March 1: Uniqueness and automorphism group of Pi_4 (day 1)

We just finished showing: for q=2 and 3, a projective plane Pi
of order n has respectively 7 hyperovals or 234 ovals O, so
4!*7=168 and 4!*234=5616 ordered hyperovals, and Aut(Pi) acts
simply transitively on them; moreover Pi is unique up to isomorphism:
for any other Pi' and O', any bijection  O --> O'  extends uniquely to
an isomorphism  Pi --> Pi'.

Can we expect this to hold in general?  Not if the automorphism group
is no larger than PGL_3(F_q): the size of this group is
(q^3-1) (q^3-q) (q^3-q^2) / (q-1)  or about q^8,  which for
large q is much smaller than the number (q+1)! or (q+2)! of
permutations of an oval, let alone that times the number of ovals.
Nevertheless it still works for n=4 (with 168 hyperovals), though
in a special way.  (For n=5, you'll see that it doesn't quite work,
but as many as 120 of the 6!=720 possible bijections lift.)

For starters, how many ordered ovals are there?  The number of
ordered n-arcs in a projective plane of order q is:

  n=1: q^2+q+1
  n=2: (q^2+q+1) (q^2+q)
  n=3: (q^2+q+1) (q^2+q) (q^2)
  n=4: (q^2+q+1) (q^2+q) (q^2) (q-1)^2
  n=5: (q^2+q+1) (q^2+q) (q^2) (q-1)^2 (q^2-5q+6)
  n=6: (q^2+q+1) (q^2+q) (q^2) (q-1)^2 (q^2-5q+6) (q^2-9q+21)

and for n=7 and higher we can't tell because we don't know how many
coincident triples of secants there are.  So this technique cannot work
(at least not in this form) past q=5.

Also, just how transitively do we expect Aut(Pi) to act?
Suppose Pi = P^2(F_q) and consider PGL_3(F_q).  It acts
2-transitively; not 3-transitively because of collinearity,
but that's the only obstruction even up to 4 points.  Indeed
for any P_1,P_2,P_3,P_4 no three of which are collinear,
and any Q_1,Q_2,Q_3,Q_4 no three of which are collinear,
there is an automorphism in PGL_3(F_q) taking P_1 to Q_1,
P_2 to Q_2, P_3 to Q_3, and P_4 to Q_4 -- and this automorphism
is unique!  More generally we have:

  Proposition:  For any field k, the group PGL_n(k) acts
  simply transitively on ordered (n+1)-tuples of points in P^(n-1)(k)
  in general linear position.

"Simply transitively" means that for any two such (n+1)-tuples
(P_0,P_1,...,P_n) and (Q_0,Q_1,...,Q_n) there is a unique group element
taking one to the other.  "General linear position" means no 3 points
on a line, no 4 on a plane, ... no n on a hyperplane.  This last
condition is actually sufficient: if we have k points on a projective
subspace of dimension k-2 then any n points containing them are on a
subspace of dimension n-2.

  Proof: It is enough to prove this for one choice of (P_0,P_1,...,P_n).
  Let P_0 be (1:1:1:...:1) and for each i=1,2,...,n let P_i be the
  (1-dimensional subspace of k^n generated by) the i-th unit vector.
  Let v_0,...,v_n be nonzero vectors in k^n corresponding to Q_0,...,Q_n.
  General linear position <==> any n of these are linearly independent
  <==> any n of these form a basis.  So there's a linear map T taking
  e_i to v_i.  This map is unique, so how do we get P_0 to go to Q_0?
  Projectively, P_1,...,P_n go to Q_1,...,Q_n iff each e_i goes to
  c_i v_i for some nonzero scalar c_i, and any choice of (c_1,...,c_n)
  gives a unique T; conversely T determines (c_1:...:c_n) [note the
  colons: these are projective coordinates].  Now v_1,...,v_n is a basis
  so  v_0 = a_1 v_1 + ... + a_n v_n  for some scalars a_1,...,a_n;
  moreover none of them are zero because then there'd be a linear
  dependence among v_0 and the other n-1 v_i's.  So choose
  (c_1:...:c_n) = (a_1:...:a_n)  and we're done.

Note that it should follow that #(PGL_n(F_q)) is the number of
ordered (n+1)-tuples of points in general linear position;
this can be checked directly (as we in effect did for q=n=3).

OK -- so we expect transitivity on n-arcs for n=4 but not beyond.
Yet a hyperoval in Pi_4 has 6 points.  How can this be?

Let's count: taking q=4 in the above n-arc counts, we find that
the number of ordered hyperovals is

  21 * 20 * 16 * 9 * 2 * 1 = 6! * 168

in particular, having chosen the first 4 points we have no choice
about the last 2 except for their order.  We can give explicit
coordinates starting from those of our proof above for the first
4 points: each of the remaining two must have nonzero coordinates,
all distinct.  Scaling the first to 1, there are only two choices:
(1:a:b) and (1:b:a), where a,b are the elements of F_4 other than
0 and 1, satisfying  ab = a+b = a^3 = b^3 = 1.  So we have the
hyperoval

  (1 : 0 : 0)
  (0 : 1 : 0)
  (0 : 0 : 1)
  (1 : 1 : 1)
  (1 : a : b)
  (1 : b : a)

and now it's clear that the remaining two points are switched by
the automorphism of P^2(F_4) that takes each coordinate to its
Galois conjugate: 0,1,a,b go to 0,1,b,a respectively.

So in fact it is reasonably to expect even for n=4 that every
bijection of hyperovals lifts to a unique isomorphism of projective
planes (provided it's true that Pi_4 is unique).  That's what we'll prove.

Before starting this, yet another way that n=4 is special.  We know
one way of getting hyperovals in an algebraic projective plane P^2(k)
when #k is even: use a conic plus its center.  If all *ordered*
hyperovals are equivalent then any point can be the center.
How can this be?  Two conics can meet in at most 4 points (Bezout).
But that's just enough for center-switching to work for #k=4.
(It also works for #k=2 but that's not as surprising...)

OK: start with any order-4 projective plane Pi and one of its
6!*168 ordered ovals  O = {p1,p2,p3,p4,p5,p6}.  This lets 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.  How to label O*?  Since we'll
show that any permutation g of O lifts uniquely to Aut(Pi), we have
a homomorphism from the permutations of O to those of O* -- and that
turns out to be an outer automorphism from S_6 to S_6 !  We'll see how
this happens starting next time.