# Trivial homology with local system

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 $$i \ge 0$$.
• 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) – lun Feb 28 at 18:29