COURSE OF PROJECTIVE GEOMETRY

§ 6: answers

*O22* Let *l* and *m* be lines in *P ^{2}*, corresponding with planes α and β through

If we take a projective transformation φ, which comes from the regular linear transformation *A*, then φ(*l*) and φ(*m*) are parallel if and only if
*A*(*s*)⊂γ, so if and only if *A* is affine as well.

Indeed, *A*(α)∩*A*(β) = *A*(α∩β) = *A*(*s*), and φ(*l*) and φ(*m*) are parallel if and only if *A*(α)∩*A*(β)
lies in γ.

*O23* The angle θ between two lines *l* and *m* in x_{3}=1 is invariant under the similarity stransformation that comes about if we first project perpendicular to
x_{3}=0, then apply in x_{3}=0 a rotation or reflection, multiplied with a scalar ρ, and finally translate over (a,b,1).

Also, if ρ=1, then the distance between two points
in x_{3}=1 is preserved under the composition of these three mappings.

*O24*

Complete quadrilateral: *P*, *A*, *G*, *C*; parallellogram: *A*, *G*, *C*; rhomb and square: *G*, *C*.

For example, take the points (0,0,1), (1,0,1), (0,1,1) and (1,1,1) in x_{3}=1, the vertices of a square.

Let *A* be the regular linear transformation of ℜ^{3} with matrix rows ((1,1,1),(0,0,1),(1,0,1)).

Then under *A*, (0,0,1) → (1,1,1), so φ(0,0,1)=(1,1,1);
likewise (1,0,1) → (2,1,2), so φ(1,0,1)=(1, 1/2, 1); likewise (0,1,1) → (2,1,1), so φ(0,1,1)=(2,1,1); likewise (1,1,1) → (3,1,2), so φ(1,1,1)=(3/2, 1/2, 1).

So the image under φ of this square isn't even a parallellogram.

Apparently, the concepts square, rhomb and parallellogram are not projective.

*O25* No. Definer a hyperbola as a conic that intersects the line at infinity L_{∞} in two points. Example: *H* with inhomogeneous equation xy=1, homogeneous equation
xy=z^{2}; the points at infinity *S*:λ(1,0,0) en *T*:λ(0,1,0) are lying on it.

Now let φ be an affine transformation, which comes from the regular linear transformation *A* of ℜ^{3}.

Then {*A*(*S*), *A*(*T*)} = *A*(*H*∩L_{∞}) = *A*(*H*)∩*A*(L_{∞}) = *A*(*H*)∩L_{∞}, so
*A*(*H*) intersects L_{∞} in two points, too. We see that an affine transformation maps each hyperbola onto a hyperbola.

So a hyperbool and an ellipse can't ever be *A*-equal.

*O26* (*A*,*B*;*C*,L_{∞}) = *A**C*/*B**C*, so:

i) (*A*,*B*;*X*,L_{∞}) smaller than 0;

ii) (*A*,*B*;*C*,L_{∞}) ∈ {4,-2};

iii) (*A*,*B*;*C*,L_{∞}) = -1.