Projective Geometry - Lecture Notes Chapter 2

2.4 The hyperbolic quadric of PG(3,F)

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 g1 and g2 be two skew lines, and denote by P a point on neither line. Then there is at most one transversal of g1 and g2 through P. If P is 3-dimensional then there is exactly one transversal of g1 and g2 through P.

Theorem 2.4.2: Let P be a 3-dimensional projective space over the division ring F. Let {g1,g2,g3} and {h1,h2,h3} be sets of skew lines with the property that each line gi meets each line hj. Then F is commutative (hence a field) if and only if each transversal g of {h1,h2,h3} meets each transversal h of {g1,g2,g3}.

Def: Let P be a 3-dimensional projective space. A non-empty set R of skew lines of P is called a regulus if:

  1. Through each point of each line of R there is a transversal of R.
  2. 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.

Theorem 2.4.3: Let P be a 3-dimensional projective space over the division ring F. Let g1,g2,g3 be three skew lines of P. Then

  1. There is at most one regulus containing g1,g2,g3.
  2. If F is non-commutative (a skewfield) then there is no regulus in P.
  3. If F is commutative (a field) then there is exactly one regulus through g1,g2,g3.
Theorem 2.4.4: Let P be a 3-dimensional projective space over the field F which is represented by homogeneous coordinates. Let
g1 = <(1:0:0:0), (0:1:0:0)>,
g2 = <(0:0:1:0), (0:0:0:1)>,
g3 = <(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 g1,g2,g3 can be described as
Q = {(a0:a1:a2:a3) | a0a3 = a1a2 with ai in F, not all ai = 0};
therefore, the coordinates of the points of Q satisfy the quadratic equation
x0x3 - x1x2 = 0.

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 a2 = a+1. The hyperbolic quadric Q consists of the points
(0:0:0:1)
(0:1:0:0)
(0:1:0:1)
(0:1:0:a)
(0:1:0:a2)
(0:0:1:0)
(1:0:0:0)
(1:0:1:0)
(1:0:a:0)
(1:0:a2:0)
(0:0:1:1)
(1:1:0:0)
(1:1:1:1)
(1:1:a:a)
(1:1:a2:a2)
(0:0:1:a)
(1:a:0:0)
(1:a:1:a)
(1:a:a:1)
(1:a:a2:1)
(0:0:1:a2)
(1:a2:0:0)
(1:a2:1:a2)
(1:a2:a:1)
(1:a2:a2: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.