y = (ax + b)/(cx + d), with non zero determinant ad-bc (*).

Equivalently.

cxy-ax+dy-b=0 (**).

Hence the graph of the function is a rectangular hyperbola. This is the only kind of quadratic curve that represents an invertible real function of the extended line to itself.

By

The main properties of a homographic relation are.

[1] They preserve the cross ratio (x

This means that the corresponding cross ratio (y

[2] They are distinguished in

[3] Involutive/non-involutive homographic relations are completely determined by prescribing arbitrarily the values to two/three arbitrary points.

[4] If a homographic relation f interchanges two points, i.e. there are (x,y) such that f(x)=y and f(y)=x, then f is involutive.

1] The first is an easy calculation.

2] Non-involutive homographies are determined by their values at three distinguished points.

This follows from ([1]) the equality of cross ratios (x

3+4] The involutive property amounts to the symmetry of the hyperbola about the first diagonal (line y=x).

This in turn is equivalent with the condition.

a+d = 0.

This is equivalent also to the lying of the center O of the hyperbola on the first diagonal.

Involutive homographies are determined by their values at two points. This is seen most easily from (**). The homographic relation obtains in this case the form.

Axy+B(x+y)+C=0 (***).

This determines the coefficients {A,B,C} (up to a multiple) uniquely by giving two pairs (x

General examples of homographies can be easily constructed by defining one-to-one maps y = F(x) of the points of a line L to the points of a line L' involving geometric constructions with intersection points.

a) of two lines.

b) of a line and a conic.

Such constructions reduce to algebraic relations of (x,y), thus defining homographic relations.

An example of a homographic relation defined geometrically is given in RectHypeRelation.html .

Assume F is an involution and there are two different, real or imaginary fixed points x

Another model of the projective line is a

If the line corresponding to m'=F(m) is orthogonal to the one corresponding to m, m' must be equal to -1/m. Thus to locate pairs (m,m') which are orthogonal we must solve the equation -(1/m)=(am+b)/(cm-a), which is equivalent to am

[1] If a is non-zero, setting (b+c)/a = 2d, we arive at equation m

[2] If a is zero, then the equation above becomes (b+c)m=0 i.e. b+c=0, and F(m)=b/(cm)=-1/m. Thus, in this case we have the special involution

The result of this short discussion is: An involution on a pencil of lines either has exactly one pair (m,m') of orthogonal lines or all pairs (m,m') are orthogonal.

D = (a+d)

- if D>0, then there are two real fixed points and the relation is called

- if D<0, then there are two imaginary fixed points and the relation is called

- if D=0, then there are two real coincident fixed points and the relation is called

Denoting by {x

This cross ratio is easily calculated.

Last expression is constant, equal to k say. Thus,

the general homographic relation is characterized by this constant k and the property (x

The particular case of

Inversely, the previous equation, solving for x', implies the homographic relation:

Note that for x = (1/(1-k))(x

In the special case of involutions equation (x

Selecting the middle O with line-coordinate (x

x' = r

This means that the homography coincides with the restriction of an inversion I(O,r) on its supporting line.

From this follows that points (x,x') are intersection points of the supporting line with the members of a circle-bundle. The circle bundle being hyperbolic (ad-bc<0), elliptic (ad-bc>0) or parabolic (ad-bc=0), which is in accordance with the naming conventions at the beginning of the paragraph.

Chasles_Steiner_Envelope.html

FourPtsAndTangent.html

GoodParametrization.html

HomographicRelationExample.html

Involution.html

Line_Homography.html

ProjectiveLine.html

RectHypeRelation.html

Schwerdtfeger, Hans

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