9

Some classmates and I were working on the following question - is the fundamental group of the Klein Bottle $K$ torsion-free? We have the following presentation: $$\pi_1(K) = \langle a,b: aba = b \rangle.$$

In trying to answer this problem, we came up with the following question that could resolve this problem, and seemed like an interesting claim in general:

Is the Fundamental Group of a space with contractible universal cover torsion-free?

We thought about it a bit and unfortunately did not find a counterexample or a proof. Is this true? If it's not, what is a good way to answer our original question about the Klein Bottle?

darij grinberg
  • 19,173
Jack Burkart
  • 694

2 Answers2

11

This is false in general; in fact every group $G$ is the fundamental group of a unique (up to weak homotopy) space with contractible universal cover, namely the classifying space $BG$, or equivalently the Eilenberg-MacLane space $K(G, 1)$. In particular, the Klein bottle is such a classifying space, as is the torus.

What's true is the following.

Theorem: If $BG$ is a finite-dimensional CW complex, then $G$ is torsion-free.

Proof. With the above hypotheses, the universal cover of $BG$ is also a finite-dimensional CW complex on which $G$ acts freely, hence for any subgroup $H$ of $G$, the quotient $BH$ of the universal cover by the induced action of $H$ is also a finite-dimensional CW complex, and in particular has finite cohomological dimension. But finite cyclic groups have infinite cohomological dimension, by the standard computation of their group cohomology. $\Box$

Qiaochu Yuan
  • 468,795
  • This is basically exactly the type of result I was wondering was true and the outline of it is appreciated - thanks! – Jack Burkart Apr 20 '16 at 22:11
  • 1
    For those unfamiliar with group cohomology, one can instead note that a model for $K(\mathbb{Z}_m, 1)$ is an infinite-dimensional lens space $L_m(\ell_1, \ell_2, \dots)$ which has a simple cellular structure (much like that of $\mathbb{RP}^{\infty}$) from which its homology can be easily calculated; see example 2.43 of Hatcher's Algebraic Topology. – Michael Albanese Jul 27 '17 at 17:18
2

No, the infinite sphere is the universal cover of $BZ/2$ the classifying space of $Z/2$.

Tsemo Aristide
  • 89,587
  • 1
    Thanks! Do you know of any hypotheses we can add to make the claim true? For example, what happens we impose the condition that the base space is a finite CW complex, such as the Klein bottle. – Jack Burkart Apr 20 '16 at 18:50