An obstruction to homogeneous manifolds being Kähler.
Throughout this paper consider the setting where \(\Gamma\) is discrete in the connected complex Lie group \(G\) and there is a homogeneous fibration \(X:G/\Gamma \to G/I:=Y\) where \(I^0\) is an abelian normal subgroup of \(G\) and the fiber \(I^0/(I^0\cap \Gamma)\) of this fibration is an abelian complex Lie group that has no noncompact holomorphic functions. Such complex Lie groups are called Cousin groups. Since \(I^0\) is normal in \(G\) set \(\widehat{G}:= G/I^0\) and \(\widehat{\Gamma}:=I/I^0\). Then \(Y=\widehat{G}/\widehat{I}\) with \(\widehat{G}\) acting almost effectively. If \(X\) is Kähler one would like to determine if \(Y\) is Kähler too. Write \(\widehat{G}=\widehat{S} \ltimes \widehat{R}\), where \(\widehat{R}\) is the radical of \(\widehat{G}\) and \(\widehat{S}\) is a maximal complex semisimple subgroup of \(\widehat{G}\). If \(Y=\widehat{G}/\widehat{\Gamma}\) were Kähler then \(\widehat{S}\cap \widehat{\Gamma}\) would be finite. Actually, the author shows that no element of infinite order in \(\widehat{\Gamma}\) is contained in a semisimple group of \(\widehat{G}\), equivalently \(\widehat{S}\cap\widehat{\Gamma}\) is finite for every semisimple subgroup \(\widehat{S}\) of \(\widehat{G}\). Another result is the following one. Assume \(\widehat{G}\) is a linear algebraic group whose radical \(\widehat{R}\) is a vector group with maximal semisimple subgroup of \(\widehat{G}\) acting on \(\widehat{R}\) by a linear representation that has no nonzero invariant vector. Then \(G/\Gamma\) does not admit a Kähler structure. A particular example of such a linear algebraic group \(\widehat{G}\) is the affine group of \(\mathbb C^n\) for \(n >1\).
scientific article; zbMATH DE number 2162472