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 $(k,\xi)\rightarrow k\text{exp}(\sqrt{-1}\xi).$
