April 21: More on M_12, the (5,6,12) Steiner system, and the affine (2,3,9) and inversive (3,4,10) planes We should develop a few more facts about the (5,6,12) Steiner design and its automorphism group before proceeding to finding Aut(M_12) in M_24. For starters, we found Aut(S_6) by the action of S_6 on totals. Here's a somewhat analogous construction of the outer automorphism of M_12. There are 66 block pairs. Any two divide the 12 points into either four triplets or 4+2+2+4. Form a graph whose vertices are the block pairs, with two adjacent iff they give four triplets. I claim that this is the triangular graph T(12), and that the outer automorphism can be constructed from the action of M_12 on the 12 maximal cliques (sometimes called the "totals" of the design for the sake of analogy with the S_6 situation). This is easy from our description last week of any three or four rows of H_12: block pairs of rows biject with pairs of columns, and two column pairs intersect iff their block pairs yield four triplets. (We can also check using the intersection triangle that our graph has degree 2 * Bin(6,3) = 20.) This also gives a bijection from triples of columns to partitions of the 12 points into four triples any two of which constitute a block; on T(12) those are triangles that are not contained in a maximal clique. (Check: there are Bin(12,3)=220 triples, and also 20*66/Bin(4,2)=220 such sets of four triples.) See below for what remains of M_12 after we specify a triple (a.k.a. M_9). What do the Bin(12,4)=495 *quadruples* of columns correspond to? Answer: the 495 quadruples of rows. It's easier to do this in the other direction: given 4 rows, we can confirm using our analysis of the previous lecture that their product has either four +1's and eight -1's or vice versa; choose the four, and check (using "one match, three mismatches") that applying the same construction to those four columns retrieves the same four rows. MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM Just as with the (5,8,24) Steiner system, the first three derived designs of the (5,6,12) are unique, and the design can be reconstructed from three kinds of ovals in the third derived design. The uniqueness is easy for the first derived design with parameters (4,5,11), because from such a design we immediately reconstruct a (5,6,12) design using the 66 block complements and the 66 hexads consisting of a block plus the 12th point -- and we already know the (5,6,12) design is unique and has a transitive automorphism group. On the other side, a (2,3,9) design is an affine plane of order 3, so it's unique too: we know that any affine plane is the complement of a line in a projective plane, and Pi_3 is unique with an automorphism group acting transitively on lines. That leaves the (3,4,10), which we've already constructed in several ways (e.g. as the algebraic inversive plane of order 3), but I don't think we've quite proved its uniqueness, let alone given the construction of the (5,6,12) design analogous to what we did for the (5,8,24) starting with Pi_4. At the end of the day several aspects of the picture are quite similar. Let P be the affine plane of order 3, which has 9 points and 12 lines. (An equivalent description: points are a 2-dimensional space over F_3; lines are sets of three distinct points that sum to zero.) An "oval" of P is a set of 4 points of P that meet each line in at most 2 points. These are exactly the ovals in Pi_3 that do not meet the line at infinity. We readily count them as we did for Pi_3: (9 * 8 * 6 * 3) / 4! = 54. They naturally partition into three classes each of size 18; we'll use one class to get the (3,4,10) design and will need all three to reach the (5,6,12). The group Aut(P) has a normal subgroup of index 6 with quotient group S_3 permuting the three classes. But naturally the details are different; e.g. here the subgroup arises from the normal subgroup V_4 of S_4, here occurring as PGL_2(F_3) ! The beginning is familiar too. Start with a (3,4,10) design D, which has 30 blocks, and derive a (2,3,9), which is P by deleting a point I. Since any three points of D determine a unique block, we must add 18 4-element subsets of P that meet no line in more than 2 points -- that is, ovals -- and also meet each other in at most 2 points. The key fact is that any oval O in P can be written uniquely as the symmetric difference of two non-parallel lines l,m. Indeed such a symmetric difference is readily seen to be an oval; each oval that arises this way does so uniquely -- its four points can be split in two other ways but both yield parallel lines; and this accounts for 9 * Bin(4,2) = 54 ovals, i.e. all of them. Let p be the intersection of l and l'. There are two other lines l',m' through p, which give rise to another oval O' disjoint from P. We claim that if O appears in D then so is O'. Indeed by the intersection triangle for a (3,4,10) Steiner system there are 3 blocks disjoint from O. These are either I+line or another oval. But there are only two lines disjoint from O, namely l' and m', and only one other oval, namely O' (else one would contain p and this one of l' and m'). So we must use O'. [On further thought, this last step was not needed: the next step doesn't use it, showing instead that we must use the four translates of O' by nonzero vectors in l' or m'; repeating this argument a few times finds all 18 translates of O and O'.] We claim that we must also the other 16 ovals that are translates of O or O'. [Recall that Aut(P) = AGL_2(F_3) = Ax+b group = stabilizer in Aut(Pi_3) of the line at infinity; the map (A,b) --> A is a homomorphism onto GL_2(F_3) with kernel F_3^2 = translations.] Proof: let q,r be points on l and m, and let s = q+r-p (this makes sense even in the affine plane, not just in F_3^2). So in coordinates we may take p=(0,0), l and m to be x=0 and y=0, and q,r,s to be (0,1), (1,0), (1,1). What's the fourth point in the unique block containing q,r,s? Not I because they're not collinear; nor (0,2) or (2,0) because then that block would meet O in 3 points; nor (1,2) or (2,1) because that's collinear with r,s or q,s respectively. So it's p and we get O' translated by (-1,-1). This argument gives four translates each of O and O', and applying it again to those translates gives the remaining ones. So we've determined all 18 ovals. We can either verify that the resulting 30 blocks constitute a (3,4,10) system, or as usual we can use the fact that we've already constructed one to show that once we've reduced to one possibility it automatically works. Likewise for the (5,6,12) design we join I + II + III to the 12 lines; each pair from I, II, III to one of the three classes of 18 ovals; each of I, II, III to the oval complements from the class associated with the complementary pair; and the empty subset of {I,II,III} with the 12 line complements on P. This gives a total of 12 + 54 + 54 + 12 = 132 blocks as expected. To see the connection with V_4 in S_4: the action of GL_2(F_3) on the line at infinity is via PGL_2(F_3) = S_4; all the translates of O come from pairs of lines through the same two points at infinity, and those of O' come from pairs through the other two points, so a choice of 18 ovals means a choice of even split of the 4 points at infinity as 2+2 -- and we know that S_4 acts on them (and thus on the extra three points I, II, III) via the map S_4 --> S_3 with kernel V_4. There's yet more to be said about the humble P [not to be confused with humble Pi, I suppose]: P can be found in any algebraic projective plane over a field where x^2+x+1 has roots, e.g. Pi_4. Indeed we've done this in characteristic 3 where x^2+x+1=(x-1)^2, and otherwise there are three distinct roots of unity 1,r,s and we can use 0, 1, -1; -1, 0, 1; 1, -1, 0; 0, r, -1; -1, 0, r; r, -1, 0; 0, s, -1; -1, 0, s; s, -1, 0. (These all satisfy x^3+y^3+z^3=0, and are the inflection points of this plane cubic, and indeed of x^3+y^3+z^3=cxyz for any c.) Note that we still can't do it over R, for instance because of Sylvester's theorem (in a finite configuration of points in RP^2, not all collinear, there are two points not collinear with any other) -- and indeed R has no sqrt(-3) or cube roots of 1. But it does work over C, so Sylvester fails over C.