A very

natural and

appealing context in which projective planes arise is that of 3-

dimensional

vector spaces over

finite fields. We shall start this node off, however, by counting the number of

subspaces of a given

dimension in a

finite-dimensional vector space over a

finite field.

**The linear algebra**

Let V be an n-dimenionsal vector space over the finite field F of order q = p^{s} for p prime and s a positive integer. Let k be an integer such that 1 ≤ k ≤ n. We begin by noting that the cardinality of V is q^{n} since every element of V may be written uniquely as a linear combination of n basis vectors, and there are q^n possibilities for the coefficients in a linear combination. Now every k-dimensional subspace of V is going to have a basis of size k. So let us count the number of ways to pick a basis of size k from V.

To begin with, we can pick our first vector in q^{n} - 1 ways since that's how many non-zero vectors there are in V. For the second vector (if we have to pick one), there will now be q^{n} - q vectors to choose from because we have eliminated the 1-dimensional vector space generated by our first vector, which consists of all of its q scalar multiples, from our consideration. Similarly, after picking the second vector, we have eliminated the q^{2} vectors in the new 2-dimensional subspace of V that we have. So there will be q^{n} - q^{2} ways of picking a third vector. So it is easy to see that there are (q^{n} - 1)(q^{n} - q)(q^{n} - q^{2})...(q^{n} - q^{k-1}) ways of picking k linearly independent vectors in V. This, however, admits a lot of repetition since we are counting the total number of bases for all the k-dimensional subspaces of V as compared to actually counting the subpaces. We can fix this easily, however, because we know by the previous argument the number of bases for any such k-dimensional subspace is (q^{k} - 1)(q^{k} - q)(q^{k} - q^{2})...(q^{k} - q^{k-1}). Thus, the number of k-dimensional subspaces (1 ≤ k ≤ n) is given by

(q^{n} - 1)(q^{n} - q)(q^{n} - q^{2})...(q^{n} - q^{k-1})/
(q^{k} - 1)(q^{k} - q)(q^{k} - q^{2})...(q^{k} - q^{k-1}).

**Some combinatorics**

Now let us restrict our attention to the case where V is a 3-dimensional vector space over F, so n = 3. Now, let us define P to be the set of all 1-dimensional subspaces of V. Similarly, we shall define B to be the set of all 2-dimensional subspaces of V. Using our formula for the number of k-dimensional subspaces of V, it is easy to see that |P| = q^{2} + q + 1 = |B|. The goal here is to force some geometric structure on V, so we are going to have to define some sort of incidence relation. P will be our set of "points", and B will be our set of "lines". So we shall say that a 1-dimensional subspace of V (an element of P, aka. a point) will be incident to a 2-dimensional subspace of V (an element of B, aka. a line) if and only if it is contained in the 2-dimensional subspace. So we have that for x ∈ P, b ∈ B, I(x,b) = 1 if x ⊂ b and 0 otherwise.

Now, suppose that we had two elements x and y of P and wanted to count the number of elements b ∈ B such that I(x,b) = 1 = I(y,b). Since the two subspaces x and y of V are distinct, we know that they intersect only at the 0 vector. In order for a 2-dimensional subspace to contain them, it would have to contain all q^{2} linear combinations of any two non-zero vectors from x and y. We know that there is at least one such 2-dimensional vector space. This means that there is a unique b ∈ B that contains both x and y. On the other hand, let b_{1} and b_{2} ∈ B, b_{1} ≠ b_{2}. The fact that b_{1} and b_{2} are distinct tells us that their interesection has dimension less than 2. We want to show that their interesction has dimension 1 and is therefore in P. Suppose that it had dimension 0. Then, if {u,v} and {w,z} were bases for b_{1} and b_{2} respectively, we would have that {u,v,w,z} would be a basis of four vectors for some subspace of V. But V has dimension 3, so the dimension of every subspace of V is at most 3. Thus we have arrived at a contradiction and we have to have that b_{1} ∩ b_{2} is a subspace of V of dimension 1.^{†}

It is easy to see that every element b ∈ B contains (q + 1) 1-dimensional subspaces of V using the formula that we developed at the beginning of this little exercise. Let x be a 1-dimensional subspace of V. Then, the number of 2-dimensional subspaces of V containing x is given by the number of choices for the other basis vector for a 2-dimensional subspace divided by the number of its choices for a second basis vector from the new 2-dimensional subspace. This is given by (q^{3} - q)/(q^{2} - q) = q+1.

So we have proved that every pair of points is contained on a unique line, which means that in this incidence system two points determine a line. Similarly, every pair of lines intersect at a unique point. This is the Elliptic Parallel Property. Since q is a power of a prime p and p ≥ 2, we have that every line in our incidence system contains at least three points. All we have to do to show that our system is a projective plane now is show the existence of four points, no three of which are collinear (contained on the same line).

Such a set of four points, no three of which are collinear is called a quadrangle in our incidence system. We shall demonstrate the existence of such a quadrangle. Since V is 3-dimensional, we know that V has a basis {u,v,w}. Let us consider the set Q = {u,v,w,u+v+w}. It is clear that any 2-dimensional subspace containing three of these guys will contain two of the original basis vectors for V. This means that the third vector will invariably be linearly independent to the other two. Thus the subspace would actually be 3-dimensional, and we have a contradiction. So we are done.

We shall denote our incidence system by PG(2,q).

**Concluding remarks**

In general, one is interested in such systems PG(n-1,q) in n-dimensional vector spaces V over fields F = GF(q). n-1 is called the projective dimension of the vector space V. The blocks, or lines, that one would be interested in using in this case would be the (n-1)-dimensional subspaces of V. These (n-1)-dimensional subspaces of V are called hyperplanes pf PG(n-1,q). The points and hyperplanes of PG(n-1,q) give us what is known as a symmetric balanced incomplete block design (SBIBD)‡ with parameters (v,k,λ) = ((q^{n}-1)/(q-1),(q^{n-1}-1)/(q-1),(q^{n-2}-1)/(q-1)). These parameters (v,k,λ) are called Singer parameters. It is important to note that the construction given here is not the only way to construct SBIBD's with Singer parameters. The designs constructed using the technique given here are called Desarguesian.

The existence question for projective planes is still open. We have just showed that there exists a projective plane of order q for every prime power q. Thus we know that there are infinitely many projective planes. There is, however, no projective plane of order 6 as a consequence of the Bruck-Ryser-Chowla theorem. The question of existence of a projective plane of order 10 was answered negatively in 1991 and involved heavy use of computer. The next unsettled case is that of projective planes of order 12. The general question is for which natural numbers n does there exist a projective plane of order n?

In 1938, the mathematician James Singer discovered that the projective geometries PG(n-1,q) have automorphism groups which act regularly on them. This discovery gave rise to the theory of difference sets. That, however, is a topic for a different node altogether! :)

† Note that the intersection of two subspaces of a vector space V is again a subspace of V.

‡ v is the number of points, k is the number of points per line, λ is the number of lines containing each pair of distinct points. In general, we also have the parameters r and b, where r is the number of lines containing each point, and b is the number of lines. However, we have a great deal of symmetry here since r = k and b = q. This is why these designs are called symmetric.