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 $$G^\mathbb C\cong {T^*G}?$$

  • $\begingroup$ 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

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 $T^*(G/H)$.


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.


Your Answer

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