COURSE OF PROJECTIVE GEOMETRY

§ 17:

*O65*

The intersection point *x*.*a* is a fixed point, so it also lies on *x ' *.

For any line *l* we choose through *C* holds: if *X* = *x*. *l*, then *X ' * = *x ' *. *l*. See *O14* for the rest.

*O66*

Let *A* be a regular linear transformation of ℜ^{3} inducing φ, with *A*(0,0,1) = (0,0,1).

Then from *A*(1,0,0) = (ρ,0,0), *A*(0,1,0) = (0,σ,0) and *A*(1,1,0) = (τ,τ,0) we find ρ = σ = τ.

Furthermore, φ maps the point (1,0,1) to the fourth harmonic with {(0,0,1),(1,0,0)} and (1,0,1); this is (-1,0,1).

So *A*(1,0,1) = ω(-1,0,1). We also have *A*(1,0,1) = *A*(1,0,0) + *A*(0,0,1) = (ρ,0,1).
So ω = 1 and ρ = -1.

Then the matrix becomes ((-1,0,0),(0,-1,0),(0,0,1)), so φ is the reflection in the point (0,0,1) in {x_{3} = 1}.

*O67*

Require for three points on λ(1,1,1) they map onto themselves, and require for two points λ(1,1,1) that the image point lies on the line through that point and through
λ(0,1,-1).
It follows that the matrix has the form ((p,0,0),(q,p+q,q),(-q,-q,p-q)).

*O68*

Check that the corresponding linear transformations have a twodimensional eigenspace and a onedimensional eigenspace.

Axis λ(-1,9,3), centre λ(-11,-2,1).

*O69*

Axis λ(1+√2,0,-1), centre λ(1,0,1-√2).

Now check for one point *X* that {*X*,*X ' * } separates the pair {*C*,*S _{X}*} harmonically.

Alternative: determine the square of the matrix and show that this is a multiple of the identity matrix.

*O70*

Start from a linear transformation *A* with *A*(0,0,1) = (β,γ,1).

If α=1 and (β,γ) ≠ (0,0), then there is no fixed point outside x_{3}=0.

The line through (0,0,1) and (β,γ,1) intersects x_{3}=0 in the centre (β,γ,0).