# Automorphism of algebraic group preserving a hyperspecial maximal compact

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...

• 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. – Peter McNamara May 26 '10 at 22:53
• 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 $G'$. – BCnrd May 26 '10 at 23:17
• 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? – blt May 27 '10 at 0:59
• @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 ${\rm{Isom}}(G', G)$. – BCnrd May 27 '10 at 1:33
• @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. – BCnrd May 27 '10 at 2:11

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.
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.