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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 $G_{\mathbb C}$?

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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 $M$.

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  • $\begingroup$ You are assuming that $M$ is compact, which is reasonable -- otherwise this is clearly false, the Poincaré upper half-space being an obvious counter-example. $\endgroup$
    – abx
    Commented Apr 29, 2017 at 16:58
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    $\begingroup$ @abx: The problem has $M$ a $G$-homogeneous space, so $M$ admits a transitive action of a compact Lie group, so is compact. $\endgroup$
    – Ben McKay
    Commented Apr 29, 2017 at 17:19
  • $\begingroup$ Oops! Of course, sorry. $\endgroup$
    – abx
    Commented Apr 29, 2017 at 17:30
  • $\begingroup$ @BenMcKay: It seems to me that your argument gives a complex Lie group $H$ ( i.e. generated by all the flows of the complex Lie algebra ) of diffeomorphisms of $M$ which contains the original $G$. So by the universal property (en.wikipedia.org/wiki/Complexification_(Lie_group)) the complexification $G_C$ covers $H$ hence acts on $M$ perhaps not faithful, but the OP do not ask for a faitful action. Thus your argument solves the question, isn't it? $\endgroup$
    – Holonomia
    Commented Apr 29, 2017 at 19:50
  • $\begingroup$ @Holonomia: yes, I think so. $\endgroup$
    – Ben McKay
    Commented Apr 29, 2017 at 20:13

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