Hello. 


Let $K = F\_q(x)$ be the rational function field and let $G = \textbf{PGL}\_{2,K}$. 
For any finite and non empty set $S$ of valuations of $K$, 
we refer to the subgroup of the adelic group $G(\Bbb{A})$: 


$$ G(A\_S) = \prod\_{p \in S} G\_p(K\_p) \times \prod\_{p \notin S} \underline{G}\_p(\mathcal{O}\_p) $$


on which for any prime $p$, $K\_p$ is the completion of $K$ at $p$, 
$G\_p = G \otimes K\_p$, $\mathcal{O}\_p$ is the ring of integers of $K\_p$ and $\underline{G}\_p$ is some matrix realization of $G\_p$ in $\textbf{GL} \otimes \mathcal{O}\_p$.  

In particular, if $S$ contains only the single point at infinity we denote $G(\Bbb{A}\_S)$ by $G(\Bbb{A}(\infty))$.  

The adelic group is decomposed into double cosets:


$$ G(\Bbb{A}) = \bigcup\_{i=1}^h G(\Bbb{A}(\infty)) x\_i G(K) $$ 


where $h=h(G)$ is finite and is called the class number of $G$. 


As the universal covering of $G$ -- namely $\textbf{SL}\_{2,K}$ -- 
admits the strong approximation property, the subgroup $G(\Bbb{A}(\infty))G(K)$ is normal in $G(\Bbb{A})$ 
and: $h(G) = (G(\Bbb{A}):G(\Bbb{A}(\infty)) G(K))$. 


Could someone please tell me what is $h(G)$ in this case ? 


Thank you, rony.



Class number of PGL_2
Let $H$ denote $SL\_2$.
By strong approximation, see <http://www.jstor.org/stable/1970924>, $H(K\_\infty)G(K)$ is dense in $H({\mathbb A})$. 
Now $H({\mathbb A}(\infty))$ contains $H(K\_\infty)$ and contains a unit-neighborhood. 
Therefore we have $H({\mathbb A})=H({\mathbb A}(\infty))H(K)$.
We can conclude the same thing for $G$, if we can show that $H(R)$ surjects onto $G(R)$ and likewise for $R={\mathbb A}, {\mathbb A}(\infty), K$.
The adelic result can be deduced from the result for fields and for $R={\cal O}\_p$.


By Hilbert's Theorem 90, for each field $k$ the Galois cohomology $H^1(k,GL\_1)$ is trivial and so the exact sequence of Galois cohomology shows that the sequence
$$
1\to GL\_1(k)\to GL\_2(k)\to PGL\_2(k)\to 1
$$
is exact, which implies the claim for fields.
To verify the claim for $R={\cal O}\_p$, we have to analyze the coordinate ring of $PGL\_2$.
First, the coordinate ring of $GL\_2$ over $R$ is
$$
A\_{GL\_2}= R[x\_1,x\_2,x\_3,x\_4,y]/(x\_1x\_4-x\_2x\_3)y-1.
$$
The coordinate ring of $PGL\_2=GL\_2/ GL\_1$ is the ring of $GL\_1$-invariants, where the action of $GL\_1$ is given by
$$
\lambda.f(x\_1,x\_2,x\_3,x\_4,y)=f(\lambda x\_1,\lambda x\_2,\lambda x\_3,\lambda x\_4,\lambda^{-2} y).
$$
The ring of invariants is generated by all monomials of the form $x\_ix\_jy$ for $1\le i,j\le 4$.
Let now $\chi\in PGL\_2({\cal O}\_p)$. 


Then $\chi$ is a homomorphism from $A\_{PGL\_2}$ to ${\cal O}\_p$.


Every such can be extended to $A\_{GL\_2}\to K\_p$ and we have to show that there exists an extension mapping to ${\cal O}\_p$.
Pick any extension and denote it by the same letter $\chi$.
For the valuation $v$ on $K\_p$ we have
$$
0\ \le\ v(\chi(x\_j^2y))=2v(\chi(x\_j))+v(\chi(y)).
$$
We are free to change $\chi(y)$ to $\chi(y)\pi^{2k}$ for any $k\in{\mathbb Z}$ if at the same time we change $\chi(x\_j)$ to $\chi(x\_j)\pi^{-k}$ and $\pi$ is a local uniformizer.
Thus we can assume $v(\chi(y))\in\{ 0,1\}$.
Then we conclude $v(\chi(x\_j))\ge 0$ for every $j$ and so $\chi$ indeed maps into ${\cal O}\_p$ as claimed.
This shows the result for ${\cal O}\_p$.


Now let's put things together. We habe ${\mathbb A}^\times={\mathbb A}(\infty)^\times K^\times$, which is due to the fact that we are dealing with the curve ${\mathbb P}^1$ which has genus zero. Let $g\in GL\_1({\mathbb A})$ then
$$
g=d(x,1)y,
$$
where $d(a,b)$ is the diagonal matrix with entries $a$ and $b$, $x$ is an idele and $y$ is in $SL\_2({\mathbb A})$. By what we have shown, we have
$$
g=d(x\_\infty x\_K,1)y\_\infty y\_k=d(x\_\infty,1)\tilde y d(x\_K,1)y\_K,
$$
where The $\infty$ indicates entries in ${\mathbb A}(\infty)$ and The $K$ indicates entries in $K$. The element $\tilde y$ is conjugate to $y\_\infty$, therefore it no longer has entries in ${\mathbb A}(\infty)$, but it still lies in $SL\_2$. Therefore, it can be decomposed again and finally we get $g=g\_\infty g\_K$.


We find that $h(G)$ is one. This however changes, if you take $K$ to be the rational function field of an arbitrary curve and ${\mathbb A}(\infty)$ the adeles which are unrestricted only at a given point. 


