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A few days ago, I asked a question about a group homology, and it was actually easy. I am continuing computing group homologies, but I am stuck on this: $H_*^{\textrm {grp}}(T, \mathbb{Z}) = \textrm{Tor} _* ^{\mathbb{Z}(T)}(\mathbb{Z}, \mathbb{Z})$ where $T=<a,b,c,d\, |\, aba^{-1}b^{-1}cdc^{-1}d^{-1}>$.

From my text it is clear that $H_0 = \mathbb{Z}$ and $H_1 = T/[T,T]=\mathbb{Z}^4$. But how could I compute for the case $* \ge 2$? I cannot use the Kunneth formula, and it does not seem to be something in my note. I think I need to provide a projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}(T)$-module(with the trivial action). If someone gives me any comment for this, then I will do the else by myself.

Thank you for your help!

S. Ha
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1 Answers1

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You have $H_2=\mathbb Z$ and $H_k=0$ for $k\geq 3$. It is the homology of the surface $S_2$ of genus $2$. The reason is that $S_2$ is $K(T,1)$: the fundamental group of $S_2$ is $T$ and the universal cover of $S_2$ is contractible. To get a free resolution of $\mathbb Z$ choose a decomposition of $S_2$ to cells (the standard one will do - with 1 2-cell, 4 1-cells and 1 0-cell, the 4 1-cells corresponding to $a,b,c,d$) and take its preimage in the universal cover (which is a disk). You get a cell decomposition of the disc, and $T$ acts freely on the these lifted cells. The resolution is $C_2\to C_1\to C_0$ where $C_k$ is the free $\mathbb Z$-module generated by the $k$-cells of the disc. After you mod out by the action of $T$, you get the cell complex corresponding to $S_2$.

user8268
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  • Thank you! So It seems that $H_* (\pi_1(S),\mathbb{Z}) = H_*(S,\mathbb{Z})$. Is it true? – S. Ha Jun 17 '14 at 11:28
  • @S.Ha yes, provided the universal cover of $S$ is contractible (which is true for any surface $S$ with the exception of the sphere and projective plane) – user8268 Jun 17 '14 at 12:07