Homology of solvable Lie groups made discrete

Question

In what follows "homology" will mean group homology, i.e. $H_*(BG^\delta;{\mathbf R})$ for the group $G$ with the discrete topology.

It is well-known how to compute the homology of abelian groups, for example for a torsion-free abelian group A one has $H_n(A)=A\wedge_{\mathbf Z}A\wedge_{\mathbf Z}\ldots\wedge_{\mathbf Z}A$, with n factors.

Can one compute the homology of solvable groups, for example of the group $$\left\{\left(\begin{array}{cc}a&b\\ 0&a^{-1}\end{array}\right)\colon a,b\in{\mathbf C}\right\}$$ or of the group $$\left\{\left(\begin{array}{cc}a&b\\ 0&a^{-1}\end{array}\right)\colon a,b\in{\mathbf k}\right\}$$ where $k\subset{\mathbf C}$ is some number field?

If $G$ is a Lie group, Milnor's Comm.Math.Helv.'83-paper gives an isomorphism $H_*(BG^\delta;{\mathbf Z}/p{\mathbf Z})=H_*(BG;{\mathbf Z}/p{\mathbf Z})$, but it's not clear to me what to do with this and what might be the analogue for the number field case.

Answer 1

The example that you mention is the semidirect product for the multiplicative group $k^\times$ acting nontrivially on the additive group $k^+$. Its homology coincides with the homology of $k^\times$. To see this you can a Hochschild-Lyndon-Serre spectral sequence. The point is that $H_i(Bk^\times;H_j(Bk^+))=0$ when $j>0$ and $i\ge 0$. For this it suffices if $k^\times $ has a subgroup $T$ such that $H_i(BT;H_j(Bk^+))=0$ for all $i\ge 0$. Choose $T$ to be infinite cyclic, generated by a rational number $c\notin\lbrace 0,1,-1\rbrace$. $c$ acts on the rational vector space $H_j(Bk^+)$ by $c^{2j}$, so that the homology of $BT$ is $H_0=coker(c^{2j}-1)=0$, $H_1=ker(c^{2j}-1)=0$, $H_i=0$ for $i>1$.