Isometries of the plane

Transkript

1 Isometries of the plane Mikael Forsberg August 23, 2011 Abstract Här följer del av ett dokument om Tesselering som jag skrivit för en annan kurs. Denna del handlar om isometrier och innehåller bevis för att en isometri måste vara en sammansättning av en ortogonal avbildning (ortogonal matris) och en translation. (parallellförflytting). I denna kurs så behöver man inte gå så här djupt utan detta dokument kan lugnt hoppas över. Den som strävar efter djupare förståelse kan dock ha viss glädje av dokumentets innehåll... 1

2 1 Introduction to isometry We begin by defining the concept of Isometry: Definition 1.1. By an Isometry of a subset of the euclidean space R n we mean a map I : R n R n which preserves length, that is x y = I(x) I(, for every pair of vectors x and y in R n. In spite of its general appearance an isometry is a rather restricted object. In fact in what follows we prove that an isometry is allways affine with a particular structure. Definition 1.2. A map G : R n R n is affine if there is a linear map L and x 0 R n such that G( = L( + x 0, for every y in R n. Consider now an isometry F which has the origin as a fix point, that is, F (0) = 0. Then we have Lemma 1.3. Suppose F is an isometry with F (0) = 0, then F is linear. Proof. Taking y = 0 in the definition of an isometry we get that F (x) = x for every x R n. Now, we have that from which we get that x y 2 = (x (x = x 2 2x y + y 2, F (x) 2 2F (x) F ( + F ( 2 = x 2 2x y + y 2, for every x and y in R n, and so we have F (x) F ( = x y. (1) This means, of course, that an isometry preserves the scalar product as well as the length. Now let e j, j = 1,..., n be the standard orthonormal basis in R n. Then by applying (1) we have that F (e j ) F (e k ) = e j e k = δ jk, which means that v j = F (e j ), j = 1,..., n is an orthonormal basis in R n. For each x, y we have again by (1) That F (x) v j = x e j = x j, F ( v j = y j, F (x + v j = x j + y j, and hence F (x + and F (x) + F ( have the same coordinates and so we conclude that F (x + = F (x) + F (. By a similiar argument we easily get that F (cx) = cf (x) and so we have proven that any isometry which leaves the origin fixed is linear. 2

3 Let us note that by elementary linear algebra we can prove that the condition (1) is equivalent to F being an orthogonal transformation; if F is expressed as a matrix A F it will be an orthogonal matrix, and as such it possesses a lot of nice properties, as is well known from linear algebra. For instance we have that from which it follows that A 1 F = AT F (det A F ) 2 = det(a F ) det(a T F ) = det(a F A T F ) = det(i) = 1 and so, any orthogonal matrix has determinant either 1 or 1. Another important consequence of A F being orthogonal is that its column vectors (which are precisly the vectors v j ) are an orthonormal basis for R n. And the converse is also true: a matrix which have orthonormal set of columnvectors is indeed an orthogonal matrix. We are now in position to prove our main theorem. Theorem 1.4. A transformation G is an isometry of R n if and only if G is an affine transformation of the form G(x) = L(x) + x 0 for every x R n where L is orthogonal. Proof. Let G be an isometry of R n. Let F (x) = G(x) G(0) for every x R n. Then it is easy to see that F is an isometry and evidently we also have F (0) = 0. By 1.3we therefore have that F is linear and by the discussion above we have also that F has to be orthogonal. Conversely, let G(x) = L(x) x 0, where L be orthogonal. Then L(x) L( = x y for every x, y R n. If we take x = y we get L(x) 2 = L(x) L(x) = x x = x 2, hence, L(x) = x. Replacing x by x y we have that L(x) L( = L(x = x y, for every x, y R n and so we have that L is an isometry. Since G(x) G( = L(x) L( we conclude that G is an isometry, which finishes the proof. 2 Examples Let us now consider the special case where n = 2. Exempel 2.1. If L is equal to the identity matrix I and x 0 = 0 then we have the identity isometry. A matrix representation is ( ) ( ) ( ) x 1 0 x I = y 0 1 y If x 0 0 then G is a pure translation and in matrix notation we get ( ( ( ) ( ) x 1 0 x x ) y y 0 3

4 Exempel 2.2. If L is an orthogonal matrix with positive determinant (= +1) then G is a pure rotation if x 0 = 0. In matrix notation we have for a rotation by an angle t: ( ( ( ) x cos t sin t x. sin t cos t) y Isometries of the above types are called direct or even isometries. Övning 2.1. Show that a nonpure rotation, i.e. a rotation followed by a translation, is a pure rotation. The following two isometries are called indirect or odd isometries. Exempel 2.3. If R where R is an orthogonal matrix with determinant equal to 1 then G is called a pure reflection. In matrix notation a reflection becomes ( ( ( ) x cos 2t sin 2t x, sin 2t cos 2t) y where t is the angle between the x-axis and the line of reflection. Övning 2.2. Prove the matrix representation of a reflection. Exempel 2.4. If G is a translation followed by a reflection in the line of translation then the result is called a glide reflection. If we express this in matrix form we get: ( ( ) ( ) ( ) x cos 2t0 sin 2t 0 x x0 +, sin 2t 0 cos 2t 0 y y 0 where t 0 = arctan y 0 /x 0. Övning 2.3. Show that a translation followed by a reflection in a line not necessarily parallell to the line of tranlation is in fact a glide reflection. Proposition 2.5. An isometry I has a fix point iff I is a rotation or a reflection. Proof. If I is a reflection or a rotation we know that it has no translation and therefore is an orthogonal matrix. Any matrix is linear and so has 0 as fix point. Assume now that I has a fixpoint; I(x) := L(x) + x 0 = x, for some orthogonal mapping L. If x 0 = 0 we have immediately that I is either a rotation or a reflection. Assume now that x 0 0. For L this means that L(x) = x x 0, from which it follows that L is not linear: L(a + b) = (a + b) x 0 L(a) + L(b) = a x 0 + b x 0 = a + b 2x 0. This is a contradiction to the fact that L was orthogonal! The only possibility to have an isometry with a fixpoint is therefore that it has no translational component and so is a pure orthogonal mapping 4

5 Consider again Theorem 1.4. With the notation of the examples it is not difficult to see that the theorem states that any isometry is one of the the above four types. Composition of two or more isometries will of course result in an isometry. In fact isometries of the plane together with composition will constitute a group, an object which we will study in the next chapter. 5