## Projective Geometry - Lecture Notes Chapter 4

### 4.6 The Klein Quadratic Set

Def:A hyperbolic quadratic set of a 5-dimensional projective space is also called a Klein quadratic set.

We define a new geometry based on a Klein quadratic set.

Let C and C* be the two equivalence classes of planes of a Klein quadratic set Q in a 5-dimensional projective space P. We define the geometry S as follows:

1. the points of S are the planes of C.
2. the lines of S are the points of Q.
3. the planes of S are the planes of C*.
4. the incidence between a line of S and either a point or plane of S is induced by the incidence in P.
5. a point 1 of S is incident with a plane 2 of S iff the planes 1 and 2 of P are not disjoint (then by 4.5.5 they intersect each other in a line of P).

Lemma 4.6.1: Let Q be a Klein quadratic set.

1. Each Q-line is on exactly one plane of each equivalence class.
2. If P is finite of order q, then each point of Q is on exactly q+1 planes of each equivalence class.

Theorem 4.6.2: The geometry S is a 3-dimensional projective space; more precisely, S is isomorphic to a 3-dimensional subspace of P.

Def: Let V be a vector space over a field F. A map q: V -> F is called a quadratic form of V if
1. q(av) = a2q(v) for all v in V and a in F.
2. the map B: V × V -> F defined by
B(v,w) = q(v+w) - q(v) - q(w)
is a symmetric bilinear form.

Examples:

Consider the 3 dimensional vector space V over the reals.

q(x,y,z) = x2 + y2 + z2
q(ax,ay,az) = (ax)2 + (ay)2 + (az)2 = a2(x2 + y2 + z2) = a2q(x,y,z).
With v = (x,y,z) and w = (x',y',z') we have
B(v,w) = (x+x')2 + (y+y')2 + (z+z')2 - (x2+y2+z2) - (x'2 + y'2 + z'2)
= 2xx' + 2yy' + 2zz'.
Clearly, B(v,w) = B(w,v) so this form is symmetric.
Let v' = (a,b,c), then
B(v+v',w) = 2(x+a)x' + 2(y+b)y' + 2(z+c)z'
= 2xx' + 2yy' + 2zz' + 2ax' + 2by' + 2cz' = B(v,w) + B(v',w).
B(dv,w) = 2dxx' + 2dyy' + 2dzz' = d B(v,w).
Thus B is linear in the first coordinate, and by symmetry is bilinear.

Consider the 3 dimensional vector space V over the reals.

q(x,y,z) = xy - z2
q(ax,ay,az) = (ax)(ay) - (az)2 = a2(xy - z2) = a2q(x,y,z).
With v = (x,y,z) and w = (x',y',z') we have
B(v,w) = (x+x')(y+y') - (z+z')2 – (xy - z2) – (x'y' - z'2)
= xy' + x'y - 2zz'.
Clearly, B(v,w) = B(w,v) so this form is symmetric.
Let v' = (a,b,c), then
B(v+v',w) = (x+a)y' + x'(y+b) – 2(z+c)z'
= xy' + x'y -2zz' + ay' + x'b – 2cz' = B(v,w) + B(v',w).
B(dv,w) = dxy' + x'dy – 2dzz' = d B(v,w).
Thus B is linear in the first coordinate, and by symmetry is bilinear.

Lemma 4.7.1: Let {v1,...,vn} be a basis of the vector space V.

1. If aij in F, then a quadratic form is defined by the following rule:
q( bi vi ) = aij bi bj.
2. Conversely, for any quadratic form q there are elements aij in F such that for all v in V we have the above form.

Def: A quadratic form is nondegenerate if q(v) = 0 and B(x,v) = 0 for all x in V implies that v = O.

Def: Let q be a quadratic form of the vector space V. The quadric of the projective space P(V) corresponding to q is the set of all points <v> of P(V) with q(v) = 0.

Examples:

By Theorem 2.4.4 the points on a regulus in a 3-dimensional projective space can be given coordinates that satisfy,

x0x3 – x1x2 = 0.
Since x0x3 – x1x2 is a quadratic form, these points lie on a quadric (the hyperbolic quadric).

The quadratic form x02 – x12 - x22 yields a quadric in the plane which is an oval. The quadratic form x02 – x12 + x22 yields a quadric in the plane which is the union of two lines.

In a 3-dimensional projective space, the quadric given by x02 - x12 – x22 – x32 is an ovoid, while the quadric given by x02 + x12 - x22 - x32 is a hyperboloid.

Lemma 4.7.2: Let q be a quadratic form of a vector space V, and let Q be the corresponding quadric in P(V). Then, if a line g contains three points of Q, each point of g lies in Q.

Def: Let q be a quadratic form of the vector space V. For a non-zero vector v in V we define

<v>perp = {x in V | B(x,v) = 0 }.

Lemma 4.7.3: Let q be a quadratic form of the vector space V, and let Q be the corresponding quadric of P(V). Then for each non-zero vector v in V we have:

1. <v>perp is a subspace of V, hence also a subspace of P(V).
2. <v>perp is either a hyperplane or equal to V.
3. Suppose that <v> in Q. Then each line of <v>perp through <v> is a tangent line of Q.
4. Suppose that <v> in Q. Then each line through <v> that is not contained in <v>perp intersects Q in precisely one further point.