An obstruction to homogeneous manifolds being Kähler

Gilligan, Bruce. "An obstruction to homogeneous manifolds being Kähler." Annales de l’institut Fourier 55.1 (2005): 229-241. <http://eudml.org/doc/116187>.

@article{Gilligan2005,
abstract = {Let $G$ be a connected complex Lie group, $H$ a closed, complex subgroup of $G$ and $X:=G/H$. Let $R$ be the radical and $S$ a maximal semisimple subgroup of $G$. Attempts to construct examples of noncompact manifolds $X$ homogeneous under a nontrivial semidirect product $G=S \ltimes R$ with a not necessarily $G$-invariant Kähler metric motivated this paper. The $S$-orbit $S/S\cap H$ in $X$ is Kähler. Thus $S\cap H$ is an algebraic subgroup of $S$ [4]. The Kähler assumption on $X$ ought to imply the $S$-action on the base $Y$ of any homogeneous fibration $X \rightarrow Y$ is algebraic too. Natural considerations allow a reduction to the case where $H=\Gamma $ is a discrete subgroup and there is a homogeneous fibration $X=G/\Gamma \rightarrow G/I =:Y$ with $I^\{\circ \}$ an abelian, normal subgroup of $G$ and the fiber $I^\{\circ \}/(I^\{\circ \}\cap \Gamma )$ a Cousin group. An algebraic condition does hold in the homogeneous manifold $Y=\hat\{G\}/\hat\{\Gamma \}$, where $\hat\{G\}:=G/I^\{\circ \}$ and $\hat\{\Gamma \}:=I/I^\{\circ \}$, namely, an element $\hat\{g\}\in \hat\{\Gamma \}$ of infinite order lying in a semisimple subgroup $\hat\{S\}$ of $\hat\{G\}$ is an obstruction to the existence of a Kähler metric on $X$. So $X$ Kähler implies $\hat\{S\}\cap \hat\{\Gamma \}$ finite.},
affiliation = {University of Regina, department of Mathematics and Statistics , Regina, S4S 0A2 (Canada)},
author = {Gilligan, Bruce},
journal = {Annales de l’institut Fourier},
keywords = {homogeneous complex manifolds; Kähler manifolds},
language = {eng},
number = {1},
pages = {229-241},
publisher = {Association des Annales de l'Institut Fourier},
title = {An obstruction to homogeneous manifolds being Kähler},
url = {http://eudml.org/doc/116187},
volume = {55},
year = {2005},
}

TY - JOUR
AU - Gilligan, Bruce
TI - An obstruction to homogeneous manifolds being Kähler
JO - Annales de l’institut Fourier
PY - 2005
PB - Association des Annales de l'Institut Fourier
VL - 55
IS - 1
SP - 229
EP - 241
AB - Let $G$ be a connected complex Lie group, $H$ a closed, complex subgroup of $G$ and $X:=G/H$. Let $R$ be the radical and $S$ a maximal semisimple subgroup of $G$. Attempts to construct examples of noncompact manifolds $X$ homogeneous under a nontrivial semidirect product $G=S \ltimes R$ with a not necessarily $G$-invariant Kähler metric motivated this paper. The $S$-orbit $S/S\cap H$ in $X$ is Kähler. Thus $S\cap H$ is an algebraic subgroup of $S$ [4]. The Kähler assumption on $X$ ought to imply the $S$-action on the base $Y$ of any homogeneous fibration $X \rightarrow Y$ is algebraic too. Natural considerations allow a reduction to the case where $H=\Gamma $ is a discrete subgroup and there is a homogeneous fibration $X=G/\Gamma \rightarrow G/I =:Y$ with $I^{\circ }$ an abelian, normal subgroup of $G$ and the fiber $I^{\circ }/(I^{\circ }\cap \Gamma )$ a Cousin group. An algebraic condition does hold in the homogeneous manifold $Y=\hat{G}/\hat{\Gamma }$, where $\hat{G}:=G/I^{\circ }$ and $\hat{\Gamma }:=I/I^{\circ }$, namely, an element $\hat{g}\in \hat{\Gamma }$ of infinite order lying in a semisimple subgroup $\hat{S}$ of $\hat{G}$ is an obstruction to the existence of a Kähler metric on $X$. So $X$ Kähler implies $\hat{S}\cap \hat{\Gamma }$ finite.
LA - eng
KW - homogeneous complex manifolds; Kähler manifolds
UR - http://eudml.org/doc/116187
ER -