2.6.3 The Shape of X
Suppose (W, S) is irreducible. IfW is neither spherical nor Euclidean, then a result of Vinberg [270, p. 1112, Lemma 15] says that the Tits coneXis strictly convex, i.e., its closure does not contain any lines through the origin. (This is obviously false in the spherical case. It is also false in the Euclidean case, where the closure of the Tits cone is a closed half-space. We have seen this in the case of the infinite dihedral group in Section 2.2.2, and the assertion in general follows from some results that we will prove in Section 10.2.2.) Our goal in this subsection is to prove the following weak form of Vinberg’s result, which is valid in the Euclidean case also; see Krammer [149, Theorem 2.1.6]
for a different proof.
Proposition 2.91.IfW is infinite and irreducible, then the Tits coneX does not contain any lines through the origin.
Our proof will be based on the following lemma:
Lemma 2.92.Suppose W is infinite and irreducible. For any x = 0 in X, there are infinitely many walls not containing x.
Proof. We may assume that the cellAcontainingxis a face of the fundamen- tal chamber C and hence that its stabilizer is WJ for someJ S. Suppose x (and hence A) is contained in all but finitely many walls. Then there is an upper bound on the gallery distance d(A, D), where D ranges over the chambers wC (w∈W). [Hered(A, D) is defined as in Exercise 1.61, and, as in that exercise, it is equal to the number of walls that strictly separate A fromD.] Equivalently, there is an upper bound ond(wA, C) forw∈W. This implies that the W-orbit of A is finite and hence that WJ has finite index
in W, contradicting Proposition 2.43.
(See also Exercise 3.83(b), where the same result is stated and proved from a combinatorial point of view.)
Proof of Proposition 2.91. SupposeX contains a pair of opposite points ±x with x = 0. Since xand −x are strictly separated by all walls that do not contain them, it follows from Theorem 2.80 that x is contained in all but
finitely many walls. This contradicts Lemma 2.92.
108 2 Coxeter Groups
Throughout this section we denote byHan arbitrary collection of linear hyperplanes in a finite-dimensional real vector spaceV. For convenience, we assume that we have chosen for each H ∈ H a linear function fH: V → R such thatH is defined by the equationfH= 0. Exactly as in Definition 1.18, we then obtain a partition of V into “cells” A, each of which is defined by equalities or strict inequalities, one for each H ∈ H. We again encode the definition of a cellAby itssign sequence σ(A) =
σH(A))H∈H. Explicitly, we defineσ(A) to beσ(x) for anyx∈A, whereσH(x) is the sign offH(x). The set of cells is a poset under theface relation defined in terms of sign sequences as in Definition 1.20.
Continuing as in Section 1.4.2, we can also work with theclosed cells A.
HereAis the set obtained by replacing the strict inequalities in the definition ofAby weak inequalities; it is also the topological closure ofAinV. We have
A=
B≤A
B .
This implies the following characterization of the partial order on cells, which does not explicitly refer toH:
B≤A ⇐⇒ B⊆A . (2.17)
We define thesupportof a cellA, denoted by suppA, to be its linear span, and we define thedimension ofAby dimA:= dim(suppA). It is immediate from the definitions that ifB < Athen dimA <dimB, sinceA is contained in at least one hyperplane that does not containB. In particular, every nonempty collection of cells has a maximal element.
Assume throughout the rest of the section that we are given a nonempty set Σof cells, and set
X :=
A∈Σ
A . We can now state our axioms.
(H0) X linearly spansV.
This axiom is harmless; if it failed, we could simply replaceV by the spanV ofX, and we could replaceHbyH:={H∩V|H ∈ H, HX}.
(H1)Σ is closed under passage to faces; equivalently,X is a union of closed cells.
The next two axioms are more serious and can be viewed as finiteness prop- erties. For any A, B ∈Σ, let S(A, B) be the set of hyperplanesH ∈ Hthat strictly separate AandB, i.e., that satisfyσH(A) =−σH(B)= 0.
(H2) For anyA, B∈Σ, the setS(A, B)is finite.
(H3) For anyA∈Σ there is a finite subsetHA⊆ H that definesA.
Here, as in Section 1.4.3, the statement means thatAis defined by the condi- tionsfH=σH(A) forH ∈ HA. It follows that each cellA∈Σis a polyhedral cone of the sort studied in Section 1.4.
One consequence of (H3) is thatAis open in its support. More precisely, we have
suppA ⊆
H∈H H⊇A
H ⊆
H∈HA
H⊇A
H , (2.18)
and A is open in the last of these spaces. Hence the latter is spanned byA, and the three spaces are equal.
Before proceeding further, we need to resolve a potential ambiguity. Given A ∈Σ, we can talk about the faces of A as defined at the beginning of this section; let’s call these theH-facesofA. But if (H3) holds, then it would seem more natural to consider theHA-faces ofA, which can in fact be intrinsically defined, without reference to the set HA (see Proposition 1.27). It is this second notion of “face” that we used in the case of the Tits cone. Fortunately, there is no conflict:
Lemma 2.93.Suppose that axiom (H3) holds. Given A ∈ Σ and HA as in (H3), the HA-faces of Aare the same as the H-faces of A.
Proof. SinceHA⊆ H, the partition ofAintoH-cells refines the partition into HA-cells. It therefore suffices to show that everyHA-face ofAis contained in anH-cell. LetB be anHA-face ofA, and consider anyH ∈ H. Suppose, for instance, thatfH ≥0 on A. ThenfH ≥0 on B; hence, sinceB is open in its support, either fH >0 on B orfH ≡0 onB. Thus everyfH has a constant
sign onB, soB is contained in anH-cell.
We turn now to the most interesting axioms, involving products and con- vexity. Given two sign sequences σ= (σH)H∈H andτ = (τH)H∈H, we define theirproduct στ to be the sign sequence given by
(στ)H =
σH ifσH= 0, τH ifσH= 0, forH ∈ H. Consider now the following two conditions:
(H4)For any two cells A, B∈Σ, there is a cellAB∈Σ such thatσ(AB) = σ(A)σ(B).
(H5) X is a convex subset ofV.
These two conditions are in fact equivalent:
Proposition 2.94.In the presence of (H1)and (H2), axioms (H4)and(H5) are equivalent to one another. When these axioms are satisfied, the product of cells can be characterized as follows: Given x∈A andy ∈B, the cell AB contains(1−t)x+ty∈ABfor all sufficiently small t >0.
110 2 Coxeter Groups
Proof. Suppose (H4) holds. Given x, y∈X, let A (resp.B) be the cell con- taining x (resp. y). We show by induction on |S(A, B)| that the open seg- ment (x, y) is contained in X. If S(A, B) = ∅, let z be the first point (i.e., the point closest tox) where (x, y) crosses a hyperplane in H. Otherwise, set z = y. Let F be the cell containing z. Then every point in (x, z) has sign sequence equal toσ(A)σ(B), so (x, z)⊆AB⊆X. We also haveF ≤AB, so F is inΣandzis inX. Ifz=y, we are done. Otherwise,S(F, B)S(A, B);
hence (x, y) = (x, z)∪ {z} ∪(z, y) ⊆ X by the induction hypothesis. This proves (H5).
Conversely, suppose that (H5) holds. GivenA, B ∈Σ, choose x∈Aand y ∈B, and considerzt:= (1−t)x+ty for 0< t≤1. By hypothesis, zt∈X for all t. If t is small enough, σH(zt) = σH(A) = 0 for all H ∈ S(A, B);
hence σH(zt) =
σ(A)σ(B)
H for such H. And if H ∈ HS(A, B), then σH(zt) =
σ(A)σ(B)
H forall t∈(0,1). There is therefore a cell inΣ with sign sequence σ(A)σ(B), and it contains zt for sufficiently smallt >0. Thus (H4) holds, as does the last assertion of the proposition.
Assume from now on thatX andΣsatisfy axioms (H0)–(H5). (In partic- ular,X could be the Tits cone associated to a Coxeter group.) It is then easy to extend toΣmost of the concepts and results of Section 1.4. We will briefly run through some of these.
(1) Any two cellsA, B∈Σ have a greatest lower boundA∩B, whose corre- sponding closed cell is the intersection A∩B.
Use a finite subset ofH that defines bothA andB, and then appeal to the corresponding fact about finite hyperplane arrangements.
(2) AchamberofΣis a cellC∈Σsuch thatσH(C)= 0 for allH ∈ H. These are precisely the maximal elements ofΣ.
Indeed, a chamber is trivially maximal. Conversely, suppose C ∈ Σ is max- imal and consider any H ∈ H. In view of (H0), there is a cell A ∈ Σ with σH(A)= 0. But thenC=CAby maximality; henceσH(C) =σH(CA)= 0, andC is a chamber.
(3) A panel is a cell P ∈Σ with exactly one 0 in its sign sequence. Equiva- lently, it is a cell in Σof dimension equal to dimV −1. Every panel is a face of at least one chamber and at most two.
(4) Two distinct chambers C, D ∈ Σ are adjacent if they have a common panel. One can now define galleries in the obvious way and prove that any two chambers C, D can be connected by a gallery; moreover, the minimal length of such a gallery is|S(C, D)|.
(5) More generally, we can consider galleries connecting two arbitrary cells A, B ∈ Σ as in Exercise 1.62. The solution to that exercise goes through
without change to show that the minimal length d(A, B) of such a gallery is |S(A, B)|. Moreover, the chambers that can start a minimal gallery from Ato B are precisely those havingABas a face. In particular, every minimal gallery from a cellAto a chamberC starts withAC.
We turn now to subcomplexes, which we didnothave occasion to consider in the setting of finite hyperplane arrangements. By a subcomplex of Σ we mean a nonempty subset Σ that is closed under passage to faces. LetΣ be a subcomplex and letX:=
A∈ΣA. Then we can studyΣby viewing it as a set ofH-cells in the linear spanV ofX, where
H:={H∩V|H ∈ H, HX} .
Viewed in this way,Σsatisfies all of the axioms of this section except possibly the (equivalent) axioms (H4) and (H5).
Definition 2.95.We say that a subcomplexΣ ofΣisconvex ifΣ satisfies (H4) and (H5), i.e., ifX :=
A∈ΣAis a convex subset ofV or, equivalently, ifΣ is a subsemigroup ofΣ.
Thus we can apply to convex subcomplexes all of the results that we have proven aboutΣ. To state one explicitly, assume for simplicity that the chambers ofΣare simplicial cones (as in the setting of Section 2.6). We make this assumption only so that we can apply the language of Section A.1.3. Then the results of this section show that Σ is a chamber complex in which any panel is a face of at most two chambers. Consequently:
Proposition 2.96.Suppose that the chambers of Σ are simplicial cones.
Then every convex subcomplexΣ ofΣ is a chamber complex in which every
panel is a face of at most two chambers.
(Of course we might have dimΣ <dimΣ, soΣ is not in general achamber subcomplex ofΣ.)
We close this section by giving some useful characterizations of con- vex subcomplexes. Note first that there is an obvious way of construct- ing convex subcomplexes of Σ using half-spaces. Namely, for any H ∈ H we have a convex subcomplex Σ+(H) (resp. Σ−(H)) consisting of the cells in Σ on which fH ≥ 0 (resp. fH ≤ 0). Further examples can be obtained from these by taking intersections. For example, the convex subcomplex Σ0(H) :={A∈Σ|σH(A) = 0}is the intersectionΣ+(H)∩Σ−(H). The fol- lowing proposition, which should be compared with Exercise 1.68, implies that every convex subcomplex can be obtained as an intersection of such “halves”
ofΣ.
LetDbe a nonempty set of chambers inΣ. We say thatDisconvex if for allC, D∈ D, every minimal gallery inΣ fromC toD is contained inD. Proposition 2.97.LetΣ be a subcomplex ofΣ. Then the following two con- ditions are equivalent:
112 2 Coxeter Groups
(i)Σ is a convex subcomplex of Σ.
(ii)Σ is an intersection of subcomplexes of the form Σ±(H)(H ∈ H).
If Σ contains at least one chamber, then (i) and (ii) are equivalent to each of the following conditions:
(iii)The maximal elements ofΣ are chambers ofΣ, and the set of chambers inΣ is convex.
(iv)Given A, C ∈Σ with C a chamber, Σ contains every minimal gallery inΣ from AtoC.
Proof. The implication (ii) =⇒ (i) is trivial, since an intersection of convex subcomplexes is a convex subcomplex. [Note that it is automatically nonempty because it contains the smallest cell, which is
H∈HH.] To prove the converse, assume first thatΣ contains a chamber, in which case we will prove (i) =⇒ (iv) =⇒ (iii) =⇒ (ii).
(i) =⇒ (iv): Consider a minimal gallery A ≤ C0, C1, . . . , Cl = C in Σ.
SetA0:=A andAi :=Ci−1∩Ci fori= 1, . . . , l. ThenCi, Ci+1, . . . , Cl is a minimal gallery from Ai to C for 0≤i≤l; henceCi =AiC. So if (i) holds andA, C∈Σ, it follows inductively thatCi∈Σ for alli.
(iv) =⇒ (iii): This is trivial.
(iii) =⇒ (ii): LetΣbe the intersection of the subcomplexesΣ±(H) that containΣ. If (iii) holds, we claim thatΣ andΣ have the same chambers.
This implies that they are equal, sinceΣ is a subsemigroup ofΣ containing a chamber, and hence every maximal cell of Σ is a chamber. To prove the claim, letC be the setC(Σ) of chambers of Σ, and setD:=C(Σ). We must show that for any C∈ CDthere is a hyperplaneH ∈ H that separatesC from D.
Choose D ∈ D at minimal distance from C, and let D, D, . . . , C be a minimal gallery fromDtoC. ThenD ∈ D, and the hyperplane/ H separating D fromD also separatesD fromC. We will show that all chambers inDare on theD-side ofH. Given E∈ D, we haved(E, D) =d(E, D)±1. The sign cannot be +, because then there would be a minimal gallery from E to D passing throughD, contradicting the convexity ofD. So the sign is−, which means that E and D are on the same side of H. This completes the proof that (iii) =⇒ (ii) and hence that all four conditions are equivalent when Σ contains a chamber.
Suppose now that Σ does not necessarily contain a chamber. To prove (i) =⇒ (ii), note that (i) implies that all maximal cells ofΣ have the same support U. [If A and B are maximal, thenA =AB and B =BA; now use the fact that AB andBA have the same zeros in their sign sequences.] Let HU := {H∩U |H ∈ H, HU}, and let ΣU be the set of elements of Σ contained inU. ThenΣU is a set ofHU-cells inU satisfying all of our axioms, andΣ is a convex subcomplex ofΣU whose maximal simplices are chambers of ΣU. Moreover, the “halves” of ΣU are the subcomplexes ΣU ∩Σ±(H) for H ∈ H, H U. By the case already treated, it follows that Σ is an intersection of such subcomplexes. This implies (ii), since
ΣU =
H⊇U
Σ0(H) =
H⊇U
Σ+(H)∩Σ−(H).
Exercise 2.98.Show that in the presence of the other axioms, (H3) can be replaced by the following apparently weaker condition:
(H3)Any chamberC∈Σ can be defined by finitely many linear inequalities of the form f >0.