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Synthetic Geometry 1.5 Finite Projective Spaces Lemma 1.5.1 Lemma 1.5.1: Let G and G Be Two Lines of a Projective 1 2 Space P

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Synthetic Geometry 1.5 Finite Projective Spaces Lemma 1.5.1 Lemma 1.5.1: Let G and G Be Two Lines of a Projective 1 2 Space P Synthetic Geometry 1.5 Finite Projective Spaces Lemma 1.5.1 Lemma 1.5.1: Let g and g be two lines of a projective 1 2 space P. Then there exists a bijection between the point sets (g ) and (g ). 1 2 Pf: If g = g the identity map is a bijection, so we may assume that 1 2 g ≠ g . 1 2 Case I: The lines g and g intersect in a point S. 1 2 Let P be a point of g and P a point of g , neither equal to S. By 1 1 2 2 Axiom 3 there is a third point P on the line P P . P is not on either 1 2 g or g (if it were we would contradict the choices of P and P .) By 1 2 1 2 the Veblen-Young axiom any line through P that contains a point X ≠ S of g intersects the line g in a uniquely determined point 1 2 ϕ(X) ≠ S. Lemma 1.5.1 Lemma 1.5.1: Let g and g be two lines of a projective 1 2 space P. Then there exists a bijection between the point sets (g ) and (g ). 1 2 Pf (cont.): Thus, the map ϕ defined by: X P 1 ϕ: X → XP ∩ g g 2 1 P is a bijection from (g )\{S} to (g )\{S}. S 1 2 g ϕ(P) By defining ϕ(S) = S, we obtain a 2 P 2 bijection from (g ) onto (g ). 1 2 Lemma 1.5.1 Lemma 1.5.1: Let g and g be two lines of a projective 1 2 space P. Then there exists a bijection between the point sets (g ) and (g ). 1 2 Pf (cont.): Case II : The lines g and g have no point in common. 1 2 Let h be some line joining a point of g with a point of g . By 1 2 the first case, there are bijections ϕ : (g ) → (h) and ϕ : (h) → (g ). 1 1 2 2 Then the composition ϕ °ϕ is a bijection from (g ) onto (g ). ❑ 2 1 1 2 Finite Projective Spaces 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 previous 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 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. Pf: By 1.4.1, P/Q is isomorphic to to any hyperplane H of P which does not pass through Q. The points on any line of a hyperplane are all contained in the hyperplane, so the order of the hyperplane, thought of as a projective space itself, is the same as the order of the projective space in which it lies. Point and Line Counts 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 (1 ≤ t ≤ d). Then: a. |U| = qt + qt-1 + ... + q + 1 = (qt+1 -1)/(q-1). In particular |P| = qd + ... + q + 1. b. The number of lines of U through a fixed point of U equals qt-1 + ... + q + 1. c. The total number of lines of U equals (qt + qt-1 + ... + q + 1)(qt-1 + ... + q + 1)/(q+1). Point and Line Counts Theorem 1.5.3: a. |U| = qt + qt-1 + ... + q + 1 = (qt+1 -1)/(q-1). In particular |P| = qd + ... + q + 1. b. The number of lines of U through a fixed point of U equals qt-1 + ... + q + 1. Pf: We prove a) and b) simultaneously by induction on t. If t =1, then U is a line of P and since all lines have the same number of points by 1.5.1, |U| = q + 1 since the order of P is q. There is only one line in U, which is the correct count given by b) when you realize that the formula there is really: i=t−1 ∑ qi = qt−1qt−2⋯q1. i=0 Point and Line Counts Theorem 1.5.3: a. |U| = qt + qt-1 + ... + q + 1 = (qt+1 -1)/(q-1). In particular |P| = qd + ... + q + 1. b. The number of lines of U through a fixed point of U equals qt-1 + ... + q + 1. Pf (cont.): Now suppose that (a) and (b) are true for projective spaces of dimension t-1. Let Q be a point of U. U/Q is a (t-1)- dimensional projective space of order q, so by induction its number of points is qt-1 + ... + q + 1. Since this is the number of lines of U through Q we have shown (b). Each of these lines contains exactly q points other than Q, and every point of U lies on exactly one of these lines, so the number of points of U is 1 + (qt-1 + ... + q + 1)q = qt + ... + q + 1. Point and Line Counts Theorem 1.5.3: c. The total number of lines of U equals (qt + qt-1 + ... + q + 1)(qt-1 + ... + q + 1)/(q+1). Pf (cont.): Let b be the number of all lines of U. Count the number of flags (Q,l) where Q is a point of U, l a line of U and Q I l. On one hand there are (qt + qt-1 + ... + q + 1)(qt-1 + ... + q + 1) such flags by counting the (# of points)(#lines through a point), and on the other hand there are (q+1)b such flags by counting (#lines)(#points on a line). As the counts must be equal, we solve for b to obtain the result. Hyperplane Counts Theorem 1.5.4: Let P be a finite projective space of dimension d and order q. Then a. the number of hyperplanes of P is qd + ... + q + 1; b. 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. Hyperplane Counts Theorem 1.5.4: Let P be a finite projective space of dimension d and order q. Then a. the number of hyperplanes of P is qd + ... + q + 1; Pf (cont.): 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. Hyperplane Counts Theorem 1.5.4: Let P be a finite projective space of dimension d and order q. Then a. the number of hyperplanes of P is qd + ... + q + 1; Pf (cont.): For each (s-2)-dimensional subspace U of H and each point P∈P\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. Hyperplane Counts Theorem 1.5.4: Let P be a finite projective space of dimension d and order q. Then a. the number of hyperplanes of P is qd + ... + q + 1; b. the number of hyperplanes of P through a fixed point P is qd-1 + ... + q + 1. Pf (cont.):(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. ❑ Finite Projective Planes Corollary 1.5.5: A finite projective plane of order q has q2 + q + 1 points and the same number of lines. Finite Projective Planes Finite Projective Planes of Small Order Order 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 # 1 1 1 1 0 1 1 4 0 ≥1 ?? ≥1 0 ?? ≥22 ≥1 ?? ≥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).
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