i have spent some time over the following problem from a problem sheet of a course on complex geometry.
Let $M=G/H$ be a homogeneous space (where $G$ is a Lie group, $H$ a closed subgroup) and let $I$ be a $G$-invariant almost complex structure on $M$. Assume that for each point $x \in G/H$ (equivalently, for the identity $e$ of $G$) we have a distinguished element $g_x$ of the stabilizer that is known to act on $T_x M$ as:
- $-\mathrm{Id}$
- $2\,\mathrm{Id}$
- as an operator with all eigenvalues $\neq 1$
(three possibilites) The problem is to determine, in each of these cases, if $I$ is always integrable.
I have given a thought to it, and the problem seems strange. I think that the answer is negative in all three cases, and since the first and the second one are particular cases of the third one, we only have to deal with them.
My questions below are to check the validity of my reasoning because the three conditions about the way $g_x$ acts look pretty arbitrary to me.
Firstly, the only connection between the existence of such an operator acting on $T_e M$ and the almost complex structure seems to be that the space $T^{1,0}_e M \subset T_e M \otimes \mathbb{C}$ must be preserved by it, but this is automatic in the first and in the second case.
Question 1: is this true?
(of course in order to satisfy the problem statement there has to exist an element $g$ in the stabilizer that acts like that)
If the answer to Q1 is positive, then it gives a lot of room for counterexamples. For example, on one hand take complex Grassmanian with its natural complex structure (example of integrability), on the other hand, keeping the same Grassmanian, define the almost complex structure by picking a general enough real subspace $V \subset T_e M \otimes \mathbb{C} \cong \mathfrak{g}/\mathfrak{h} \otimes \mathbb{C}$ such that $V \cap \bar{V} = \{0\}$ and such that the preimage of $V$ in $\mathfrak{g} \otimes \mathbb{C}$ is not a Lie subalgebra.
Question 2: does this indeed give a non-integrable almost-complex structure?