1
$\begingroup$

Let $G$ be a simple group over $k=\mathbb{C}$, $A=k[[t]]$, $K=k((t))$, and consider the group $G(K)$. This group is a split reductive group over a local field, and therefore the results of Bruhat and Tits in their paper "Groupes reductifs sur un corps local II" apply: for every parahoric subgroup $\mathcal{P} \subset G(K)$ there exists a smooth group scheme $\mathcal{G}$ over $\text{Spec}(A)$ such that $\mathcal{G}(K) \cong G(K)$ and $\mathcal{G}(A) \cong \mathcal{P}$.

My question is about these group schemes for hyperspecial maximal parahorics. The maximal parahoric subgroups correspond to vertices of the affine Dynkin diagram of $G$. A maximal parahoric is hyperspecial if it corresponds to a hyperspecial vertex of the diagram, and a vertex is hyperspecial (in the case of $G(K)$) if removing it gives a diagram isomorphic to the diagram of $G$. In this case the group schemes are reductive, and correspond to Chevalley group schemes.

My question is: are all the hyperspecial group schemes isomorphic? My understanding is that Chevalley group schemes should be determined by their generic fibers, which is the same for all parahoric group schemes.

If they are not isomorphic, what am I missing about the uniqueness of Chevalley groups?

If they are isomorphic, why bother distinguishing between them?

An example for $G=\text{SL}(2)$ would be much appreciated!

Improve this question
$\endgroup$
5
  • $\begingroup$ They are not just isomorphic, but $G(K)$-conjugate! From my point of view, the reason to distinguish them is not as groups, but for the way that they fit into the Bruhat--Tits picture: they correspond to different (but $G(K)$-conjugate) points in the BT building. $\endgroup$
    – LSpice
    Commented Feb 5, 2015 at 19:33
  • $\begingroup$ (Hmm, since they don't sit in a superstructure, I guess I mean that the $G(K)$-conjugacy of the associated building points induces an isomorphism of the underlying schemes.) $\endgroup$
    – LSpice
    Commented Feb 5, 2015 at 19:49
  • $\begingroup$ Could you explain a bit more? Not all hyperspecial parahorics are conjugate in $G(K)$, but they will be conjugate over a finite extension $L$ of $K$. Are you saying conjugation in $G(L)$ will induce a $K$-isomorphism of the group schemes? $\endgroup$
    – Michael Schuster
    Commented Feb 6, 2015 at 16:44
  • $\begingroup$ @MartinSchuster, I'm sorry, you are right. There are at most two orbits under $G_{\text{ad}}(K)$ (not necessarily $G(K)$ as I originally said), but I don't see any obvious reason that these two orbits should give rise to the same schemes. $\endgroup$
    – LSpice
    Commented Mar 2, 2015 at 0:08
  • 1
    $\begingroup$ (Oops, sorry; I misspelt your name.) On further reflection (no pun intended), since we've got a split group, I return to my original position: all (hyper)special points in the (reduced) building are $G_{\text{ad}}(K)$-conjugate; this is stated in \S2.5 of Tits's Corvallis article. $\endgroup$
    – LSpice
    Commented Mar 2, 2015 at 0:22

0

Reset to default

You must log in to answer this question.