Sporadic and related groups. Lecture 15 Matrix representations of sporadic groups.

Transcription

1 Sporadic and related groups. Lecture 15 Matrix representations of sporadic groups.

2 Section 1 fields of characteristic zero Several of the sporadic groups were contstructed as matrix groups in characteristic zero. The Conway groups acting on the Leech Lattice is the prime example, but I've spent considerable time on that already. Rudvalis, J 3 and the Monster were also initially constructed this way.

3 Rudvalis 2.Ruv has a 28-dimensional representation over the Gaussians consisting of A + B.i where A and B are rationals and i 2 = -1. Vectors now come in fours {V, -V, iv, -iv} The starting group was 2 6 U 3 (3).2 which acts monomially.

4 Gaussians are quite nice Being a non-real field quadratic over the rationals, any representation fixes, in twice the dimension, a positive definite integral quadratic form, so discrete implies finite. Also Z(i) is a P.I.D. So the Rudvalis group is the stabilizer of a (Gaussian) lattice which is not much different from an oridinary (integral) lattice.

5 Geometry arises. There are 4 x 4060 vectors of smallest norm in this lattice, corresponding to the rank 3 graph on 4060 points that Rudvalis originally noticed. These can be constructed using 2 6 U 3 (3) and hence the graph and the lattice constructed together.

6 3.J 3 can be 18 x 18 matrices. These matrices need sqrt(5) and sqrt(-3) so only exist in a quartic extension of the rationals. Although an early construction used this, it was quite messy. I will not say any more about it.

7 The Monster Originally constructed in dimensions, in this case it is the size of the representation that gave trouble. The construction is not completely hopeless though, since there is a fixed algebra product as well as the quadratic form. I believe it has all been done by hand

8 Section 2 - modular Matrices in characteristic zero have many useful properties, But computationally they tend to be slow to work with. (Though actually I'm not sure there is any fundamental reason why this is so). Anyway, for practical work is is usually faster to work over finite fields.

9 PID implies integers. In case you are not aware, if a finite group has a representation over a field (in characteristic zero) whose algebraic integers form a PID, then the representation can be written over those integers. This is true theoretically, but also an effective algorithms are known.

10 This means you can reduce modulo a prime. The major work of looking for maximal subgroups of the sporadic groups (much of it the work of R. A. Wilson) is usually done with matrices modulo p, since they can be worked with much faster. P=2 is a common case (the fastest) but P=3 often makes it easier to see the 2-locals.

11 Rational integral modular So given the rational matrices (e.g. in the literature), if the integers form a PID it is algorithmic to convert them into integer matrices and then reduce them modulo a suitable prime. But many representations only exist over one finite field, so they must be constructed in that field.

12 First example is J 1 How would we make the 7 dimensional representation of this group modulo 11 today? (Well we wouldn't we'd just download it) The (modular) character values tell us that it is irreducible on L 2 (11) L 2 (11) is not so hard to make mod 11! And an A 5 in L 2 (11) commutes with an extra involution in J 1. Hence a plan!

13 Making J 1 as 7 x 7 matrices mod 11. So we start off by typing in a 2- dimensional representation of 2.L 2 (11), are two easy generators We could make the action on homogenous degree 6 polynomials x 6 x 5 y x 4 y 2 x 3 y 3 x 2 y 4 xy 5 y 6 And work out what the matrices do.

14 Next step Find an A 5 in L 2 (11). If you are using a computer, a few random tries will soon find it. By hand it is not too hard either. And then the A 5 will be found to be (characters again) the direct sum of a 4 and a 3 dimensional representation. The additional element we want is the involution that negates the 4-space and fixes the 3 space that A 5 acts on.

15 Diagram for J 1 construction. J 1 / \ 2 x A L (11) 5 2 \ / A 5

16 Diagram for J 4 construction. J 4 / \ x2S U (11) 4 3 \ /

17 What is the best way to make a group? There are two considerations. Firstly you need to be able to make the correct representation of at least one and ideally both the groups. Secondly you don't want there to be too many cases to try at the end. How many cases are there?

18 Number of cases Let L be the centralizer (in the full matrix group) of the Left-hand group And R be the centralizer of the Right And H the centralizer of the intersection. Then the cases to consider are the double coset representatives h i such that H = Σ L.h i.r

19 Why is this? You can consider one side (R, say) to be fixed, and you must fix the intersection. By there is no point conjugating R by something that commutes with R, since it doesn't change anything. And if you conjugate by something that commutes with L, you have just conjugated everything, which is also pointless.

20 This can give uniqueness. If, in a construction, the number of cases works out as 1, you know without doing it that there can be at most one thing that works. In some cases the ordinary character table provides a construction with a unique case, implying that there is only one group like that. Uniqueness of J 4 was done that way.

21 3.O'Nan in 45 dimensions mod 7 This was a fairly messy construction, but the matrices were fairly small so it was quite easy. This representation has a cubic form a trilinear map (x,y,z) field that is G- invariant. Ryba was able to show that the stabilizer of this form was 3.O'Nan.

22 Remark about O'Nan group There is only one class of involution in the O'Nan group, and its centralizer is 4.L 3 (4).2. Hence for every involution there is a unique cyclic-4 associated with it. Involutions have square-roots So whenever you have a (small) group you can wonder what the square-roots generate. Most unusual situation.

23 45 is a common dimension. 3.McL also has a 45 dimensional representation, this time modulo 5. It's character looks very similar to the O'Nan group character - except for the field. And M 24 has a 44+1 looking rather similar too.

24 As if L 3 (2) is a Weyl group The 3-dimensional representation of 2 x L 3 (2) over the complex numbers has 21 reflections giving 42 roots. So it looks like you might make several sporadic groups as if they were Chevalley groups with this as the Weyl group. But no-one (as far as I know) has made any meaningful comments along these lines.

25 And on to Lyons. From the 5-modular characters of McL and 2A 11, it looks as if the Lyons group might have a 111 x 111 representation modulo 5, as for 3.McL and for 2.A 11. And by now it is not hard to think of a plan. Make 2.A 11, find 2.A 8 in that, and go up to 3x2.A 8 to generate the group. Few dozen cases.

26 Wilson and Jansen In a short visit to Aachen, Wilson (who knows how to work with matrices) and Jansen (who knows how to work with character tables) found two or three new and unsuspected representations of sporadic groups this way. Try to fit small characters of a couple of subgroups together, and if it looks plausible, just make it and see if it works.

27 The computer did the work Indeed on two separate occasions they were already in the pub by the time the construction had been completed by the computer. But the matrices were still there in the morning.

28 Probably worth summarizing. Group Dimension Field J J 3 9 GF(4) Ly O'Nan 45 7 J

29 The Baby Monster This has a 4371 dimensional representation over the complex numbers Which, mod 2, is This was made (by Wilson) via the Fischer group F 23. Nobody knew what the module structure was for F 23, but Wilson boldly hoped that it was the direct sum of the irreducibles. He was lucky - it is!

30 So he made the 4370 This representation of BM is much easier to work with, and enabled investigations that would not have been possible using permutations on 13,571,955,000 points. What about the Monster?

31 Matrices of this scale are starting to be reasonable. This laptop could (with suitable software that does not quite exist) multiply dense matrices of this size in about an hour or two. The new St. Andrews machine could do so (with suitable software that is rather further from existence) in a minute or so. No-one has done this. Not yet anyway.

32 To find subgroups of M Wilson works mod 3 (so that H = Co 1 is available) and has an additional generator T normalizing the M 24 in that. And using subroutines can operate on any vector b any element of H and also by T. He has to organize his work so that this is all that he needs to do. Investigating a field while tied to a post. He seems to like that post!

33 A basic pair of vectors He has two particular vectors and has proved that the only element of the monster that stabilizes both is the identity. Hence if he has a word (in H and T) that looks like the identity he can test it. In fact, he almost never needs to prove it it is sufficient to be right most of the time. The recent proof that L 2 (41) is a subgroup of M is an example of this sort of work.

34 Look for L 2 (41) in M Your generators are H= Co 1 and T. Let z be the central involution of H Find a random element x (in H and T) such that <x,z> is Find out which 20 in H normalizes the element of order 41 (10 first). Look through all dihedral 20.2 (still in H) and see if it generates L 2 (41) with x.

35 Theme... Finding the elements that in fact have the properties you want is the hard bit. Checking that they have those properties once you have them is usually much easier.