Projective Geometry - Lecture Notes Chapter 1

1.5 Finite Projective Spaces

Lemma 1.5.1: Let g1 and g2 be two lines of a projective space P. Then there exists a bijection between the point sets (g1) and (g2).

Def: A projective space P is finite if its point set is finite.

(Note that in exercise 31 you show that if dim(P) > 1 then P is finite if and only if its line set is finite.)

By virtue of the above lemma, there exists an integer q such that every line has q+1 points on it. This q is called the order of the finite projective space P. Note that q 2.

Lemma 1.5.2: Let P be a finite projective space of dimension d 2 and order q. Then for each point Q of P the quotient geometry P/Q has order q.

Theorem 1.5.3: Let P be a finite projective space of dimension d and order q, and let U be a t-dimensional subspace of P (1t d). Then:

  1. |U| = qt + qt-1 + ... + q + 1 = (qt+1 -1)/(q-1).
    In particular |P| = qd + ... + q + 1.
  2. The number of lines of U through a fixed point of U equals qt-1 + ... + q + 1.
  3. The total number of lines of U equals
    (qt + qt-1 + ... + q + 1)(qt-1 + ... + q + 1)/(q+1).

Theorem 1.5.4: Let P be a finite projective space of dimension d and order q. Then

  1. the number of hyperplanes of P is qd + ... + q + 1;
  2. the number of hyperplanes of P through a fixed point P is qd-1 + ... + q + 1.

Pf: (a) We prove the result by induction on the dimension. If d = 1, P is a line and the hyperplanes are points. Since P has order q, there are q + 1 points in P. If d = 2, P is a plane and the hyperplanes are lines. By Theorem 1.5.3 (c), the number of lines in P is (q2 + q + 1)(q + 1)/q+1 = q2 + q + 1.
   Now, suppose that the assertion is true for all finite projective spaces of dimension s-1. Note that we may assume that s-1 > 2 since the result has been established for dimensions 1 and 2. Let P be a finite projective space of dimension s, and H a hyperplane of P. By the dimension formula, every hyperplane H, meets H in a subspace of dimension s - 2 (which is > 1). Thus, any hyperplane of P (H) is spanned by an (s-2)-dimensional subspace of H and a point outside of H.
   For each (s-2)-dimensional subspace U of H and each point PP\H, the subspace <U,P> is a hyperplane (of P) containing (qs-1 + ... + q + 1) - (qs-2 + ... + q + 1) = qs-1 points outside of H. Since there are qs points of P outside of H, there are q = qs/qs-1 hyperplanes H through U. By induction, there are qs-1 + ... + q + 1 hyperplanes of H (since dim(H) = s-1), which are (q-2)-dimensional subspaces of P contained in H. Therefore, the total number of hyperplanes of P is 1 + q(qs-1 + ... + q + 1) = qs + ... + q + 1.
(b) Let P be a point of P, and H a hyperplane not through P. Any hyperplane of P through P meets H in a hyperplane of H. By (a) there are qd-1 + ... + q + 1 of these.

Corollary 1.5.5: A finite projective plane of order q has q2 + q + 1 points.

Finite Projective Planes of Small Order

Order23456789 1011121314151617 181920
No.11110114 01??10??221 ??1??
Notes:

1. The only known projective planes have orders that are prime powers.

2. (Bruck-Ryser Theorem) If q 1 or 2 mod(4) and a projective plane of order q exists, then q is the sum of two integral squares (one of which may be 0).

3. The order 10 was eliminated by a massive computer search.