# Finite CW complex with finite non-abelian fundamental group and higher homologies zero

I want to build a finite CW complex such that $$\pi_1$$ is non-abelian and $$H_i$$ are zero for $$i\geq 2.$$ From Hatcher for a given group G, one can create an example of a 2-complex $$X_G$$ with $$\pi_1(X_G)=G.$$ I also checked from Mayer-Vietoris that if $$G$$ is cyclic such complex won't have any higher homology for $$i\geq 2.$$ I tried to take $$G=S_3,$$ the symmetric group of order 6 and from Mayer-Vietoris I get $$H_2$$ is $$Z.$$ I believe this was a correct calculation. Or is there a way to get the groups $$G$$, with $$G$$ non-abelian, such that we can get $$\pi_1 =G$$ and $$H_i=0$$ for $$i\geq 2.$$

Any reference or idea to create such an example? Or is there a way to claim such a finite complex can't exist?

• Is there anything wrong with a bouquet of circles? Aug 4, 2021 at 13:54
• I edited the question. I meant finite non-abelian group in $\pi_1.$ Aug 4, 2021 at 14:02
• In fact there is a 2-dimensional simplicial complex whose fundamental group is $2I$, the "binary icosahedral group", which is nonabelian and perfect and order 120, and whose homology groups $H_i$ are all zero for all $i \geq 1$. This can be constructed by identifying appropriate twists of antipodal faces on a dodecahedron (and triangulating each pentagon), but to the modern eye the fastest description is probably "This is the punctured Poincare homology sphere".
– mme
Aug 4, 2021 at 14:15
• Note that if $G = \pi_1(X)$ is perfect, you are asking for a finite acyclic space $X$ with fundamental group $G$. There's a lot of literature about acyclic spaces which could be relevant. Aug 4, 2021 at 14:32

Theorem. Let $$G$$ be a group. There exists a finite 3-complex $$X_G$$ with $$\pi_1 X_G = G$$ and $$H_i X_G = 0$$ for $$i > 1$$ if, and only if, $$G$$ is finitely presentable and has second group homology $$H_2(G) = 0$$.

The more interesting question to me is whether this is possible for a finite 2-complex, and Jens Reinhold's answer gives us the first step there. This is possible for the group $$2I$$ mentioned in my comment above.

Lemma: Let $$X$$ be any CW complex. Then the cokernel of the Hurewicz map $$\pi_2 X \to H_2 X$$ is isomorphic to $$H_2(\pi_1 X)$$. (Proof: attach cells to make $$X$$ into a $$K(\pi_1 X, 1)$$; doing so kills off precisely $$\pi_2 X$$ inside of $$H_2 X$$.) This is exercise 23 in Hatcher section 4.2.

Proof of theorem. Pick any finite presentation $$P$$ of $$G$$ and construct the presentation complex $$X_P$$. This is a finite complex with $$\pi_1 X_P = G$$, and with $$H_2(X_P)$$ a free abelian group $$\Bbb Z^k$$ on a finite number of generators. Because the cokernel of $$\pi_2(X_P) \to H_2(X_P)$$ is precisely $$H_2(\pi_1 X_P) = H_2(G) = 0$$, it follows that $$\pi_2(X_P) \to H_2(X_P) = \Bbb Z^k$$ is surjective. Now choose $$k$$ maps $$\rho_i: S^2 \to X_P$$ so that these give a basis of the second homology group, and set $$X_G = X_P \cup_{i=1}^k D^3,$$ attaching these 3-cells along the $$\rho_i$$. The resulting complex has $$H_i X_G = 0$$ for $$i > 1$$ by a Mayer-Vietoris argument, while $$\pi_1 X_G = \pi_1 X_P \cong G$$ by the van Kampen theorem.

• Notice that by use of simplicial approximation one may assume that $X_G$ is a simplicial complex, not merely a CW complex.
– mme
Aug 4, 2021 at 16:09
• The lemma is also a straightforward application of the Serre spectral sequence to the fibration $\tilde X\to X \to B\pi_1X$ (where $\tilde X$ is the universal covering so that we can identify the Hurewicz map with $H_2\tilde X\to H_2 X$). Aug 5, 2021 at 6:42

Take any group $$G$$ (non-abelian if you like) that has a presentation $$G = \langle g_1, \dots, g_s \ | \ r_1, \dots, r_{s} \rangle$$ with the same number of generators and relations.

Form a CW-complex $$X$$ with one $$0$$-cell, $$s$$ 1-cells (representing the generators $$g_i$$) and $$s$$ $$2$$-cells that represent the relations. Then $$\pi_1(X) = G$$ and the cellular chain complex that computes $$H_{\ast}(X)$$ looks as follows: $$\mathbb Z^{s} \xrightarrow{\partial} \mathbb Z^s \xrightarrow{0} \mathbb Z$$ The differential $$\partial$$ can of course be calculated easily by abelianizing the relations $$r_i$$. If $$G$$ is finite, then $$H_1(X;\mathbb Q) = G^{\text{ab}} \otimes \mathbb Q = 0$$, hence the induced map $$\partial \otimes \mathbb Q$$ is surjective, hence injective, hence also $$\partial$$ itself is injective, hence $$H_2(X;\mathbb Z) = 0$$. Thus, $$X$$ is as desired.

It remains to undertstand which finite groups admit such a presentation. In line with mme's great comment above, $$2I$$ is such an example.

• math.stackexchange.com/questions/3273061/… gives more examples Aug 4, 2021 at 16:24
• As a remark, every fundamental group of an oriented 3-manifold admits such a presentation, obtained from a Heegaard splitting. One could also directly take the 2-complex to be homotopy equivalent to $Y \setminus p$.
– mme
Aug 4, 2021 at 21:02

One way to construct a space with given $$\pi_1$$ and without higher homotopy/homology is with a covering $$X \to X/G$$ for some free action of $$G$$ on a contractible $$X$$ which will then be isomorphic to $$\pi_1$$. For example $$X=\mathbb{R}^2$$ is the universal covering space of the Klein bottle. The Klein bottle is a finite CW-complex, its fundamental group is the non-abelian group $$G=\langle x,y \mid xyx^{-1} = y^{-1} \rangle$$, and its homotopy and homology groups are trivial from degree two upwards.

• +1 I learned something from this! After the edit, I suppose it doesn't quite answer the question, since the OP wants $\pi_1(X)$ to be finite. As explained here, if $G$ has any torsion, then any $K(G,1)$ must have cells in arbitrarily large dimension, and so $K(G,1)$ is not finite. Aug 4, 2021 at 14:22
• I have edited my question. I meant for a finite non-abelian group in $\pi_1.$ Aug 4, 2021 at 14:27
• And seemingly the higher homology groups of $K(G,1)$, i.e. group homology of $G$, are not trivial in general?
– Z. M
Aug 4, 2021 at 14:45