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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?

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    $\begingroup$ Is there anything wrong with a bouquet of circles? $\endgroup$
    – Tim Campion
    Aug 4 at 13:54
  • $\begingroup$ I edited the question. I meant finite non-abelian group in $\pi_1.$ $\endgroup$
    – piper1967
    Aug 4 at 14:02
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    $\begingroup$ 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". $\endgroup$
    – mme
    Aug 4 at 14:15
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    $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Aug 4 at 14:32
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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.

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  • $\begingroup$ Notice that by use of simplicial approximation one may assume that $X_G$ is a simplicial complex, not merely a CW complex. $\endgroup$
    – mme
    Aug 4 at 16:09
  • $\begingroup$ 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 $H_2\tilde X\to H_2 X$). $\endgroup$ Aug 5 at 6:42
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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.

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    $\begingroup$ math.stackexchange.com/questions/3273061/… gives more examples $\endgroup$ Aug 4 at 16:24
  • $\begingroup$ 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 $Y \setminus p$. $\endgroup$
    – mme
    Aug 4 at 21:02
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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.

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    $\begingroup$ +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. $\endgroup$
    – Tim Campion
    Aug 4 at 14:22
  • $\begingroup$ I have edited my question. I meant for a finite non-abelian group in $\pi_1.$ $\endgroup$
    – piper1967
    Aug 4 at 14:27
  • $\begingroup$ And seemingly the higher homology groups of $K(G,1)$, i.e. group homology of $G$, are not trivial in general? $\endgroup$
    – Z. M
    Aug 4 at 14:45

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