Let $G$ be a compact Lie group and let $M$ be a $G$-homogeneous manifold. Suppose that $M$ is endowed with a complex structure invariant by the action of $G$. Denote by $G\_{\mathbb C}$ the complexification of $G$.


Are there known conditions so that the action of $G$ on $M$ can be extended to a holomorphic action of $G\_{\mathbb C}$?



Holomorphic extension of an action by a compact Lie group on a complex homogeneous manifold
I think this should work for a connected group $G$. The Lie algebra action extends, by multiplying by $J$, to a complex Lie algebra action. By compactness of $M$, these vector fields are all complete. So some covering group of the complexification acts, for example the universal covering group $\tilde{G}\_{\mathbb{C}}$ acts. Hochschild, **The Structure of Complex Lie Groups**, chapter XVII, §5, proves that if $G$ admits a faithful finite dimensional representation, then $G$ injects into its complexification. For example, all compact Lie groups admit faithful finite dimensional representations. The group $G$ is then the maximal compact subgroup of $G\_{\mathbb{C}}$; Broecker and tom Dieck, Representations of Compact Lie Groups, p. 153, prop. 8.3. Every connected Lie group retracts to its maximal compact subgroup. In particular, the fundamental group of $G$ and of $G\_{\mathbb{C}}$ are the same, represented by loops in $G$.
The fundamental group of $G\_{\mathbb{C}}$ embeds into the universal covering group $\tilde{G}\_{\mathbb{C}}$. We need to see which elements of it act trivially, so that we can see which quotient of $\tilde{G}\_{\mathbb{C}}$ the action descends to. The fundamental group of the original compact group sits as a subgroup of $\tilde{G}$, but acts trivially on $M$. So the same is true for the fundamental group of $G\_{\mathbb{C}}$: $\pi\_1(G\_{\mathbb{C}}) \subset \tilde{G}\_{\mathbb{C}}$ acts trivially on $M$. Hence $G\_{\mathbb{C}}$ acts on the complex homogeneous space $M$.


