A space $X$ (by which I mean a CW complex) is acyclic if its reduced singular homology $\tilde H_\ast(X;\Bbb Z)$ is trivial in all degrees.

A discrete group $\pi$ is said to be acyclic if its classifying space $B\pi$ is acyclic.

A space $X$ is aspherical if its universal cover is contractible. A space $X$ is non-aspherical if its universal cover isn't contractible.

Question: Does there exist a non-aspherical, acyclic space $X$ whose fundamental group $\pi$ is also acyclic?

  • $\begingroup$ What is the distinction between $\pi$ being acyclic and being perfect? I mean, I see the logical distinction; but are there examples to show that they are actually different concepts? $\endgroup$ – Jeff Strom Nov 28 '17 at 20:05
  • $\begingroup$ I think so, take the binary icosahedral group. $\endgroup$ – John Klein Nov 28 '17 at 21:38
  • 3
    $\begingroup$ @JeffStrom Take a look at the Wiki page on perfect groups. In particular, $A_5$ and the groups of elementary infinite matrices over a ring $E(R)$ are both perfect and non-acyclic. For $E(R)$ its $H^2(\mathbb Z)$ is equal to $K_2(R)$. $\endgroup$ – Anton Fetisov Nov 28 '17 at 21:42

Yes, such things exist. Take any finitely presented infinite acyclic group $G$, for example, Higman's group. It is a theorem by Kervaire (''Smooth homology spheres and their fundamental groups'') that for each $n \geq 5$, there is an integral homology sphere $M^n$ with fundamental group $G$. Consider $X=M-\ast$. The integral homology of $X$ is trivial, the fundamental group of $X$ is $G$. Suppose that $X$ were aspherical. Observe that $M\simeq X \cup_f D^n$ is obtained by attaching an $n$-cell. Since $\pi_{n-1} (X)=0$, $f$ is nullhomotopic, and hence $M \simeq S^n \vee X$. It follows that the universal cover of $M$ has the homotopy type of $\tilde{X}$ with infinitely many $S^n$'s attached (since $G$ is infinite). So $H_n (\tilde{M};\mathbb{Z}) \neq 0$, a contradiction, because $\tilde{M}$ is a noncompact $n$-manifold.

  • $\begingroup$ Johannes, it just occurred to me: why is $\pi_{n-1}(X)$ trivial? $\endgroup$ – John Klein Nov 28 '17 at 21:37
  • $\begingroup$ I just realized you meant to write "aspherical" instead of acyclic. I fixed your text. It's good now. $\endgroup$ – John Klein Nov 28 '17 at 21:45
  • $\begingroup$ Yes, indeed. I incorporated your change. $\endgroup$ – Johannes Ebert Nov 28 '17 at 21:58

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