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I was reading a paper in which the authors use the fact that any compact simply-connected homogeneous symplectic manifold has non-zero Euler characteristic. They prove it by quoting a theorem by Kostant which implies that the manifold is symplectomorphic to a coadjoint orbit of a semisimple group, then state that compact coadjoint orbits of semisimple groups have non-zero Euler characteristic.

I am looking for a more direct proof of that fact. Do you know some?

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    $\begingroup$ can you please refer the paper you are reading? $\endgroup$ – Anubhav Mukherjee Nov 26 '18 at 20:30
  • $\begingroup$ Sure! The paper is "Homogeneous symplectic manifolds with Ricci-type curvature", by M. Cahen, S. Gutt, J. Horowitz and J. Rawnsley. $\endgroup$ – Valentino Nov 26 '18 at 21:05
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Let your manifold be $X=G/H$. First of all, since it is simply connected, we can write it as $K/U$ where $K$ and $U=K\cap H$ are compact in $G$ (Montgomery’s theorem, 1950). Next, since $K/U$ is homogeneous symplectic, one knows that $U$ is the centralizer of a torus $S\subset K$.1) In particular $U$ contains any maximal torus containing $S$, i.e. $U$ is an equal rank subgroup of $K$. And finally, one knows that equal rank subgroups satisfy $χ(K/U)\ne0$: e.g. Samelson (1958), or Mostow (2005).


1) That is clear, with $S$ the closure of $\exp(\mathbf Rx)$, if we already know that $X\simeq$ the (co)adjoint orbit of some $x\in\mathfrak k^*\simeq\mathfrak k$. But it can also be proved a priori : Borel–Weil (1954, Thm 1), or in more detail Matsushima (1957, Thm 1).

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    $\begingroup$ For someone like me who does not know anything about coadjoint orbits the proof linked in the footnote is quite satisfying because it explains how the assumptions "simply-connected" and "symplectic" are used. $\endgroup$ – Igor Belegradek Nov 27 '18 at 16:21
  • $\begingroup$ Thank you, Francois Ziegler! But there is still an issue: Borel-Weil's theorem requires the group to be semisimple. How can we prove that that can be supposed? $\endgroup$ – Valentino Nov 29 '18 at 0:25
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    $\begingroup$ @Valentino That reduction is done in Matsushima (now added), also Bourbaki (2005, Exerc. IX.6.13 iii) $\endgroup$ – Francois Ziegler Nov 29 '18 at 11:35

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