Feb. 17 -- affine and inversive planes

There are two topics remaining in chapter 1 that we'll cover
(and several that we won't).  One is the construction of the
star examples of design theory: the Steiner systems whose
automorphism groups are the sporadic Mathieu groups (or, for
the (3,6,22) system, a group containing M_{22} with index 2).
This is described on pages 22-24, but requires various notions
that are postponed to later chapters (e.g. chapter 6 for the
uniqueness of Pi_4).  We'll get to them later in the course.
The other topic is our concern today (pages 13-15).


AFFINE PLANES

Recall that P^2(k) is obtained from the affine plane k^2
by adjoining a line at infinity.  It's called an "affine plane"
because there's no choice of origin -- imagine an unmarked blackboard
or flat sheet of paper extended out to infinity for the familiar case
of k=R, the real numbers.  Of course this means that we can recover k^2
by starting from P^2(k) and *removing* a line and all its points.
This is still a Steiner system: any two points in the affine plane
determine a unique line.  This follows just from the axioms of a
projective plane, so we don't need any algebra on k to check it.
If  \D = (X, \B)  is a combinatorial projective plane of order q,
and B is any of its lines, then the _residual_ \D^B := {X \ B, \B \ {B}}
is a finite affine plane.  (More precisely, the collection of blocks
of \D^B isn't quite \B \ {B}, because a block other than B needn't be
a subset of X \ B; the blocks are  B' \ B  for blocks B' of \B other
than B itself.)  This \D^B is a 2-(q^2, q, 1) design with q^2+q blocks,
called an _affine plane_ of order q.  If the projective plane was
algebraic then we recover the affine plane k^2, an "algebraic affine
plane" (which the book will call AG(2,q), a notation I'm deprecating).

We can try this for any design \D and block B, but for \D^B to be
itself a design we must have each B'\B of the same size, and if that's
to be true for _all_ B then \D must be a square design.  Conversely,
the residual of a square 2-(v,k,\lam) design is a 2-(v-k, k-\lam, \lam)
design.  Warnings:

(i) Different choices of B may yield non-isomorphic residuals \D^B
with the same parameters.

(ii) \D^B could have repeated blocks  But then two blocks of \D meet in
at least k-\lam points, whence k <= 2\lam.  But since k(k-1)=v\lam,
this means v \leq 2(k-1)  [the book has the slightly weaker bound 2k-1,
which is good enough], and this cannot hold for both a design and its
complement.  So either \D or its complement \D@ (whichever has the
smaller blocks) has a residual design without repeated blocks.

Writing  (v-k, k-\lam, \lam) =: (v^B, k^B, \lam^B)  we solve for
\lam = \lam^B,  k = k^B + \lam^B,  v = v^B + k^B + \lam^B.
Then the condition  k(k-1) = (v-1)\lam  on the parameters of
the square design \D becomes

  k^B (k^B + \lam^B - 1) = v^B \lam^B

(p.13; check this manipulation, and that it works for the parameters
(v^B, k^B, \lam^B) = (q^2, q, 1) of a finite affine plane of order q).

A 2-design satisfying these constraints is said to be quasi-residual.
Not every such design need be the residual of an actual square 2-design,
but nicely enough the case of affine planes (with \lam=1) always works!
See p.14, Prop. 1.40.  The proof: in such a design, any two lines
meet in either 0 or 1 points.  If the design is \D^B then two lines are
disjoint iff they were obtained from lines that meet at a point of B.
To reconstruct \D, we need an equivalence relation || on affine lines L
such that L||L' iff L||L' came from projective lines through the same
point of B.  So we've seen L||L' iff L=L' or L,L' are disjoint.
We say that L,L' are _parallel_ in this case.  Reflexivity and
symmetry are clear, but why is this transitive?

Lemma: for every line L and point p of a finite affine plane
there's a unique line L' containing p and parallel to L.
(That is, "Playfair's axiom" holds for finite projective planes.)

Pf.: if p in L then L'=L works, and no other line does because then
two distinct parallel lines would meet.  (We use n>1 here; if n=1
there's only one choice for L' anyhow...)  If p not in L, there are
q+1 lines through p, of which q meet L (one for each point of L),
so a unique line not meeting it.

Corollary: || is an equivalence relation.

Pf.: The only nontrivial case is three distinct lines L,L',L" with
L||L'||L".  If L and L" were not parallel they would meet at some point
p which has two parallels to L', namely L and L", contradiction.

Thus: Prop. 1.40 -- a quasi-residual Steiner 2-design (=affine plane)
is residual; thus there's a finite projective plane of order q
iff there is a finite affine plane of the same order.


INVERSIVE PLANES

Remarkably k^2 always extends to a Steiner 3-design; this gives us
our second infinite examples of Steiner 3-designs, and again with a
3-transitive automorphism group.  The necessary condition (1.33)
[p.11: k+1 must divide b(v+1)] is always satisfied by a combinatorial
affine plane [q+1 | (q^2+q)(q^2+1)] but we already know that such
integrality conditions are rarely sufficient.

Again the construction is nicely motivated by a classical geometrical
picture over the real numbers.  The (2-)sphere S can be considered a
3-design whose blocks are the circles (not just "great circles").
Stereographic projection from the north pole P identifies this with
the classical inversive plane: the blocks become Euclidean circles
as well as lines ("circles through infinity" -- NB this is a different
compactification of R^2, with just one point at infinity rather than
a whole line).  It's called the "inversive plane" because inversions
are automorphisms, taking any block (circle of the sphere) to another
block.  You've seen that S is also the "Riemann sphere" C u {infinity},
which we now know is also \P^1(C), the "projective line over C".
The circles are just the images of its subset \P^1(R) under PGL_2(C).
This works just as well over a finite field k = F_q, since there is
a quadratic extension k' = F_{q^2} to take the place of C: the images
of \P^1(k) under PGL_2(k) are the blocks, and the blocks through the
point at infinite (0:1) are just the affine k-lines on k' which may be
identified with k^2 by an arbitrary choices of k-basis for k'.

The same construction works for any d>1 to give a 3-(q^d+1,q+1,1)
design whose points are the points of the projective line over
a field of q^d elements and whose derived design is affine d-dimensional
space over a field of q elements.  Your first problem on the next
problem set is to prove that this actually works.

The book notes another description via "ovoids" in P^3(k).  This
construction uses more about quadratic forms over finite fields
than I can assume here.  If you know or want to learn about this
then you might do your final project on that construction, and maybe
the Suzuki-Tits alternative for q=8,32,128,... noted on page 16.

Mild warning: the 3-(q^d+1,q+1,1) designs coming from P^1(F_{q^d})
are sometimes called "spherical geometries".  Indeed over R
there are inversive spaces of all dimensions d, each of which
gives an infinite Steiner 3-design -- namely, the d-sphere with
its circles -- and the derived 2-design is R^d with its lines.
But once d>2 there's no directly analogous construction in the
finite case: R has no algebraic extension of degree d, while
quadratic forms in d+2>4 variables over a finite field don't work
because they have isotropic lines whereas the Euclidean sphere
contains no Euclidean lines.