COURSE OF PROJECTIVE GEOMETRY

§ 7: answers

*O27*

a) Each plane through the normal of a plane α is perpendicular to α.

b) Two lines *l* and *m* through the origin in ℜ^{3}are perpendicular if and only if *m* belongs to the plane α whereof *l* is a normal.

c) If *l* is a normal of α and *m* a normal of β, then *l* lies in β if and only if *m* lies in α.

d) If *n* runs through the plane γ, then the plane whereof *n* is a normal runs through the set of planes through the normal of γ.

*O28* Beneath we have drawn three hyperbolic lines through *P* (in the circle model of the hyperbolic plane) that don't intersect *l*.

*O29* First the definitions of sinh and cosh:
sinh(x) = (exp(x)-exp(-x))/2, cosh(x) = (exp(x)+exp(-x))/2.

The points (cosh(t),sinh(t)) run through one half of the hyperbola x^{2}-y^{2}=1, like (cos(t),sin(t)) run through the ellipse (circle) x^{2}+y^{2}=1.

Furthermore, d/dx sinh(x) = cosh(x) and d/dx cosh(x) = sinh(x), like d/dx sin(x) = cos(x) and d/dx cos(x) = -sin(x).

Now let f(x)=cosh(x) (x not negative). Because f ' isn't negative, and f(0)=1, f is at least 1 and a bijection from [0,∞) onto [1,∞). The inverse mapping is called arccosh.

Now we have to show:
|p_{3}q_{3}-p_{1}q_{1}-p_{2}q_{2}| ≥ √(p_{3}^{2}-p_{1}^{2}-p_{2}^{2})
√(q_{3}^{2}-q_{1}^{2}-q_{2}^{2}) and ... = ... ⇔ *p* = *q*.

Define a symmetrical bilinear form by *p*.*q* := p_{3}q_{3}-p_{1}q_{1}-p_{2}q_{2}.

(λ*p*+*q*).(λ*p*+*q*) is positive if λ*p*+*q* is lying inside *k*, 0 if λ*p*+*q* is lying on *k*, and negative if
λ*p*+*q* is lying outside *k*. So if we choose two distinct fixed hyperbolic points *p* and *q*, then
λ^{2}*p*.*p*+2λ*p*.*q*+*q*.*q* has two distinct zeros, and then the discriminant is positive.

*O30* Let *P* = λ(0,0,1) and *Q* = λ(a,0,1). Then d(*P*,*Q*) = arccosh(1/√(1-a^{2})) → ∞ als a ↑ 1.

*O31* Consider *P*^{2} with in it the circle x^{2}+y^{2}=z^{2}. Let *X*=(x_{o},y_{o},z_{o}).

The *polar line * of *P* is the line with equation xx_{o}+yy_{o}=zz_{o}.

We have: *P* is on the polar line of *Q* ⇔ x_{P}x_{Q}+y_{P}y_{Q}=z_{P}z_{Q} ⇔
*Q* is on the polar line of *P*.

Furthermore: if *X* lies on the circle, then the polar line of *X* is tangent to the circle in *X*.

So, in the picture, *P* lies on the polar lines of both intersection points of *l* and the circle, so *l* is the polar line of *P*.

Hence, the equation of *l* is xx_{P}+yy_{P}=zz_{P}.

Now let *n* be a hyperbolic line "through *P*". Then the equation of *n* has the form ax+by=cz where ax_{P}+by_{P}=cz_{P}.

Since *l* has projective coordinates (x_{P},y_{P},-z_{P}), and *n* has projective coordinates (a,b,-c), whilst
cz_{P}-ax_{P}-by_{P}=0, we find *l* ⊥ *m*.