Is there a nontrivial example of an acyclic group $G$ such that its corresponding Eilenberg space $K(G,1)$ is homotopy equivalent to a finite CW-complex ?
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3$\begingroup$ It would be helpful to say what you mean by "nontrivial". If it means "nontrivial group" I would move the adjective accordingly. $\endgroup$– YCorCommented Feb 5, 2020 at 22:46
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2 Answers
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The Higman group with presentation $$\langle{a,b,c,d}\mid{aba^{-1}b^{-2}},~bcb^{-1}c^{-2},~cdc^{-1}d^{-2},~ dad^{-1}a^{-2}\rangle$$ is perfect, and the 2-complex associated to this presentation has Euler characteristic 0. Hence this complex is acyclic. It is in fact aspherical, but it may be simpler to observe that Higman's group is also an iterated generalized free product with amalgamations $(A*_{\langle{b}\rangle}B)*_{F(a,c)}(C*_{\langle{c}\rangle}D)$, where $A,B,C$ and $D$ are copies of the Baumslag-Solitar group $BS(1,2)$, generated by $\{a,b\}$, $\{b,c\}$, $\{c,d\}$ and $\{d,a\}$, respectively. We may assemble a 2-dimensional Eilenberg-Mac Lane complex for the Higman group in a similar way.
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I presume by "acyclic" you are referring to homology with $\mathbb{Z}$ coefficients. There are many such examples.
For instance, you can take two elements $u,v$ in the free group $F_2$ of rank 2 that satisfy the $C'(1/6)$ small-cancellation condition, and also such that $u,v$ together generate the abelianisation $\mathbb{Z}^2$. Explicit examples are easy to construct.
The small-cancellation condition then implies that the corresponding presentation complex $X$ is aspherical, and the assumption about the abelianisation implies that $X$ has the homology of a point. But $\pi_1X$ is an infinite hyperbolic group, in particular non-trivial.
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7$\begingroup$ If explicit examples are easy to construct, it might be worth constructing some explicit examples for the interested readers. $\endgroup$ Commented Feb 5, 2020 at 22:24
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10$\begingroup$ @AndréHenriques Sure, take $F_2 = \langle a,b\rangle$. A suitable pair of elements is $a(aba^{-1}b^{-1})^5$ and $b^6a^2b^3a^{-1}b^{-8}a^{-1}$. I believe the longest common piece is $a^2b$, and $3$ is less than one sixth of the length of each word, which is cyclically reduced. Counting exponent sums, we see that they generate the abelianization. $\endgroup$ Commented Feb 6, 2020 at 3:39
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$\begingroup$ @AndréHenriques -- that would be a great thing to do, but unfortunately I don't have time. If you click through to the wikipedia page, you'll see that small-cancellation conditions are quite explicit, and easy to fulfill, as Rylee's example shows. Someone who's genuinely interested in answering the question won't have any difficulty building an infinite family of examples of presentations using this recipe (though it is a little trickier to show that you get infinitely many different isomorphism types). $\endgroup$– HJRWCommented Feb 6, 2020 at 14:11