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Let me take this question again from the top.

I would like to know what a special parahoric subgroup is.

I think this is a "real" question, though not an especially good one -- it indicates my complete lack of expertise in this area, but it is certainly a question of interest to research mathematicians (who else?).

The reason that I want to know -- and that I would prefer a short, relatively simple answer rather than an invitation to read the original paper of Bruhat and Tits -- is that I have been asked to write a MathReview for a paper using this concept. Now, perhaps I shouldn't have agreed to review this paper, and I didn't, exactly, but it was sent to me about three months ago so it seems a little late to object. Anyway, the introduction is clear, so I think my final product will be a reasonable specimen of the form. It's just that, for my own benefit, when I write sentences like

"The stabilizer of a self-dual periodic lattice chain is a parahoric subgroup (or, in some cases, contains a parahoric subgroup of index 2, which one recovers by intersecting with the kernel of the Kottwitz homomorphism). The author restricts to subsets $I$ for which the parahoric so obtained is special in the sense of Bruhat-Tits theory."

it would be nice if I understood a little better what that meant.

I am looking ideally for a quick answer that increases my knowledge at least a little bit together with a place to read up on it when I get the time / inclination to deepen my knowledge.

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Should it be closed because not really a question? –  Mark Sapir Oct 4 '10 at 18:40
    
Have you explained to your friend how to use MO? :-) –  Robin Chapman Oct 4 '10 at 18:43
    
@Mark: Why is it not really a question? –  Pete L. Clark Oct 4 '10 at 18:50
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@Mark: yes, I'm afraid you've dragged it out of me: I don't know what a special parahoric subgroup is. And to think, I call myself a man... –  Pete L. Clark Oct 4 '10 at 21:32
2  
@Pete: I am happy that now you know. See, MO is better than Google and Wikipedia combined –  Mark Sapir Oct 4 '10 at 23:33
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1 Answer 1

up vote 10 down vote accepted

I'd recommend first that you and your friend spend more time with Tits :), "Reductive groups over local fields", from the Corvallis volume (free online, last time I checked). Undoubtably there are other references, like papers of Prasad-Raghunathan mentioned by Greg Kuperberg, and any paper that treats Bruhat-Tits theory for unitary groups. I'll try to provide a background/basic treatment here.

You're certainly used to Bruhat-Tits theory for $SL_2$, or at least for $PGL_2$, over $Q_p$, where one encounters the $(p+1)$-regular tree. As you know, $PGL_2(Q_p)$ acts on this tree, and the stabilizer of a point is a maximal compact subgroup that is conjugate to $PGL_2(Z_p)$, and the stabilizer of an edge is an Iwahori subgroup. This I assume is a familiar picture.

To understand the "special point" subtleties, you should first think about a group like $G=SU_3$ -- the $Q_p$-points of a quasisplit form of $SL_3$, associated to an unramified quadratic field extension $K/Q_p$ with $p$ odd. Let $S$ be a maximal $Q_p$-split torus in $G$. Then $S$ has rank one, though the group $G$ has absolute rank two. It follows that the Bruhat-Tits building for $SU_3$ is again a tree, though not as simple as the $SL_2$ case. In fact, in this unramified situation, the building can be seen as the fixed points of the building of $G$ over $K$ (which is the building of $SL_3$), under the Galois involution.

Now, one must think about the relative roots of $G$ with respect to $S$ -- i.e., decompose the Lie algebra of $G$ with respect to the adjoint action of $S$. There are four eigenspaces with nontrivial eigenvalue -- these are the root spaces for the relative roots which I'll call $\pm \alpha$ and $\pm 2 \alpha$. The root spaces ${\mathfrak g}_{\pm \alpha}$ are two-dimensional.

Now, let $A$ be the apartment of the building associated to $S$ -- $A$ is a principal homogeneous space for the one-dimensional real vector space $X_\bullet(S) \otimes_Z R$. After choosing a good base point (a hyperspecial base point, using the fact that $K/Q_p$ is unramified), $A$ may be identified with $X_\bullet(S) \otimes_Z R$ and the affine roots are functions of the form $\pm \alpha + k$ and $\pm 2 \alpha + k$, where $k$ can be any integer.

Let $h$ be a generator of the rank 1 $Z$-module $X_\bullet(S)$, so that $\alpha^\vee = 2 h$, and $A = R \cdot h$. The affine roots are given by: $$[\pm \alpha + k](r h) = r + k, [\pm 2 \alpha + k](r h) = 2r + k.$$ The vanishing hyperplanes of these affine roots are the points: $$r h : r \in \frac{1}{2} Z.$$ These are the vertices of the building, contained in the apartment $A$.

Now consider a vertex $nh$, where $n$ is an integer. The affine roots $\pm (\alpha - n)$ and $\pm (2 \alpha - 2n)$ vanish at the vertex $n$. The gradients of these affine roots are the roots $\pm \alpha$ and $\pm 2 \alpha$. These are all of the roots in the original (relative) root system. That's why these vertices are hyperspecial vertices.

On the other hand, consider a vertex $(n + \frac{1}{2}) h$, where $n$ is an integer. The affine roots vanishing at this vertex are $\pm (2 \alpha - 2n - 1)$. The gradients of these affine roots are the roots $\pm 2 \alpha$. These are not all of the original roots, but all original roots are proportional to these roots. You can see how this phenomenon requires the setting of a non-reduced root system to happen. These "half-integral" vertices are special points, since the original root system does not occur in the system of gradients, but it does up to proportionality. At these special (but not hyperspecial) points, the Bruhat-Tits group scheme underlying the parahoric has special fibre with reductive quotient isomorphic to $PGL_2$ (I think... or is it $SL_2$) over the residue field. At the hyperspecial points, the group would be a quasisplit $SU_3$ over the residue field.

If it's not clear from above, a special point in the building occurs where the set of gradients of affine roots vanishing at that point is equal, modulo proportionality, to the set of relative roots. That's the general definition.

Hope this helps - see Tits for more.

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Thanks, Marty, this is very helpful. –  Pete L. Clark Oct 4 '10 at 23:31
    
Dear Marty, You have helped more than one person with this explanation. Thanks! –  Emerton Oct 5 '10 at 3:41
    
Glad to hear it! I hope more people read Tits's article in Corvallis too - it's a real gem, and full of great hard-to-find (outside Bruhat-Tits) examples. I only wish there were more pictures in it. –  Marty Oct 5 '10 at 5:29
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