Let $K$ be a nonarchimedean local field and $G$ a (connected) reductive group over $K$, so that $G(K)$ carries a natural topology. An element $g \in G(K)$ is compact if it is contained in a compact subgroup of $G(K)$ (equivalently, if $g$ is contained in a compact open subgroup of $G(K)$). Let $G(K)^o$ be the (normal and open) subgroup of $G(K)$ generated by the compact elements. Let $Z$ be the center of $G(K)$. I am looking for proofs or references for the following facts:

  1. $G(K)/G(K)^o$ is a finitely generated free abelian group.
  2. The image of $Z$ in $G(K)/G(K)^o$ is of finite index.
  3. $Z \cap G(K)^o$ is a maximal compact subgroup of $Z$.

Answers handling any subset of these claims are very welcome.


2 Answers 2


It is easy to see that if $G$ is abelian (hence a torus, since $G$ is reductive), then $G(K)$ modulo maximal compact is a free abelian group, of rank equal to the $K$-rank of $G$, namely the dimension of the maximal $K$-split torus.

If $G$ is semi-simple and simply connected, and is isotropic, then $G(K)$ is generated by unipotent elements (which are compact elements); therefore, $G(K)=G(K)^0$. If $G$ is anisoptropic, then $G(K)$ is compact, and hence $G(K)/G(K)^0$ is trivial.

If $G$ is semi-simple and not simply connected, then it is known that $G(K)/G(K)^+$ is compact and abelian (and finite if $Char (K)=0$), where $G(K)^+$ is the subgroup which is the image of the $K$-rational points of the simply connected cover of $G$ into $G(K)$. It can be shown (I think it is in Bruhat-Tits) that there is a compact subgroup which maps onto $G(K)/G(K)^+$. So the semi-simple case can also be handled.

Since any connected reductive group is a product (almost direct) of a central torus and a semi-simple group, the answer to all the questions follows from the torus case.

  • $\begingroup$ Why is a semi-simple, simply connected, isotropic $G$ generated by unipotent elements? $\endgroup$ Oct 13, 2014 at 21:35
  • 1
    $\begingroup$ This is called the "Kneser-Tits conjecture" for local fields and has been proved by various people; there is a proof (by Raghunathan and Prasad) by reducing this to groups of $K$-rank one. Note also that Kneser-Tits is false for many fields; it is a relatively recent result due to Gille (I think) that KT is true for number fields. $\endgroup$ Oct 13, 2014 at 23:29
  • 2
    $\begingroup$ @Venkataramana: Of course, one really intends to impose the additional (harmless in practice) assumption that if $G$ is "absolutely simple" (or at least $K$-simple) for Kneser-Tits, to avoid the silliness of $K$-anisotropic direct factors. $\endgroup$
    – user27920
    Oct 14, 2014 at 2:31
  • $\begingroup$ @user52484: Yes, thank you for pointing this out; otherwise you may have a product of anisotropic and isotropic simply connected groups, for which KT cannot hold. $\endgroup$ Oct 14, 2014 at 12:34

It should be that $G(K)^0$ is the group of $g \in G(K)$ such that $|\chi(g)| = 1$ for all $K$-rational characters $\chi:G \rightarrow {\rm{GL}}_1$.

Once this is shown, it follows that $G(K)/G(K)^0$ is a subgroup of the torsion-free finitely generated group of $K$-rational cocharacters of the $K$-torus $T := G/\mathscr{D}(G)$ (since ${\rm{Hom}}({\rm{X}}_K(T),K^{\times}/O_K^{\times}) = {\rm{X}}_{K\ast}(T)$), the image of $Z$ in there has finite index because the maximal central $K$-torus of $G$ has finite index in $Z_G$ and maps isogenously on $T$ (and a $K$-isogeny between $K$-tori induces a finite-index inclusion between $K$-rational cocharacter groups).

To likewise see #3, the isogeny of the maximal central $K$-torus onto $T$ reduces the task to checking that for any $K$-torus $S$ the quotient of $S(K)$ modulo its unique maximal compact subgroup is a finitely generated free abelian group. If $S' \subset S$ is the maximal $K$-anisotropic subtorus, so $S/S'$ is $K$-split, the fact that ${\rm{H}}^1(K,\cdot)$ is finite on $K$-tori (by local class field theory; see Milne's book on arithmetic duality theorems) reduces the problem to the easy case of split tori.

So it remains to prove that $G(K)^0$ has the asserted description at the start. For any $K$-rational character $\chi$ on $G$, clearly $|\chi|$ is trivial on all compact elements and hence is trivial on $G(K)^0$. Thus, we need to show that if $g \not\in G(K)^0$ then there exists such a $\chi$ for which $|\chi(g)| \ne 1$. Consider the $K$-isogeny $f:G \rightarrow T \times G'$ where $T = G/\mathscr{D}(G)$ and $G'$ is the quotient of $G$ modulo its center. Since $f$ is a finite morphism, the induced map on $K$-points (whose image can be very "thin" in positive characteristic!) is at least topologically proper with finite fibers. Thus, $G(K)^0$ is the full preimage of $(T \times G')(K)^0$, and by design the $K$-rational character groups of $G$ and $T \times G'$ coincide. This allows us to replace $G$ with $T \times G'$, so now we just need to check separately the case of tori and the case of (adjoint) semisimple $G$.

The case when $G$ is a torus is an easy consequence of the (non-trivial!) finiteness of degree-1 cohomology for the maximal anisotropic subtorus (basically the exact sequence consideration mentioned above), so now we assume $G$ is semisimple and just have to check that $G(K)^0 = G(K)$. (We even saw above that it suffices to replace such $G$ with its isogenous adjoint central quotient.) Let $S$ be a maximal $K$-split torus.

We first treat two special cases: either $S = 1$ (i.e., $G$ is $K$-anisotropic) or $S$ is a maximal $K$-torus (i.e., $G$ is $K$-split). In the first case $G(K)$ is compact; the simplest proof is due to G. Prasad (see his paper Elementary proof of a theorem of Bruhat-Tits-Rousseau... in Bulletin SMF 110). In the second case the Bruhat decomposition and compactness of unipotent $K$-points reduces the task to analyzing points in $S(K)$. But we could have passed to the adjoint case, so then ${\rm{X}}(S)$ has a basis given by a base for a positive system of roots and hence we can pass to the rank-1 split semisimple case. The case of ${\rm{SL}}_2$ is easy (generated by unipotent $K$-points) and for $G = {\rm{PGL}}_2$ the Bruhat decomposition and consideration of the image of ${\rm{SL}}_2(K)$ reduces the task to the specific element $g = {\rm{diag}}(\pi, 1) \bmod K^{\times}$. For the standard Weyl element $w$ that comes from ${\rm{SL}}_2(K)$ we see that $gw$ has order 2 in $G(K)$, so $g = (gw)w^{-1} \in G(K)^0$.

Consider the general semisimple case with $S \ne 1$ (so $S$ is not central in $G$). By Theorem 7.2 in the Borel-Tits IHES paper on reductive groups, there is a split connected semisimple $K$-subgroup $H \subset G$ for which $S$ is a maximal $K$-torus and $\Phi(H,S)$ is the set of non-multipliable roots in the relative root system $\Phi(G,S)$. In particular, $W(H,S) = W(G,S)$, so $N_H(S)(K)$ acts transitively on the set of minimal parabolic $K$-subgroups of $G$ containing $S$ (via bijection of the set of such with the set of positive systems of roots). By the settled split semisimple case, $H(K) = H(K)^0 \subset G(K)^0$. Hence, we just need to consider $g$ that normalizes and hence is contained in a minimal parabolic $K$-subgroup $P \supset S$. But $P = Z_G(S) \ltimes U$ with $U = \mathscr{R}_u(P)$ a $K$-split unipotent $K$-group, so $U(K) \subset G(K)^0$, so we just need to consider $g \in Z_G(S)(K)$.

The derived group $G' := \mathscr{D}(Z_G(S))$ is $K$-anisotropic with $S \times G' \rightarrow Z_G(S)$ a central $K$-isogeny, so $G'(K)$ is compact and hence lies in $G(K)^0$ while $S(K) \subset H(K) \subset G(K)^0$. If $n$ is the order of $S \cap H$ then every element of $Z_G(S)(K)$ has $n$th power in $S(K)\cdot H(K) \subset G(K)^0$, and we have also shown that $G(K)/G(K)^0$ is a quotient of $Z_G(S)(K)/Z_G(S)(K)^0$. By dimension induction (since $\dim Z_G(S) < \dim G$) this latter quotient is a finitely generated abelian group, so $G(K)/G(K)^0$ is finitely generated, abelian, and killed by $n$, so it is a finite abelian group.

This method gives that in general $G(K)/G(K)^0$ is a finitely generated commutative group whose torsion part corresponds to those $g \in G(K)$ such that $|\chi(g)|=1$ for all $\chi:G \rightarrow {\rm{GL}}_1$. So one needs a better way to work with central isogenies to eliminate this finite discrepancy (e.g., to somehow link things back to the simply connected case without creating a finite discrepancy).

  • $\begingroup$ In the third paragraph from below, why does it suffice to consider $g$ that normalize a minimal $K$-parabolic subgroup $P \supset S$? $\endgroup$ Oct 13, 2014 at 21:28
  • $\begingroup$ @QuestionMark: I should replace the last two sentences of that paragraph with the following, which I hope is clearer: Since $N_H(S)(K)$ represents $W(G,S)$ and $U(K) \subset G(K)^0$ for a minimal parabolic $K$-subgroup $P = Z_G(S) \ltimes U$ containing $S$ (with $U = \mathscr{R}_u(P)$ a $K$-split unipotent $K$-group), the Bruhat decomposition for $(G,S,P)$ allows us to reduce to considering $g \in Z_G(S)(K)$. $\endgroup$
    – user27920
    Oct 14, 2014 at 2:29
  • $\begingroup$ Thanks. Perhaps you meant that $N_H(S)(K)/Z_H(S)(K)$ rather than $N_H(S)(K)$ represents $W(G, S)$? $\endgroup$ Oct 14, 2014 at 3:15
  • $\begingroup$ @QuestionMark: By "represents" I just meant "maps onto", so I think we're on the same page. $\endgroup$
    – user27920
    Oct 14, 2014 at 3:46

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