## Kähler structure on a complex reductive group

Let $G$ be a complex reductive group, and $K$ a maximal compact subgroup (such that $K_{\mathbb{C}}=G$). By the polar decomposition theorem one has that, as manifolds, $G\cong T^*K$. The inherited symplectic structure is compatible with the complex structure, making $G$ into a Kähler manifold.

On the other hand $G$ is a smooth affine variety, and therefore inherits a Kähler structure from any embedding in an affine space. The ring of regular functions of $G$ is described by the algebraic Peter-Weyl theorem, and affine embeddings are of course just given by choices of generators.

Can one obtain the Kähler structure coming from $T^*K$ by any of these affine embeddings?

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 How about any embedding? – Reimundo Heluani Oct 19 2011 at 23:29 My formulation was slightly ambiguous, but that was the question I had intended. I've rephrased the question - thanks for the comment. – Johan Oct 20 2011 at 8:32 I meant something different and I might be wrong, but given such an embedding, G also inherits a Kähler structure coming from affine space as well. – Reimundo Heluani Oct 20 2011 at 10:08

Isn't the answer no in the very simplest case? If $K$ is the circle group, then the Kähler structure on the cotangent bundle makes it metrically a cylinder $R \times S^{1}$. I believe this cylinder cannot be isometrically embedded in $C^n$ (apply the maximum modulus principal to the derivative of the map).

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 Ooops, my bad, didn't see the exponential when I tried this example. Should remove my answer. – Reimundo Heluani Oct 20 2011 at 17:56 Thanks Peter and Reimundo, the problem is much clearer to me know. Indeed it seems like the Kähler structure coming from the polar decomposition (which needs the choice of an invariant metric) is never compatible with an affine embedding. In the paper "Phase Space Bounds for Quantum Mechanics on a Compact Lie Group" by Brian Hall this Kähler structure is discussed a bit more, he shows it provides the unique "adapted complex structure" on $T^*K$ determined by the choice of the metric on $K$. Reimundo: no need to apologize, your answer made me think of something else, I will write it below. – Johan Oct 20 2011 at 23:15

Let me give a simple argument until we think a better answer. Using Peter-Weyl you can choose an embedding of $G \subset \mathbb{A}^n$ such that $K$ is Lagrangian (using real representations for example), since $G = K_\mathbb{C} \simeq T^* K$ you obtain that $K$ is also Lagrangian in this manifold. Now you can use Weinstein's Lagrangian Neighborhood theorem (which gives a Lagrangian in a neighborhood of $K$, but perhaps this can be deformed by using that $K$ is maximal compact?).

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This is meant to be a comment to Reimundo's answer, but as it runs longer than comments allow I am posting it as an answer.

In the general situation when you think of $G$ as a symplectic manifold, and $K$ a Lagrangian submanifold, it is often possible to make the local symplectomorphism guaranteed by Weinstein's Lagrangian Neighbourhood theorem quite explicit.

Suppose $G$ is complex reductive and $K$ an maximal compact subgroup. Consider the following two (left) actions of $K$ on $G$: $\mathcal{L}_k(g)=kg$ and $\mathcal{R}_k(g)=gk^{-1}$, for $k\in K$ and $g\in G$. Suppose $G$ has a symplectic form $\omega$, for which both actions, $\mathcal{L}$ and $\mathcal{R}$ are Hamiltonian, with moment maps $\mu_{\mathcal{L}}$ and $\mu_{\mathcal{R}}$. Since the actions commute the moment map for one is invariant for the other. Since the actions are free $K$ is Lagrangian and both moment-maps map onto open subsets in $\mathbb{k}^*$.

Consider now $G$ as a $K$-principal bundle by means of the action $\mathcal{L}$. This bundle is of course locally trivial, and it is not hard to see that one can in fact use $\mu_{\mathcal{R}}$ as the quotient map. Moreover this gives a symplectomorphism from $G$ to $K\times \mu_{\mathcal{R}}(G)\subset T^*K$. If $\mu_{\mathcal{R}}(G)$ contains $0$ this provides you with the local symplectomorphism guaranteed by the Lagrangian neighbourhood theorem; if $\mu_{\mathcal{R}}$ is surjective it gives a global symplectomorphism $G\cong T^*K$. This is the case for the Kähler structure provided by the polar decomposition and a choice of a metric, but also for at least a fair amount of the Kähler structures coming from affine embeddings.

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