Last update: 12 May 2014
This is an excerpt of the lecture notes Discrete complex reflection groups by V.L. Popov. Lectures delivered at the Mathematical Institute, Rijksuniversiteit Utrecht, October 1980.
Now we move on to proofs. The greater part of these proofs relies heavily on the fact that for an infinite the subgroup is sufficiently "massive".
In order to clarify the last statement we need the following principal fact:
Theorem. Let be an infinite Then
Assume that Then is a monomorphism.
It follows from the discreteness of that is a torus (here bar denotes the closure and denotes the connected component of unity). (See [Bou1972], Ch. III §4, ex 13a.)
Then and Let also and let be the subspace of determined by The subspaces and are Definitely, (indeed, if not, then and hence which is a contradiction with
The set is dense in Therefore is dense in Hence (because is dense in
We claim that there exists a point with for every
Consider an arbitrary point and an operator such that Take with We have where But the restriction of to is nondegenerate. Hence there exists a vector such that So, we have Put We have We shall prove that is a point as wanted.
Let us first check that if Write and We need to prove that But is commutative, so Hence We have now: So, i.e. But hence It follows from that This establishes the claim.
Now we can prove for arbitrary Indeed, write as before. We have: It follows from that is Therefore we have Moreover, if then and, by the claim, Therefore, and hence i.e. But the image of is hence Therefore i.e. and we are done.
Now we consider two subgroups of resp. is the subgroup generated by those reflections for which resp.
The subgroup is finite. Indeed, identifying and by means of we have for any reflection and hence for any So is a discrete subgroup of a compact group hence finite.
We claim now that is infinite. Before proving this, we shall show how to finish the proof of the theorem if the statement holds. Thus, assume is an infinite
Note that Using the above arguments and constructions we obtain a torus a subgroup and a decomposition Also, we have and But therefore whence So and It follows now that and for every
Let be a reflection. Then and in view of we have: (root line of Therefore and acts on trivially. But is generated by reflections; therefore acts trivially on which contradicts the existence of such that and
It remains to show that Let us assume that this is not the case. Then has a fixed point By construction, We shall show that and commute. This then yields that is a finite group (for and clearly generate which is a contradiction.
Let and be reflections. We have
But is a reflection with mirror
Hence is either in or in If this element is in then i.e., which is a contradiction. Therefore, and We also have and (for otherwise which is absurd).
Consequently, there is a unique line on and This line lies in and is orthogonal to Hence i.e.
It follows from this theorem that is "big enough":
Theorem. Let be an infinite irreducible Then is a lattice of rank if and of rank or if (here
Let Then is an invariant subspace of the This subspace is nontrivial according to the previous theorem. Therefore because of irreducibility (see the theorem in 1.4), and the assertion follows from the equality
Let As above, we have Hence Therefore But is a complex subspace of and hence either or i.e. If then hence i.e.
|2)||if then is a crystallographic group iff|
1) follows from the fact that is contained in a compact group and has an invariant lattice of a full rank.
Let be a finite and be the set of mirrors of all reflections from Let
Then it is easy to see that the subgroup of generated by all reflections with is a cyclic group. Let be the order of this group.
Theorem. Let be a generating system of reflections of such that the order of is equal to for every Then each reflection is conjugate to for certain and
Let be the of in (it follows from that acts on Let be the product of linear equations of the mirrors in normalized by the condition Then is a character of i.e. a homomorphism of into the multiplicative group of The group is generated by We have also Therefore there exists a with But iff is the orbit of see [Spr1977]. Therefore for a certain and the statement follows.
Theorem. Let be an Then for any reflection there exists a reflection with
be the set of all reflections of Then
is a generating system of reflections of Let
be the of in
Then there exists a number with
Let be such that We have:
Let be a subgroup of We have When is a semidirect product of and We have the following criterion:
Theorem. Let Then:
|a)||is a semidirect product iff there exists a point such that induces an isomorphism of the stabilizer of with|
|b)||For every finite group and every subgroup of there exists a unique group (up to equivalence) such that and is a semidirect product of and|
Proof is left to the reader.
The point a from part a) of this theorem is called a special point of see [Bou1968].
Using this theorem we can clarify the structure of infinite in a number of important cases:
Theorem. Let be a group generated by reflections. Assume that and that is an essential group (i.e. generated by reflections. Then is a semidirect product of and
Let be a generating system of reflections of Then there exists a reflection such that see Section 3.2. We have because is essential. Therefore more precisely, this intersection is a single point
We have now that is a surjective map, hence an a isomorphism. Thus we are done in view of the theorem above.
The conditions of this theorem are always fulfilled if and is an infinite irreducible In general this is not the case if though "in most cases", it is.
Let and let be a group as in the title. Set Then by 3.1.
is a of Let us consider the restriction of to We claim that this restriction has only real values. Indeed, defines an euclidean structure on such that is orthogonal with respect to this structure. But because of irreducibility. Hence, there exists a canonical extension of determined up to a hermitian scalar product, say on Thus we have two hermitian structures, and on They are proportional because of irreducibility of Taking restrictions to we have: In other words, is a real form of and the restriction of the action of to gives an irreducible real finite (and itself is the complexification of this group). It follows from the classification that this group is generated by reflections, see 1.5. Therefore is generated by reflections, too. It follows from the theorem above (section 3.3) that is a semidirect product. Let be a special point of Then is a real form of It is clear that is The restriction of to is a real form of Therefore, this restriction is an affine Weyl group and is its complexification. This completes the proof of the theorem in Section 2.2.