Let $A$ be a local domain. We let $T=T(A) $ be the subgroup of $\mathrm{SL}_{2}$ consisting of diagonal matrices and $V$ be the subgroup of unital matrices of $\mathrm{SL}_{2}$; i.e.
$V:=\left\{\left( \begin{array}{cc} 1 & a \\ 0 & 1 \\ \end{array} \right) \;|\; a\in A\right\}$.
$T:=\left\{\left( \begin{array}{cc} u & 0 \\ 0 & u^{-1} \\ \end{array} \right) \;|\; u\in A^{\times}\right\}$.
We have that $T$ acts on $V$ by conjugation so $V$ is a $T$-module.
This action is non-trivial, so I am trying to calculate $H_{p}(T, V)$ with $p\in\{0, 1,2,3\} $
For the zero homology, it is well known that $H_{0}(T, V)=V_{T}$ i.e. the co-invariants.
In this case I have to find all the matrices in $V$ such that the conjugation by matrices in $T$ are trivial.
For the $H_{0}(T, V)=V_{T}$ , since $T\cong A^{\ast}$ and $V\cong A$, through this identifications, we get that $A^{\ast}$ acts on $A$ by multiplication of a square unit.
Thus for $V_{T}$ I need to check the elements $ua-a=(u^{2}-1) a$ with $u$ a unit in $A$ and $a\in A$.
If $u^{2}-1$ is unit for some $u\in A^{\ast}$ then $V_{T}=0$.
From this I found that if the residue field of the local domain (it can be a local ring) $A$ has at least four elements, then $H_{0}(T, V)=V_{T}$ is equal to zero.
For the next homology $H_{1}(T, V)$, I have a non trivial action in this case, so I tried the next example:
For $A=\mathbb{Z}_{p}$ with $p$ an odd prime we have that
$$\mathbb{Z}_{p}^{\times}\cong \mathbb{F}_{p}^{\times}\times U_{1}$$,
where $U_{1}=1+p\mathbb{Z}_{p}$. In other words, we have the following group extension that splits
$$0\rightarrow U_{1}\rightarrow A^{\times}\rightarrow \mathbb{F}_{p}^{\times}\rightarrow 0$$
Now we have the following Hochschild-Serre spectral sequence associated to this extension
$$E^{2}_{p,q}=H_{p}(\mathbb{F}_{p}^{\times},H_{q}(U_{1},A))\Rightarrow H_{p+q}(A^{\times},A)$$
Since the group extension splits, we have that
$$H_{1}(A^{\times},A)=H_{0}(\mathbb{F}_{p}^{\times},H_{1}(U_{1},A))\times H_{1}(\mathbb{F}_{p}^{\times},H_{0}(U_{1},A))$$
For the homology $H_{1}(U_{1},A)$, it is calculated using the complex
$$A\otimes B_{2}\rightarrow A\otimes B_{1}\rightarrow A\otimes \mathbb{Z}[U_{1}]$$
where $B_{n}$ it the normalized bar resolution over $\mathbb{Z}[U_{1}]$ and the tensors are taken over $\mathbb{Z}[U_{1}]$. Now for $a\in A$ and $u\in U_{1}$, $a\otimes [u]\in \mathrm{Ker}(A\otimes B_{1}\rightarrow A\otimes \mathbb{Z}[U_{1}])$ if and only if $(u^{2}-1)a=0$. Since $A$ is an integral domain, it follows that $u=\pm1$ or $a=0$. Since $-1\not\in U_{1}$, it follows that $\mathrm{Ker}(A\otimes B_{1}\rightarrow A\otimes \mathbb{Z}[U_{1}])=\{0\}$ and thus $H_{0}(\mathbb{F}_{p}^{\times},H_{1}(U_{1},A))=0$.
Now we have that $H_{0}(U_{1},A)=A/I$ where $I$ is the ideal generated by $u^{2}-1$ with $u\in U_{1}$. Since $u^{2}-1$ is not a unit in $A$ for all $u\in U_{1}$. We have that $I\not=A$. From the Hensel lemma, we get that $H_{0}(U_{1},A)=\mathbb{F}_{p}$. Therefore I need to calculate this homology $H_{1}(\mathbb{F}_{p}^{\times},\mathbb{F}_{p})$. i am confused about the action here of $\mathbb{F}_{p}^{\times}$ on $\mathbb{F}_{p}$ in this case. Is it multiplication by a square of a unit?
Thank you for your time!
