# Fundamental group of a compact Riemannian manifold whose universal covering is not compact

This might be a silly question, but my algebra and topology are horrible.

Let $$(M,g)$$ be a compact Riemannian manifold with boundary, and $$\Pi_1(M)$$ its fundamental group. If its universal covering $$(M^*,g^*)$$ is not compact, does there exist a subgroup of $$\Pi_1(M)$$ isomorphic to $$\mathbb{Z}$$?

Thanks in advance for any help.

Shockingly, as far as I can tell this appears to be an open problem.

First of all, this is a purely group-theoretic question: the fundamental group of a compact manifold is finitely presented, and conversely every finitely presented group is the fundamental group of a smooth closed manifold, in every dimension $$n \ge 4$$. We can add boundaries by cutting out a small hole, which doesn't affect the fundamental group for $$n \ge 3$$. Furthermore the universal cover is noncompact iff the fundamental group is infinite, and Riemannian metrics always exist on smooth manifolds. So your question is equivalent to:

Does every infinite finitely presented group have an element of infinite order? Equivalently, does there exist an infinite finitely presented group all of whose elements have finite order?

According to this MO thread from 2011 this is open! Famously, it's known that there exist such groups which are finitely generated, by the Golod-Shafarevich theorem; examples include the Grigorchuk group and (much harder) Tarski monsters. This is related to the literature on the Burnside problem.

For positive results with further hypotheses on the group see this MO thread. The answer should be yes for compact manifolds in dimensions $$n \le 3$$, I think. If the universal cover is contractible (for example if $$M$$ is hyperbolic) then $$\pi_1$$ is even torsion-free.

• Yes, it is a well-known open problem. And yes, the answer is known in dimension 3: The fundamental group $G$ of a compact 3-manifold splits as a free product of a torsion-free group and finitely many finite groups $G_i$. Each finite subgroup of $G$ is conjugate to a subgroup of some $G_i$. Commented Aug 10 at 10:03
• Thanks for your help! Good news, I have answer now. Bad news, it's not what I was expecting.
– Jhin
Commented Aug 12 at 12:50