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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.

Simon Parker
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  • You might try Einstein Manifolds by Besse; the chapter on homogeneous Einstein-Kähler manifolds discusses coadjoint orbits, and if memory serves, the book aimed to be self-contained. (Even if I'm mis-remembering, the bibliography is extensive.) – Andrew D. Hwang Jan 03 '16 at 15:39

3 Answers3

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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?

Tsemo Aristide
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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.

Malkoun
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  • Thank you very much for the detailed discussion. Can this be generalized to certain non-compact Lie groups? Most notably, the quotient Sp(2N,R)/U(N) is isomorphic to the co-adjoint orbit and it's a Kähler manifold constructed in this way. The point is that if one restricts the Killing form onto the orthogonal complement of the stabilizer group U(N), it actually becomes definite and thereby defines the positive definite metric, we need (moreover, it's compatible with the symplectic form). However, I don't know if this applies to all/most non-compact semi-simple Lie groups!? – LFH Oct 28 '19 at 13:06
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    @LFH, I cannot find the proof in Prof. Bryant's notes anymore. What I would do is the following. Assume you have a Cartan involution, go through the proof, and see what assumptions you need. I am sorry if this is not very helpful. If you can find the proof in Prof. Bryant's notes, then I may be able to help more concretely. – Malkoun Oct 28 '19 at 15:28
  • Thank you so much for your response. I was trying to read the notes of Christmas, but still wasn't quite able to extract the required assumptions. I found some other notes on the issue of seeing when co-adjoint orbits are Kähler. I had written essentially the same question before (https://physics.stackexchange.com/questions/508652/inner-product-on-group-theoretic-coherent-states-and-anti-commutator-of-lie-alge?noredirect=1#comment1146201_508652), but this was more contrived and I didn't make a connection to co-adjoint orbits. If you have any suggestions, let me know! – LFH Jan 22 '20 at 14:21
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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