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.
As we have seen in Section 2.3, one of the ingredients of the description of infinite complex crystallographic is the lattice We shall show in this chapter how one can find all the invariant lattices of full rank for an arbitrary fixed finite
We use the following notation:
a finite essential
a lattice,
a fixed generating system of reflections of
the set of all root lines of
Definition. is called the root lattice associated with
If then is called a root lattice.
Theorem. is a lattice and
Proof. | |
It is clear that is so let us prove the assertion about ranks. We can assume that because is essential. Put If then therefore because Hence i.e. So, is non-singular. But and Therefore |
Corollary. If (hence then for every If then for every
Proof. | |
We may take in the previous proof. As this proof shows that the first assertion follows from the evident inequality The second assertion can be proved similarly by means of reduction to the real form. |
It appears that one can reconstruct from the
Theorem.
Proof. | |
Let and then there exists such that for certain (because every reflection in is conjugate to the a power of some see Section 3.2). Let It is easy to see that is invariant, hence and |
The problem of finding of all lattices can be solved in two steps: 1) description of all root lattices; 2) description of all K-invariant lattices with a fixed associated root lattice. We shall first show how to solve the second problem.
Definition. Clearly, is a subgroup of It is more convenient to use another description of
Let be a natural map and be the set of points fixed by Then Therefore, It is clear that is and
Theorem. is a lattice and
Proof. | |
Let us first show that is a lattice. If it is not a lattice, then there exists a vector such that for every (because a closed subgroup of Then for every and Therefore as is a lattice and, hence, because is an essential group. Therefore is a lattice. is a euclidean space with respect to Let be the orthogonal complement of in this space. Let and Then Therefore hence (again because is essential). Thus This completes the proof. |
Cohomological meaning of
We have an exact sequence of groups It gives the exact cohomological sequence But because is essential, and because is divisible. Therefore
Now we can explain how to find all lattices with a fixed root lattice.
Theorem. Let be a fixed root lattice in Then for every lattice in the following properties are equivalent:
a) | is a lattice and |
b) | and |
Proof. | |
a) b). Let Then Therefore and b) a). Let Then where is the natural map, and Hence is and therefore is also If then (because |
We have already seen that for irreducible one can take if and or if It appears that if then there exists a good constructive way to find by means of
Theorem. Let and be a lattice in Let also (Recall from Section 4.1 that is nonsingular). Then:
a) | |
b) | If then |
Proof. | |
a) Let then, by definition, for all hence and b) Let and Then But and because of linear independence of Therefore, by definition of we have |
This theorem is useful in practice because one can explicitly describe the operator its matrix with respect to the basis is
What to do if and
It is not difficult to see that one can take in such a way that generate a subgroup of which is also irreducible (the numbering in Table 1 has this property).
Example: Therefore the problem may be solved as follows: find all lattices and select those lattices among them which are invariant under
We shall show how this can be done in Section 4.8, after we have explained (in Section 4.7) how to describe invariant root lattices.
By now we want to discuss several general properties of invariant lattices and explain the significance of root lattices in the theory of infinite
Theorem. Let be a finite linear irreducible group and be a nonzero lattice in Then if and or if Moreover, if is an and then is the complexification of the Weyl group of a certain irreducible root system.
Proof. | |
as in Section 3.1 and 3.4. |
Corollary. If and has an invariant lattice of rank then has an invariant lattice of rank
Proof. | |
is the complexicification of a Weyl group, let be a lattice of rank in the corresponding real form of which is invariant under this Weyl group. Then, for every a lattice is and has rank |
It should be noted that if and is a Weyl group then a root lattice of rank is (up to similarity) a lattice of radical weights and is a lattice of weights, see [Bou1968]. The matrix of with respect to a basis of simple roots, is, in this case, the Cartan matrix.
Theorem. Let be irreducible, be a nonzero lattice and Then:
a) |
and
if
and if |
b) | depends only on but not on the choice of the generating system of reflections. |
c) |
is divisible by
if is divisible by if |
d) | if and if |
e) | If and are the invariant factors of a matrix of the endomorphism of defined by then, Specifically, do not depend on the choice of the generating system of reflections. |
f) | if and if |
Proof. | |
If then, by considering the corresponding real form, we reduce the problem to the case If then It is known that in this case for every one has and (here is considered as a linear operator of a real vector space Now, a) follows from the fact that a basis of is an of We have also: But whence if or if The assertions b), c), d), e), f) follow from these equations. |
Corollary.
a) | If then i.e. any invariant lattice is a root lattice. |
b) | Let and suppose is a prime number. Then or |
Remark. One can calculate directly from the graph of i.e. where runs through the permutations of degree and for a decomposition of as a product of cycles.
Examples.
a) type Write for the graph Simple cyclic products are and
We have Assume, by induction, that if Then,
b) type Let us consider the generating system of reflections given by the graph
The list of all nonzero simple cyclic products is
It is easy to see that
Remark. If and is a lattice of radical weights with Weyl group then are invariant factors of a Cartan matrix and is isomorphic to the centre of a simple simply connected Lie group corresponding to
We shall now show that if there exists a nonzero lattice then there also exists an infinite with and
Theorem. Let be a finite linear and be a nonzero lattice in Then the semidirect product of and is an iff
Proof. | |
Let be the semidirect product of and Let be a special point. We identify and Then the set of all reflections in is is a reflection of and Let be a subgroup of generated by this set. Then Also because Moreover, if and is the root line of a reflection then Therefore we have But every reflection in lies in (because it has the form where is a reflection and see Section 1.2). Therefore, is generated by reflections, and we have Hence, iff |
Corollary. The infinite for which is generated by reflections are exactly the groups where is a nonzero root lattice.
We shall prove now the theorem from Section 2.5 and part of the theorem from Section 2.9.
Proof. | |
a) b) is already proved in Section 4.3. b) a) is trivial. c) b) is also trivial: such a lattice is b) c). Let be an invariant lattice of rank Then the semidirect product of and is a crystallographic because of the previous theorem. b) d). Let us first prove that coincides with the ring with unity generated over by all cyclic products. We have Indeed, clearly hence The reverse inclusion follows from the fact that Tr is an additive function. But generate the ring (here is a generating system of reflections of Therefore, the monomials generate as a We have: and it easily follows from this equality that Therefore, and we are done. By now, let be a lattice of rank It follows from the equality that for every But see Section 4.1. Therefore, is an integral element of a certain purely imaginary quadratic extension of denote this extension by Let we shall show that for every being irreducible, there exists for certain Let Then and But hence Therefore the ring lies in the maximal order of d) e). This is proved in [Vin1971-2], Lemma 1.2. d) a). Let where is the maximal order of a certain purely imaginary quadratic extension of being irreducible and being integrally closed, one can again apply Lemma 1.2 from [Vin1971-2] and obtain that is defined over Therefore there exists a of such that is an isomorphism. But is a Dedekind ring and is a of rank without torsion. Therefore, is isomorphic to a direct sum of fractional ideals of the field see [CRe1961]. Let be these ideals. Then there exists a of such that But is a lattice of rank in and for every fractional ideal of there exists with Hence is also a lattice of rank in e) d). Let be defined over a purely imaginary quadratic extension of Then for every hence But, is an integral algebraic number. Therefore lies in a maximal order of d) f). Using Table 1, we can easily find those for which lies in the maximal order of a certain purely imaginary quadratic extension of It is convenient to use the following necessary condition on each generator of this ring is an integral element of a certain purely imaginary quadratic extention of iff and It appears aposteriori that this condition is also sufficient. After a certain amount of calculations we obtain the list of f). Example. Let type The graph of is We have The previous condition for the gives: Therefore and in these cases lies in a maximal order of a certain purely imaginary extension of |
We assume now that We shall first describe (up to similarity) all invariant root lattices of rank when (it follows from Section 4.5 that only these lattices are of interest to us).
Theorem. Let be an irreducible finite linear and let be a set of lattices of rank and In order for to be a root lattice with it is necessary, and, if also sufficient, that:
a) | |
b) | and for every such that |
c) | For every and such that and one has |
Proof. | |
Assertion a) follows from the formula and the fact that is generated over by cyclic products. The "necessary part" of b) follows from the invariance of the lattice. Let us prove the "sufficient part". If then is in fact a direct sum of hence is a lattice. This lattice is invariant under for every if then it follows from b); if then it follows from a). Hence, is Now let us prove that b) c). b) c). One obtains the proof by applying to both sides of c) b). One obtains the proof by applying to both sides of |
Corollary. Let be a nonzero lattice. If then and determine each other uniquely by the formulas
Proof. | |
We have by c): (we assume that The index of the left lattice in the right lattice is Therefore the inclusions are in fact equalities. Applying we get Again, the inclusions are in fact equalities. |
If we also need to know for a lattice, and to select those for which
a) | is |
b) |
Theorem. In particular, if there is a number such that
Proof. | |
We have Assume that i.e. If then, applying to the both sides of this relation, we obtain i.e. Vice versa, if for some then applying we obtain: i.e. In order to prove the second assertion, let us apply to We obtain and, if then Thus the inclusion is in fact an equality; hence |
Let us assume now that
An algorithm for constructing root lattices of full rank: case 1.
We shall only consider here groups from the theorem in Section 1.6 with the property: every two nodes of the graph of (see Table 1) can be connected by a path of edges such that the absolute value of the weight of each edge equals
These groups are: type type and
Algorithm:
Let be an arbitrary lattice in of rank which is invariant under
Let:
For let us consider an arbitrary path of edges from to for which the absolute value of the weight of each edge equals Let We claim that:
a) | is a lattice of full rank. |
b) | does not depend on the construction (i.e. on the choice of the paths). |
c) | each invariant lattice is obtained in this way. |
Proof. | |
The assertions b) and c) follow from the corollary above. Let us prove a). We check the conditions a) and b) of the theorem proved at the beginning of this section. The condition a) is clearly fulfilled, so we need only check b). We have and Let Then (thanks to the construction of But Hence so that Let us consider We have Therefore, But the restriction of to has a trivial kernel, because Therefore, Using the same arguments we obtain also |
Therefore we only need to find all such that is in the ring of integers of the field it generates. It is well known how to do it (see [BSh1963]). We obtain, after checking, that in fact if then (in the cases under consideration) is the maximal order in its fraction field. Hence, up to similarity, we have
Example.
type The graph is
The path we choose connecting and will be and the path, connecting and will be
Here and
Hence, Therefore Checking all cases as done in this example, one obtains exactly those lattices that are in the column with of Table 2.
An algorithm for constructing root lattices of full rank: case 2.
We shall consider now the remaining irreducible finite i.e. the groups
We see that the graph of in these cases is a chain. Taking a suitable numbering, we can assume that are the only nonzero (the numbering in Table 1 has this property).
Algorithm.
Let be a lattice in We have Let us take an arbitrary lattice between these two lattices (such a lattice exists, e.g. has this property): And so on: Let We claim that is an invariant lattice and that any invariant lattice is of this form.
Proof. | |
Let us check the conditions a) and c) of the theorem proved in the beginning of this section. We need in fact only check c), because a) is obvious. We have Therefore, |
It appears that, as in the case 1, if
Example.
type The graph is We have and Therefore, We have
But Hence, Thus, we have only two possibilities: either or
One can check that, using this algorithm, one obtains, the lattices that are in the column with of Table 2 as well as other lattices in the cases types and As a matter of fact, these last lattices are not included in Table 2 because they do not give new (i.e. the semidirect product of and the corresponding lattice is equivalent to one of the groups from Table 2). We omit the check of these last assertions.
Now we shall briefly consider the case
These are the groups: We shall explain the approach by several examples.
Examples.
a) type or The graphs of these groups are
We see that type cf. Table 1. Using the previous theory, one checks that, up to similarity, there exists only one root lattice of full rank, i.e. But for we have see the example b) in Section 4.4. It follows from the corollary in Section 4.4 that, up to similarity, is a unique lattice of full rank. But must have an invariant lattice because of the theorem in Section 2.5. Therefore this lattice has to be (of course, one can check that is a lattice also by means of a straightforward computation).
b) type The graph is We see that type One can check that there exists only one (up to similarity) root lattice of full rank:
So, to describe all lattices of full rank we need to find all such that:
a) | (with respect to |
b) | (with respect to |
c) |
The matrix of with respect to the basis has the form We have see Section 4.4. Therefore the coefficients of lie in and (cf. Section 4.4), the order of is equal to It follows from these facts that It is not difficult to see that where It is readily checked that are representatives of different nonzero elements of Therefore every lattice of full rank has to be similar to one of the lattices: Using the equalities one can straightforwardly verify that all the lattices are
The same considerations can be carried out for other groups from the list above, and the description of allˆ(up to similarity) lattices of full rank will be obtained. We leave this to the reader (the most complicated case is type