4.4. LINEAR ISOMETRIES
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4.3.3. Consider a plane P that does not pass through the origin. Let ˛ be a unit normal vector to P and let x0 be a point on P . Find a formula (in terms of ˛ and x0) for the reflection of a point x through P . Such a reflection through a plane not passing through the origin is called an affine reflection.
4.3.4. Here is a method to determine all the products of the symmetries of the square tile. Write J for Jr , the reflection in the .x; y/–plane.
(a) The eight products ˛J , where ˛ runs through the set of eight rotation matrices of the square tile, are the eight nonrotation matrices. Which matrix corresponds to which nonrotation symmetry?
(b) Show that J commutes with the eight rotation matrices; that is, J ˛ D ˛J for all rotation matrices ˛.
(c) Check that the information from parts (a) and (b), together with the multiplication table for the rotational symmetries, suffices to compute all products of symmetries.
(d) Verify that the sixteen symmetries form a group.
4.3.5. Another way to work out all the products, and to understand the structure of the group of symmetries, is the following. Consider the matrix 20 1 03
S D Jc D 1 0 0
4
5 : 0 0 1 (a) The set D of eight diagonal matrices with ˙1’s on the diagonal is a subset of the matrices representing symmetries of the square tile. This set of diagonal matrices is closed under matrix multiplication. Show that every symmetry matrix for the square tile is either in D or is a product DS , where D 2 D.
(b) For each D 2 D, there is a D0 2 D (which is easy to compute) that satisfies SD D D0S, or, equivalently, SDS D D0. Find the rule for determining D0 from D, and use this to show how all products can be computed. Compare the results with those of the previous exercise.
4.4. Linear Isometries
The main purpose of this section is to investigate the linear isometries of three–dimensional space. However, much of the work can be done without much extra effort in n-dimensional space.
Consider Euclidean n-space, Rn with the usual inner product hxjyi D
Pn i D1 xi yi , norm jjxjj D hx; xi1=2, and distance function d.x; y/
D jjx yjj. Recall that transposes of matrices and inner products are
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4. SYMMETRIES OF POLYHEDRA related by hAx; yi D hx; At yi for all n–by–n matrices A, and all x; y 2 Rn.
Definition 4.4.1. An isometry of Rn is a map W Rn ! Rn that preserves distance, d. .a/; .b// D d.a; b/, for all a; b 2 Rn.
You are asked to show in the Exercises that the set of isometries is a group, and the set of isometries satisfying .0/ D 0 is a subgroup.
Lemma 4.4.2. Let be an isometry of Rn such that .0/ D 0. Then h.a/; .b/i D ha; bi for all a; b 2 Rn.
Proof. Since .0/ D 0, preserves norm as well as distance, jj.x/jj D jjxjj for all x. But since ha; bi D .1=2/.jjajj2 C jjbjj2 d.a; b/2/, it follows that also preserves inner products.
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Remark 4.4.3. Recall the following property of orthonormal bases of Rn: If F D ffi g is an orthonormal basis, then the expansion of a vector x with respect to F is x D Pi hx; fi ifi . If x D Pi xi fi and y D Pi yi fi , then hx; yi D Pi xi yi .
Definition 4.4.4. A matrix A is said to be orthogonal if At is the inverse of A.
Exercise 4.4.3 gives another characterization of orthogonal matrices.
Lemma 4.4.5. If A is an orthogonal matrix, then the linear map x 7! Ax
is an isometry.
Proof. For all x 2 Rn, jjAxjj2 D hAx; Axi D hAt Ax; xi D hx; xi D jjxjj2.
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Lemma 4.4.6. Let F D ffi g be an orthonormal basis. A set fvi g is an orthonormal basis if and only if the matrix A D Œaij D Œhvi ; fj i is an orthogonal matrix.
Proof. The i th row of A is the coefficient vector of vi with respect to the orthonormal basis F . According to Remark 4.4.3, the inner product of vi and vj is the same as the inner product of the i th and j th rows of A.
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Hence, the vi ’s are an orthonormal basis if and only if the rows of A are an orthonormal basis. By Exercise 4.4.3, this is true if and only if A is orthogonal.
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Theorem 4.4.7. Let be a linear map on Rn. The following are equivalent: (a) is an isometry.
(b) preserves inner products.
(c) For some orthonormal basis ff1; : : : ; fng of Rn, the set f.f1/; : : : ; .fn/g is also orthonormal.
(d) For every orthonormal basis ff1; : : : ; fng of Rn, the set f.f1/; : : : ; .fn/g is also orthonormal.
(e) The matrix of with respect to some orthonormal basis is orthogonal.
(f) The matrix of with respect to every orthonormal basis is orthogonal.
Proof. Condition (a) implies (b) by Lemma 4.4.2. The implications (b) ) (d) ) (c) are trivial, and (c) implies (e) by Lemma 4.4.6. Now assume the matrix A of with respect to the orthonormal basis F is orthogonal.
Let U.x/ be the coordinate vector of x with respect to F ; then by Remark 4.4.3, U is an isometry. The linear map is U 1 ı MA ı U , where MA means multiplication by A. By Lemma 4.4.5, MA is an isometry, so is an isometry. Thus (e) H)(a). Similarly, we have (a) H)(b) H)(d) H)(f) H)(a).
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Proposition 4.4.8. The determinant of an orthogonal matrix is ˙1.
Proof. Let A be an orthogonal matrix. Since det.A/ D det.At /, we have 1 D det.E/ D det.At A/ D det.At / det.A/ D det.A/2:
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Remark 4.4.9. If is a linear transformation of Rn and A and B are the matrices of with respect to two different bases of Rn, then det.A/ D det.B/, because A and B are related by a similarity, A D VBV 1, where V is a change of basis matrix. Therefore, we can, without ambiguity, define the determinant of to be the determinant of the matrix of with respect to any basis.
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4. SYMMETRIES OF POLYHEDRA Corollary 4.4.10. The determinant of a linear isometry is ˙1.
Since det.AB/ D det.A/ det.B/, det W O.n; R/ ! f1; 1g is a group homomorphism.
Definition 4.4.11. The set of orthogonal n–by–n matrices with determinant equal to 1 is called the special orthogonal group and denoted SO.n; R/.
Evidently, the special orthogonal group SO.n; R/ is a normal sub group of the orthogonal group of index 2, since it is the kernel of det W O.n; R/ ! f1; 1g.
We next restrict our attention to three-dimensional space and explore the role of rotations and orthogonal reflections in the group of linear isometries.
Recall from Section 4.3 that for any unit vector ˛ in R3, the plane P˛ is fx W hx; ˛i D 0g. The orthogonal reflection in P˛ is the linear map j˛ W x 7! x
2hx; ˛i˛. The orthogonal reflection j˛ fixes P˛ pointwise and sends ˛ to ˛. Let J˛ denote the matrix of j˛ with respect to the standard basis of R3. Call a matrix of the form J˛ a reflection matrix.
Let’s next sort out the role of reflections and rotations in SO.2; R/.
˛
˛
Consider an orthogonal matrix . Orthogonality implies that
ˇ
ı
ˇ
ˇ
˛
is a unit vector and
D ˙ : The vector can be written
ı
˛
ˇ
cos./
as for some angle , so the orthogonal matrix has the form sin. /
cos./
sin. /
cos./
sin. /
or
. The matrix sin. /
cos. / sin. /
cos. /
cos./
sin. /
R D sin./ cos./ is the matrix of the rotation through an angle , and has determinant equal
cos./
sin. /
1
0
to 1. The matrix equals R sin. /
cos. / J , where J D
0
1
is the reflection matrix J D JOe . The determinant of R
2
J is equal to 1.
Now consider the situation in three dimensions. Any real 3–by–3 ma trix has a real eigenvalue, since the characteristic polynomial is cubic with real coefficients. A real eigenvalue of an orthogonal matrix must be ˙1 because the matrix implements an isometry.
Lemma 4.4.12. Any element of SO.3; R/ has C1 as an eigenvalue.
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Proof. Let A 2 SO.3; R/, let be the linear isometry x 7! Ax, and let v be an eigenvector with eigenvalue ˙1. If the eigenvalue is C1, there is nothing to do. So suppose the eigenvalue is 1. The plane P D Pv orthogonal to v is invariant under A, because if x 2 P , then hv; Axi D hAv; Axi D hv; xi D 0. The restriction of to P is also orthogonal, and since 1 D det./ D . 1/.det.jP /, jP must be a reflection. But a reflection has an eigenvalue of C1, so in any case A has an eigenvalue of C1.
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Proposition 4.4.13. An element A 2 O.3; R/ has determinant 1 if and only if A implements a rotation.
Proof. Suppose A 2 SO.3; R/. Let denote the corresponding linear isometry x 7! Ax. By the lemma, A has an eigenvector v with eigenvalue 1. The plane P orthogonal to v is invariant under , and det.jP / D det. / D 1, so jP is a rotation of P . Hence, is a rotation about the line spanned by v. On the other hand, if is a rotation, then the matrix of with respect to an appropriate orthonormal basis has the form
21
0
0
3
0
4
cos. / sin. /5 ;
0
sin. /
cos. / so has determinant 1.
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Proposition 4.4.14. An element of O.3; R/ n SO.3; R/ implements either an orthogonal reflection, or a reflection-rotation, that is, the product of a reflection j˛ and a rotation about the line spanned by ˛.
Proof. Suppose A 2 O.3; R/ n SO.3; R/. Let denote the corresponding linear isometry x 7! Ax. Let v be an eigenvector of A with eigenvalue ˙1. If the eigenvalue is 1, then the restriction of to the plane P orthogonal to v has determinant 1, so is a reflection. Then itself is a reflection.
If the eigenvalue is 1, then the restriction of to P has determinant 1, so is a rotation. In this case is the product of the reflection jv and a rotation about the line spanned by v.
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4. SYMMETRIES OF POLYHEDRA
Exercises 4.4 4.4.1.
(a) Show that j˛ is isometric.
(b) Show that det.j˛ / D 1.
(c) If is a linear isometry, show that j˛ 1 D j.˛/.
(d) If A is any orthogonal matrix, show that AJ˛ A 1 D JA˛ .
(e) Conclude that the matrix of j˛ with respect to any orthonormal basis is a reflection matrix.
4.4.2. Show that the set of isometries of Rn is a group. Show that the set of isometries satisfying .0/ D 0 is a subgroup.
4.4.3. Show that the following are equivalent for a matrix A: (a) A is orthogonal.
(b) The columns of A are an orthonormal basis.
(c) The rows of A are an orthonormal basis.
4.4.4.
(a) Show that the matrix R2 J D R JR is the matrix of the reflection in the line spanned by cos :
sin
Write J D R2 J .
(b) The reflection matrices are precisely the elements of O.2; R/ with determinant equal to 1.
(c) Compute J J .
1
2
(d) Show that any rotation matrix R is a product of two reflection matrices.
4.4.5. Show that an element of SO.3; R/ is a product of two reflection matrices. A matrix of a rotation–reflection is a product of three reflection matrices. Thus any element of O.3; R/ is a product of at most three reflection matrices.
4.5. The Full Symmetry Group and Chirality
All the geometric figures in three-dimensional space that we have con sidered — the polygonal tiles, bricks, and the regular polyhedra — admit reflection symmetries. For “reasonable” figures, every symmetry is implemented by by a linear isometry of R3; see Proposition 1.4.1. The rotation
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group of a geometric figure is the group of g 2 SO.3; R/ that leave the figure invariant. The full symmetry group is the group of g 2 O.3; R/ that leave the figure invariant. The existence of reflection symmetries means that the full symmetry group is strictly larger than the rotation group.
In the following discussion, I do not distinguish between linear isome tries and their standard matrices.
Theorem 4.5.1. Let S be a geometric figure with full symmetry group G and rotation group R D G \ SO.3; R/. Suppose S admits a reflection symmetry J . Then R is an index 2 subgroup of G and G D R [ RJ .
Proof. Suppose A is an element of G n R. Then det.A/ D 1 and det.AJ / D 1, so AJ 2 R. Thus A D .AJ /J 2 RJ .
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For example, the full symmetry group of the cube has 48 elements.
What group is it? We could compute an injective homomorphism of the full symmetry group into S6 using the action on the faces of the cube, or into S8 using the action on the vertices. A more efficient method is given in Exercise 4.5.1; the result is that the full symmetry group is S4 Z2.
Are there geometric figures with a nontrivial rotation group but with no reflection symmetries? Such a figure must exhibit chirality or “handedness”; it must come in two versions that are mirror images of each other.
Consider, for example, a belt that is given n half twists (n 2) and then fastened. There are two mirror image versions, with right–hand and left– hand twists. Either version has rotation group Dn, the rotation group of the n–gon, but no reflection symmetries. Reflections convert the right–handed version into the left–handed version. See Figure 4.5.1.
Figure 4.5.1. Twisted band with symmetry D3.
There exist chiral convex polyhedra with two types of regular polygo nal faces, for example, the “snubcube,” shown in Figure 4.5.2 on the next page.
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4. SYMMETRIES OF POLYHEDRA Figure 4.5.2. Snubcube.
Exercises 4.5 4.5.1. Let G denote the full symmetry group of the cube and R the rotation group. The inversion i W x 7! x with matrix E is an element of G n R, so G D R [Ri. Observe that i2 D 1, and that for any rotation r, ir D ri.
Conclude that G Š S4 Z2.
4.5.2. Show that the same trick works for the dodecahedron, and that the full symmetry group is isomorphic to A5 Z2. Show that this group is not isomorphic to S5.
4.5.3. Show that the full symmetry group of the tetrahedron is S4.
4.5.4. What is the full symmetry group of a brick?
4.5.5. What is the full symmetry group of a square tile?
4.5.6. What is the full symmetry group of a tile in the shape of a regular n–gon?
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