# Does the graded face poset of a BN-pair admit a EL-labeling?

Let $G$ be a finite group with a BN-pair of rank $n$. Let $B$ be the associated Borel subgroup.
Let $P$ be the poset of proper right cosets (i.e. $Kg$ with $K \in [B,G)$ and $g \in G$).
Let $\hat{P}:= P \cup \{\hat{0}, \hat{1}\}$ be the bounded extension of $P$.

All the following questions are very closely related:

Question 1.1: Is $\hat{P}$ Cohen-Macaulay?
Question 1.2: Does $\hat{P}$ admits a dual EL-labeling?

Question 2.1: Is the order complex $\Delta(P)$ equal (or equivalent) to the building of the BN-pair (as an abstract simplicial complex)?
Question 2.2: Is the geometric realization $|\Delta(P)|$ homotopic to a wedge of $r \ge 1$ spheres $\mathbb{S}^{n-1}$.
Question 2.3: What is $r$? How to compute it?

Edit (17/11/2016)

We are not expert on these questions, but we are also not completely ignorant. All these questions (except perhaps Questions 1.2 and 2.3) should be obvious to experts, and we need their clarification.

About Question 2.1:
The definition of the building associated to a BN-pair (as written in [2, Chapter V], [3, Section 15.5] and [4, Section 5.7]) looks like to the definition of $\Delta(P)$ in [6, Section 1.1], but we are not sure that they are equivalent. One doubt comes from the fact that $\Delta(P)$ is defined from the poset of proper cosets, whereas, in  and , the building seems to be defined for the poset of all the cosets (in , it is with proper cosets). Anyway, is the equivalence immediate, or does it need a proof? Any reference?

About Question 1.1:
In [2, Remark 3 p94] it is written: << If you know what a Cohen-Macaulay complex is, then you can easily deduce from our study of the homotopy type of a building that every building is a Cohen-Macaulay complex. >> This result is written without proof, so it should be obvious to expert also. Anyway, is it explicitly proved in some other reference?
We are using the definition of Cohen-Macaulay poset of : it is a property of the order complex of all the open intervals of the poset. Assuming that Question 2.1 has a positive answer, then, $\Delta(P)$ is a building, so is a Cohen-Macaulay complex. Can we deduce that $\hat{P}$ is a Cohen-Macaulay poset?

About Question 1.2:
It is known that a poset admitting a (dual) EL-labeling is Cohen-Macaulay, but the converse is false. So this question seems relevant. Is it an open problem?

About Question 2.2:
By [2, Theorem 2 p 93], it has a positive answer, if Question 2.1 has so.

About Question 2.3:
Let $G$ be a semisimple algebraic group defined over some field $k$. In [5, Section 4, (ii')], Tits states that << A spherical building has the homotopy type of a bouquet of spheres. Furthermore, the number $N$ of these spheres is easily determined; for instance, if the ground field $k$ is finite of characteristic $p$, $N$ is the $p$-contribution to the order of $G$. >>
Is there a proof of this statement somewhere? Is it true for any finite group with a BN-pair? Or should we assume $G$ of Lie type? and if so how to compute $r$ (denoted $N$ by Tits) beyond the Lie type?

References
$$ Nicolas Bourbaki, Lie groups and Lie algebras. Chapters 4–6, Elements of Mathematics, 2002.
$$ Kenneth S. Brown, Buildings, 1989.
$$ Roger W. Carter, Simple groups of Lie type, 1972.
$$ Paul Garrett, Buildings and classical groups, 1997.
$$ J. Tits, On buildings and their applications, Proceedings of the ICM (Vancouver), 1975.
$$ Michelle L. Wachs, Poset topology: tools and applications, 2007.
$$ Anders Björner, Adriano M. Garsia, and Richard P. Stanley, An introduction to Cohen-Macaulay partially ordered sets, 1982.

Edit (18/11/2016)

About Question 2.1:
According to the notation of [Section 1.1, 6], from a simplicial complex $\Delta_0$, we can make a poset $P(\Delta_0)$ called the face poset of $\Delta_0$, and from a poset $P_0$, we can make a simplicial complex $\Delta(P_0)$ called the order complex of $P_0$. The result is (stated without proof in [Section 1.1, 6]): $$|\Delta_0| \simeq |\Delta(P(\Delta_0))|$$ i.e. the geometric realization are homeomorphic. Now take $[H,G]$ a boolean interval of finite groups, $P_0:=C(H,G)$ the poset of proper right cosets $Kg$, and $P_0^*$ the dual of $P_0$ (reversed order). Then $P_0^*$ is naturally the face poset of a simplicial complex, which can be called $P^{-1}(P_0^*)$. If $H$ is also a Borel subgroup coming from a BN-pair, then $P^{-1}(P_0^*)$ is the simplicial complex structure of the building associated to the BN-pair. By applying the result above, it follows that $$|P^{-1}(P_0^*)| \simeq |\Delta(P(P^{-1}(P_0^*)))| = |\Delta(P_0^*)| \simeq |\Delta(P_0)|$$ The fact that some references consider all the cosets whereas other references consider only proper cosets, does not matter because $G$ would map to the empty simplex.
So we have a positive answer to Question 2.1.

Edit (19/11/2016)

About Question 1.1:
Russ Woodroofe has points out to us the reference answering this question:
Anders Björner, Some combinatorial and algebraic properties of Coxeter complexes and Tits buildings, Adv. in Math. 52 (1984), no. 3, 173–212.

So now, all the questions have been answered, except Questions 1.2 and 2.3 (in general).

• I am not sure what a dual EL-labeling is (I am clearly not an expert for these things) -- does that just mean that the dual complex has an EL-labeling? In any case, it is known that thin buildings of classic type (i.e. apartments) are EL-shellable. Also, building are shellable (as shown in Björner's paper pointed out to you by Russ Woodroofe). (Correction: I gave the wrong reference for the EL-shellability of apartments) – Max Horn Nov 19 '16 at 20:20
• @MaxHorn: A poset $P$ admits a dual EL-labeling if its dual $P^*$ (i.e. order reversed) admits a EL-labeling. – Sebastien Palcoux Nov 19 '16 at 20:32
• Re: Wachs' assertion that $|\Delta_0| \cong |\Delta(P(\Delta_0))|$, this is true because $\Delta(P(\Delta_0))$ is the barycentric subdivision of $\Delta_0$ (as an abstract simplicial complex). Each face of $\Delta_0$ becomes a point of $P(\Delta_0)$; interpret this point as the barycenter of the face it came from. This interpretation makes the faces of the new complex $\Delta(P(\Delta_0))$ to be simplices spanned by barycenters of faces contained in the same maximal face of $\Delta_0$. This is the barycentric subdivision. – benblumsmith Feb 3 '17 at 3:40
• Minor correction to my previous comment: in the second-to-last sentence, replace "...contained in the same maximal face..." with "...contained in the same full flag..." – benblumsmith Feb 3 '17 at 17:23
• Regarding Question 2.3: Björner makes the homopy-type calculation in Theorem 4.1 of the same paper ("Some combinatorial and algebraic properties of Coxeter complexes and Tits buildings"). See also Lemma 3.1. It's a certain sum of indices. – Russ Woodroofe Jan 30 '19 at 12:38

## Regarding question 1.2

I do not know an answer to this, nor whether it has been studied. However, it is known that apartments (i.e. Coxeter complexes) are EL-shellable.

UPDATE: OK, I am not so sure anymore this is known. I thought the following reference do the right things; but they deal with the Bruhat poset of a Coxeter group, not the Coxeter complex resp. the poset of parabolic subgroups. I am terribly sorry for the confusion. Here are the references in any case:

• M. Dyer, “Hecke algebras and shellings of Bruhat intervals,” Compositio Math. 89(1), 91–115 (1993).

Moreover,

• P. H. Edelman, “The Bruhat order of the symmetric group is lexicographically shellable,” Proc. Amer. Math. Soc. 82(3), 355–358 (1981) [doi:10.2307/2043939].
• R. A. Proctor, “Classical Bruhat orders and lexicographic shellability,” J. Algebra 77(1), 104–126 (1982) [doi:10.1016/0021-8693(82)90280-0].
• A. Björner and M. Wachs, “Bruhat order of Coxeter groups and shellability,” Adv. Math. 43(1), 87–100 (1982) [doi:10.1016/0001-8708(82)90029-9].

## Regarding question 2.2

This is the Solomon-Tits theorem, for which there are multiple proofs in the literature. For example, see Chao Ku, MR 1459131 A new proof of the Solomon-Tits theorem, Proc. Amer. Math. Soc. 126 (1998), no. 7, 1941--1944..

## Regarding question 2.3

This follows from the Solomon-Tits theorem plus some knowledge about finite groups of Lie-type.

Specifically, each apartment of a spherical building $\Delta$ of rank $n$ (i.e. a Coxeter complex) is (or rather: its geometric realization is) an $(n-1)$-sphere. Fix a single chamber $c$ (i.e. a maximal simplex), set the of apartments containing $c$ cover the whole building, and the building is homotopic to the wedge of these apartments / spheres.

In particular, the number $r:=r(\Delta)$ you are looking for is equal to the number of apartments through $c$, which is equal to the number of chambers opposite $c$.

To compute $r$ for a finite building $\Delta$, let's first consider the case where your group $G$ is a finite group of Lie type, defined over a field with characteristic $p$. Then essentially, $r$ is equal to the $p$-part of the order of $G$. For a more specific statement, see e.g. Corollary 24.11 in

• Gunter Malle and Donna Testerman (2011). Linear Algebraic Groups and Finite Groups of Lie Type. Cambridge University Press.

What happens here is that whenever you have a group with a split BN-pair (i.e. $B=U\ltimes T$, where $T$ is a maximal torus), then the unipotent radical $U$ of the Borel subgroup $B$ acts sharply transitively on the apartments through the chamber corresponding to $B$. In the finite case, then $U=O_p(B)$ (the largest normal $p$-subgroup of $B$). One can turn this into a formula which, however, depends on the type of the building and size of field of definition. If for example your group is of type $A_n$, i.e. $G:=SL_{n+1}(q)$ with $q=p^k$ and $p$ prime, then $U$ is a $p$-Sylow subgroup of $G$. W.l.o.g. it is the subgroup of strict upper triangular matrices, and thus its size (the number $r$ you are looking for) is $q^{n(n-1)/2}$.

Now in general, you can form direct products of buildings to get new buildings, whose rank is the sum of the rank of the factors. On the group side, this corresponds to almost direct products (basically, a central extension of a direct product; think about taking two copies of $SL_{n+1}(q)$ with non-trivial center, then form a product where you glue them together along the center. That is, form the direct product, then factor out the subgroup $\{ (z,z) \mid z\in Z(SL_{n+1}(q)\}$.

Anyway: In a product $\Delta:=\Delta_1\times\cdots\times\Delta_k$ of buildings, we have $r(\Delta) = r(\Delta_1)\cdots r(\Delta_k)$. So one can "easily" reduce to the irreducible case. Note that this something one freuqently does when dealing with problems in buildings / coxeter groups / Lie-type groups: Focus on the irreducible case, and then the general case usually follows in an "obvious" way from this, modulo relatively simple but tedious details.

Note that of course you can form products in cross characteristic, and then the simple trick of taking the $p$-part of the order of $G$ cannot work anymore. Also, you can get factors which are not of algebraic nature, as you could take a product of multiple groups of rank 1 (i.e., Moufang sets), in particular, of sharply 2-transitive groups. The resulting group can have arbitrarily large rank, depending on how many factors you take, an is not of Lie type.

Of course in this case, it's still not hard to compute the number $r$ once you have a decomposition of the group (resp. the building) into its factors: In a Moufang set $X$, all points in $X\setminus\{c\}$ are opposite $c$. Also, there are irreducible buildings of rank 2 which are not of "pure" algebraic type: These "Moufang polygons" have been classified here:

• Jacques Tits and Richard Weiss (2002). Moufang Polygons. Springer Berlin Heidelberg, Berlin, Heidelberg
• What is $O_p(B)$? In its paper above, Tits states that for $G$ a semisimple algebraic group over a finite field of characteristic $p$, $r$ is the $p$-contribution of $|G|$. This is what you wrote about $A_n$. Is it true for the other Chevalley groups and their twist? Then, how to compute $r$ if $G$ is not of this form (this is what I mean by "beyond the Lie type")? – Sebastien Palcoux Nov 19 '16 at 20:31
• << A finite groups with a BN pair of rank at least 3 is of Lie type >>. Don't you need to assume the BN-pair to be irreducible? What does this additional assumption means for $G$? – Sebastien Palcoux Nov 19 '16 at 20:58
• A finite group $G$ with a BN-pair in general, is not associated to a particular prime $p$, isn't it? So the formula "$p$-contribution of $|G|$" should not work in general. – Sebastien Palcoux Nov 19 '16 at 21:05
• If $G$ has a general BN-pair with $B$ the Borel subgroup, is it true that what you wrote implies that $G = G_1 \times \dots \times G_s$ and $B=B_1 \times \dots \times B_s$ with $G_i$ admitting an irreducible BN-pair of Borel $B_i$ ? – Sebastien Palcoux Nov 19 '16 at 21:27
• You would have to ask him, obviously, but my guess would be that there are countless factoids one could insert into any given mathematical paper; in the end, you have to draw the line somewhere. Computing the precise number is hardly the main focus of the Solomon-Tits theorem either. – Max Horn Nov 20 '16 at 20:43