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It is known that any connected compact Lie group $G$ is a finite quotient of the product of a compact simply connected semisimple Lie group $\tilde{G}$ and a torus $\mathbb{T}^n$ (see for example Chapter V Theorem 8.1 in Brocker). In other words, the universal covering map $\tilde{G}\times \mathbb{R}^n\to G$ factors through some $\tilde{G}\times\mathbb{T}^n$.

More generally, consider a compact globally symmetric space $M$. Then the universal cover of $M$ is the product of a simply connected globally symmetric space $\tilde{M}$ of compact type and a Euclidean space $\mathbb{R}^n$. My question is, is it still true that the universal covering map $\tilde{M}\times \mathbb{R}^n\to M$ factors through some $\tilde{M}\times \mathbb{T}^n$ so that $\tilde{M}\times \mathbb{T}^n\to M$ is a finite Riemannian cover? Any help is appreciated. Thank you!

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  • $\begingroup$ If $G$ is the isometry group of $M$, then one can deform the given metric to the one that splits off a torus a finite cover in the way you describe, and so that $G$ still acts by isometries after the deformation, see arxiv.org/abs/math/0302221. Perhaps in your case one can show that the deformation must be constant. $\endgroup$
    – Igor Belegradek
    Commented May 16, 2018 at 18:57
  • $\begingroup$ To clarify, you are asking for an isometric splitting $\tilde{M}\times \mathbb{T}^n$? $\endgroup$
    – Ian Agol
    Commented May 17, 2018 at 9:13
  • $\begingroup$ For not necessarily isometric splitting the answer is yes: Any compact manifold of nonnegative Ricci curvature splits of a torus factor (by Cheeger-Gromoll splitting theorem). But in this case the splitting should be isometric, I think. $\endgroup$
    – Igor Belegradek
    Commented May 17, 2018 at 15:03
  • $\begingroup$ @IanAgol Yes. I think the answer is yes. Now I have a sketch but need to check the details. Write $M=U/K$, and let $\mathfrak{u},\mathfrak{k}$ be the Lie algebras of $U,K$ respectively. Write $\mathfrak{u}=\mathfrak{c}+[\mathfrak{u},\mathfrak{u}]$ where $\mathfrak{c}$ is the center of $\mathfrak{u}$, and then $\mathfrak{k}=\mathfrak{c}\cap\mathfrak{k}+[\mathfrak{k},\mathfrak{k}]$. $\endgroup$
    – shrinklemma
    Commented May 17, 2018 at 16:25
  • $\begingroup$ @IanAgol If $\mathfrak{u}=\mathfrak{k}+\mathfrak{m}$ denotes the Cartan decomposition, then the torus factor should come out of $\mathfrak{c}\cap\mathfrak{m}$ and the simply connected factor from the pair $([\mathfrak{u},\mathfrak{u}],[\mathfrak{k},\mathfrak{k}])$. $\endgroup$
    – shrinklemma
    Commented May 17, 2018 at 16:25

1 Answer 1

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See Theorem A here:

https://www.ams.org/journals/proc/2001-129-12/S0002-9939-01-06008-7/S0002-9939-01-06008-7.pdf

Your $M$ is compact globally symmetric so any geodesic is contained in a compact flat. So you get a covering as you wanted.

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