Let $K$ be a connected compact Lie group, $K^{\mathbb{C}}$ be complexified Lie group of $K$.

Denote $Z(k)$ by the centralizer of k∈K and $Z^{\mathbb{C}}(k) $ be the complexified Lie group of $Z(k)$ .

Q: Does the homogeneous spaces $K^{\mathbb{C}}/{{Z(k)}^{\mathbb{C}}}$ have a natural Kähler or sympletic structure?

My interpretation is following:
at the point $[Z(k)]\in K/Z(k)$ of $K/Z(k)$, ~~$T^*_{[Z(k)]}(K/Z(k))\cong ({\text{Lie}K})^*/{({\text{Lie}{Z(k)}})^*}$,

so $T^*{(K/Z(k))}\cong T^*K/T^*{Z(k)}$,

Combined with $T^*{K}\cong K^{\mathbb{C}} $ and $T^*{Z(k)}\cong {Z(k)}^{\mathbb{C}} $.

We can get $ K^{\mathbb{C}}/{Z(k)}^{\mathbb{C}}\cong T^*{(K/Z(k))} $.

Hence $K^{\mathbb{C}}/{Z(k)}^{\mathbb{C}}$ have a canonical sympletic form.

Is it true??? \

to Henrik Winther:

In fact, Fix $\langle,\rangle$ an Ad-invariant inner product on the Lie algebra $\mathfrak{k},$ Then $$T^*K\cong TK \cong K\times \mathfrak{k}\cong K^{\mathbb{C}} ,$$

where the first isomorphism comes from the inner product, the second one by left invariance, and the last one via the polar decomposition $P: K\times \mathfrak{k}\rightarrow K^{\mathbb{C}} $ defined by $(k,\xi)\rightarrow k\text{exp}(\sqrt{-1}\xi).$


I interpret "natural" to mean "invariant" in this question. I'm going to assume the action is almost effective. Let's consider left invariant Hermitian structures. If such structures exist on a homogeneous space $G/H$, the isotropy representation $\rho$ of the isotropy algebra $\mathfrak{h}$ is a unitary representation. In particular, the image $\rho(\mathfrak{h})\subset\mathfrak{gl}(\mathfrak{g}/\mathfrak{h})$ is contained in $\mathfrak{u}(\mathfrak{g}/\mathfrak{h})$.

Since multiplication by $i$ takes skew-Hermitian matrices to Hermitian ones, this will fail when $\mathfrak{h}$ is the complexification of something, provided $\rho(\mathfrak{h})$ is nonzero.

This answers negatively the case when $Z(k)$ has positive dimension.

  • $\begingroup$ I am apparently not allowed to comment on the question, so this goes here: What is the isomorphism $T^\ast K\simeq K^\mathbb{C}$? $\endgroup$ – Henrik Winther Sep 23 '14 at 12:46

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