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May 4, 2017 at 13:24 vote accept Max Reinhold Jahnke
May 1, 2017 at 12:31 comment added Holonomia @Max Reinhol Jahnke: you have to regard the Lie algebra of the group $G$ as a finite dimensional Lie algebra of vector fields of $M$. If $X_1,\cdots,X_k$ are generators of such Lie algebra then consider the vector fields $JX_1, \cdots JX_k$. Then the 2k vectors fields $X_1,\cdots,X_k, JX_1,\cdots,JX_k$ generates a finite dimensional complex Lie algebra. By Palais theorem,using flow completeness, you have that a complex Lie group $H$ acts on $M$. Then by the universal property of the complexification (en.wikipedia.org/wiki/Complexification_(Lie_group)) you get that $G_C$ acts on $M$.
May 1, 2017 at 7:43 history edited Ben McKay CC BY-SA 3.0
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May 1, 2017 at 1:07 comment added Max Reinhold Jahnke Dear @BenMcKay, I'm having a hard time understanding your answer. I understand that you use Palais' theorem to extend the action to a covering of the complexified group, but what guarantee that you can define it on the complexification? Can you be more specific on what results you are using or indicate some references? Anyway, thank you very much.
Apr 30, 2017 at 18:18 comment added Ben McKay @MaxReinholJahnke: No. The action is $X+iY \in \mathfrak{g} + i \mathfrak{g} \mapsto \left.\frac{d}{dt}\right|_{t=0} \exp(t(X+JY)m).$.
Apr 30, 2017 at 18:11 comment added Max Reinhold Jahnke Just to clarify, you are suggesting that I consider the Lie algebra action $X + iY \in \mathfrak g + i \mathfrak g \mapsto \frac{d}{dt} |_{t=0} (\exp(tX)m) + J\left(\frac{d}{dt} |_{t=0} (\exp(t Y)m)\right)$?
Apr 30, 2017 at 18:02 comment added Ben McKay @MaxReinholJahnke: I should have said that you use $J$ rather than $i$, the almost complex structure induced by the complex structure. Then you get a $J$-invariant collection of real vector fields. You then use $J$ as the complex structure on that Lie algebra of vector fields.
Apr 30, 2017 at 16:09 comment added Max Reinhold Jahnke Thank you for your answer, @BenMcKay. Is the first extension due to Palais theorem? If so, I understand why you are using that the vector fields are all complete. If I'm not mistaken, the Lie algebra action you are considering is, for $X + iY \in \mathfrak g + i \mathfrak g$ you associate the smooth vector field on $M$ given by $\frac{d}{dt} |_{t=0} (\exp(tX)m) + i \frac{d}{dt} |_{t=0} (\exp(tY)m)$ for each $m \in M$. But the version of the theorem I known use the flow on M to construct the action, but the flow only exist for real vector fields. Is there a complex version of theorem?
Apr 29, 2017 at 20:13 comment added Ben McKay @Holonomia: yes, I think so.
Apr 29, 2017 at 19:50 comment added Holonomia @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?
Apr 29, 2017 at 17:30 comment added abx Oops! Of course, sorry.
Apr 29, 2017 at 17:19 comment added Ben McKay @abx: The problem has $M$ a $G$-homogeneous space, so $M$ admits a transitive action of a compact Lie group, so is compact.
Apr 29, 2017 at 16:58 comment added abx 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.
Apr 29, 2017 at 16:45 history answered Ben McKay CC BY-SA 3.0