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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.

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up vote 9 down vote accepted

Let $H$ denote $SL_2$. By strong approximation, see, $H(K_\infty)G(K)$ 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.

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Thank you very much, it was very helpful ! rony. – Rony Bitan Nov 15 '11 at 15:30
I found that the proof was not finished. I added a few lines. – Anton Nov 16 '11 at 9:27
Dear Anton, when you write "We can conclude the same thing for $G$, if we can show that $H(R)$ surjects onto $G(R)$ and likewise for $R=A,A(\infty),K$", you probably mean not $H=SL_2$, but $GL_2$. – Mikhail Borovoi Nov 16 '11 at 18:53
The assertion that $GL_2(\mathcal{O}_p)$ surjects onto $PGL_2(\mathcal{O}_p)$ seems to follow from Lang's theorem and Hensel's lemma. – Mikhail Borovoi Nov 16 '11 at 20:01
@Mikhail: you are right, the first lines of my proof don't reflect what I do then... – Anton Nov 17 '11 at 9:58

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