Let $m,n$ be integers. The Baumslag Solitar Group if defined by
$$G=G_{m,n}=\langle a,b: ba^{m}b^{-1}=a^{n}\rangle $$
This group acts naturally on $\mathbb{R}^{2}$ by multiplication and I want to compute the Cohomology groups $H^{k}(G_{1,2},\mathbb{R}^{2})$ for $k=0,1,2,…$
My Approach:
The matrices $A=\left(
\begin{array}{cc}
1 & 1 \\
0 & 1 \\
\end{array}
\right)$ and $B=\left(
\begin{array}{cc}
2 & 0 \\
0 & 1 \\
\end{array}
\right)$ make a copy of $G=G_{1,2}$. By definition
$$H^{0}(G,\mathbb{R}^{2})=\{x\in \mathbb{R}^{2}:g\cdot x=x, \ \text{for all} \ g\in G\}$$
It is easy to see by a simple calculation that $H^{0}(G,\mathbb{R}^{2})=0$. On the other hand $H^{1}(G,\mathbb{R}^{2})$ is defined by
$$H^{1}(G,\mathbb{R}^{2})=\frac{\operatorname{Der}(G,\mathbb{R}^{2})}{ \operatorname{Ider}(G,\mathbb{R}^{2})}$$
And…. again… by a simple calculation $H^{1}(G,\mathbb{R}^{2})=0$
In order to compute $H^{2}(G,\mathbb{R}^{2})$ I want to use the identity
$$H^{2}(G,\mathbb{R}^{2})=H^{1}(G, \operatorname{Hom}_{\mathbb{Z}}(I[G], \mathbb{R}^{2}))$$
My quetions are:
1.- If $G=\langle A,B\rangle $ how to compute the group ring $\mathbb{Z}[G]$?
2.- If $G=\langle A,B\rangle $ how to compute the augmentation ideal $I[G]$?
3.- Is there an easy way to compute $H^{2}(G,\mathbb{R}^{2})$?
4.- How to compute $H^{k}(G,\mathbb{R}^{2})$ for $k>2$?
Best Answer
In fact, $H^2(G, {\mathbb R}^2)=0$. The simplest way to see this is to use Euler characteristic with coefficients in the ${\mathbb R}G$-module $M$. It is a general fact that $$ \chi(G, M)= \dim_{\mathbb R}(M)\cdot \chi(G), $$ where $\chi(G)=\chi(K(G,1))$. For all Baumslag-Solitar groups $G$, the natural presentation complex is $K(G,1)$, from which it follows that $\chi(K(G,1))=0$. Now, in your example, you already know that $$ H^0(G,{\mathbb R}^2)=H^1(G,{\mathbb R}^2)=0. $$
Since $\chi(G, M)=0$ where $M={\mathbb R^2}$, it follows that $H^2(G,{\mathbb R}^2)=0$ as well: Since presentation complex is a 2-dimensional $K(G,1)$, all BS groups have cohomological dimension 2, hence, for every ${\mathbb Z}G$-module $M$, $$ H^i(G, M)=0, i\ge 3. $$ Thus, for your module $M$, you get: $$ 0=\chi(G, M)= \dim H^0(G,M) - \dim H^1(G, M) + \dim H^2(G, M)=H^2(G, M). $$