# Compact simply-connected homogeneous symplectic manifold

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?

• can you please refer the paper you are reading? – Anubhav Mukherjee Nov 26 '18 at 20:30
• Sure! The paper is "Homogeneous symplectic manifolds with Ricci-type curvature", by M. Cahen, S. Gutt, J. Horowitz and J. Rawnsley. – Valentino Nov 26 '18 at 21:05

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).