Lecture 13: Isometries

Lecture 13: Isometries 13 Isometries 13.1 Review Last time, we discussed the orthogonal matrices On , which are matrices which preserve the dot product, which is a measure of length. We found that if we looked at the orthogonal matrices which had determinant 1, SOn , they actually turned out to be rotations in 2-space and 3-space! The rest of the orthogonal matrices O3 can be obtained from SO3 by multiplying a rotation matrix by   −1  1 ; 1 the result will be a refection over some axis. As a result, all length-preserving 3×3 matrices are rotations or refections. 13.2 Isometries Without any prior knowledge, we might assume that there are many diferent types of length-preserving mappings, called isometries. We found that for linear mappings, the isometries were the orthogonal matrices, and two or three dimensions, they were rotations or refection. What are the possibilities for isometries that are not linear? Guiding Question Orthogonal matrices are the linear mappings that preserve distance. What are the other possibilities for distance-preserving mappings that are not necessarily linear? An isometry from Rn to Rn is a length-preserving mapping. Defnition 13.1 A function f : Rn −→ Rn is an isometry if |f (u) − f (v)| = |u − v| for all u, v ∈ Rn . Let’s take a look at two key examples. Example 13.2 For a matrix A ∈ On , the linear transformation Rn −→ Rn #» x 7→ A #x» is an isometry. Example 13.3 Translation by a vector #v» ∈ Rn a is an isometry: t #» b Rn Rn −−→ #» #» x 7→ #x» + b a This is not a linear transformation! How crazy can an isometry be? The answer, fortunately or unfortunately, is not very. In fact, these two examples and their compositions turn out to be the only isometries. Lecture 13: Isometries Theorem 13.4 #» #» #» #» n Every isometry f is of the form t #» b ◦ A, for A ∈ On and b ∈ R . So f ( x ) = A x + b . Despite the fact that preserving distance does not appear to be a very strong condition on f , it turns out that it is equivalent to the very strong condition that it basically has to be linear, combined with a shift. It boils down to the following lemma. What form do the isometries that fx the origin take? The answer is that they must be linear. Lemma 13.5 If f : Rn → Rn is an isometry such that f (0) = 0, it must be a linear transformation.a a It must respect the additive and scalar multiplicative structure on Rn . Proof. We must show that f preserves sums and scalar products. First, we see that a dot product can be #» written in terms of 0 and distances:
#» #u» · #v» = 1 �| #u» − #» 0 |2 + | #v» − 0 |2 − | #u» − #v»|2 . 2 As a result, the following equation also holds:
1 � #» #» #» f ( #u») · f ( #v») = |f ( u ) − f ( 0 )|2 + |f ( #v») − f ( 0 )|2 − |f ( #u») − f ( #v»)|2 . 2 Because f is an isometry, |a − b| = |f (a) − f (b)|. Setting f (0) = 0 gives the equation ⃗u · ⃗v = f (⃗u) · f (⃗v ), so it must be the case that since f preserves lengths, f also preserves the dot product. • The sum can be expressed using a dot product again. For #z» = #» x + #» y, ( #z» − #x» − #y») · ( #z» − #x» − #y») = 0, and so #» z · #» z + #» x · #» x + #» y · #» y − 2 #x» · #» z − 2 #y» · #» z + 2 #x» · #» y = 0. Now, since we know that addition is determined in some complicated way from dot product, since f fxes the dot product, it must fx addition as well. 44 So f (z) = f (x) + f (y). • A similar reasoning gives us the scaling product: f (cx) = cf (x). Despite the fact that the only piece of information is that f preserves distances and maps the origin to itself, it is enough to play around algebraically to fnd out that f must be linear. This rules out lots of crazy functions that you could imagine could be isometries. Proof of Theorem 13.4. Now, we can prove the original theorem. Given f : Rn −→ Rn , there is some vector b ∈ Rn such that f (0) = b. Then t−b ◦ f is an isometry that fxes 0. Thus, there is some linear transformation A such that t−b ◦ f = A, and this implies that f = tb ◦ A, since tb is the inverse of t−b . From the defnition of an isometry, it is easily seen that the composition of two isometries is an isometry. Given that isometries are all of the same restrictive form, it is not surprising that they form a group. 44 If we had some other crazy invented operation determined from the dot product, f must also fx that! Lecture 13: Isometries Defnition 13.6 The group of isometries is f Mn := {isometries Rn − → Rn } ⊆ Perm(Rn ).a a Any bijective function on Rn permutes the vectors in Rn , since it maps each vector in Rn to exactly one vector in Rn , which is potentially itself. Clearly, translations, which are isomorphic to (Rn , +), form a subgroup of Mn , since t #b» +t b#»′ = t #b»+ b#»′ . Orthogonal matrices On also form a subgroup of Mn . Note that the composition of an orthogonal matrix with a translation is #» A ◦ t #» b = tA b ◦ A, since A(x + b) = Ax + Ab. In particular, a translation and an orthogonal matrix do not commute with each other. Consider the projection π :Mn −→ On tb ◦ A 7−→ A. It is a group homomorphism, since (tb ◦ A) ◦ (tb′ ◦ A′ ) = tb+Ab′ ◦ AA′ . Also, π is surjective, and the kernel is ker(π), which are translations. Thus, the subgroup of translations is normal inside Mn . 13.3 Isometries in 2-space Now that we have an understanding of isometries in general, let’s narrow it down to an analysis in two dimensions. Guiding Question For n = 2, what do isometries look like? The following defnition is an intuitive extension of the idea of orientation for linear mappings. Defnition 13.7 An isometry x 7→ Ax + b is orientation-preserving if det(A) = 1, and orientation-reversing if det(A) = −1. In two dimensions, isometries can be classifed into one of four types. Lecture 13: Isometries Theorem 13.8 Every isometry on R2 is 1. Translation 2. Rotation around a point pa 3. Refection across a line L b 4. Glide refection — frst, refect across a line L, then translate by some vector b parallel to Lc a It is no longer required that p is the origin, since the isometry does not have to be a linear transformation the line L may or may not pass through 0; the isometry is not necessarily linear. c We will see diagrams next week which have glide refections in their symmetry group! b Again, The frst two are orientation-preserving; the last two are orientation-reversing. By composing with translations, it is possible to essentially change coordinate systems. For example, consider rotations and refections. Let f be an isometry, say a rotation around the origin. Then, tp f t−p is a new isometry that fxes p, instead of the origin, since it is applying f but after shifting coordinates by p. 45 Similarly, letting f be a refection across any line, we can represent it in new coordinates as a refection across a line through the origin. Proof. We split the proof up into two cases depending on whether f is orientation-preserving or reversing. • Case I. Consider an orientation-preserving isometry f (x) = Aθ x + b. 1. If Aθ = I2 , the identity, then f = tb , which is possibility 1 in the theorem. 2. Otherwise, if Aθ ̸= I2 , we want to fnd a fxed point p such that f (p) = p. Since Aθ has no fxed vectors, pA (1) ̸= 0, and so Aθ − I2 has a trivial kernel, and so A − I2 is invertible. Then the equation (A − I2 )p = −b has a unique solution p = (A − I2 )−1 (−b), and then f (p) = Ap + b = p. So t−p Atp = Aθ , 45 When we apply t−p , we shift p to 0, then we use f to rotate around 0, and lastly use tp to shift 0 back to p. Lecture 13: Isometries since it fxes 0. This corresponds to the second possibility: rotation around a point p. • Case II. Let f be an orientation-reversing isometry. Then f = tb ◦ A, where A is refection across a line L. First, change the origin to b/2. Then t−b/2 f tb/2 = t−b/2 tb Atb/2 = tb/2 tAb/2 A = tm A, where m = b+Ab 2 . Since b and Ab are refections over a line L, m, the average, must lie on that line. 3. If m = 0, it is a refection. 4. If m ̸= 0, it is a glide refection. 46 Again, the same idea from Case I applies. Shifting to a new coordinate system gives us either a refection or a glide refection. 46 In our proof, we try to be slicker about it, but if we are uncomfortable with that, we know f is just Ax + b, and we could simply crunch through lots of sines and cosines to force f into one of the four forms in Theorem 13.8.