Let $F$ be a $p$-adic field with residue field $k$ and let $G$ be a connected reductive group over $F$. Let us assume that $G$ is simply connected as an algebraic group over an algebraic closure of $F$. If I am not mistaken, it will insure that maximal compact subgroups and maximal parahoric subgroups of $G(F)$ are the same thing.

Given a maximal compact subgroup $K$ of $G(F)$, one can ask whether it is special or hyperspecial. This is defined in terms of the affine building of $G$ ; and there is a characterization of hyperspecial subgroups in terms of integral models of $G$.

On the other hand, given a parahoric subgroup $K$ in $G(F)$, one can form its pro-p radical $K^+$ (I also saw it called pro-unipotent radical, but I assume it is the same thing). The quotient $\mathcal K := K/K^+$ can be seen as the group of $k$-rational points of a reductive group over the finite field $k$. Therefore we call it the finite reductive quotient of $K$. Usually, when $G$ is a classical group then $\mathcal K$ is a product of two classical groups of the same type as $G$.

Both the adjectives "special", "hyperspecial" and the finite reductive quotient only depend on the conjugacy class of $K$.

When $K$ is a maximal parahoric subgroup, is there a way to detect when $K$ is special or hyperspecial solely in terms of conditions on the finite reductive quotient $\mathcal K$ ?

It's a naive question and I don't know what to expect. It just came to my mind while looking at the case of unitary groups. That is, if $E/F$ is an unramified quadratic extension, $V$ is a nondegenerate hermitian space over $E/F$ of dimension $n$ and $G = \mathrm U(V)$. Then, finite reductive quotients of maximal parahoric subgroups look like $\mathrm U_{t}(k) \times \mathrm U_{n-t}(k)$ where $1\leq t \leq n$ is some integer. Unless I am mistaken, we have

  • When $n$ is odd, then $t$ must be odd and the corresponding parahoric is special for $t = 1$ and $t = n$. It is hyperspecial when $t = n$.
  • When $n$ is even and $G$ is not unramified, then $t$ must be odd and the corresponding parahoric is special for $t = 1$ and $t = n-1$.
  • When $n$ is even and $G$ is unramified, then $t$ must be even and the corresponding parahoric is special for $t = 0$ and $t = n$. In fact, they are hyperspecial in both cases.

Thus, it looks like the parahorics are special when one summand of the finite quotient is "as small as possible". Furthermore they are hyperspecial when this summand actually vanishes.

As a sidenote, maybe not related to the problem above, let me denote by $K_t$ a parahoric subgroup having finite quotient $\mathrm U_t(k) \times \mathrm U_{n-t}(k)$ where $t$ is odd or even depending on the cases we described. They form a system of representatives for each conjugacy class of maximal parahoric subgroups of $G$ (the number of such classes being precisely the Witt index of $V$). In particular, when $n$ is even the parahorics $K_t$ and $K_{n-t}$ are not conjugate in $G$ even though their finite reductive quotients are isomorphic.

But I noticed that it is "fixed" when we consider unitary similitude groups $G = \mathrm{GU}(V)$ instead. In this case, we still have maximal parahoric subgroups $K_t$ with finite quotient isomorphic to $\mathrm{G}(\mathrm{GU}_t(k)\times \mathrm{GU}_{n-t}(k))$, however when $n$ is even then the parahorics $K_t$ and $K_{n-t}$ are conjugate in $G$.

  • $\begingroup$ Practically by definition, $x$ being a special point is the same as the Weyl group of the $p$-adic group is the same as the Weyl group of the reductive quotient. Is that the sort of answer you have in mind? $\endgroup$
    – LSpice
    Oct 30, 2021 at 3:36

1 Answer 1


(I write this answer in quite a haste, so there will probably be some inaccuracies. Sorry for that and, please, let me know, when you find them.)

First, I think, that if the group is not almost simple, then the question makes few sense. For example, each of the split groups $G_1 = Sp_4(F)$ and $G_2 = {\rm SL}_2(F) \times {\rm SL}_2(F)$ have a parahoric subgroup of the same type, namely ${\rm SL}_2(\mathcal{O}_F) \times {\rm SL}_2(\mathcal{O}_F)$ (where $\mathcal{O}_F$ are the integers of $F$), obviously with the same finite reductive quotient. For $G_1$, the corresponding parahoric is non-special, whereas for $G_2$ it is hyperspecial.

For simplicity, I will even assume that $G$ is absolutely almost simple over $F$, which means that the absolute Dynkin diagram is connected. We have the affine Dynkin diagram $\widetilde \Delta$ of $G(F)$, whose vertices are in 1-1 bijection with the vertices in a fixed base alcove of the Bruhat--Tits building of $G(F)$. Alternatively, the affine Dynkin diagram can be obtained from the usual one by adding one vertex. Conjugacy classes of maximal parahorics $K = K_x$ of $G(F)$ are in 1-1 bijection with vertices $x$ of $\widetilde \Delta$ (use that $G$ is simply connected). The Dynkin diagram of the finite reduction quotient $\mathcal{K}_x = K_x/K_x^+$ is then just $\widetilde \Delta \backslash \{x\}$ (with all edges beginning/ending in $v$ removed). Thus the number of simple factors of $\mathcal{K}_x$ is equal to the number of connected components of $\widetilde \Delta \backslash \{x\}$. For example, if you remove the middle point in (affine) type $E_6$ (but also in some other cases) it is possible for $\widetilde \Delta \backslash \{x\}$ to have $3$ connected components, so that there are three factors.

Now, $x$ is special, iff $\widetilde \Delta \backslash \{x\}$ equals $\Delta$, the (usual, not affine) Dynkin diagram of $G$. To answer the question, observe (by looking at the finite and affine Dynkin diagrams, which can be found for example in wikipedia) that $\widetilde \Delta \backslash \{x\}$ will tend to be of type $\Delta$ when $x$ tends to be an extremal vertex of $\widetilde \Delta$. For example, if $G = Sp_4$, i.e. with $\Delta$ of type $C_2$ and $\widetilde \Delta$ being $\cdot = \cdot = \cdot$, then $\widetilde \Delta$ without the left or the right vertex will be of type $C_2$ (so these are hyperspecial), whereas $\widetilde \Delta$ without the middle vertex is of type $A_1 \times A_1$ (and it is not hyperspecial).

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    $\begingroup$ Even in the absolutely simple case we can have coincidences of parahorics: $\operatorname{SL}_3(\mathcal O_F)$ is a parahoric subgroup of both $\operatorname{SL}_3(F)$ and $\mathsf G_2(F)$, for example. Surely "solely in terms of $\mathcal K$" should really be interpreted as "solely in terms of $\mathcal K$, for a given $G$" or "solely in terms of $\mathcal K$, in terms of $G$" (which is indeed how your answer proceeds, by comparing the Dynkin diagrams)? $\endgroup$
    – LSpice
    Oct 30, 2021 at 3:39
  • $\begingroup$ Thank you for the detailed explanations, this is precisely what I was looking for. I understand a little bit more what is going on behind my observations. To learn more about the material covered in your answer, I assume there is no going around persevering with Tits' works ; I shall become more familiar with it. @LSpice Absolutely, when writing my question I was implicitely assuming "for a fixed $G$". Nonetheless your examples are enlightening $\endgroup$
    – Suzet
    Oct 30, 2021 at 7:07

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