When does a Shimura variety have contractible universal cover? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:02:31Z http://mathoverflow.net/feeds/question/67699 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67699/when-does-a-shimura-variety-have-contractible-universal-cover When does a Shimura variety have contractible universal cover? Dan Petersen 2011-06-13T20:05:38Z 2011-06-14T07:35:14Z <p>Disclaimer: I know very little about Shimura varieties.</p> <p>Some Shimura varieties have a contractible universal covering space, for instance $A_g$ itself. Are there any nice necessary and/or sufficient conditions implying this?</p> http://mathoverflow.net/questions/67699/when-does-a-shimura-variety-have-contractible-universal-cover/67740#67740 Answer by ACL for When does a Shimura variety have contractible universal cover? ACL 2011-06-14T07:26:23Z 2011-06-14T07:35:14Z <p>Connected Shimura varieties are quotients $S=X/\Gamma$, where $X$ is a Hermitian symmetric space without compact factor and $\Gamma$ is a discrete subgroup acting properly discontinuously on $X$. If $\Gamma$ acts without fixed points, then $X\to S$ is a universal covering. (In the general case, the universal covering of $S$ is $X/\Gamma_1$, where $\Gamma_1$ is the subgroup of $\Gamma$ generated by stabilizers of fixed points on $X$.)</p> <p>It is known that $X=G/K$, where $G$ is a semisimple adjoint Lie group and $K$ is a maximal compact subgroup. The Iwasawa decomposition $G=KAN$ implies that $X$ is diffeomorphic to $AN$, hence to a real vector space. In particular, $X$ is contractible.</p>