Kähler metrics on the coadjoint orbits of a compact Lie group

Question

Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}$. It is well-known that each orbit for the coadjoint representation of $G$ on $\mathfrak{g}^*$ carries a canonical symplectic structure, known as the Kirillov-Kostant-Souriau symplectic form.

Moreover, I've read at a few different places that the coadjoint orbits are also Kähler manifolds:

Theorem. Let $G$ be a compact Lie group, $\mathcal{O}$ a coadjoint orbit and $\omega$ its Kirillov-Kostant-Souriau symplectic form. Then, there exists a unique $G$-invariant Kähler metric on $\mathcal{O}$ that is compatible with $\omega$.

For example, this result is mentioned in Robert Bryant's lecture notes An Introduction to Lie Groups and Symplectic Geometry on page 150, and at the beginning of this paper by Kronheimer.

However, I didn't find any proof of that theorem. Does someone know how to prove it or can point a good reference?

According to Bryant, it is "not hard" to prove it "using roots and weights". But I wasn't able to do so.

Answer 1

Suppose that $G$ is compact and semi-simple, let $c\in {\cal G}^*$ the dual of ${\cal G}$, consider ${\cal G}_c$ the stabilizer of $c$, and ${\cal,G}'_c$ its orthogonal for the Killing metric, you can identify the tangent space at $c$ of the coadjoint orbit to ${\cal G}'_c$. If you restrict -Killing to ${\cal G}'_c$ you obtain a metric on the adjoint orbit, is this metric does not define the Kahler structure with the Konstant Kirillov Souriau form?

Answer 2

well, I had a look at prof. Bryant's notes. I understand what he means that it is not hard to do it, using roots and weights. I know you have asked for a full proof, but instead, I will indicate how to modify the construction of the Kahler metric on coadjoint orbits of $\mathfrak{U(n)}$, to have it work on a coadjoint orbit of an arbitrary compact semisimple Lie group.

The Adjoint/adjoint invariant symmetric bilinear form can be replaced by minus the Killing form (minus to make it positive-definite). The diagonalization by a unitary transformation "becomes" the statement that given an element $x$ in $\mathfrak{g}$, there exists a $g \in G$ (depending on $x$), such that $Ad_g(x) \in \mathfrak{t}$, where $\mathfrak{t}$ is some (fixed) chosen Cartan subalgebra.

Instead of $U(n)_{\xi}$, replace it with $G_{\xi}$, the stabilizer of $\xi$ in $G$. The $1$-form with values in $\mathfrak{g}$, namely $g^{-1}dg$, is to be replaced by the left-invariant Maurer-Cartan form on $G$.

Finally, the 2 formulae at the end (in Prof. Bryant's notes), should be modified by replacing the sum with a sum over a choice of positive roots, and replacing $\xi_j - \xi_k$ by a positive root acting on an element in the Cartan subalgebra which is in the orbit of $\xi$ (in some sense, a "diagonalization" of $\xi$, and such an element is unique up to the action of the Weyl group on the Cartan subalgebra).

That's probably what Prof. Bryant meant. There are some details that need to be checked of course, but in any case, I hope this helps.

Answer 3

In fact if $M$ be a compact Kähler manifold, then its symplectic quotient is also Kähler manifold. So, because $T^∗G$ is Kähler manifold, so coadjoint orbit is also Kähler due to the following reason,

We have this useful decomposition $G^{\mathbb C}\cong G\times \mathfrak g^{*}\cong T^*G$

If we take $\mu:T^*G\to \mathfrak g^*$ as momentum map then $\mu^{-1}(\lambda)=G$, so

$$\mu^{-1}(\lambda)/G_\lambda\cong G/G_\lambda\cong \mathcal O_\lambda$$

so we need to take $M=T^*G$

As far as I know this method is from Mostow