27 '14 at 19:12
  • $\begingroup$ I am looking for finding $M$ such that the reduced space $S$ be EXACTLY complexified coadjoint orbit $G^\mathbb C/(G_\mu)^\mathbb C $and not coadjoint orbit . But my primary idea is that we must consider moment map for hyper-kahler manifolds, because $M=T^*G^\mathbb C$ has hyperkahler structure but it is just an idea and I don't think it be correct. Here if $G/H$ be homogeneous space then the space $G^{\mathbb C}/H^{\mathbb C}$is called complexified homogeneous space $G/H$ AND FOR COADJOINT ORBIT $T^*(G/G_\mu)\cong G^{\mathbb C}/G_\mu^{\mathbb C} $ $\endgroup$ – user21574 Apr 27 '14 at 19:23
  • $\begingroup$ en.wikipedia.org/wiki/Hyperk%C3%A4hler_quotient $\endgroup$ – user21574 Apr 27 '14 at 19:25
2
$\begingroup$

I found the answer of my question. This question is well known, but I didn't know this fact.

Consider the right action of the Lie subgroup $H$ to $G$ : $(g,h)\to gh$, $g\in G$, $h\in H$. If we identify $\mathfrak h\cong \mathfrak h^*$ we get the moment map $\mu:T^*G\to \mathfrak h$, $\mu(g.\zeta)=\text{pr}_\mathfrak h \zeta$, where $\zeta\in \mathfrak g$. Here $\text{pr}_\mathfrak h \zeta$ denotes the orthogonal projection with respect to invariant scalar product $<,>$

Then $$\mu^{-1}(0)/H\cong T^*(G/H)\cong G^{\mathbb C}/H^{\mathbb C}$$

I learned this fact from a mathematician, but still I have problem to show

$$\mu^{-1}(0)/H\cong T^*(G/H).$$

I guess we need to pass to the Springer resolution on $T^*(G/H)$.

$\endgroup$
1
$\begingroup$

Maybe the following two papers help, which concentrate on cotangent bundle reduction (even in the singular case).

  • MR2408270 Hochgerner, Simon Singular cotangent bundle reduction & spin Calogero-Moser systems. Differential Geom. Appl. 26 (2008), no. 2, 169–192.

  • MR2241438 Hochgerner, Simon; Rainer, Armin Singular Poisson reduction of cotangent bundles. Rev. Mat. Complut. 19 (2006), no. 2, 431–466.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy