Before considering the finite geometries, we need to examine an important development in post-Euclidean geometry. Starting with the perspective drawings of the Renaissance artists, geometers began toying with the idea that parallel lines might meet "at infinity". To legitimatize this concept the subject of projective geometry was developed. Projective geometry is a slight extension of Euclidean geometry in which no parallel lines exist (i.e., every pair of lines meet at a unique point.) To accomplish this in plane geometry, we introduce a new point on each line of a family of parallel lines - because these lines now meet in the new point they are no longer parallel. This new point, since it can not exist in the original plane is called a point at infinity. For each family of parallels we introduce a distinct point at infinity and then collecting all the points at infinity, we introduce a new line consisting entirely of these points and call it the line at infinity. This construction gives us the Real Projective Plane which is an extension of the Euclidean plane. Since we have only added things to the original plane, we have not lost any of the geometry of that plane which can be recaptured by tossing out the infinite points and line. It turns out that many statements and theorems are simplified when one views them as statements about this extended plane. The advantages of working with the Projective Geometry are so great that a number of geometers have expounded the statement that "Projective Geometry is all geometry."

- Every pair of distinct points is incident with a unique line,
- Every pair of distinct lines is incident with a unique point, and
- There exist 4 points no three of which are incident with the same line.

If the sets or *L* are finite then the projective plane is called a **finite projective plane**. As
an example of a finite projective plane (in fact the smallest possible example), let =
{1,2,3,4,5,6,7} and let *L* consist of the following sets:

**Theorem VIII.1.1** -* If one line of a projective plane is incident with only a finite number of
points, say 1 + n, then
*

- every line is incident with 1 + n points,
- every point is incident with 1 + n lines,
- there are n
^{2}+ n + 1 points in , and - there are n
^{2}+ n + 1 lines in*L*.

*Proof*: Let be the line incident with n + 1 points and ' be any other line. Let Q be a point
not on either line (Q must exist, for if it didn't, i.e., all points lie on one or the other of these
two lines, then axiom 3 would be violated). Q and each, in turn, of the n+1 points on
determine n+1 distinct lines incident with Q (why are they distinct?). If another line, distinct
from these n+1 were incident with Q then it would have to meet by axiom 2 in one of its
n+1 points and so there would be two points (one on and Q) which would have two distinct
lines incident with them contradicting axiom 1. Thus Q has exactly n+1 lines incident with it.
Now these n+1 lines through Q each must meet ' at a distinct point (why are they distinct?).
If ' had another point incident with it, then this point together with Q would determine one
more line through Q, which as we have already seen is impossible. So, ' has exactly n+1
points incident with it, and since ' was arbitrary, all lines are incident with exactly n+1
points proving part a).

Let P be an arbitrary point, and any line which is not incident with P. Since by part a),has exactly n+1 points, these points determine n+1 lines through P. Ther e can be no other lines through P, for any such would have to intersect at a point other than the n+1 already accounted for. Thus every point is incident with exactly n+1 lines.

Let P be an arbitrary point. By axiom 1, every point in determines a line together with
P which obviously passes through P. Hence, all the points of lie on the n+1 lines incident
with P. Each of these lines is incident with n points other than P, so there are n(n+1) of them
and including P we get n^{2} + n + 1 points in .
Let be an arbitrary line. By axiom 2, every line of meets at a point on . So,
all
the lines of are incident with the n+1 points of . Each of these points is incident with n
lines other than , so there are n(n+1) of them and including we get n^{2} + n + 1 lines of .

The same theorem could have been proved with the assumption that there existed a point that was incident with exactly n+1 lines (exercise).

In light of the above theorem, we define the **order** of a finite projective plane to be the
number n, i.e., one less than the number of points on a line. [The reason for defining it this
way will be made clearer later]. Our example then is a projective plane of order 2.

**Theorem VIII.1.2** - *A projective plane of order n is a 2-(n ^{2}+n+1,n+1,1) design.*

*Proof*: Let the points be the varieties and the blocks are the set of points incident with a line,
for each line. By Thm. VIII.1.1 c), there are n^{2}+n+1 varieties, and by a) of the same theorem
the block size is n+1. That every pair of points (varieties) appear together in exactly one
block is axiom 1.

**Theorem VIII.1.3** - *A projective plane of order n is equivalent to a complete set of MOLS of
order n.*

*Proof*: The construction of the projective plane of order n from a complete set of MOLS of
order n was given in VII.2.4.

Let be a projective plane of order n. Select any line of and arbitrarily label with
the digits 1,...,n each line which passes through a point of , for each point of . Now select
two points of . The lines which pass through these points will be used to index rows and
columns of the latin squares, so label one of the points R and the other C. The n^{2} points of
intersection of the lines through R and C are associated with pairs of numbers, the number of
the line through R and the number of the line through C. Now, for each point of other than
R or C, we will form a Latin square in the following way: If P is the point on , we have
already labelled all the lines through P other than . The n^{2} points of intersection of the R
and C lines must all lie on the n labelled lines through P. In the cell of the square
corresponding to one of the intersection points we place the label of the line through P which
passes through this point. It is easy to see that the square produced this way is a Latin square
of order n. This procedure is repeated for each of the points of , giving n-1 Latin squares.

To see that they are mutually orthogonal, consider two such squares and suppose that when superimposed there are two cells containing (a,b). Since the two cells of the first square received the label a, the two points which correspond to the cells must have been on the same line (labelled a) going through a point of . Since these same cells have the label b in the second square, the two points must also be on the line labelled b passing through a different point of . This is impossible by axiom 1, so we see that these squares must be mutually orthogonal.

By Construction III.1.1 and the above theorem we see that projective plane of order n always exists if n is a prime or prime power, i.e., the order of a finite field. By Thm. VIII.1.2, we may apply Thm. VII.1.4 to obtain the following result.

**Theorem VIII.1.4** - [Bruck-Ryser Thm.] *If n 1 or 2 mod 4 then a projective plane of order n does not exist unless n is the sum of two integral squares.*

*Proof*: By Thm. VIII.1.2 we know that a projective plane of order n is a symmetric BIBD
with v = n^{2}+n+1 , k = n+1 and = 1. Since n^{2}+n+1 is always odd, Thm. VII.1.4 implies that
a necessary condition for the existence of a projective plane of order n is that the equation

This theorem implies that there does not exist a projective plane of order 6, a result we could have inferred from Thm. VIII.1.3, since no pair, let alone a complete set, of MOLS of order 6 exists. Notice that, since 10 = 9 + 1, this theorem says nothing about the existence of a projective plane of order 10. An extremely long computer search has finally proved that this plane does not exist (see note at the end of the chapter). The smallest open case is n = 12 which the Bruck-Ryser theorem do es not cover and then n = 18, which since 18 = 9 + 9, the theorem says nothing about. The only known orders are primes and prime powers, although alternate constructions exist which give projective planes which are not isomorphic to those produced by the Latin square construction. Instant fame and recognition will be the reward of the mathematician who settles the question of the existence of projective planes of composite order.

Let K be a k-arc in the projective plane of order n. The lines of fall into three
classes with respect to K. Lines which intersect K in two points are called **secant lines**, those
that meet K in only one point are called **tangent lines** and those that have no point in
common with K are called **exterior lines**. We can easily count the numbers of these types of
lines. There is a secant line for each pair of points of K, so there are k(k-1)/2 secant lines
altogether, k-1 passing through each point of K. A point of K will therefore have n+1 - (k-1)
= n - k + 2 tangent lines through it, and so the total number of tangent lines is k(n-k+2). All
other lines of are then exterior
lines so there are n^{2} + n + 1 - k(k-1)/2 - k(n-k+2) of them.
Some other easy counting arguments give the following result.

**Theorem VIII.2.1** - *If n is odd, no hyperoval can exist in a projective plane of order n.
If n is even, every oval in a projective plane of order n can be extended to a hyperoval
in a unique way.*

*Proof*: Suppose n is odd and K is an (n+2)-arc. By the above remarks, there are n+2-1 =
n + 1 secants passing through each point of K, i.e., every line through a point of K is a secant
line. Consider a point P which is not in K. The lines through P either intersect K, in which
case they must contain two points of K or else they are exterior lines. Since all the points of
K are joined to P in pairs, n + 2 must be even which is impossible if n is odd.

Now suppose that n is even and let K be an oval. Since k = n + 1, there is exactly one tangent line through each point of K and n + 1 tangents in all. Let A and B be points of K, and let X be any other point on the secant they determine. Since K contains n - 1 points other than A and B, and n is even, there must exist at least one tangent line which passes through X. Thus there are at least n + 1 distinct tangent lines, one through each point on this secant line, but since that is the total number of tangent lines, through each point on a secant line there passes exactly one tangent line. Now consider the intersection point of two tangent lines (which exists by axiom 2), through this point no secant line can pass and since it must be connected to each of the points of K, its n + 1 lines must all be tangent lines. Since this point is not on any secant line, it can be joined to K to produce an hyperoval.

In the even order case, the point at which all the tangent lines to an oval meet is called
the **knot** or **nucleus** of the oval. This strange behavior of tangent lines in the even order case
does not have an analog in continuous geometry.

If n is odd, it is easy to see what the code C looks like. If we take all the rows of A which contain a 1 in a specific position (i.e., all the lines through a given point) and add them, we get a vector with a 0 in the specified position and 1's in all other components, since n+1 is even. These vectors generate all code words of even weight, just add together these special vectors that have 0's in the positions you want to have 1's in. Since C can not contain any code words of odd length (verify - a parity argument), this must be C, i.e., the s/space consisting of all even weight vectors.

If n is even the situation is more complex. We shall only investigate one case, that is n 2 mod 4.

**Theorem VIII.3.1** - *If n 2 mod 4 then the incidence matrix A of a projective plane of
order n generates a code C with dimension ½(n ^{2} + n + 2).
*

*Proof*: We shall only prove half of this theorem (actually the easier half). Let C* be the code
C extended by adding a parity check bit to each code word. The code C* is a s/space of a
vector space of dimension n^{2} + n + 2 and we have that dim(C) dim(C*). Since n is even,
every pair of code words in C* are orthogonal, so C* (C*)^{perp} . Then, since the sum of the
dimensions of a s/space and its orthogonal compliment equals the dimension of the space, we
have that dim(C) dim(C*) ½(n^{2} + n + 2). The inequality in the other direction can also be
shown.

We note that the conclusion of Thm. VIII.3.1 implies that the extended code C* is self-orthogonal, i.e., it is its own orthogonal complement. The code of the above theorem has other interesting properties of geometric significance.

**Theorem VIII.3.2** - *The code C of Thm. VIII.3.1 has minimum weight n + 1 and every
vector of minium weight is a line in the projective plane. Also, the vectors of C of weight n +
2 are precisely the hyperovals of the projective plane.*

*Proof*: Let v 0 be a code word with weight d. In C* every vector corresponding to a line of
the projective plane has a parity check of 1 since n + 1 is odd. If d is odd, then in C* v
would also have a parity check of 1. Since C* is self-orthogonal, v must have at least one 1 in
common with every line. If d is even, then every line through a point of v must meet v in a
second point. In this second case, since there are n + 1 lines through any point, we must have
d n + 2. In the first case, since there are n + 1 lines through any point and all lines meet v,
we have d(n+1)n^{2} + n +
1 which
implies that dn + 1. Thus, since the lines do have
weight n + 1, this is the minimum weight of C.

Now suppose v has weight n + 1 (which is odd) and is not a line. Then since (n+1)^{2} > n^{2}
+ n + 1, some line must meet v in at least 3 points. Let be such a line, if there is a point
on which is not on v then every linethrough this point must meet v in at least one
distinct point. Thus v would have at least n + 3 points contradiction.

Suppose now that v has weight n + 2. Every line meets v in an even number of points. Let be a line and suppose v and have 2a points in common. Each of the n lines through one of these 2a points meets v at least once more. Therefore, 2a + n n + 2, i.e., a = 0 or a = 1, so v is a hyperoval.

Let K be an hyperoval, we must show that the points of K when written as a vector v, is
a member of the code. Let S be the set of secant lines of K. Each of the n^{2} - 1 points not in v
is on exactly ½(n+2) (which is even since n 2 mod 4) lines of S. Each of the points of v is
on n + 1 lines of S (which is odd). Thus the sum of the lines of S will give v.

These coding theory results have lead to the recent disproof of the existence of a projective plane of order 10. The extended code C* of such a hypothetical plane would be a (112,56)-self dual code. A result due to F.J. MacWilliams, relates the weights of a code to the weights of its dual code. These relations can be used to disprove the existence of a code. In the particular case in hand, all the relations would be determined if the number of code words for each weight up to 16 could be determined. We know that there is one code word of weight 0 (the zero vector), and no code words of weights 1 - 10 by Thm. VIII.3.2, and 111 code words of weight 11 by the same result. It is not hard to show that all the code words of C* have weight divisible by 4, so there are no code words in C of weights 13 or 14. This leaves just the words of weights 12,15 and 16 to be determined. MacWilliams, Sloane and Thompson assumed that there were code words of weight 15. It was found that if this were the case, the projective plane would have to contain a particular configuration of 15 lines. A computer search showed that starting from this configuration it was impossible to construct a projective plane of order 10. Thus there can be no code words of weight 15. Furthermore, Lam, Thiel, Swiercz and McKay, have shown that there are no hyperovals in a plane of order 10, and so no code words of weight 12. The only remaining case is that of code words of weight 16. This work was finally completed in 1989 after a considerable amount of CPU time on a Cray Computer (one of NSA's) with the result that there are no code words of weight 16 and so there is no projective plane of order 10.

**Dembowski**,* Finite Geometries*, Springer-Verlag, Berlin, 1968.

However, this is not a text, rather a compilation of research results with most of the proofs ommitted but referenced. Some texts in the area are

**Hughes & Piper**, *Projective Planes*, Springer-Verlag, New York, 1973.

**Stevenson**, *Projective Planes*, Freeman, San Francisco, 1972.

**Karteszi**, *Introduction to Finite Geometries*, North-Holland, Amsterdam, 1976.

**Hirschfeld**, *Projective Geometries over Finite Fields*, Clarendon Press, Oxford, 1979.

**Pickert**, *Projektive Ebenen*, Springer-Verlag, Berlin, 1975.

Two excellent introductory texts are

**Room & Kirkpatrick**, *Miniquaternion Geometry*, Cambridge Univ. Press, 1971.

**Albert & Sandler**, *An Introduction to Finite Projective Planes*, Holt, Rinehart and Winston,
New York, 1968.

For the coding theory connections see any of the references in that chapter, but especially the
book by **Cameron and van Lint**.

The role that ovals play in the structure of projective planes will be the subject of my next book, but until then there are only scattered references to them in the literature.