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 $$G^\mathbb C\cong {T^*G}?$$
Best Answer
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 $T^*(G/H)$.