f(t) = (at+b)/(ct+d).

Here P

Inversely, given the two lines {AX, AY}, the fixed point O, and the Moebius function f(t)=(at+b)/(ct+d), introduce a coordinate t on AX and for each point P

To prove this take the coordinates-origin at O, and consider the (variable) vector u = OB

sOB

From there, denoting by J(Z) the pi/2-rotated of Z and by (..., ...) the inner product we get.

s(OB

Analogously s' = (A,Ju

If L is described parametrically by B

s = 1/(mt+n) and s' = 1/(m't+n').

Since the cross-ratio is (P

To prove the inverse consider a fixed line M through O and take the trace of the line-bundle at A.

A(X,Y,B

B

Taking inner products with J(m) we get.

(A,J(m)) = r(P

Taking a parametrization along A: P

Hence B

Putting A' = A/b, u'

This shows that B

This is essentially the first part of the previous exercise. But another argument results from the reduction of the cross ratio of bundles to ratios of sines (see Harmonic_Bundle.html ). (A,C,B,D) = (sin(v)/sin(v')):(sin(w)/sin(w')), where v = angle(OA,OB), v'=angle(OA,OD), w=angle(OC,OB), w'=angle(OC,OD).

In fact, by the assumed constancy of angles the second ratio sin(w)/sin(w') is constant.

Since the cross ratio is the same for all secants of the line bundle at O.

O(A,B,C,D), we can replace L with a line L' orthogonal to the fixed OA at A. Then the sin(v) is expressed through t = AB' (B' the trace of OB on L') through its tangent which is tan(v) = t/OA = t/a (putting a = |OA|). This implies sin

Besides v'= v+w+w' = v+w'', where w''=w+w' implies sin(v') = sin(v)cos(w'')+cos(v)sin(w'').

Thus the ratio (sin(v)/sin(v')) has the form t/(tcos(w'')+asin(w'')) and this proves the claim.

Define the additional intersection points as shown in the figure. Introduce also a line coordinate t = AB along line AX.

By section-2 it suffices to show that the cross ratio f(t) = (O,D,E,F) is a Moebius function. Considering the bundle of lines at C: C(O,D,E,F) and taking its tracce on line AX this ratio transfers to (I,B,A,F), which by section-3 has the desired form of a Moebius function of t. Hence D moves on a line.

Harmonic_Bundle.html

Harmonic.html

HomographicRelation.html

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