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Let $Z$ be a compact, connected, orientable (Edit: as Misha point out) and locally Riemannian symmetric space. As a complete, simple connected, locally symmetric space is a global symmetric space. We can write $Z=\Gamma \backslash G/K$. One of such $(G,K)$ is $G=\mathrm{Iso}(\widetilde{Z})$,$K$ is the stablizer of a point in $\widetilde{Z}$.

My question is, if $\widetilde{Z}$ is of non compact type, can we always choose $(G,K,\Gamma)$ such that

  1. $G,K$ is connected.
  2. $\Gamma\subset G$ closed discrete?

If not, under what natural condition, $Z$ is this type.

One of naive choice is to take $G=\mathrm{Iso}^0(\widetilde{Z})$, but it is to always true that $\Gamma\subset \mathrm{Iso}^0(\widetilde{Z})$.

Edit: I pose this question because in papers of H. Moscovici and R.J. Stanton http://link.springer.com/article/10.1007%2FBF01393895 http://link.springer.com/article/10.1007%2FBF01232263 where proved the result for this type of space(maybe I missed something) but state the theorem for locally symmetric space.

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1 Answer 1

up vote 2 down vote accepted

No, take $Z$ to be a non-orientable compact connected hyperbolic surface.

Edit: Here is an easy orientable example. Let $S$ be a non-orientable hyperbolic surface. Then take $Z$ to be the orientable 2-fold cover of $S\times S$. The point is is that fundamental group of $Z$ (regarded as a deck-transformation group) still contains non-holomorphic automorphisms of the bidisk. I am quite sure that there are even stably parallelizable examples of similar to this, but I do not see a sufficient motivation for constructing them.

I looked briefly at one of the papers you have links to: They are just sloppy in their formulations. Whenever they say "locally symmetric space" you should just read "quotient $\Gamma\backslash X$, where $\Gamma< G^0$''. The correct statement is that for a lattice (actually, any subgroup) $\Gamma$ in the Lie group $G=Isom(X)$, the intersection $\Gamma \cap G^0$ is a finite-index subgroup of $\Gamma$, where $G^0$ is the identity component of $G$.

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Thank you for the answer. But if we suppose that $Z$ is orientable or even more is spin. Can this be true? –  shu May 15 '13 at 16:00
    
For orientable manifolds this is still false. I know examples in dimension 6. These are compact quotients of complex 3-d ball by nonholomorphic orientation-preserving discrete isometry groups. I would have to think about spin condition, but I am sure it is not enough either. Even triviality of the tangent bundle is probably not enough (but this will be harder to arrange). –  Misha May 15 '13 at 20:24
    
Yes, I do agree. –  shu May 17 '13 at 14:28

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