We demonstrate in this appendix the properties of the bilinear form associated with the geometric
realisation of a Coxeter group
. Recall that the matrix
Main theorem:
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 [116], 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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