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Let $G$ be the group generated by $a_i,b_j$, $i,j=1,2,3$ and the following relations: $$b_ib_j=b_jb_i, a_ib_j=b_ja_i, a_1a_2=b_3a_2a_1, a_2a_3=v_1a_3a_2, a_3a_1=v_2a_1a_3$$ I want to comute the homology of $H_nG=H_n(G,\mathbb{Z})$ where $\mathbb{Z}$ is considered as a trivial $G$-module. I have made some attempts of applying spectral sequences (Lyndon-Hochschild-Serre). Here I chose $N=\langle b_1,b_2,b_3\rangle$ as the normal (central) subgroup of $G$. However I could not manage to compute the differentials. Does anyone know how to do this (either by spectral sequences or any other way)?

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I assume your $v_1,v_2$ is really $b_1,b_2$. – Chris Gerig Dec 18 '11 at 22:06
What DO you know about your group? (is it finite? is it torsion-free?etc) – Igor Rivin Dec 18 '11 at 22:35
@Igor, you may want to remove your comment. What if your student sees it here? – Mark Sapir Dec 18 '11 at 22:47
@Mark: you taunt me cruelly:) My question was really: there are a lot simpler questions to answer before you reach for spectral sequences, so has the OP asked them? (given your answer, it appears that no, (s)he has not). – Igor Rivin Dec 18 '11 at 22:57
@Igor: You are right. If the OP cannot recognize nilpotent group by its natural presentation, he/she should not know about spectral sequences. It is like sex education in kindergarten. – Mark Sapir Dec 18 '11 at 23:12
up vote 10 down vote accepted

If $v_i=b_i$, your group is the free nilpotent group of class 2 and rank 3. Its homology is known:

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Yes, $v_1,v_2$ is really $b_1,b_2$. The paper seems very helpful. By a quick calculation it seems like the second homology (Schur multiplier) of the groups $G(r)$ defined here is $\mathbb{Z}^{\frac{1}{3}(r^3-r)}$. Thanks! – xonort Dec 19 '11 at 13:13

I'll try to give an idea on how the computation can be done using the LHS spectral sequence (SS). I start computing cohomology, for cohomology admits a cup product in the SS that is quite useful. Let be $C := Z(G) = \langle b_1,b_2,b_3 \rangle \cong \mathbb{Z}^3$ and $Q := G/C \cong \mathbb{Z}^3$.

Note that $H^\ast(C) = \Lambda_{\mathbb{Z}}(x_1,x_2,x_3)$ and $H^\ast(Q) = \Lambda_{\mathbb{Z}}(y_1,y_2,y_3)$.

Since the extension $$ 1 \to C \to G \to Q \to 1\hspace{20pt}(\ast)$$ is central, by universal coefficients, $$E_2 = H^\ast(Q;H^\ast(C)) = H^\ast(Q) \otimes H^\ast(C) = \Lambda_{\mathbb{Z}}(x_1,...,y_3)$$ with $\deg x_i = (0,1)$, $\deg y_j = (1,0)$. Note that $d_2$ is a derivation, i.e. $$d_2(a \cdot b) = d_2(a)\cdot b + (-1)^{i+j}a \cdot d_2(b) \hspace{20pt}(\ast\ast)$$ for $a \in E_2^{i,j}, b \in E_2^{k,l}$. Since $d_2(y_j) = 0$ (for positional reasons), $d_2$ is completely determined by its values on the $x_i's$.

Now there's an important formula for $d_2^{0,1}$. It depends on the extension class from $(\ast)$. A good reference is [Neukirch, Schmidt, Wingberg: Cohomology of Number Fields, Theorem 2.1.7]. Using the formula, I found
$$d_2(x_1) = y_1y_2,\hspace{2pt} d_2(x_2) = y_1y_3,\hspace{2pt} d_2(x_3) = y_2y_3.$$ Using $(\ast\ast)$ one can compute the $E_3$-term. In my (hasty) computation, I obtained $$E_\infty = E_3 = \begin{array}{cccc} 0 & 0 & \mathbb{Z}^3 & \mathbb{Z} \newline 0 & \mathbb{Z}^6 & \mathbb{Z}^8 & 0 \newline 0 & \mathbb{Z}^8 & \mathbb{Z}^6 & 0 \newline \mathbb{Z} & \mathbb{Z}^3 & 0 & 0 \end{array}$$ For positional reasons, $E_3 = E_\infty$ and since all groups are free abelian, it follows $$H^i(G) = \begin{cases} \mathbb{Z} & i=0,6 \newline \mathbb{Z}^3 & i=1,5 \newline \mathbb{Z}^8 & i = 2,6 \newline \mathbb{Z}^{12} & i=3 \end{cases}$$ Finally, by duality, $H_i(G) = H^i(G)$.

Perhaps someone can compare the result with the paper referenced by Mark Sapir. Thanks.

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Thanks, I will check this out in more detail. Does this mean that all differentials are either injective or surjective? – xonort Dec 19 '11 at 13:16
If there are no errors, yes. – Ralph Dec 19 '11 at 20:13

Trivially $H_0G=\mathbb{Z}$. And it's well known that $H_1G=G/[G,G]$. And we have Hopf's formula $H_2G=R∩[F,F]/[F,R]$ for presentation $G=F/R$. So "low" dimensions are taken care of. Computer programs (GAP,HAP) can compute group homology for a ton of finite groups and certain infinite groups, but I haven't used it in 3 years and am not sure if it works in your case.

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Thanks, I've tried Hopf's formula, but do me it seems hard to do concrete calculations with it by hand (I reallt don't like to apply computer programs). – xonort Dec 19 '11 at 13:24

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