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Disclaimer: I know very little about Shimura varieties.

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?

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Always, I imagine, although I'm hardly an expert. My intuition is that the universal cover should be negatively curved, and so contractible. –  Donu Arapura Jun 13 '11 at 20:24
    
Universal covers of Shimura varieties are Hermitian symmetric spaces of noncompact type, which are each in turn holomorphically diffeomorphic to some bounded open simply connected subset of $\mathbf{C}^n$ equipped with its Bergman metric. –  David Hansen Jun 13 '11 at 20:39
    
I think the universal cover depends on the level structure; if there is torsion in the corresponding discrete subgroup then it need not be a Hermitian symmetric space of noncompact type. For example, $\mathcal{A}_1= \mathbf{A}^1$ (but this is still contractible...). –  ulrich Jun 14 '11 at 6:32
    
On the other hand, if you take the universal cover of $\mathcal A_1$ in the orbifold sense you get $\mathfrak H_1$, so everything works out fine. Maybe I should have specified that this is what I was interested in, sorry. –  Dan Petersen Jun 14 '11 at 6:39
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1 Answer

up vote 9 down vote accepted

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 $X$.)

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.

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Thanks!${}{}{}$ –  Dan Petersen Jun 14 '11 at 12:59
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