Sporadic and related groups. Lecture 11 Matrices over finite fields J 4

Sporadic and related groups. Lecture 11 Matrices over finite fields J 4

Original aim of the meat-axe. Find the (degrees of the) 13 irreducible representations of M 24 mod 2. Gordon James found 12 of them theoretically in his PhD thesis (~1972) Next step - work directly with matrices. Tensor representations to make bigger ones Chop the result up into irreducibles. The 13 th degree is 1242 (~1977)

The algorithms are not new. Most of them are just linear algebra programs. Add. C = A + B Tensor. C = A [X] B Transpose. B = A' Invert. B = A -1 Multiply. C = A x B Nullspace. Find B such that B.A = 0

Two slightly more specialized programs Split given (two) generators and a vector, make a basis for the smallest subspace invariant under the generators and containing that vector. Then go on to compute the action of the generators on that subspace and on the quotient. Standard Basis. Spin up one vector in a canonical way.

These are all easy. When I went to Montreal in 1982, I took these 8 programs on punched cards. They only worked mod primes. I think they totalled about 200 lines of FORTRAN for all eight of them! These programs are not rocket science!

Better performance versions For the main research work, the 1981 FORTRAN version would have been too slow and used too much memory. For the previous five years we had worked in IBM 370 assembler, mainly mod 2. Now we work in c, with occasional pentium assembler. Larger fields are no problem either.

Multiply All the main algorithms work with rows, suitably packed in memory. Mod 2 we put each entry in a single bit. For other small primes there are various fascinating tricks, but as the prime rises it basically becomes more and more competitive to use the computer s Add instruction. Scalar multiplication is less important we can multiply each row by all scalars before we start if the field is small. This also generalizes to grease - computing certain rows in advance.

Grease When multiplying large matrices over small fields, A = B x C say, choose a dimension d (e.g. 3), and group the rows of C into sets of 3 rows. Make all the vectors in the space spanned by each group of rows. This reduces the number of scalar multiplications to zero, and the number of adds by a factor of d.

Grease codes There is actually a coding-theory problem here which seems untouched. Find a good set of vectors to make in a space such that every vector in the space is the sum of, say, three of them. The Golay co-code from M 24 is very good for this purpose.

Semi echelon form 1 3 2 5 5 2 3 2 3 The first entry in every 0 0 0 1 4 2 3 1 1 row is 1, and below it is 0 1 4 0 3 5 2 2 7 all zeros 0 0 0 0 0 0 1 0 4 0 0 1 0 5 5 0 6 2 The rows are obviously linearly independent. A vector that is in the space is readily expressed as a linear combination of these rows A new row is readily appended after cleaning out the Pivots

Row Nullspace We input the matrix A and set B = identity. Do identical row operations to A and B to put A into semi-echelon form. For every zero row of A that is made, output the corresponding row of B.

Grease can be used in Gaussian Elimination also It does make it quite complicated and hard to implement, so I will not describe it in detail here Just to say that you have chunks of rows that you put into full echelon form (zeros above the pivots also). This is work-in-progress.

Tensor Product 1 2 5 6 5 6 10 12 3 4 [X] 7 8 = 7 8 14 16 15 18 20 24 21 24 28 32 ( A [X] B ) x (C [X] D) = (A x C) [X] (B x D) Given two representations of G, tensoring them gives another representation.

Split (1 of 2) First get a semi-echelon form basis X for the invariant subspace by starting with the X as the single input vector, repeatedly multiply rows of X by the generators, reducing the result using X and appending if the end product is non-zero. This is usually called spinning up the vector.

Split (2 of 2) Then multiply the rows of X again by the generators, and express the result in terms of the basis for X, thereby computing the rows of the action of the generators on the subspace. Then compute the action on the quotient by completing the echelon form to a nonsingular matrix (adding rows all zero except for a 1 in a non-pivot column).

Null-vectors are good candidates for Split. Given two generators A and B for a (reducible) representation. Compute the nullspace of (A+B+AB). Take a vector of this (if there is one) and use split to see if it lies in an invariant subspace. It is quite likely that it does! If that doesn t work, try a null vector of A+B+AB+ABA, or BA+ABA+BBABA or...

Why is a null-vector good If there is an invariant subspace, that means that there is a basis so that the matrices look like. X * But a random element of the group 0 Y algebra (such as A+B+AB) gives X a random matrix. Over a small field, a random matrix often has a single null-vector. If so, from all the (not very many) null-vectors of the whole matrix, one of them lies in the invariant susbpace. Hence, over a small field, it will soon work.

Mod 2 It is possible to compute the nullity of a large random matrix mod 2. For large dimension the result is Non-singular 26% Nullity 1 52% Nullity 2 or more 22% Hence mod 2 it is very easy indeed to find elements of the group algebra with null-vectors in an invariant subspace if there is one.

Norton s Irreducibility Test Suppose you have a representation and find an element of the group algebra of nullity 1. Suppose the null vector is not in an invariant subspace. Suppose further that the null-vector of the transpose of the group algebra element is not in an invariant subspace of the transposed generators Then the representation is irreducible.

Equivalent representations Two representation ρ(g) and σ(g) of G are equivalent iff there is a matrix X such that X -1.ρ(g).X = σ(g) for all g in G. Can we find X? Yes we can. Find an element of the group algebra of nullity 1 and let v ρ and v σ be its null vectors in the two representations.

Standard basis. The single null vector v (e.g. of A+B+AB) is a standard vector. We can then form a matrix whose rows are images of v in some canonical way. We use v, v.a, v.b, v.aa, v.ab etc. in that order provded they are linearly independent (if they are not, we discard them). The result conjugates the representation into a standard basis.

Finding X We find Y and Z that conjuagate ρ(g) and σ(g) into standard basis. And then set X = Y.Z -1 If anything works, that will.

Building groups Given a subgroup H in a representation thought to be a restriction from G. Look at a subgroup K of H whose normalizer in G is bigger. Make (using standard basis) all the matrices that normalize K in the required way and look to see if you have generated G by H and this new matrix. Keep one that works and chuck the rest.

Bilinear forms Problem - Given an irreducible representation that is self-dual, find the bilinear form. Solution put the representation into a standard basis, transpose and invert the generators and find the matrix that conjugates that back (using standard basis) The resulting matrix is the bilinear form.

Original Objective number 2 In those days (1977) J 4 had been conjectured, but not proven, to exist. It might have a representation of degree 112 mod 2. We therefore set about making matrices. Start with 111 representation of U 3 (11) mod 2 (made by pratting about with 3x3 matrices over GF(121), acting on 1332 projective vectors, and chopping up the 1332 x 1332 matrices mod 2).

from 111 to 112 for U3(11) We actually made 1.110 for U3(11). Then transpose inverse gives us 110.1 Standard base puts the 110 in the same basis. Then had to guess the 1x1 matrix - only two possibilities for each generator, so we tried all four cases.

From U3(11) to J 4 Find a subgroup 11 1+2.40 in U 3 (11) Extend to a subgroup 11 1+2.5x2S 4 This is done by using standard base to conjugate generators of 11 1+2 into other generators of the same 11 1+2. There were quite a few possible cases to try. 48, if I remember correctly.

First time none of them worked This was because we had tried to make the D 12 subgroup of L 2 (11) instead of the A 4 which we needed - all in the automorphism group 11 2.(5x2.L 2 (11)).2 of 11 1+2. I'm sure you can see that this incredibly error-prone!

Second time it worked And we had generators in 112 dimensions for a group that looked a lot like J 4. Half way there! Now we just needed to prove it.

1978 Problem. We have 112 x 112 matrices mod 2 that in fact generate J 4. We wish to prove it... to recognize the group generated as J 4. And hence prove that J 4 exists. How can we do anything?

What can we do? We can multiply two matrices. We can find the order of a matrix As we later realized, we can find the characteristic polynomial of a matrix, which helps with recognizing conjugacy classes. Not much!

Involutions An involution is just an element of order two. They are important because two involutions generate a dihedral group. If t 1 and t 2 are both involutions, and the order of t 1 t 2 is n, then <t 1,t 2 > is order 2.n - the symmetry group of the n-ogon.

First find an involution t. This is easy in J 4. Half the elements have even order. So suppose we find an element X of order 22, say... Then its 11 th power is an involution. t = X 11 (There are groups, like L 2 (2 n ) where almost all elements have odd order)

Find the centralizer of t Conjugate t to make other involutions t' = (y -1.t.y) What is the order of t.t'? If it is even 2.m, then (t.t') m is an involution in the centre of <t,t'> so in particular commutes with t. We can collect elements commuting with t

Bray Trick Much later, John Bray noticed that even if t'.t=y -1.t.y.t has odd order 2m+1, we still have something that commutes with t... namely (y -1.t.y.t) m.y -1 since (y -1.t.y.t) m.y -1.t. y.t.(y -1.t.y.t) m = 1 (y -1.t.y.t) m.y -1. t = (t.y -1.t.y) m.t.y -1 = t.(y -1.t.y.t) m.y -1

Bray's trick is better When you find the product of two conjugate involutions being of odd order, the resulting centralizing element is evenly distributed - it is essentially a random element of the centralizer. This is better than the even case, where it can sometimes be hard to finish off the centralizer.

Anyway - to summarize You can find the centralizer of an involution. But you haven't proved that it is the whole centralizer, so we can't use this to prove that J 4 exists. Now what?

Need to prove G is not O 112 (2) We can find the matrix that conjugates our group to the transpose-inverse. And hence find the fixed quadratic form. Hence we know that G - the group generated by the matrices, is contained in O 112 (2). We need to show that G is smaller than this.

This part was done 4 times One method was to find an invariant subspace in the exterior square (action on quadratic polynomials modulo symmetric ones). 112 2- = 6216 = 4995 + 1221

Method 1 We then define a relationship on pairs of vectors in 112 if their wedge is in the 4995 subspace. And define J 4 to be the stabilizer of the relationship. We already have a lot of matrices And showed that this is all of them.

Other 4995 method S. P. Norton managed to show more theoretically, using the 4995 result, that the group generated was J 4. This was the first published proof.

Other methods Conway described an alphabettheoretical way of showing that a certain set of matrices was closed under multiplication by generators. He used double coset decomposition, and then aimed to find the necessary proofs that various words of one form also had an expression in another.

Yet other methods Conway's idea was modified in David Benson's PhD thesis to a method of showing that the generators preserved a certain set of vectors in the 112 space. Other proofs of existence have since been found.

Recognition Over the last dozen years or so, there has been a growing project to recognize an arbitrary matrix group over a finite field. Given a bunch of matrices, name the group that they generate. Uses classification!

This is a large project. Aim... to put into GAP and Magma enough code and data, so that if a user enters a bunch of matrices over a finite field, the computer can answer questions about it... composition factors, Sylow subgroups, conjugacy classes etc. etc. etc. I can only hope to give you the flavour.

Aschbacher's theorem Given a bunch of generators over a finite field, the group must fall into at least one of about nine classes Reducible Reducible if you extend the field Field is too big Tensor product Monomial... etc. etc. etc.

Last case An almost simple group in a representation that is irreducible, in the correct field, not a tensor product etc. etc. etc. These are the groups you need to be able to recognize individually. The ATLAS project continues. We need to collect more and more data about these groups

Initial Analysis Any finite group has the following structure A maximal normal solvable subgroup. On top of which comes the direct product of some non-abelian simple groups On top of which come some automorphisms of some of these groups And then a permutation group.

Simple group recognition. The heart of this procedure is to recognize the simple groups in the second layer. We don't just want their names though... we want to find the standard generators. Then we can use the encyclopedia to find things we want.

Encyclopaedia? QMUL Atlas website... Getting into GAP and Magma Lots and lots of information Permutations, matrices, recognizers, class-list finders, etc. etc. etc.

Example - M 22 Standard generators of M 22 are a, b where a has order 2, b is in class 4A, ab has order 11 and ababb has order 11. 1. Find any element of order 8. Its square is a 4Aelement, y say, and its fourth power is a 2A-element, x say [Probability of success on each attempt is 1/8] 2. Find a conjugate a of x and a conjugate b of y such that ab has order 11 and ababb has order 11. [Probability of success on each attempt is 64/1155 (approx. 1/18)]

Example - M 22 M22 = a, b a 2 = b 4 = (ab) 11 = (ab 2 ) 5 = [a,bab] 3 = (ababab 1 ) 5 = 1 (on the standard generators)

Straight Line program for A 7 mu 1 2 3 mu 3 2 4 mu 3 4 5 mu 3 3 6 pwr 7 5 7 mu 6 7 8 iv 8 9 mu 9 4 10 mu 10 8 2

Hence, if you give me M 22 In any way you like, I'll do the black box to find standard generators, then test the presentation. Now I just have to find the original generators in my nice M 22. This part of the problem can also be dealt with by knowing the nice M 22 very well.

Presentation for J 4 We can now recognize J 4 by standard generators and a presentation too. J 4 = x, y, t x 2 = y 3 = (xy) 23 = [x, y] 12 = [x, yxy] 5 = (xyxyxy 1 ) 3 (xyxy 1 xy 1 ) 3 = (xy(xyxy 1 ) 3 ) 4 = t 2 = [t, x] = [t, yxy(xy 1 ) 2 (xy) 3 ] = (yt yxy 1xyxy 1x ) 3 = ((yxyxyxy) 3 tt (xy)3y(xy)6y ) 2 = 1