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.
When it is no longer true in general that an infinite irreducible complex crystallograph is the semidirect product of and We shall explain here how one can find the corresponding extensions of by in this case.
We assume that
We recall that the structure of the extension is given by a of where with values in see Section 2.4.
Let and Taking as an origin, we can identify and then consists of the elements
First of all, one can assume, upon replacing by a suitable cocycle homologuous to that We assume that the order of is equal to see Section 3.2.
Let be the reflections with We assume, as usual, that the group generated by is irreducible. We have hence Then we have and we are done.
We shall assume now that
Therefore is defined by only one vector Moreover, one can take because there exists a reflection in
Therefore, given an invariant lattice of full rank, one has to solve the following problems
Find those for which
|a)||the cocycle of with values in given by condition satisfies for any relation of (i.e. for every element of see Section 2.4).|
|b)||Condition a) holds and the corresponding group is an|
|1)||If then a) b).|
|2)||If for every relation then b)|
1) Let We know that But is a root lattice for and the condition shows that the semidirect product of and lies in This semidirect product is an see Section 4.5. We also have that contains the reflections
Let be an arbitrary element, Then is a product of reflections in some order. Therefore, multiplying by and by in a suitable order, one can obtain an element of the form But the elements are also products of reflections. Therefore, multiplying by reflections, one can obtain But Therefore is a product of reflections.
2) We have Therefore in fact, defines a of with values in Let be the group defined by with and It is an because of 1). Also we have a group defined by with and
Let be a reflection. By section 3.2, we know, there exists with for certain and This is also a reflection, hence But We have By definition of we have Hence, also Therefore, or, in other words, This means that It follows now from that hence and
The following simple observation is very useful in practice because it gives strong restrictions on the choice of
Theorem. Let a) be satisfied and let be such that Then
It follows from the definition of that and We have now: and we are done.
The graph is The vectors are given in Table 2. Note that
There exists only one (up to similarity) lattice of full rank
The presentation (i.e. the sets of generators and relations) of is known, see [Cox1974].
The relations are: It follows from that
Therefore and hence and (mod But Therefore, we may assume that
It only remains to check whether or not for this particular This has to be done by straightforward computations: and so on (one only needs to consider the relations which involve This check shows that gives a cocycle of indeed, and, hence, defines an with It can be proved in a straightforward manner that this cocycle is not a coboundary, i.e. is not a semidirect product.
The same considerations can be given for each with and, as a result, one obtains Table 2. It appears a posteriori that in all cases either is a root lattice or for every relation of (therefore, for an is always a root lattice). By performing similar straightforward computations in the remaining cases, a proof of the theorems in Section 2.8. is obtained.
As for the proof of the theorem from Section 2.9, the first part of it is given in Section 4.6; the second part (about minimality of follows from [BSh1963]. The third part follows from Table 1.