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!

  • $\begingroup$ 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. $\endgroup$ 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
    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. $\endgroup$ 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 $\mathfrak{k}=\mathfrak{c}\cap\mathfrak{k}+[\mathfrak{k},\mathfrak{k}]$. $\endgroup$ 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 $([\mathfrak{u},\mathfrak{u}],[\mathfrak{k},\mathfrak{k}])$. $\endgroup$ May 17, 2018 at 16:25

1 Answer 1


See Theorem A here:


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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