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Let $F$ be a local non-archimedean field and let $G$ be a connected reductive algebraic group defined over $F$. Let $G_{der}$ denote the algebraic derived group of $G$; this is connected and semisimple. Let $Z$ be the center of $G$ and let $Z^0$ denote its algebraic connected component of the identity. Then $G = Z^0G_{der}$; in fact the map $Z^0 \times G_{der} \to G$ is a central isogeny.

My questions arise from considering the restriction of complex representations of $G(F)$ to $G_{der}(F)$. We note that $Z^0 \times G_{der}$ and $G$ are connected and have isomorphic Lie algebras.

Question 1: Does it follow that $Z(F)G_{der}(F)$ is a subgroup of finite index in $G(F)$?

For example, with $G=GL_n$ the index is $F^\times/(F^\times)^n$. If true in general, I would love a reference.

Question 2: Let $C$ be the commutator subgroup $G(F)$. Are there bounds (in terms of the Coxeter number and cardinality of the residue field, in particular) on the index of $C$ in $G_{der}(F)$?

For example, let $G(F)$ be the group of quaternions of norm 1 over a p-adic field with residue field of odd cardinality $q$. Then $G=G_{der}$, being the algebraic group $SL_2$. Here $C=G(F)\cap(1+P)$ (where $P$ is the maximal ideal of the integer ring of the algebra of quaternions); it has index $q+1$ in $G(F)=G_{der}(F)$.

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You are assuming char($F$) is 0. The exact sequence $1 \rightarrow \mu \rightarrow Z \times G' \rightarrow G \rightarrow 1$ ($G'$ the derived group, $\mu$ finite over $F$ and central) induces a connecting homomorphism $G(F) \rightarrow {\rm{H}}^1(F,\mu)$ with kernel $Z(F)G'(F)$, so the index in Question 1 divides the size of ${\rm{H}}^1(F,\mu)$. Finiteness of Galois cohomology of finite Galois modules for $p$-adic fields is proved in Serre's "Galois cohomology". –  user30180 Jun 8 '13 at 3:06
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For Question 2, the commutator $G \times G \rightarrow G$ factors through $G \times G \rightarrow \widetilde{G}$ into the simply connected cover $\widetilde{G}$ of the derived group $G'$ (since $G^{\rm{ad}} = \widetilde{G}^{\rm{ad}}$), so if $\widetilde{G}(F)$ is perfect then by the Kneser-Bruhat-Tits vanishing theorem the index is exactly the size of ${\rm{H}}^1(F,\mu)$ where $\mu = \ker(\widetilde{G} \rightarrow G')$. Any simply connected semisimple $H$ is a direct product of $F$-simple factors, and $H(F)$ is perfect for $F$-simple simply connected $F$-isotropic $H$ by a theorem of Tits. –  user30180 Jun 8 '13 at 3:21
    
Tate's Euler characteristic and local duality results for Galois cohomology of finite Galois modules over $p$-adic fields will provide bounds on the size of ${\rm{H}}^1(F,\mu)$ in the preceding comments. –  user30180 Jun 8 '13 at 5:03
    
To Ayanta: Thanks very much! I would accept your first comment as an answer to my first question, if you'd post it as such. For the second comment: since $G\times G \to G$ is not a homomorphism, I don't see how to recover $[G(F),G(F)]$ using Galois cohomology (or from your argument). –  Monica Jun 10 '13 at 9:11

1 Answer 1

up vote 3 down vote accepted

This is an answer to the 2nd question, since the OP now understands an answer to the first one. Let $G$ be a connected reductive group over a field $k$, and let $G'$ be the connected semisimple derived group, and $\widetilde{G} \rightarrow G'$ the simply connected central cover of $G'$. Writing $G^{\rm{ad}} := G/Z_G$ to denote the adjoint central quotient of $G$ (i.e., quotient by scheme-theoretic center, or equivalently the image of ${\rm{Ad}}_G:G \rightarrow {\rm{GL}}(\mathfrak{g})$), note that the natural maps $$\widetilde{G}^{\rm{ad}} \rightarrow {G'}^{\rm{ad}} \rightarrow G^{\rm{ad}}$$ are all isomorphisms.

The commutator map $c:G \times G \rightarrow G$ lands inside $G'$ and it is invariant under $Z_G \times Z_G$-translation on either factor of the source, so by the universal property of the quotient morphism of schemes $$G \times G \rightarrow (G \times G)/(Z_G \times Z_G) = (G/Z_G) \times (G/Z_G) = G^{\rm{ad}} \times G^{\rm{ad}}$$ it follows that $c$ factors uniquely through this quotient morphism via a $k$-scheme map $$G^{\rm{ad}} \times G^{\rm{ad}} \rightarrow G'.$$ But this can be applied just as well to the semisimple $\widetilde{G}$ in place of $G$, and so since the map $\widetilde{G} \rightarrow G$ induces an isomorphism between adjoint central quotients we arrive at a map $$G^{\rm{ad}} \times G^{\rm{ad}} = \widetilde{G}^{\rm{ad}} \times \widetilde{G}^{\rm{ad}} \rightarrow (\widetilde{G})' = \widetilde{G}$$ with the property that its composition with $\widetilde{G} \rightarrow G$ on the target and its pre-composition with $G \times G \rightarrow G^{\rm{ad}} \times G^{\rm{ad}}$ on the source recovers the commutator map $c$ of $G$.

In particular, we have obtained a $k$-scheme map $$G \times G \rightarrow \widetilde{G}$$ whose composition with $f:\widetilde{G} \rightarrow G$ is the commutator. Passing to $k$-points (finally!), we conclude that $(G(k),G(k))$ is contained in $f(\widetilde{G}(k))$. Hence, in cases where $\widetilde{G}(k)$ is its own commutator subgroup then it follows that $(G(k),G(k)) = f(\widetilde{G}(k))$. In general $f$ factors through the $k$-subgroup $G' \subset G$, and we have an exact sequence of groups $$\widetilde{G}(k) \rightarrow G'(k) \rightarrow {\rm{H}}^1(k,\mu)$$ where $\mu = \ker(\widetilde{G} \twoheadrightarrow G')$, with this exact sequence surjective on the right for cases where ${\rm{H}}^1(k,\widetilde{G}) = 1$.

As for any simply connected group, $\widetilde{G}$ is a direct product of its $k$-simple factors $G_i$, each of which is the Weil restriction of an absolutely simple and simply connected semisimple group $G'_i$ over a finite separable extension $k_i$ of $k$. Thus, for group-theoretic questions about the rational points of $\widetilde{G}$ (such as perfectness) one can focus on the absolutely simple case at the cost of passing to a finite separable extension of $k$. Note in particular that the $k$-simple factor $G_i$ of $\widetilde{G}$ is $k$-isotropic if and only if $G'_i$ is $k_i$-isotropic (since finite separable Weil restriction induces a bijection between sets of maximal tori, and has no effect on isotropicity or not of a given torus).

By a theorem of Kneser (in char. 0) and Bruhat-Tits (in general), if $k$ is a non-archimedean local field and $H$ is a connected semisimple $k$-group that is simply connected then ${\rm{H}}^1(k,H) = 1$. Moreover, for such $k$ it is a theorem of Tits that a connected semisimple $k$-group which is isotropic, absolutely simple, and simply connected has perfect group of rational points. (Google "Kneser-Tits conjecture" for more general $k$.)

The upshot is that if $k$ is a non-archimedean local field and all $k$-simple factors of $G'$ (or equivalently, of $G^{\rm{ad}}$) are $k$-isotropic then $(G(k),G(k))$ is equal to the image of $\widetilde{G}(k)$ in $G'(k)$, so if moreover ${\rm{char}}(k)=0$ then this image is open with finite index equal to the size of ${\rm{H}}^1(k,\mu)$ for the finite kernel $\mu = \ker(\widetilde{G} \twoheadrightarrow G')$ that "is" a finite Galois module when in characteristic 0. The size of this cohomology group is then provided by Tate's results in local duality (including his Euler characteristic formula), expressed in terms of several ingredients: the size of $\mu(k)$, the number of $k$-rational characters of $\mu$, and $|\!|\#\mu|\!|_k$ (normalized absolute value for the $p$-adic field $k$, with $\#\mu$ the order of the Galois module $\mu$).

What if $G'$ (or equivalently, $\widetilde{G}$) has $k$-simple factors that are anisotropic? We know exactly what the $k$-simple connected semisimple anisotropic groups are over non-archimedean local fields are, namely central quotients of ${\rm{R}}_{k'/k}({\rm{SL}}_1(D'))$ for central division algebras $D'$ over finite separable extensions $k'$ of $k$. That is exactly the case which you addressed in the question, so it is precisely the anisotropic $k$-simple simply connected cases for which the group of rational points is not perfect (and hence a more direct analysis is required, probably best done case-by-case depending on what you may need, whereas the "everywhere isotropic" case has a clean uniform answer as described above).

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Thanks so much for the detailed explanation! –  Monica Jun 11 '13 at 7:42

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