Analytical Geometry. 2.4 The Hyperbolic Quadric of PG(3,F)

Transcription

1 Analytical Geometry 2.4 The Hyperbolic Quadric of PG(3,F)

2 Transversals Def: A set of subspaces of a projective space are skew if no two of them have a point in common. A line is called a transversal of a set of skew subspaces if it intersects each in exactly one point. Lemma 2.4.1: Let P be a projective space. Let be two skew lines, and denote by P a point on neither line. Then there is at most one transversal of through P. If P is 3-dimensional then there is exactly one transversal of through P. Pf: Assume that there are two transversals through P. These two lines span a plane which contains, a contradiction.

3 Transversals Lemma 2.4.1: Let P be a projective space. Let be two skew lines, and denote by P a point on neither line. Then there is at most one transversal of through P. If P is 3-dimensional then there is exactly one transversal of through P. Pf (cont.): Now suppose that P is 3-dimensional. The plane <P, must intersect the line g 2 in some point Q (since the plane is a hyperplane). The line PQ lies in the plane and so intersects the line. PQ is thus a transversal through P.

4 The 16 Point Theorem Theorem 2.4.2: Let P be a 3-dimensional projective space over the division ring F. Let { } and {h 1 } be sets of skew lines with the property that each line g i meets each line h j. Then F is commutative (hence a field) if and only if each transversal g of {h 1 } meets each transversal h of { }. Pf: Consider this diagram: h 1 h 2 h 3 <v 2 + v 2 g 2 <v 3 <v 4 + v 3 <v 3 + v 4

5 The 16 Point Theorem Theorem 2.4.2: Let P be a 3-dimensional projective space over the division ring F. Let { } and {h 1 } be sets of skew lines with the property that each line g i meets each line h j. Then F is commutative (hence a field) if and only if each transversal g of {h 1 } meets each transversal h of { }. Pf(cont.): h1 h 2 h 3 g 2 <v 2 <v 3 <v 4 + v 3 + v 2 h 2 = <v 2 + av 4. And h 3 = <v 3 + v 4 <v 2 + v 4 +v 2 +v 3 +v 4 <b(v 1 +v 3 ) + c(v 2 +av 4 ) = <d(v 1 +v 2 ) + e(v 3 +v 4 ). Equating the coefficients: b = d c = d b = e ca = e b = c = d = e and a = 1.

6 The 16 Point Theorem Theorem 2.4.2: Let P be a 3-dimensional projective space over the division ring F. Let { } and {h 1 } be sets of skew lines with the property that each line g i meets each line h j. Then F is commutative (hence a field) if and only if each transversal g of {h 1 } meets each transversal h of { }. Pf(cont.): h1 h 2 h 3 <v 2 + v 2 h + bv 2 g 2 g <v 3 <v 4 <v 3 + v 4 + v 3 <v 2 + v 4 +v 2 +v 3 +v 4 + av 3 Claim: g meets h iff ab = ba.

7 Pf(cont.): h1 h 2 h 3 g 2 g The 16 Point Theorem <v 2 <v 3 <v 4 + v 3 + av 3 + v 2 <v 3 + v 4 <v 2 + v 4 +v 2 +v 3 +v 4 h + bv 2 <v 2 + av 4 is on h 2, +v 2 + a(v 3 +v 4 ) is on h 3. Since v 1 + av 3 = v 1 + v 2 + a(v 3 +v 4 ) (v 2 + av 4 ), + av 3 is on the line determined by those two points, and since there is only one transversal to h 2 and h 3 through +av 3 this line must be g. So g = + av 3, v 2 + av 4 and similarly, h = + bv 2, v 3 + bv 4.

8 The 16 Point Theorem Theorem 2.4.2: Let P be a 3-dimensional projective space over the division ring F. Let { } and {h 1 } be sets of skew lines with the property that each line g i meets each line h j. Then F is commutative (hence a field) if and only if each transversal g of {h 1 } meets each transversal h of { }. Pf(cont.): If there is a point in the intersection of the subspaces g and h, there are scalars x,y,z and w such that: x(v 1 +av 3 ) + y(v 2 +av 4 ) = z(v 1 +bv 2 ) + w(v 3 +bv 4 ). By the independence of the vectors, we can equate coefficients: x = z, y = zb, xa = w and ya = wb. But then, xba = zba = ya = wb = xab. If x = 0, then z = y = w = 0 and there is no point of intersection. Otherwise, we get ba = ab. Conversely, if ab = ba, then + av 3 + bv 2 + bav 4 is a common point of g and h.

9 Reguli Def: Let P be a 3-dimensional projective space. A nonempty set R of skew lines of P is called a regulus if: a. Through each point of each line of R there is a transversal of R. b. Through each point of a transversal of R there is a line of R. The set R' of all transversals of a regulus R is itself a regulus, called the opposite regulus of R. If P has finite order q then any regulus consists of exactly q+1 lines.

10 Existence of Reguli Theorem 2.4.3: Let P be a 3-dimensional projective space over the division ring F. Let be three skew lines of P. Then a. There is at most one regulus containing. b. If F is non-commutative (a skewfield) then there is no regulus in P. c. If F is commutative (a field) then there is exactly one regulus through. Pf: (a) Each point of is on exactly one transversal of. Let R' be the set of these transversals. Any regulus containing, g 2 and must therefore have R' as its set of transversals. Since contains at least three points, there are at least three transversals in R', h 1, h 2 and h 3.

11 Existence of Reguli Theorem 2.4.3: Let P be a 3-dimensional projective space over the division ring F. Let be three skew lines of P. Then a. There is at most one regulus containing. b. If F is non-commutative (a skewfield) then there is no regulus in P. c. If F is commutative (a field) then there is exactly one regulus through. Pf(cont.): (a) Let P be a point on h 1 that is not incident with, g 2 or. Any regulus R containing, g 2, has the property that the line of R through P is necessarily a transversal of h 1, h 2, h 3 through P. Since there can be only one such line, the lines in R are uniquely determined. (b) and (c) follow directly from the 16-point theorem.

12 The Hyperbolic Quadric of PG(3,F) Theorem 2.4.4: Let P be a 3-dimensional projective space over the field F which is represented by homogeneous coordinates. Let = <(1:0:0:0), (0:1:0:0), g 2 = <(0:0:1:0), (0:0:0:1), = <(1:0:1:0), (0:1:0:1) be three skew lines. Then the set Q of points on the uniquely determined regulus R through, g 2, can be described as Q = {(a 0 :a 1 :a 2 :a 3 ) a 0 a 3 = a 1 a 2 with a i F, not all a i = 0}; therefore, the coordinates of the points of Q satisfy the quadratic equation x 0 x 3 - x 1 x 2 = 0.

13 The Hyperbolic Quadric of PG(3,F) Theorem 2.4.4: Let P be a 3-dimensional projective space over the field F which is represented by homogeneous coordinates. Let = <(1:0:0:0), (0:1:0:0), g 2 = <(0:0:1:0), (0:0:0:1), = <(1:0:1:0), (0:1:0:1) be three skew lines. Then the set Q of points on the uniquely determined regulus R through, g 2, can be described as Q = {(a 0 :a 1 :a 2 :a 3 ) a 0 a 3 = a 1 a 2 with a i F, not all a i = 0}; and the coordinates of the points of Q satisfy the quadratic equation x 0 x 3 - x 1 x 2 = 0. Pf: Using the notation in Theorem 2.4.2, let v 1 = (1:0:0:0), v 2 = (0:1:0:0), v 3 = (0:0:1:0) and v 4 = (0:0:0:1). The points of Q have the form + av 3 + bv 2 + abv 4 (from the proof of 2.4.2) for arbitrary choices of a and b in F. In homogeneous coordinates, such a point is (1:a:b:ab) and so satisfies x 0 x 3 - x 1 x 2 = 0.

14 The Hyperbolic Quadric of PG(3,F) Theorem 2.4.4: Let P be a 3-dimensional projective space over the field F which is represented by homogeneous coordinates. Let = <(1:0:0:0), (0:1:0:0), g 2 = <(0:0:1:0), (0:0:0:1), = <(1:0:1:0), (0:1:0:1) be three skew lines. Then the set Q of points on the uniquely determined regulus R through, g 2, can be described as Q = {(a 0 :a 1 :a 2 :a 3 ) a 0 a 3 = a 1 a 2 with a i F, not all a i = 0}; and the coordinates of the points of Q satisfy the quadratic equation x 0 x 3 - x 1 x 2 = 0. Pf(cont.): Now suppose that a point P has coordinates (a 0 :a 1 :a 2 :a 3 ) which satisfy x 0 x 3 - x 1 x 2 = 0. If a 0 0, we can also write the coordinates of P as (a 0 :a 1 :a 2 : a 1 a 2 /a 0 ) = (1:a 1 /a 0 :a 2 /a 0 :(a 1 /a 0 )(a 2 /a 0 )) and so is a point of Q. If a 0 = 0, then either a 1 or a 2 must be 0. W.l.o.g. assume that a 1 = 0. Then P = (0:0:a 2 :a 3 ). If a 2 = 0 then P = <v 4, otherwise P I g 2. So in all cases P in Q.

15 The Hyperbolic Quadric of PG(3,F) Q is called the hyperbolic quadric of PG(3,F). Example: Consider PG(3,4). The field GF(4) = {0,1,a,a+1} where a 2 = a+1. The hyperbolic quadric Q consists of the points: (0:0:0:1) (0:0:1:0) (0:0:1:1) (0:0:1:a) (0:0:1:a 2 ) (0:1:0:0) (1:0:0:0) (1:1:0:0) (1:a:0:0) (1:a 2 :0:0) (0:1:0:1) (1:0:1:0) (1:1:1:1) (1:a:1:a) (1:a 2 :1:a 2 ) (0:1:0:a) (1:0:a:0) (1:1:a:a) (1:a:a:1) (1:a 2 :a:1) (0:1:0:a 2 ) (1:0:a 2 :0) (1:1:a 2 :a 2 ) (1:a:a 2 :1) (1:a 2 :a 2 :a) The points along each row are collinear, and the 5 lines so formed are a regulus. The points in each column are also collinear (check these statements) and these 5 lines form the opposite regulus.