The projectification of the euclidean plane extends it to the projective plane by adding a new line. The euclidean plane E is identified with the plane x3 = 1, of R3. Every line [x1,x2,x3] of R3 passing through the origin and not being parallel to that plane (i.e. not having x3 = 0) intersects E on a point (x1, x2, 1). We identify (x1,x2, 1) with this element [x1,x2,x3] of the projective plane PR2.
This establishes an identification of the euclidean plane with a part E' of the projective plane. The complement F' = PR2 - E' consists of all points of the form [x1,x2,0]. This set is a projective line of PR2. The projectification adds to E a set of points F which is isomorphic to line F'. F is called the line at infinity and consists of points at infinity which correspond to points of the projective plane of the form [x1,x2,0].
The correspondence is given by means of directions of parallel lines of E. We consider the class of all lines parallel to a non zero vector (v1,v2) as the point at infinity identified with the element [v1,v2,0] of line F'.
Creating this model of the projective plane we have the benefit of embedding all familiar shapes of the euclidean plane into the projective plane. Then we can use the tools of the projective plane to study properties of coincidence of lines and points or even more general intersection questions, to which the projective plane is better suited than the euclidean plane.
The reason for the later is that all pairs of distinct lines in the projective plane intersect at a point. In the projectification model the intersection point coincides with an ordinary point, if the lines intersect inside the euclidean plane. If they don't, then their image in the euclidean plane is that of two parallel lines, thus defining the same point at infinity, which is then considered as their intersection. An ordinary line L and the line at infinity intersect always at the point at infinity defined by the direction of L.
The equations defining curves of the euclidean plane transfer to equations of the projective plane by replacing the coordinates x1, x2, correspondingly with x1/x3, x2/x3. Thus, line ax1 + bx2 + c = 0 ==> a(x1/x3) + b(x2/x3) + c = 0 ==> ax1 + bx2 + cx3 = 0. Similarly a conic represented by equation ax12 + bx22 + cx1x2 + dx1 + ex2 + f = 0 ==> a(x1/x3)2 + b(x2/x3)2 + c(x1/x3)(x2/x3) + d(x1/x3) + e(x2/x3) + f = 0. ==> ax12 + bx22 + cx1x2 + dx1x3 + ex2x3 + fx32 = 0. The inverse procedure consists of setting x3 = 1 to an equation p(x1,x2,x3) = 0 of the projective plane, to obtain the corresponding equation p(x1,x2, 1) = 0 of the euclidean plane.
Applying a projective transformation to points of the projectification is of particular interest. Ordinary points (x,y) must extend to (x,y,1). Points at infinity though are also represented by pairs (x,y), considered now as defining directions of lines (therefore in this case (x,y) has to be non-zero). In that case the point is extended to (x,y,0). A projective transformation, represented by an invertible matrix A gives then the image point through matrix multiplication:
If (x,y) is meant to define a point at infinity. The image point (x'/z', y'/z') is meant to be an ordinary point. In the case z'=0, A maps the point at infinity (x,y) to (x',y'), considered also as a point at infinity (identified with [x',y',0]).
Exercise Consider a projectivity F, defined by a matrix A, and for each point (x,y) construct the two points A and B, corresponding to the above discussion. Show that line AB passes from a fixed point (namely C = F([0,0,1])).
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Projectification of the euclidean plane ![[0_0]](Projectification_files/Projectification_0_0_0.png)
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