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, I tried by calculating the homology of the complex

$ A\otimes B_2\xrightarrow{1\otimes d} A\otimes B_1\xrightarrow{1\otimes d} A\otimes \mathbb{Z}A^{\ast}$

where the tensor is taken over $\mathbb{Z}A^{\ast}$. The module $B_1$ is the free $G$ module on the symbols $[a_1]$ for $a_1\in A^{\ast}$ and $a_1\not=1$. If we let $[]\in\mathbb{Z}G$ be the identity, then

$$d([a_1]) = a_1[] - []$$.

Similarly, $B_2$ is the free $G$ module on $[a_1|a_2]$ where neither are the identity, and

$$d([a_1|a_2]) = a_1[a_2] - [a_1a_2] + [a_1]$$.

Here $[a_1a_2] = 0$ if $a_1a_2 = 1$ in the group $A^{\ast}$.

We have that $a\otimes u\in Ker(1\otimes d)$ if and only if $0=u\cdot a-a=u^{2}a-a=(u^{2}-1)a$. Since $A$ is a domain we have that $a=0$ or $u=\pm 1$. From this I got that $Ker(1\otimes d)=\langle a\otimes u[-1]\rangle$. I was trying to check if $a\otimes [-1]$ is a boundary element and if the size of the residue field might influence that.

  I was trying to find some literature and I only found cases when the action is only trivial. Any guideline, idea or corections might be helpful.

Thank you for your time!

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!

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.

$V:=\left\{\left( \begin{array}{cc} 1 & a \\ 0 & 1 \\ \end{array} \right) \;|\; a\in A\right\}$.

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.

Now I am trying to find the homology of $H_{0}(T,H_{2}(V,\mathbb{Z}))=H_{0}(A^{\ast},H_{2}(A,\mathbb{Z}))=(A\bigwedge_{\mathbb{Z}}A)_{A^{\ast}}$ which are the $A^{\ast}$- coinvariants of $A\bigwedge_{\mathbb{Z}}A$ where the actions is given by

$u\cdot(a\wedge b)=u^{2}a\wedge u^{2}b$.

I am trying to find out when $(A\bigwedge_{\mathbb{Z}}A)_{A^{\ast}}$ is zero, so I need to check if the ideal $I$ generated by the elements

$u^{2}a\wedge u^{2}b - a\wedge b$ $(1)$ .

It has a unit.

I was trying some cases for $A$, for instance the case of $p$- adic integers and if $u$ is unit and also and integers the condition $(1)$ reduces to

$(u^{4}-1)(a\wedge b)$

Thus I need to check if there is a unit $u$ and also and integers such that $u^{4}-1$ is a unit. When $p\geq 7$ is true since $2^{4}-1$ is unit. For the cases $p=2,3,5$ , $u^{4}-1$ is not a unit.

For the local ring $\mathbb{Z}/4\mathbb{Z}$ the ideal $I$ is zero. So I have not found the importance of the residue field for these cases.

Thank you for your time!

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, I tried by calculating the homology of the complex

$ A\otimes B_2\xrightarrow{1\otimes d} A\otimes B_1\xrightarrow{1\otimes d} A\otimes \mathbb{Z}A^{\ast}$

where the tensor is taken over $\mathbb{Z}A^{\ast}$. The module $B_1$ is the free $G$ module on the symbols $[a_1]$ for $a_1\in A^{\ast}$ and $a_1\not=1$. If we let $[]\in\mathbb{Z}G$ be the identity, then

$$d([a_1]) = a_1[] - []$$.

Similarly, $B_2$ is the free $G$ module on $[a_1|a_2]$ where neither are the identity, and

$$d([a_1|a_2]) = a_1[a_2] - [a_1a_2] + [a_1]$$.

Here $[a_1a_2] = 0$ if $a_1a_2 = 1$ in the group $A^{\ast}$.

We have that $a\otimes u\in Ker(1\otimes d)$ if and only if $0=u\cdot a-a=u^{2}a-a=(u^{2}-1)a$. Since $A$ is a domain we have that $a=0$ or $u=\pm 1$. From this I got that $Ker(1\otimes d)=\langle a\otimes u[-1]\rangle$. I was trying to check if $a\otimes [-1]$ is a boundary element and if the size of the residue field might influence that.

I was trying to find some literature and I only found cases when the action is only trivial. Any guideline, idea or corections might be helpful.

Thank you for your time!