PROJECTIVE GEOMETRY COURSE

§ 21: *Introduction starting from analytic geometry*

Let *l* and *m* be two lines in ℜ^{3} through __0__ , with direction vectors __l__ en __m__.

Let *A* be a regular linear transformation of ℜ^{3} with *A*(__l__) = __m__ and *A*(__m__) = __n__ (__n__ not in the plane
spanned by __l__ and __m__).

On the base __l__, __m__, __n__, the matrix has the form

In coordinates with respect to __l__, __m__, __n__, an arbitrary plane through *l* has equation β y' + γ z'= 0 and vector representation u(1,0,0) + v(0,-γ,β).
*A* maps this plane onto the plane through *m* with vector representation u(0,1,0) + v(ρβ, σβ, -γ+τβ) and equation
(γ-τβ)x'+ ρβz'= 0.

We find the geometric locus of the intersection lines of the original plane and its image plane by eliminating β and γ. We get (see O81): x'y' + τx'z' = ρ(z')^{2}.

Now if we take coordinates (x,y,z) with respect to the natural base, via the formula

we find (ax+by+cz)(dx+ey+fz) + τ(ax+by+cz)(gx+hy+iz) = ρ(gx+hy+iz)^{2}.

This is the equation of an arbitrary conic.

This justifies the following projective definition of a conic: a conic is the set of intersection points *x*.φ(*x*), where φ is a projective transformation of
*P*^{2} and *x* runs through a pencil of lines.

Notice that if __n__ does lie in the plane of __l__ and __m__, we find three points *L, M, N* in *P*^{2} with φ_{A}(*L*) = *M*,
φ_{A}(*M*) = *N*, whilst *L*, *M*, *N* are collinear.
According to the dual of O40, φ_{A|L} is then a perspectivity, so the conic degenerates and becomes two lines (*LM* and the axis of the perspectivity).
We also get degenerations if *L* = *M* (determine the kind of degeneration; distinguish between φ = id, φ hyperbolic, elliptic or parabolic).

*Proposition*: The conic k: x^{t}Ax = 0 is degenerated if and only if det(A) = 0.

*Proof*:

1. Suppose k is degenerated.

If k is one point, the equation has the form (px+qy+rz)^{2} + (ux+vy+wz)^{2} = 0. Check that in this case det(A) = 0.

Otherwise, there exist two distinct points *P* and *Q* such that the whole line *PQ* is lying on k. Then (p+aq)^{t}A(p+aq) = 0 for all a, so p^{t}Ap = p^{t}Aq =
q^{t}Ap = q^{t}Aq = 0. Then the points λAp and λAq are both lying on the lines with projective coordinates λp and λq respectively, so Ap = μAq.
Then A(p-μq) = 0 whilst p ≠ μq, so det(A)=0.

2. Suppose det(A)=0.

Then Im A is a plane through O or a straight line through O. The vectors Ap with p on k are then lying in one plane. But if k is non-degenerated, Ap is normal vector of the plane through O and the tangent
to k in *P*. So k must be degenerated.

*Remark* : There exist empty conics as well, like x^{2}+y^{2}+z^{2} = 0. For the last one we have A=E_{3}, so det(A) ≠ 0; we call it non-degenerated.

*O81* Eliminate β and γ from β y' + γ z' = (γ - τ β) x'+ ρβ z' = 0 (β, γ ≠ 0).

*O82* Prove, starting from a_{1 1}x_{1}^{2} + a_{2 2}x_{2}^{2} + a_{3 3}x_{3}^{2} +
2 a_{1 2}x_{1}x_{2} + 2 a_{1 3}x_{1}x_{3} + 2 a_{2 3}x_{2}x_{3} = 0, that through five points there goes in general
exactly one conic (if these five points are freely situated, there is exactly one non-degenerated conic that goes through them).