Ohrimenko+ / Barnsley. Superfractals

Figure 2.52 This diagram represents some bundles of lines in three dimensions, through the origin, before and after a linear transformation is applied. It is supposed that the lines are clustered around three orthogonal directions in which the linear transformation rescales space by constant factors, as in Theorem 2.5.4. The red lines on the left lie in a direction that is stretched by the transformation, and the same applies to the black lines. The blue lines lie in a direction that is compressed.

Furthermore, all except one of these lines intersect the line {(x, y) 0 : y = 1}. Accordingly, each line through O in the plane 0 is represented by its point of intersection with {(x, y) 0 : y = 1}. This leaves only the line (1, 0, 0) RP2 as so far unrepresented, and in fact it is represented by the point ∞ R2 L∞. In this way RP2 is represented using R2 R {∞} = R2 L∞. The description of projective transformations, in the resulting new coordinate system, is exactly the one that we gave in the earlier subsection entitled ‘The dance of the points’.

Specifically, the connection between the projective transformations P : RP2 →

for all X R2 L∞. See also Figure 2.53.

z

l = (l1, l2, l3) RP2 l3 ≠ 0

Figure 2.53 Construction of the mapping T : RP2 → R2 L ∞ . The plane 1 represents R2.

E x e r c i s e 2.7.21 Verify that P = T ◦ P ◦ T −1.

E x e r c i s e 2.7.22 Show that if P : R2 L∞→R2 L∞ is a projective transformation such that P(L∞) = L∞ then P|R2 , the restriction of P to R2, is an affine transformation.

E x e r c i s e 2.7.23 Show that if P : R2 L∞→R2 L∞ is such that it maps two distinct points on L∞ to two distinct points on L∞ then P(L∞) = L∞.

E x e r c i s e 2.7.24 Show that a set S R2 L∞ is a straight line iff the set of straight lines defined by the set of points T −1(S) RP2 defines a plane in R3. If S is the straight line in R2 L∞ defined by lx + my + n = 0 with n =0, what is the equation for the set of points in R3 that lie in the plane defined by T −1(S)?

E x e r c i s e 2.7.25 (i) Let L denote the set of straight lines in X = R2 L∞. Show that we can define an invertible mapping Z : X → L by

Z (X ) = T {l RP2 : l R3; l T −1(X )} ,

for all X X, where means ‘is perpendicular to’ and T −1(X ) is treated as a line in R3. Describe the mapping Z −1 : L → X.

p X then Z (l1), Z (l2) X are two distinct points that lie on the line Z ( p) L;

(b) if p1, p2 X, with p1 = p2, lie on the line l then Z ( p1), Z ( p2) L are two distinct lines that intersect at the point Z (l) X.

The mapping Z constructed in Exercise 2.7.25 is an example of a duality transformation. It can sometimes be used to convert the objects in a theorem that concerns straight lines, points and intersections in the projective plane into new objects, thereby yielding a new theorem.

Representation of the projective plane on a sphere and on a disk

Another way of representing R2 L∞, or equivalently RP2, that reveals a natural topology for the projective plane is to describe each point l RP2 as the pair of points for which the corresponding line l R3 intersects the surface S of the sphere

2

of radius 1 having its centre at O. In particular, a natural metric dRP2 on RP is obtained by defining dRP2 (l, l ) to be the shortest distance, on S, between the pair

of points that represents l and and the pair of points that represents l . The natural

topology of (RP2, dRP2 ) is the identification topology on (S, deuclidean ) induced the centre to the same

by mapping pairs of points lying on the same line through point. Clearly (RP2, dRP2 ) is a compact metric space.

The behaviour of a projective transformation P : R2 L∞ → R2 L∞ may

be thought of in terms of the action, on the sphere S, of the corresponding linear transformation P : R3 → R3, as illustrated in Figure 2.54. It is clear, from this point of view, that any projective transformation is continuous with respect to the

metric dRP2 .

In place of using two points on the spherical shell S to represent a single point of RP2 we can use just one of the points, say the one on the upper hemisphere. The only slight difficulty is that points on the equator, that is, on the intersection of S with the plane z = 0, are double points. So we must omit exactly half this circle. Then we see that we can represent RP2 by the set of points

S+ = {(x, y, z) : x2 + y2 + z2 = 1, z > 0}

{(x, y, 0) : x2 + y2 = 1, y > 0} {(1, 0, 0)}. Notice that we can define unique coordinates for S+ by using the points of

D+ := {(x, y) R2 : (x, y, z) S+ for some z R3}.

The set D+ is just the orthogonal projection of S+ onto the x y-plane; it consists of the interior of the unit circle centred at O together with half the unit circle. The corresponding invertible mapping T : R2 L∞ → D+ is illustrated in

Figure 2.54 The action of a projective transformation P on a picture P may be expressed in terms of how the corresponding linear transformation P : R3 → R3 acts on the sphere S. The plane 1 defined by z = 1 represents the space R2 in which P lies. First P is transformed into two pictures T (P) and T (P) by central projection onto S. Then a linear transformation P : R3 → R3 is applied to the sphere and the two pictures on it, to yield two pictures on an ellipsoid. Finally, either of these pictures is projected back onto 1. It is always possible to choose the linear transformation P in such a way that the final result, back on 1, is P(P).

Figure 2.55 Construction of the invertible mapping T : R2 L ∞ → D+ described in the text, in Equations (2.7.9)–(2.7.11). Where are T (L ∞ ) and T (∞)?

for all (x, y) D+ such that x2 + y2 < 1. Calculate T −1 35 , 45 .

In Figure 2.56 we show the result of applying the transformation T : R2 L∞ → D+, defined in Equations (2.7.9)–(2.7.11), to pictures of periodic tilings of R2 by square tiles, where each tile has a white border and a black square in the middle. The sides of the tiles and of the black squares run parallel to the coordinate axes. Images of the straight lines formed by the boundaries of the tiles that are parallel to the x-axis seem to meet at a single point on D+. This meeting point is actually T (∞). Similarly, images of lines parallel to the y-axis meet at T (0). Remember that the disk that represents D+ possesses only half its circular boundary.

Figure 2.56 Here three regular arrays of square tiles, with their sides parallel to the coordinate axes, have been transformed by T : R2 → D+ to produce three pictures. The different pictures correspond, from left to right, to successively larger tiles. Why do the images of parallel lines of tiles, in the plane, seem to converge to the same meeting point on D+ ?

We readily find that T : R2 → D+ maps the straight line y = c, c R, into the ellipse

x2 + 1 + 1 y2 = 1. c2

This family of ellipses meets at the point (1, 0) D+. You should be able to spot illustrations of parts of this family of ellipses in Figure 2.56.

E x e r c i s e 2.7.27 Show that T : R2 L∞ → D+ maps the straight line given by

lx + my + n = 0

into the conic section

(l2 + n2)x2 + 2lmx y + (m2 + n2)y2 − n2 = 0.

Make a sketch of some of these ellipses for l and m fixed and several values of n.

In the left-hand panel of Figure 2.57 a regular array of pixels has been mapped by T onto D+. The right-hand panel shows the result of applying T ◦ L to the same array, where L is the linear transformation L(x, y) = (2x, 2y). Notice how the line at infinity, represented by the boundary of the disk, remains fixed, the major axes of certain families of ellipses point to the same places and the picture material is squeezed out towards the boundary.

In place of looking at how a projective transformation P : R2 L∞ → R2 L∞ acts upon a picture P that has its domain in R2 we can instead look at how

the conjugate transformation P : D+ → D+ defined by

Figure 2.57 The left-hand panel shows the result of mapping an array of pixels onto D+ . The right-hand panel shows the result of the same mapping after the dimensions of the domains of the pixels have been doubled.

acts upon a picture whose domain lies in D+. Specifically we find, for all (x, y) D+, that

P(x, y) =

(ax + by + eF(x, y)) sgn(gx + hy + j F(x, y))

(ax +by +eF(x, y))2 +(cx +dy + f F(x, y))2 +(gx +hy + j F(x, y))2 1/2 ,

(cx + dy + eF(x, y)) sgn(gx + hy + j F(x, y))

(ax +by +eF(x, y))2 +(cx +dy + f F(x, y))2 +(gx +hy + j F(x, y))2 1/2 ,

where

+

F(x, y) = 1 − x2 − y2,

the underlying linear transformation is that given in Equation (2.7.2) and the function sgn is defined by

sgn(x) =

+1 when x ≥ 0, −1 when x < 0.

Examples of the transformation P : D+ → D+, applied directly to pictures

with domain D+, are illustrated in Figures 2.58–2.60. In each case the original picture is on the left, the transformed picture is on the right and the underlying linear transformation is the same, namely

Figure 2.58 The picture P on the left represents, on D+ , a regular array of pixels. The picture on

the disk which represents D+ , but in the right-hand panel they are squeezed towards a smooth curve lying mainly in the interior of the disk.

Figure 2.59 A transformation P : D+ → D+ , conjugate to a projective transformation acting on R2

L ∞ , is applied to a picture whose domain is D+ . Notice the lovely stretching lines and how the part of the picture on the boundary of D+ is mapped to two sides of an internal smooth curve.

We can see how a projective transformation acts on a picture P, espe-

cially in relation to L∞, by comparing the pictures P, T (P), P(T (P)) and

the relationship of T (P), P(T (P)) and the boundary D+.

Figure 2.60 Here a projective transformation, represented as acting directly on D+ , is applied to a picture, on the left, of a texture of foliage and branches. The result, on the right, is a very different looking kind of texture.

For example, in Figure 2.61 we illustrate the effect of the projective transformation associated with the linear transformation

acting on a picture P, in the top left panel, of nine cartoon trees. The tree at the centre is located in the vicinity of the origin. The top right panel shows the picture T (P); notice how the central tree is not much deformed but the other

trees are squeezed against the boundary of D+. The picture P(T (P)) is shown

in the bottom right panel; the effect of P has been to reflect the picture T (P)

about a horizontal line and then to displace the result, so that it seems to have slid off the disk across the top boundary of D+ and to have reappeared, with the orientation reversed, from across the bottom boundary. The bottom left panel

the picture and slid from L∞ back into view, with reversed orientation, at the bottom.

Figures 2.62 and 2.63 illustrate exactly the same sequence of transformations, but applied to different pictures. Each picture emphasizes different aspects of the same transformations. For example, notice how in Figure 2.62 the fish picture T (P) seems to nearly fill D+, while the flowers in the picture P in Figure 2.63 are transformed by P to be closer together, no longer separated by the birds.

Figure 2.61 A picture P, top left, is mapped onto a disk, D+ , at the top right. Then a transformation of the form given in Equation (2.7.12) is applied to produce the picture at the bottom right. This picture is mapped back onto the euclidean plane, yielding, at bottom left, a projective transformation P(P) of the original picture P. Notice how the pictures on the right look something like the pictures on the left wrapped and stretched over a hemispherical shell – you can almost see the convexity of the hemisphere. The line at infinity corresponds, in the pictures on the right, to the boundaries of the disks.

The cross-ratio

Projective transformations do not in general preserve lengths, ratios of lengths or angles. But they always preserve cross-ratios.

D e f i n i t i o n 2.7.28 The cross-ratio of a sequence of four distinct collinear points A, B, C, D R2 L∞ is a unique real number, which may be computed as follows. If A, B, C, D R2 then write

A = a · (e1, e2) + (t1, t2), B = b · (e1, e2) + (t1, t2),

C = c · (e1, e2) + (t1, t2), D = d · (e1, e2) + (t1, t2),

where a, b, c, d R, (e1, e2) R2, (t1, t2) R2 and e12 + e22 = 1;