# Is there a "spherical building" for a reductive group over a Henselian local ring?

Let $$A$$ be a Henselian local ring and let $$G$$ be a split reductive $$A$$-group. I'm interested in some notion of a "building of parabolic subgroups" for the group scheme $$G$$.

In my specific cases of interest $$A$$ is nonreduced, something like the finite ring $$\mathbb Z/p^n\mathbb Z$$.

Background: Using results from SGA 3 (if I understand correctly), the classification of parabolic subgroups is much the same over $$A$$ as over a field. For instance:

• every Borel subgroup contains a split torus,
• the Borel subgroups that contain a given split torus are a torsor under the Weyl group, and
• the parabolic subgroups that contain a Borel subgroup are in bijection with subsets of a basis of the root system.

We can therefore define a "(spherical) building of parabolic subgroups" $$B(G)$$ of $$G$$ in the usual way, where facets are (proper) parabolic subgroups and apartments correspond to split maximal tori, as the sets of parabolics that contain the torus.

So after waving our hands and using a major black box (SGA 3), it seems that we have defined some interesting building-like object in great generality. In fact, it seems like $$B(G)$$ just is a building on the nose: if we glance at the three building axioms (see e.g. Abramenko and Brown, Def. 4.1), $$B(G)$$ seems to satisfy them all.

The bad news: there is a classification of spherical buildings of rank $$\geq3$$: they arise from semisimple algebraic groups over a field.

So it seems very unlikely that $$B(G)$$ can be a building after all: it would have to magically agree with the building of some other group $$G'$$ over a field. Looking at the case where $$A$$ is finite, I struggle to imagine how this could be. (See below.) But these buildings are also hard for me to get my hands on.

In fact, here is a proof, using the classification theorem, that $$B(G)$$ cannot be a spherical building in the case where (say) $$A=\mathbb Z/p^n\mathbb Z$$ and $$\text{rank}(G)\geq3$$. Let $$X$$ be the scheme of Borel subgroups, the maximal facets of $$B(G)$$. Using the Bruhat decomposition, and the fact that its strata are affine, one can see that

1. $$p\nmid|X(\mathbb F_p)|$$ and
2. the fibers of the reduction map $$X(A)\to X(\mathbb F_p)$$ have cardinality $$p^{nD}$$, so that
3. $$|X(A)| = p^{nD}\cdot|X(\mathbb F_p)|$$,

where $$D=\dim X_{\mathbb F_p}$$. Hence the number of maximal chambers of $$X$$ is a multiple of a power of $$p$$, but this doesn't happen for the building of a reductive group over $$\mathbb F_p$$ (or $$\mathbb F_q$$).

Questions: (1) (Why) does $$B(G)$$ fail to satisfy the building axioms? (2) Has this simplicial complex $$B(G)$$ been studied before?

Stray thought: We have to be careful in working with group schemes or non-algebraically closed fields to differentiate between schemes and their sets of rational points. When $$A=\mathbb F_2$$ the definition I gave of an apartment is wrong if we interpret "contains a torus" on the level of rational points: a split torus over $$\mathbb F_2$$ does not have very many rational points.

• The spherical building has been studied, among others, notably by Curtis, Lehrer, and Tits, in Spherical buildings and the character of the Steinberg representation. They work over a field, though. I'm not sure if this building has been previously studied over a ring. (I'm making this a comment because it's not clear to me from your question whether you didn't know about this case, or knew about it and were wondering specifically about coefficients in a ring.) Sep 14, 2022 at 17:51
• Is it obvious in your case that the object over $\mathbb Z/p^n\mathbb Z$ for split reductive $G$ is anything other than the object over $\mathbb Z/p\mathbb Z$ for the split reductive group of the same type? Sep 14, 2022 at 17:56
• @LSpice Thanks for the reference, that's a nice source. I am indeed aware of the case over a field, and specifically wonder about the case over a ring. (I stuck to the Henselian local case because it is relevant to me, close to the field case, and avoids many complications that arise over a general base scheme.) Sep 14, 2022 at 17:59
• @LSpice We should get genuinely more here. In the case of $\text{GL}_2$, the Borels are parameterized by $\mathbb P^1$. The projective line has more points over $\mathbb Z/p^n\mathbb Z$ than over $\mathbb F_p$, for one, because the limit of these as $n$ grows is $\mathbb P^1(\mathbb Z_p) = \mathbb P^1(\mathbb Q_p)$. Sep 14, 2022 at 18:03

Working over $$A = \mathbb Z/p^2\mathbb Z$$, consider $$B_1 = \begin{pmatrix} * & * \\ 0 & * \end{pmatrix}$$ and $$B_2 = \operatorname{Int}\begin{pmatrix} 1 & 0 \\ p & 1 \end{pmatrix}B_1$$. Then $$B_1 \cap B_2$$ consists of all matrices $$\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$$ where $$a \equiv d \pmod p$$. In particular—here I rely on the hope that algebraic groups over rings behave closely enough to algebraic groups over fields that my understanding carries over—$$B_1 \cap B_2$$ does not contain any maximal torus in $$\operatorname{GL}_{2, A}$$, so $$B_1$$ and $$B_2$$ do not lie in any common apartment.
• @DavidSchwein, if you're interested, my thought process was: as you outlined in your edit, $\newcommand\Z{\mathbb Z}\DeclareMathOperator\GL{GL}\newcommand\B{\mathcal B}\B(\GL_{2, \Z/p^2\Z})$ can't be a building because it has power-of-$p$ times too many chambers, so I thought it must be some sort of structure fibred over $\B(\GL_{2, \Z/p\Z})$ with fibres of $p$-power size. There's a natural map $\B(\GL_{2, \Z/p^2\Z}) \to \B(\GL_{2, \Z/p\Z})$, so I thought I'd look at its fibres and see if they violated any building axioms. Sep 15, 2022 at 16:55