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 g_{1} and g_{2} be two skew lines, and denote by P a point on neither line. Then there is at most one transversal of g_{1} and g_{2} through P. If P is 3dimensional then there is exactly one transversal of g_{1} and g_{2} through P.
Theorem 2.4.2: Let P be a 3dimensional projective space over the division ring F. Let {g_{1},g_{2},g_{3}} and {h_{1},h_{2},h_{3}} 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},h_{2},h_{3}} meets each transversal h of {g_{1},g_{2},g_{3}}.
Def: Let P be a 3dimensional projective space. A nonempty set R of skew lines of P is called a regulus if:
 Through each point of each line of R there is a transversal of R.
 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 3dimensional projective space over the division ring F. Let g_{1},g_{2},g_{3} be three skew lines of P. Then
 There is at most one regulus containing g_{1},g_{2},g_{3}.
 If F is noncommutative (a skewfield) then there is no regulus in P.
 If F is commutative (a field) then there is exactly one regulus through g_{1},g_{2},g_{3}.
Theorem 2.4.4: Let P be a 3dimensional projective space over the field F which is represented by homogeneous coordinates. Let
g_{1} = <(1:0:0:0), (0:1:0:0)>,
g_{2} = <(0:0:1:0), (0:0:0:1)>,
g_{3} = <(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_{1},g_{2},g_{3} can be described as
Q = {(a_{0}:a_{1}:a_{2}:a_{3})  a_{0}a_{3} = a_{1}a_{2} with a_{i} in 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.
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)
(0:1:0:0) (0:1:0:1) (0:1:0:a) (0:1:0:a^{2})
 (0:0:1:0) (1:0:0:0) (1:0:1:0) (1:0:a:0) (1:0:a^{2}:0)
 (0:0:1:1) (1:1:0:0) (1:1:1:1) (1:1:a:a) (1:1:a^{2}:a^{2})
 (0:0:1:a) (1:a:0:0) (1:a:1:a) (1:a:a:1) (1:a:a^{2}:1)
 (0:0:1:a^{2}) (1:a^{2}:0:0) (1:a^{2}:1:a^{2}) (1:a^{2}:a: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.