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Let $M$ be compact $G$-homogeneous manifold, equipped with the equivariant complex structure, when $G$ is a semi-simple algebraic group. The obvious example is every flag manifold. In that case, all non-vanishing cohomology classes lives in $\Omega^{(k,k)}$, for some $k \leq $dim$M$. Does that hold generally?

Sketch of a possible proof: Every cohomology class has an $G$-equivariant representative. Hence, we must just to look for equivariant elements in $ {\cal T}^{(p,q)}(M)$. But by a "rep theory argument", such elements only exists if $p=q$, hence, we have non-vanishing classes only in $\Omega^{(k,k)}$.

Is proof correct? (Does it hold its water!) I would like help for the "fleshing-out" of the "rep theory argument".

A lot of thank you's!

P.S. Do the cohomology rings of such manifolds have a presentation analogous to Schubert calculus of flag manifolds?

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  • $\begingroup$ I guess your argument above needs that the stabliser of the action of G on M is also a semi-simple algebraic group. If this true, then your argument does indeed hold water. $\endgroup$
    – Tomasz Köner
    Commented May 26, 2015 at 13:00

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In general the answer is no. Check Bott's paper Homogeneous vector bundles, where he computed the Hodge numbers of the complex homogeneous space SU(3) (as a C-space classified by H.-C. Wang, it is the quotient of SL(3,C) by a complex subgroup). Geometrically it is the total space of a $T^2$-bundle over full flags in $\mathbb{C}^3$. It is non-Kahler.

I should mention that there are typos in Bott's paper. The correct numbers are $h ^{0,0} = h^{0,1} = h^{1,1} = h^{1,2} = h^{3,2} = h^{3,3} = h^{4,3} = h^{4,4} = 1$.

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  • $\begingroup$ Also look at: Griffiths, Phillip A. Some geometric and analytic properties of homogeneous complex manifolds. I. Sheaves and cohomology. Acta Math. 110 1963 115–155. 22.70 (32.32) $\endgroup$
    – Ben McKay
    Commented Apr 27, 2016 at 15:36
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In general, the answer to your question is "no". For example, let $\mathbb{C}$ act on the standard square two-torus by translations. Clearly, this action preserves the complex structure. All the cohomology classes are represented by invariant forms, and in particular in dimension 1 there are forms that represent cohomology classes that are of type $(1,0)$ and $(0,1)$. I am assuming by "equivariant" elements you mean invariant elements.

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  • $\begingroup$ In the question $G$ is a semi-simple algebraic group, which does not seem to be the case for your "counter-example". $\endgroup$
    – Daniel Loughran
    Commented May 25, 2015 at 21:22
  • $\begingroup$ Is the action of the $\mathbb C$ homogeneous? $\endgroup$
    – Falertu Vatilski
    Commented May 26, 2015 at 8:51
  • $\begingroup$ Yes. The torus is just $\mathbb{C}$ mod a lattice. I was trying to think of another counterexample. All you would need would be to mod out a semi-simple algebraic group by a lattice subgroup, and then the result would have non trivial fundamental group and thus nontrivial $H_1$. But I really don't have much experience with semi-simple group actions. I know that would work for compact group actions. $\endgroup$
    – Ken Richardson
    Commented May 31, 2015 at 21:14

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