MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer
Thanks!${}{}{}$ – Dan Petersen Jun 14 '11 at 12:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.