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A Proof of Some Important Properties of the Bilinear Form

We demonstrate in this appendix the properties of the bilinear form B associated with the geometric realisation of a Coxeter group ℭ. Recall that the matrix

( ( ) ) -π-- (Bij) = − cos m ij
has only 1’s on the diagonal and non-positive numbers off the diagonal. Recall also that a vector v is said to be positive if and only if all its components vi are strictly positive, vi > 0; this is denoted v > 0. Similarly, a vector v is non-negative, v ≥ 0, if and only if all its components v i are non-negative, vi ≥ 0. Finally, a vector is non-zero if and only if at least one of its component is non-zero, which is denoted v ⁄= 0. Our analysis is based on reference [116Jump To The Next Citation Point]. We shall assume throughout that B is indecomposable.

Main theorem:

  1. The Coxeter group ℭ is of finite type if and only if there exists a positive vector vi > 0 such that ∑ j Bijvj > 0.
  2. The Coxeter group ℭ is of affine type if and only if there exists a positive vector vi > 0 such that ∑ j Bijvj = 0.
  3. The Coxeter group ℭ is of indefinite type if and only if there exists a positive vector vi > 0 such that ∑ j Bijvj < 0.

These cases are mutually exclusive and exhaust all possibilities.

Proof: The proof follows from a series of lemmata. The inequalities v ≥ 0 define a convex cone, namely the first quadrant Q. Similarly, the inequalities Bv ≥ 0 define also a convex cone KB. One has indeed:

u,v ∈ KB ⇒ λu + (1 − λ )v ∈ KB ∀λ ∈ [0,1].

Note that one has also

v ∈ KB ⇒ λv ∈ KB ∀ λ ≥ 0

and kerB = {v|Bv = 0} ⊂ K B. There are three distinct cases for the intersection K ∩ Q B:

  1. Case 1: K ∩ Q ⁄= {0 }, K ⊂ Q B B.
  2. Case 2: KB ∩ Q ⁄= {0 }, KB ⁄⊂ Q.
  3. Case 3: KB ∩ Q = {0 }.

These three distinct cases correspond, as we shall now show, to the three distinct cases of the theorem. To investigate these distinct cases, we need the following lemmata:

Lemma 1: The conditions Bv ≥ 0 and v ≥ 0 imply either v > 0 or v = 0. In other words

KB ∩ Q ≡ {v|Bv ≥ 0} ∩ {v|v ≥ 0} ⊂ {v|v > 0} ∪ {v = 0}.

Proof: Assume that v ≥ 0 fulfills Bv ≥ 0 and has at least one component equal to zero. We shall show that all its components are then zero. Assume v = 0 i for i = 1,⋅⋅⋅ ,s and v > 0 i for i > s. One has 1 ≤ s ≤ n (with no non-vanishing component vi if s = n). From Bv ≥ 0 one gets ∑n ∑n (Bv )i = j=1Bijvj = j=s+1Bijvj ≥ 0. Take i ≤ s. As j > s in the previous sum, one has Bij < 0 and thus ∑ nj=s+1Bijvj ≤ 0, which implies ∑ nj=s+1 Bijvj = 0. As vj > 0 (j > s), this leads to Bij = 0 for i ≤ s and j > s. The matrix B would be decomposable, unless s = n, i.e. when all components vi vanish.

Lemma 2: Consider the system of linear homogeneous inequalities

∑ λ ≡ a v > 0 α αi i i

on the vector v. This system possesses a solution if and only if there is no set of numbers μα ≥ 0 that are not all zero such that ∑ α μαaαi = 0.

Proof: This is a classical result in the theory of linear inequalities (see [116], page 47).

We can now study more thoroughly the three cases listed above.

  Case 1: KB ∩ Q ⁄= {0 }, KB ⊂ Q
  Case 2: K ∩ Q ⁄= {0 } B, K ⁄⊂ Q B
  Case 3: KB ∩ Q = {0 }

Case 1: KB ∩ Q ⁄= {0 }, KB ⊂ Q

In that case, one has

Bv ≥ 0 ⇒ v > 0 or v = 0

by Lemma 1. Furthermore, K B cannot contain a nontrivial subspace W since w ∈ W implies − w ∈ W, but only one of the two can be in Q when w ⁄= 0. Hence ker B = 0, i.e., detB ⁄= 0 and

Bv = 0 ⇒ v = 0.

This excludes in particular the existence of a vector u > 0 such that Bu < 0 or Bu = 0.

Finally, the interior of KB is non-empty since B is nondegenerate. Taking a non-zero vector v such that Bv > 0, one concludes that there exists a vector v > 0 such that Bv > 0. This shows that Case 1 corresponds to the first case in the theorem. We shall verify below that Bij is indeed positive definite.

Case 2: KB ∩ Q ⁄= {0 }, KB ⁄⊂ Q

KB reduces in that case to a straight line. Indeed, let v ⁄= 0 be an element of KB ∩ Q and let w ⁄= 0 be in KB but not in Q. Let ℓ be the straight line joining w and v. Consider the line segment from w to v. This line segment is contained in KB and crosses the boundary ∂Q of Q at some point r. But by Lemma 1, this point r must be the origin. Thus, w = μv, for some real number μ < 0. This implies that the entire line ℓ is in K B since v ∈ K ⇒ λv ∈ K B B for all λ > 0, and also for all λ < 0 since w ∈ KB.

Let q be any other point in KB. If q ∕∈ Q, the segment joining q to v intersects ∂Q and this can only be at the origin by Lemma 1. Hence q ∈ ℓ. If q ∈ Q, the segment joining q to w intersects ∂Q and this can only be at the origin by Lemma 1. Hence, we find again that q ∈ ℓ. This shows that K B reduces to the straight line ℓ.

Since v ∈ KB ⇒ − v ∈ KB, one has Bv = 0 ∀v ∈ KB. Hence, Bv ≥ 0 ⇒ Bv = 0, which excludes the existence of a vector v > 0 such that Bv < 0 (one would have B (− v) > 0 and hence Bv = 0). Furthermore, there exists v > 0 such that Bv = 0. This shows that Case 2 corresponds to the second case in the theorem. We shall verify below that Bij is indeed positive semi-definite. Note that det B = 0 and that the corank of B is one.

Case 3: KB ∩ Q = {0 }

In that case, there is a vector v > 0 such that Bv < 0, which corresponds to the third case in the theorem. Indeed, consider the system of homogeneous linear inequalities

∑ − Bujvj > 0, vj > 0. j

By Lemma 2, this system possesses a solution if and only if there is no non-trivial μ α ≡ (μi, ¯μi) ≥ 0 such that ∑ iμi(− Bij) + ¯μj = 0.

Consider thus the equations ∑ iμi(− Bij) + ¯μj = 0 for μ α ≥ 0, or, as Bij is symmetric, ∑ B μ = ¯μ j ij j i. Since ¯μ ≥ 0 i, these conditions are equivalent to ∑ B μ ≥ 0 j ij j (if ∑ B μ ≥ 0 j ij j, one defines ¯μi through ∑ j Bijμj = ¯μi), i.e., μ ∈ KB. But μi ≥ 0, i.e., μ ∈ Q, which implies μi = 0 and hence also μ¯i = 0. The μα all vanish and the general solution μ ≥ 0 to the equations ∑ iμi(− Bij) + ¯μj = 0 is accordingly trivial.

To conclude the proof of the main theorem, we prove the following proposition:

Proposition: The Coxeter group ℭ belongs to Case 1 if and only if B is positive definite; it belongs to Case 2 if and only if B is positive semi-definite with detB = 0.

Proof: If B is positive semi-definite, then it belongs to Case 1 or Case 2 since otherwise there would be a vector w > 0 such that Bw < 0 and thus Bijwiwj < 0, leading to a contradiction. In the finite case, B is positive definite and hence, detB ⁄= 0: This corresponds to Case 1. In the affine case, there are zero eigenvectors and det B = 0: This corresponds to Case 2.

Conversely, assume that the Coxeter group ℭ belongs to Case 1 or Case 2. Then there exists a vector w such that Bw ≥ 0. This yields (B − λI)w > 0 for λ < 0 and therefore B − λI belongs to Case 1 ∀λ < 0. In particular, det(B − λI ) ⁄= 0 ∀λ < 0, which shows that the eigenvalues of B are all non-negative: B is positive semi-definite. We have seen furthermore that it has the eigenvalue zero only in Case 2.

This completes the proof of the main theorem.


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