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Suppose that $K/\mathbb{Q}_l$ is a finite extension, with ring of integers $\mathcal{O}_K$. Suppose $\mathcal{G}/K$ is a (linear) algebraic group (connected+reductive), and $\Gamma\subset \mathcal{G}(K)$ is a hyperspecial maximal compact subgroup. This just means (I believe) that we can find $\tilde{\mathcal{G}}/\mathcal{O}_K$ with $\tilde{\mathcal{G}}_K=\mathcal{G}$ and $\tilde{\mathcal{G}}(\mathcal{O}_K)=\Gamma\subset\tilde{\mathcal{G}}(K)$.

Suppose we have an automorphism $\alpha$ of $\mathcal{G}/K$ preserving $\Gamma=\tilde{\mathcal{G}}(\mathcal{O}_K)$. Can we necessarily extend $\alpha$ to an automorphism of $\tilde{\mathcal{G}}$? If not, are there any nice conditions under which we can? For instance, what if $\mathcal{G}$ is simple or even simple and simply connected?

Edit: added connected, reductive hypotheses on $\mathcal{G}$, which were there in my head, but I forgot to write them...

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  • $\begingroup$ I believe that there are more conditions than mentioned in the text that we must require in order to call a subgroup hyperspecial, namely that the special fibre of G tilde is connected reductive. $\endgroup$
    – Peter McNamara
    Commented May 26, 2010 at 22:53
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    $\begingroup$ You want $\mathcal{G}$ conn'd reductive. Let $R$ be henselian (e.g., complete) dvr with frac. field $K$, and $G$ and $G'$ smooth affine $R$-groups with conn'd reductive fibers. Consider if $K$-isom $G'_K \simeq G_K$ extends to $R$-isom provided it carries $G'(R)$ into $G(R)$; for residue field $k$ finite it is your question, but this makes sense in general. If $R$ strictly henselian (i.e. $k$ sep. closed), then yes: see 1.7 of Bruhat-Tits II. For more general $k$ (e.g., finite), should look deeper into Bruhat-Tits theory, especially if you wish to avoid $R$-split hypotheses on $G$ and $G'$. $\endgroup$
    – BCnrd
    Commented May 26, 2010 at 23:17
  • $\begingroup$ Lemma 6.2 of Snowden-Wiles seems to say that in the case where $\mathcal{G}$ is simply connected+semisimple, if you send $\mathcal{G}(\mathcal{O}_K)$ into itself then you also send $\mathcal{G}(\mathcal{O}_L)$ into itself for any $L/K$ tamely ramified. This seems to mean that you send $\mathcal{G}(\mathcal{O}_K^{unr})$ into itself, and hence by Brian's comment you at least get an automorphism at least of $\mathcal{G}_{\mathcal{O}_K^{unr}}$. Is that correct? $\endgroup$
    – blt
    Commented May 27, 2010 at 0:59
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    $\begingroup$ @blt: your deduction is correct, and then Galois descent via generic fiber pushes the automorphism down to $O_K$, so you win. Note that proof of S-W Lemma 6.2 uses things deep into B-T II. Since you are only using unramified (and not general tame) extensions, it's natural to wonder if the building technology in full is needed, or if one can make a more direct argument by doing the $R$-split case "by hand" and then exploiting that reductive $R$-groups split over a finite etale extension. For example, it could be studied in terms of (quotients of?) the Isom-scheme ${\rm{Isom}}(G', G)$. $\endgroup$
    – BCnrd
    Commented May 27, 2010 at 1:33
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    $\begingroup$ @blt: if $G, H$ are smooth affine $R$-gps with conn'd reductive fibers and $H \rightarrow G$ is isogeny then $H(R)$ is full preimage of $G(R)$ in $H(K)$ (use "valuative criterion" for properness in easy case of finite maps), so by usual business with max'l central torus and s.c. central cover of "derived group" [in relative setting over $R$] the original question over general $R$ reduces to separate case of tori and s.c. ss gps. Tori are easy since split over finite etale cover and the split torus case is obvious, so question over general $R$ does reduce to the s.c. ss case. $\endgroup$
    – BCnrd
    Commented May 27, 2010 at 2:11

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As noted, there seem to be some "reductive"s missing from the question. Here's what is known: let $R$ be a Henselian discrete valuation ring with field of fractions $K$, and let $R^{\prime}$ be the integral closure of $R$ in the maximal unramified extension $K^{\prime}$ of $K$; a smooth affine scheme $X$ over $R$ defines a scheme $X_{K}$ over $K$ and a subset $X(R^{\prime})$ of $X_{K}(K^{\prime})$; the functor $X\rightsquigarrow(X_{K},X(R^{\prime}))$ is fully faithful.

The proof of this is fairly easy (cf. 1.7.3 of .Bruhat, F., and Tits, J., Groupes reductifs sur un corps local II, Publ. Math. IHES, 60, 1984).

As stated this fails without "affine and smooth".

In general you can't replace $X(R^{\prime})$ with $X(R)$ (because $X(R)$ may be empty). Perhaps if $X_{k}(k)$ is Zariski dense in $X_{k}$ it's OK ($k$=residue field).

Added: As blt points out, Lemma 6.2 of Snowden and Wiles, arXiv:0908.1991v3, states that, when $K$ is a finite extension of $\mathbb{Q}_{\ell}$, $X$ is a simply connected semisimple group, and the map on the generic fibre is an automorphism, if $X(R)$ maps into $X(R)$ then $X(R^{\prime})$ maps into $X(R^{\prime})$. Thus, for simply connected semisimple groups, the answer is YES, and as BCnrd points out, that implies that the answer is YES for all reductive groups.

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