Let $X$ be the classifying space of the Higman group $G$. It is well known that $G$ is an acyclic group $$H_{\ast}(X;\mathbb{Z})=H_{\ast}(pt;\mathbb{Z}).$$

Now, suppose that $\mathcal{M}$ is a local system on the space $X$ such that

$$H_{i}(X;\mathcal{M})=0, \textrm{ for all $0\leq i$}.$$

Does such local system $\mathcal{M}$ on $X$ exist (other then $\mathcal{M}= 0$)


For $X = BG$ local systems on $X$ can be identified with $G$-modules, and homology with the derived tensor product $-\otimes^L_{\mathbb ZG}\mathbb Z$, i.e. $H_i(X;M) \cong \operatorname{Tor}^i_{\mathbb Z G}(M,\mathbb Z)$. One way to see this is to take the definition $H_i(X;M):= H_i(\mathcal S_*(\widetilde X)\otimes_{\mathbb Z\pi_1(X)} M)$, where $\pi_1(X)$ acts on (singular) chains on the universal cover $\widetilde X$ via deck transformations, and replace $\mathcal S_*(\widetilde X)$ with the cellular complex $C_*(\widetilde X)$ of the CW structure of the realization of the nerve of the groupoid $G//G$; then $C_*(\widetilde X)\otimes_{\mathbb Z G} M = \dots\to \mathbb Z[G^2]\otimes M\to \mathbb Z[G]\otimes M\to M$ is the bar complex computing group homology.

Let $G$ be an arbitrary group, and let $M = \operatorname{ker}(\mathbb Z G\to \mathbb Z)$ be the reduced group algebra, so that we have a short exact sequence $$ 0\to M\to \mathbb ZG\to\mathbb Z\to 0 $$ of $\mathbb ZG$-modules. This gives rise to a long exact sequence of Tor-groups, in particular a boundary operator $\partial:\operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z,\mathbb Z)\to \operatorname{Tor}^i_{\mathbb Z G}(M,\mathbb Z)$. Its kernel is the image of the map $0 = \operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z G,\mathbb Z)\to \operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z,\mathbb Z)$, so it is always injective; its cokernel is the kernel of the map $\operatorname{Tor}^{i}_{\mathbb Z G}(\mathbb Z G,\mathbb Z)\to \operatorname{Tor}^{i}_{\mathbb Z G}(\mathbb Z,\mathbb Z)$, which is an isomorphism for $i = 0$ and has the zero group as its codomain for $i > 0$, so it is always injective, so $\partial$ is always surjective and thus always an isomorphism.

In your example of the Higman group, we know that $H_i(X;\mathbb Z)$ is $\mathbb Z$ concentrated in degree $0$, so that $\partial$ is an isomorphism with the zero group, so that $H_i(X;M) = 0$ for all $i \ge 0$.

  • 2
    $\begingroup$ Thanks for all the details! Do you think that if we add the condition that $M$ is finitely generated as $\mathbb{Z}G$-module, then the conclusion will be that $M=0$ ? Since the kernel of $\mathbb{Z}G\rightarrow \mathbb{Z}$ is not finitely generated (hope I am not wrong) $\endgroup$ – lun Feb 28 at 18:29
  • $\begingroup$ @Bertram Arnold what is the map from singular chain complex to the cellular one (or the way around) that allows you to replace one by the other? $\endgroup$ – user51223 Feb 29 at 2:22
  • $\begingroup$ In general, there is of course no explicit direct map. In this case the CW structure is the canonical one on the geometric realization of a simplicial set, so that there is a canonical map from the cellular complex to the singular chains sending an n-cell to the non-degenerate simplex it corresponds to. This is a chain homotopy equivalence, and any homotopy inverse does the job. $\endgroup$ – Bertram Arnold Feb 29 at 9:41

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