Suppose that $F$ is a nonarchimedean local field, and that $G_1$, $G_2$ are connected, simply connected algebraic groups over $F$. Suppose moreover $G_1$ and $G_2$ are semisimple. Suppose $H$ is a hyperspecial maximal compact subgroup of $G_1\times G_2$. Is $H$ necessarily a product $H_1\times H_2$ where the $H_i$ are hyperspecial maximal compact subgroups of the $G_i$?

Edited: I had originally written that the Lie algebras of the $G_i$ were semisimple, which is not such a sensible thing to write. I have changed this to say the $G_i$ themselves semisimple.

Note: Brian Conrad's answer, below, shows we need assume nothing more than connected reductive hypotheses on $G_1$ and $G_2$.


1 Answer 1


To allow all characteristics, the semisimplicity requirement on the Lie algebras should be replaced with the requirement that the $G_i$ are semisimple as $F$-groups. (In positive characteristic the Lie algebras of connected semisimple groups can often fail to be semisimple.) In fact, the simply connected and semisimplicity hypotheses are stronger than necessary; connected reductive is sufficient. Taking into account the definition of "hyperspecial maximal compact subgroup" (of the group of $F$-rational points), the question in this modified form immediately reduces to a more "algebraic" assertion having nothing to do with local fields, as follows.

Consider a reductive group scheme $\mathcal{G}$ (with connected fibers, following the convention of SGA3) over a normal noetherian scheme $S$, and let $G$ be its fiber over the scheme $\eta$ of generic points of $S$. Assume $G$ decomposes as a direct product $G = G_1 \times G_2$ with $G_i$ necessarily connected reductive over $\eta$. Then does there exist a pair of reductive closed $S$-subgroup schemes (with connected fibers) $\mathcal{G}_i$ in $G_i$ such that $\mathcal{G}_ 1 \times \mathcal{G}_2 = \mathcal{G}$ compatibly with the given identification on generic fibers? Below is a proof that the answer is ``yes''.

First observe that such $\mathcal{G}_i$ are necessarily unique if they exist, being the Zariski closures of the $G_i$ in $\mathcal{G}$ (since $S$ is reduced). In view of this uniqueness, by descent theory it follows that to prove the existence we may work \'etale-locally on the base. (This step tends to ruin connectedness hypotheses, since $S' \times_S S'$ is generally cannot be arranged to be connected, even if $S$ is connected and $S' \rightarrow S$ is a connected \'etale cover.) Hence, we now assume that $\mathcal{G}$ is $S$-split in the sense that it admits a (fiberwise) maximal $S$-torus $\mathcal{T}$ that is $S$-split. We may also work separately over each connected component of $S$, so we can now assume $S$ is connected (as we will make no further changes to $S$ in the argument).

The maximal $\eta$-torus $T := \mathcal{T}_ {\eta}$ in $G$ uniquely decomposes as $T = T_1 \times T_2$ for necessarily $\eta$-split maximal $\eta$-tori $T_i$ in $G_i$. By the classification of split pairs (i.e., fiberwise connected reductive group equipped with a split maximal torus) over any base scheme in terms of root data, to give a direct product decomposition of $(\mathcal{G}, \mathcal{T})$ is the same as to decompose its root datum into a direct product (i.e., direct product of the $X$ and $X^{\vee}$ parts, and corresponding disjoint union for the $R$ and $R^{\vee}$ parts). By connectedness, the normal $S$ is irreducible (i.e., $\eta$ is a single point), so the root data of $(\mathcal{G}, \mathcal{T})$ and its generic fiber $(G,T)$ are canonically identified. QED


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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