Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am aware of the following question: Definitions of Reductive and Semisimple Groups

So let me phrase a precise question:

Is there a standard technique by which one can translate the unitary/smooth admissible representation theory of semisimple algebraic groups over a local field with the representation theory of reductive algebraic groups?

E.g. many results for $GL(n,F)$ follow from results for $SL(n,F)$ modulo its center. Schur's lemma implies that the restiction of an irreducible representation to the center must be a character.

Is there an algebraic semisimple group $G'$ associated to any reductive $G$, which plays the same role? E.g. such that $G'(F) \cdot Z(F)$ is cocompact in $G(F)$, where $Z$ is the centrum of $G$?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

The derived group $G'$ of $G$ always works for your second question (i.e., $G'(F)Z(F)$ is closed and cocompact in $G(F)$). Indeed, by using local class field theory and Kneser-Bruhat-Tits we know that ${\rm{H}}^1(F,H)$ is finite for any connected reductive $F$-group $H$, so ${\rm{H}}^1(F,G')$ is finite. If $X \rightarrow Y$ is any smooth map of smooth $F$-schemes then $X(F) \rightarrow Y(F)$ is an $F$-analytic submersion, so it has open image. Thus, the image of $G(F)$ in the commutative $(G/G')(F)$ is open, and it has finite index since ${\rm{H}}^1(F,G')$ is finite. In other words, $G(F)/G'(F)$ is a finite-index open subgroup of $(G/G')(F)$, so $G(F)/G'(F)Z(F)$ is compact if and only if the map $Z(F) \rightarrow (G/G')(F)$ induced by the surjection of $F$-tori $Z \rightarrow G/G'$ has compact cokernel.

Thus, it suffices to show that for any surjective map $T' \rightarrow T$ between $F$-tori (even inseparable), the map $T'(F) \rightarrow T(F)$ has closed image with compact cokernel. The maximal compact subgroup of $T(F)$ is $$T(F)^1 = \cap_{\chi\in {\rm{X}}_F(T)} \ker |\!|\chi|\!|_F$$ where $\chi$ varies through the $F$-rational cocharacters of $T$ and $|\!| \cdot |\!|_F$ is the normalized absolute value. In other words, $T(F)^1$ is the group of $t \in T(F)$ such that $\chi(t) \in O_F^{\times}$ for all such $\chi$. It is harmless to pass to quotients by maximal compact subgroups, which is to say that it is equivalent to show that the map $T'(F)/T'(F)^1 \rightarrow T(F)/T(F)^1$ has closed image with finite index.

But $T(F)^1$ is always open in $T(F)$ with the discrete cokernel $T(F)/T(F)^1$ naturally isomorphic to the $F$-rational cocharacter group ${\rm{X}}_{\ast,F}(T)$, so we are reduced to proving that ${\rm{X}}_{\ast,F}(T') \rightarrow {\rm{X}}_{\ast,F}(T)$ has image with finite index. Since these cocharacter groups are finitely generated $\mathbf{Z}$-modules and any surjection of $F$-tori admits an $F$-rational section in the isogeny category, we're done.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.