March 3: Uniqueness and automorphism group of Pi_4 (day 2) Recall: we began with any order-4 projective plane Pi and one of its 6!*168 ordered ovals O = {p1,p2,p3,p4,p5,p6}. This let us identify 15 of the lines: the 15 secants s_{i,j}. That leaves 15 points and 6 passant lines to account for. Each of the 15 points is on 3 secants and 2 passants. This immediately tells us that the passant lines form a hyperoval O* in the dual projective plane Pi*, whose secants are the 15 points off O. Consider one of these points q. It is contained in 6/2=3 secants of O, and thus 5-3=2 passant lines. Now the secants determine a partition of O into three pairs. As noted already, a partition of a six-element set into three pairs is called a _syntheme_ in this context. For any n, the number of partitions of a 2n-element set into n pairs is 1 * 3 * 5 * 7 * ... * (2n-1) = (2n)! / (2^n n!) [proofs: i) fix an element: it has 2n-1 possible partners; for each choice we have to count pairings of the remaining 2(n-1) elements, etc.; ii) standard double-counting; or iii) all are equivalent under S_{2n} and the stabilizer has order 2^n n! -- yes, this is more-or-less equivalent to (ii)]. This count is 1, 3, 15, 105, 945, ... for n=1,2,3,4,5,... . For n=2 this was already seen to give the exceptional homomorphism S4 --> S3. For n=3 it will lead us to the exceptional (outer) homomorphism S6 --> S6: the count is still small enough to coincide with the number Bin(2n,2) of pairs. We begin by observing that 15 is just big enough to account for all the points not on O (clearly the syntheme determines the point). So for any syntheme, the corresponding three secants are coincident. This is the last time that can happen: already 105 is too large for us to hope that each 1-factor of an oval in Pi_7 yields four secants that meet at a point. (As you'll see in the present homework, even for Pi_5 it doesn't quite work even though there are still only 6 points on the oval.) Getting back to Pi_4: we've now accounted for all 21 points, namely the 6 points of the oval and one point for each of the 15 synthemes; we've also accounted for 15 of the 21 lines, namely the secants (one of each of the 15 pairs). So we need only figure out how to group the 15 syntheme points into 6 passant lines, each of which will contain exactly two of them. We've already noted that the passant line forms an oval O* in Pi*; it's secants are the points where two passants intersect -- necessarily syntheme points -- so we're in effect seeking to identify synthemes in O with pairs in O*. Now the key point is that two syntheme points q,q' are on a passant line <==> the unique line through q and q' is not a secant <==> the synthemes associated to q and q' have no pair in common. Well a passant line l contains 5 points that determine 5 synthemes, each of which consists of 3 pairs, and all 3*5=15 of these pairs are distinct. But there are only Bin(6,2) pairs to go around. So each of the six passant lines l determines a partition of the pairs into 5 synthemes -- what we'll call a _total_, short for "synthematic total" (and the text calls a "factorization", short for "1-factorization"). Lemma: there are six totals. Proof: count triples (t,s,s') where t is a total and s,s' two of its synthemes. Given t there are 5*4=20 choices for (s,s'). We can check that any two disjoint synthemes lie in a unique total. (There are 4 other synthemes disjoint from both s and s'; one has a pair in common with each of the other three, but the other three are pairwise disjoint.) On the other hand, given s there are 4*2=8 choices for s', so the number of (s,s') pairs is 15*8=120. Hence there are 120/20=6 totals, QED. So we have a unique choice for completing our (re)construction of Pi from O: there's one passant line for each of the 6 totals. To check that this works, we can either verify the axioms directly, or note that the algebraic Pi_4 is already known to satisfy them and we just showed that starting from any of its 168 hyperovals we get our labeling of all 21+21 vertices and edges. We can thus formulate a theorem for projective planes of order 4 analogous to what we've done for order 3 and 2: let Pi, Pi' be projective planes of order 4, and O, O' hyperovals on them; then every bijection O --> O' extends uniquely to an isomorphism Pi --> Pi'. Taking Pi'=Pi we also find that Aut(Pi) has size 6!*168 = 120960. Identifying Pi with P^2(F_4) -- which we can do now that we've shown all these projective planes are isomorphic -- we also deduce that every automorphism is either a projective linear transformation (i.e. an element of PGL_3(F_4)) or such a transformation composed with the nontrivial automorphism of F_4. Now what about Pi'=Pi* and O'=O*? This gives an isomorphism between the symmetric groups of permutations of O and O*, both S_6. But it's not an inner automorphism! We noted already that for any group G the group of inner automorphisms is G/Z(G) [where Z(G) = center of G]; for G=S_n with n>2, the center is trivial and conjugation by g is just relabeling the objects that S_n permutes. In particular an inner automorphism of S_n preserves the cycle structure of each element. But here we can see explicitly that this is not the case. Recall that we have projective coordinates (1 : 0 : 0) (0 : 1 : 0) (0 : 0 : 1) (1 : 1 : 1) (1 : a : b) (1 : b : a) for the points in O. Using these coordinates, O* consists of the six lines (y:x:x), (x:y:x), (x:x:y) where x and y are distinct nonzero elements of F_4. Now a cyclic permutation of the three coordinates has cycle structure 3,1,1,1 on O -- to see that it fixes (1:a:b), note that b=a^2 and a^3=1, so (1:a:b) = (1:a:a^2) = (a:a^2:a^3) = (a:a^2:1) = (a:b:1), and likewise for (1:b:a); but there are no fixes points in O* so the cycle structure of the action on O* is 3,3. We'll see that this is essentially the unique way to get an outer automorphism of any S_n, in that n must equal 6 and the outer automorphism must be the one relating O and O* up to conjugation (= relabeling of O, or equivalently of O*). And we'll exploit the strongly regular graph T(n) -- recall that this graph has vertices = pairs taken from an n-element set, and edges = pairs that intersect in one element. We'll prove that Aut(T(n)) = S_n for n>4, and that for n other than 6 any automorphism of S_n induces an automorphism on T(n) via the action on involutions.