We continue to assume that (W, S) is an arbitrary Coxeter system with S finite. In Section 2.5 we constructed a representation of W on the vector spaceV =RS. We introduced a setCofchambersinV∗, determined by a setH ofwalls. Here we carry the chamber geometry further and show that a great
deal of what we did in Chapter 1 for finite reflection groups extends to general Coxeter groups. The main results first appeared in an unpublished paper of Tits [240]. Published accounts later appeared in Bourbaki [44], Vinberg [270], and Humphreys [133]. We have marked this subsection as optional because, while it contains an extremely useful tool for the study of Coxeter groups, it is not really needed in the rest of this book.
2.6.1 Cell Decomposition
We assume here that the reader is familiar with the elementary geometry of polyhedral sets defined by finitely many linear equalities and inequalities, as developed in Section 1.4. In particular, we will make use of the fact that such a set has well-defined faces. These can be determined using a collection of defining equalities and inequalities as in Definition 1.20, and they can also be characterized intrinsically (Proposition 1.27).
We apply this first to the fundamental chamber C, which is a simplicial cone. It has one faceAfor each subsetJ ⊆S, defined by−, es= 0 fors∈J and−, es>0 for s∈SJ. We use these faces and theW-action to define the cells that will be of interest to us.
Definition 2.79.The transforms wA (w∈W, A ≤C) will be called cells.
TheTits cone X is defined to be the union of all the cells. Equivalently, X =
w∈W
wC .
Note that every cell is a polyhedral set of the sort discussed above, defined by finitely many linear equalities and inequalities.
Theorem 2.80.The cone X is convex. For any x, y ∈X, the line segment [x, y] crosses only finitely many walls and is contained in a finite union of cells. Moreover:
(1)C is a strict fundamental domain for the action ofW onX.
(2)The stabilizer of any x∈C is the standard subgroup of W generated by Sx:={s∈S|sx=x}.
(3)For each cellA and wallH, Ais contained either in H or in one of the open half-spaces determined by H.
Proof. To prove the first part of the theorem, we may assume x ∈ C and y ∈ wC for some w ∈ W. Then [x, y] crosses only finitely many walls by Lemma 2.63, since any wall that it crosses separates C from wC. We will prove by induction on l(w) that [x, y] is contained in a finite union of cells (and hence, in particular, it is contained inX). Letz be the point such that [x, z] =C∩[x, y]; see Figure 2.6. Then [x, z] is contained in the union of the faces of C, so it is enough to show that [z, y] is contained in a finite union of cells. We may assumey=z. For eachs∈S, we havez∈U+(s) orz∈Hs, and
104 2 Coxeter Groups
y∈wC C
x z
Fig. 2.6.Proof of convexity.
there must be at least oneswithz∈Hs andy∈U−(s); otherwise, we could move a positive distance fromztowardywithout leavingC. Lemma 2.58 now implies that w =sw with l(w) = l(w)−1, and then [z, y] = s[z, sy] with sy∈wC, so we are done by the induction hypothesis.
We now proceed with (1)–(3). For (1) and (2) we must show that ifwx=y withx, y∈C, thenx=yandw∈ Sx. We argue by induction onl(w), which may be assumed >0. Writew=sw withl(w) =l(w) + 1. Thenwx=sy.
The left side is in U+(s) by Lemma 2.58, and the right side is in U−(s). So wx=sy∈Hs, and hencewx=y∈Hs. We now have x=y andw ∈ Sx by the induction hypothesis, and finallyw=sw ∈ Sxbecauses∈Sy =Sx. (3) We may assumeA≤C andH =wHs. We know thatAis contained in at least one of the two closed half-spaces bounded by H, since C is in one of the open half-spaces. So we must show that if A meets H then A is contained inH. In other words, iftis the reflectionwsw−1whose fixed-point set is H, we must show that if tx= x for some x ∈ A then tx = x for all x∈A. This follows immediately from (2), since Sx corresponds to the walls ofC containingx, which are the same as the walls ofC containingA.
It follows from (3) that every cellAhas a well-defined sign sequenceσ(A), generalizing the sign sequences for chambers, where now the possibility σH(A) = 0 is allowed. Moreover, A is defined by the equalities and inequal- ities corresponding to the signs. [It suffices to check this when Ais a face of the fundamental chamber, in which case the result is trivial.] In particular, distinct cells are disjoint. It follows easily that the face relation, which makes sense a priori because each cell is defined by finitely many linear equalities and inequalities, has the usual interpretation:
B≤A ⇐⇒ B⊆A ⇐⇒ σ(B)≤σ(A),
where the ordering on sign sequences is the same as in Definition 1.20.
Remarks 2.81.(a) Since the cells are disjoint, the statement that every closed line segment in X is contained in only finitely many cells can be strengthened: Every closed line segment meets only finitely many cells.
(b) Even though H is infinite, it still determines a partition ofV∗ into sets determined by sign sequences, exactly as in Section 1.4. For lack of a better
name, we call these setsH-cells. What we have just shown, then, is that the cells of the Tits cone are in factH-cells. One must be careful in what follows to distinguishcells as in Definition 2.79 from the more generalH-cells.
(c) Although each cell of the Tits cone is determined by finitely many of the hyperplanes inH, this is not necessarily true of arbitraryH-cells. The reader might find it instructive to find all the H-cells if W is the infinite dihedral group (Figure 2.2).
We can now carry the theory further.
Proposition 2.82.
(1)Given cells AandB, there is a (unique) cell ABsuch that σH(AB) =
σH(A) ifσH(A)= 0,
σH(B) ifσH(A) = 0. (2.16) For anyx∈Aandy∈B, we have(1−)x+y ∈ABfor all sufficiently small >0. The product(A, B)→ABmakes the set of cells a semigroup.
(2)X is the entire spaceV∗ if and only ifW is finite.
Proof. (1) This is proved as in Section 1.4.6, the essential point being that a line segment in X crosses only finitely many walls.
(2) If W is finite then we saw in the previous section that W can be identified with a finite reflection group acting onV∗, soX=
wC =V∗ by Chapter 1. Conversely, suppose X =V∗. Then−C is contained inX. Since
−Cis obviously anH-cell, it follows that−Cis a cell ofX, hence a chamber
ofX, soW is finite by Proposition 2.69.
Remark 2.83.Using the fact that the chambers are simplicial cones, one can show as in Section 1.5.8 that the posetΣof cells is a simplicial complex, whose vertices are the cells that are rays. Moreover, one can easily check, as in Sec- tion 1.5.9, that this simplicial complex is isomorphic to the posetΣ(W, S) of standard cosets, ordered by reverse inclusion. This simplicial complex, called theCoxeter complex associated to (W, S), will be the main object of study in the next chapter.
2.6.2 The Finite Subgroups of W
As an application of the Tits cone, we will use it to analyze the finite subgroups of W. As above, let Σ denote the set of cells inX. Fix A∈ Σ, and let HA
be the set of walls containingA. ThenHA determines a partition ofV∗ into HA-cells. This is coarser than the partition intoH-cells, i.e., every HA-cell is a union ofH-cells. As in Exercise 1.45, we can prove the following:
Lemma 2.84.Let Σ≥A be the set of cells of X having A as a face, and let ΣA be the set of HA-cells that meet X. For any cell B ∈Σ≥A, let f(B) be the HA-cell containingB. Thenf:Σ≥A→ΣA is a bijection.
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Proof. On the level of sign sequences,f just picks out the components ofσ(B) corresponding to the hyperplanes inHA. It is 1–1 because the remaining com- ponents ofσ(B) are the same as those ofσ(A). To prove that f is surjective, start with anHA-cellB that meetsX, choose a cellB ofX contained inB, and form the productAB. Then a consideration of sign sequences shows that
f(AB) =B.
Lemma 2.85.Let WA be the stabilizer of A. Then WA is finite if and only if HA is finite.
Proof. If WA is finite, then it contains only finitely many reflections, so HA
is finite. Conversely, if HA is finite, then there are only finitely many HA- cells, and henceΣAis finite. Lemma 2.84 now implies thatΣ≥Acontains only finitely many chambers. So WA is finite, since it acts simply transitively on those chambers. [Given chambers D, E ≥A, we know that there is a unique w ∈ W such that wD =E. Then wA and A are W-equivalent faces ofE;
hencewA=Aby Theorem 2.80.]
Lemma 2.86.Let Xf be the set of points x ∈ X whose stabilizer Wx is finite. Givenx∈Xf andy∈X withx=y, the half-open line segment[x, y) is contained inXf. In particular,Xf is convex.
Proof. In view of Lemma 2.85, Xf consists of the points x∈X such thatx is contained in only finitely many walls. The result now follows from the fact
that [x, y] crosses only finitely many walls.
We can now prove the main result of this subsection.
Proposition 2.87.Every finite subgroup of W is contained in a finite para- bolic subgroup.
Proof. Note thatXf isW-invariant and that by Theorem 2.80, the stabilizers Wx forx∈Xf are precisely the finite parabolic subgroups ofW. So our task is to show that every finite subgroupW ofW fixes a point ofXf. The latter being convex by Lemma 2.86, we can prove this by averaging: Start with an arbitraryx∈Xf, and then
w∈Wwxis a point ofXf fixed byW. Remark 2.88.See Bourbaki [44, Section V.4, Exercise 2(d)] or Brink and Howlett [49, Proposition 1.3] for other proofs of the proposition.
Exercises
2.89.LetA be a cell and WA its stabilizer. If WA is finite, show that every HA-cell meetsX, so the mapf of Lemma 2.84 is a bijection fromΣ≥Ato the set of allHA-cells.
2.90. (a) Show that Xf is open inV∗and hence is the interior ofX inV∗. (b) Show that the action ofW onXf is proper. Since the stabilizersWxfor
x∈Xf are finite by definition, the content of this is that everyx∈Xf
has aWx-invariant neighborhoodU such thatwU ∩U =∅ifw /∈Wx.
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.