For an algebraic group that cannot be embedded into $GL_n$, is there a nice definition for congruence subgroup? Do we just define it as the compact open subgroup of $G(A_f)$, where $A_f$ is the finite adele?
1 Answer
Even though every linear algebraic group (understood to mean affine of finite type) can be embedded into ${\rm{GL}}_ n$, if we change the embedding then the notion of "congruence subgroup" may change (in the sense of a "group of integral points defined by congruence conditions"). So a more flexible notion is that of arithmetic subgroup, or $S$-arithmetic for a non-empty finite set of places $S$ (containing the archimedean places), in which case we can make two equivalent definitions: subgroup commensurable with intersection of $G(k)$ with a compact open subgroup of the finite-adelic points, or subgroup commensurable with intersection of $G(k)$ with $S$-integral points of some ${\rm{GL}}_ n$ relative to a fixed closed subgroup inclusion of $G$ into ${\rm{GL}}_ n$ (i.e., if $\mathcal{G}$ denotes the schematic closure in ${\rm{GL}}_ n$ over $\mathcal{O}_ {k,S}$ of $G$ relative to such an embedding over $k$, we impose commensurability with $\mathcal{G}(\mathcal{O}_ {k,S})$).
Personally I like to view the ${\rm{GL}}_ n$-embedding as just a quick-and-dirty way to make examples of flat affine integral models (of finite type) over rings of integers (via schematic closure from the generic fiber). However, in some proofs it is really convenient to reduce to the case of ${\rm{GL}}_ n$ and doing a calculation there. It is not evident when base isn't a field or PID whether flat affine groups of finite type admit closed subgroup inclusions into a ${\rm{GL}}_ n$ (if anyone can prove this even over the dual numbers, let me know!), so in the preceding definition it isn't evident (if $\mathcal{O}_ {k,S}$ is not a PID) whether the $\mathcal{G}$'s obtained from such closure account for all flat affine models of finite type over the ring of $S$-integers. In other words, we really should be appreciative of ${\rm{GL}}_ n$.
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$\begingroup$ This answer has been revised from the initial one to correct some misstatements. Not sure if it merits being selected as the answer anymore. If it is "unselected", that's fine. $\endgroup$– BCnrdCommented Mar 15, 2010 at 3:05