We demonstrate in this appendix the properties of the bilinear form associated with the geometric realisation of a Coxeter group . Recall that the matrix
These cases are mutually exclusive and exhaust all possibilities.
Proof: The proof follows from a series of lemmata. The inequalities define a convex cone, namely the first quadrant . Similarly, the inequalities define also a convex cone . One has indeed:
Note that one has also
and . There are three distinct cases for the intersection :
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 and imply either or . In other words
Proof: Assume that fulfills and has at least one component equal to zero. We shall show that all its components are then zero. Assume for and for . One has (with no non-vanishing component if ). From one gets . Take . As in the previous sum, one has and thus , which implies . As (), this leads to for and . The matrix would be decomposable, unless , i.e. when all components vanish.
Lemma 2: Consider the system of linear homogeneous inequalities
on the vector . This system possesses a solution if and only if there is no set of numbers that are not all zero such that .
Proof: This is a classical result in the theory of linear inequalities (see , page 47).
We can now study more thoroughly the three cases listed above.
In that case, one has
by Lemma 1. Furthermore, cannot contain a nontrivial subspace since implies , but only one of the two can be in when . Hence , i.e., and
This excludes in particular the existence of a vector such that or .
Finally, the interior of is non-empty since is nondegenerate. Taking a non-zero vector such that , one concludes that there exists a vector such that . This shows that Case 1 corresponds to the first case in the theorem. We shall verify below that is indeed positive definite.
reduces in that case to a straight line. Indeed, let be an element of and let be in but not in . Let be the straight line joining and . Consider the line segment from to . This line segment is contained in and crosses the boundary of at some point . But by Lemma 1, this point must be the origin. Thus, , for some real number . This implies that the entire line is in since for all , and also for all since .
Let be any other point in . If , the segment joining to intersects and this can only be at the origin by Lemma 1. Hence . If , the segment joining to intersects and this can only be at the origin by Lemma 1. Hence, we find again that . This shows that reduces to the straight line .
Since , one has . Hence, , which excludes the existence of a vector such that (one would have and hence ). Furthermore, there exists such that . This shows that Case 2 corresponds to the second case in the theorem. We shall verify below that is indeed positive semi-definite. Note that and that the corank of is one.
In that case, there is a vector such that , which corresponds to the third case in the theorem. Indeed, consider the system of homogeneous linear inequalities
By Lemma 2, this system possesses a solution if and only if there is no non-trivial such that .
Consider thus the equations for , or, as is symmetric, . Since , these conditions are equivalent to (if , one defines through ), i.e., . But , i.e., , which implies and hence also . The all vanish and the general solution to the equations is accordingly trivial.
To conclude the proof of the main theorem, we prove the following proposition:
Proposition: The Coxeter group belongs to Case 1 if and only if is positive definite; it belongs to Case 2 if and only if is positive semi-definite with .
Proof: If is positive semi-definite, then it belongs to Case 1 or Case 2 since otherwise there would be a vector such that and thus , leading to a contradiction. In the finite case, is positive definite and hence, : This corresponds to Case 1. In the affine case, there are zero eigenvectors and : This corresponds to Case 2.
Conversely, assume that the Coxeter group belongs to Case 1 or Case 2. Then there exists a vector such that . This yields for and therefore belongs to Case 1 . In particular, , which shows that the eigenvalues of are all non-negative: 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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