Chapter 27: Symmetry Groups
MAT301H1S Lec5101 Burbulla
Week 11 Lecture Notes
Winter 2020
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups
Chapter 27: Symmetry Groups
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups What is an Isometry?
n n Definition: The function T : R −→ R is called an isometry if it n preserves distances. That is, for all ~x, ~y ∈ R ,
kT (~x) − T (~y)k = k~x − ~yk.
As you may have seen in your previous linear algebra course, T is an isometry if and only if T preserves lengths: kT (~x)k = k~xk. We 2 3 2 will only concern ourselves with isometries in R and R . In R there are only four possibilities: 1. T is a rotation of θ around a point, usually chosen to be the origin. 2. T is a reflection in a line, usually a line passing through the origin. 3. T is a translation. 4. T is a glide reflection.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups
The Four Types of Isometries of R2
cos θ − sin θ 1. T (~x) = R (~x) with [R ] = . θ θ sin θ cos θ cos θ sin θ 2. T (~x) = F (~x) with [F ] = , if the reflection θ θ sin θ − cos θ is in the line y = m x with slope m = tan(θ/2). 2 3. T (~x) = ~x + ~a, for some vector ~a ∈ R .
A glide reflection is a translation parallel to a line, followed by a reflection in that 4. line; or equivalently, a reflection in a line followed by a translation parallel to the line. The line is called the glide-axis.
Recall: the matrices for rotations and reflections are orthogonal, A−1 = AT , and that the order of every reflection is 2.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups
What is The Symmetry Group of a Figure in Rn?
n Definition: let F be a set of points in R . The symmetry group n of the figure F is the set of all isometries T of R that carry F onto itself: T (F ) = F . The group operation is function composition. There are two cases: 1. if F is finite in extent, then the only possible symmetries of F are isometries which can be represented by an orthogonal matrix A such that A−1 = AT . For example, rotations, reflections, and improper rotations. In this case, the symmetry group of F will be finite. 2. if F is infinite in extent, then the symmetry group of F may include one or more translations, in which case the symmetry group of F will be infinite. We will limit ourselves to finite symmetry groups in this chapter, and look briefly at some infinite symmetry groups in the next and last chapter, Chapter 28.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Finite Symmetry Groups in the Plane
Theorem 27.1: the only finite symmetry groups in the plane are Zn and Dn. (In the figure below, Cn is the same as Zn.) Proof: see the book. Note: a symmetry group consists solely of rotational symmetries if there is no axis of symmetry in the figure.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups Example 1
As we have mentioned previously, for n ≥ 3, Dn is the symmetry group of a regular n-gon. For example, if n = 4 : The four rotations are rep- And the four reflections are repre- resented by the matrices sented by the matrices
1 0 1 0 [R ] = , [F ] = , 0 0 1 0 0 −1
0 −1 0 1 [R ] = , [F ] = , 90 1 0 90 1 0
−1 0 −1 0 [R ] = , [F ] = , 180 0 −1 180 0 1
0 1 0 −1 [R ] = . [F ] = . 270 −1 0 270 −1 0
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Figure for Example 1 y-axis x = 0 y = −x y = x @ −1, 1) @ (1, 1)( r @ r @ @ @ y = 0 @ x-axis @ @ @ (−1, −1) @ (1, −1) @ r r
Figure: A square, centered at the origin, in the xy-plane.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups
Example 2: D1 and D2
The groups D1 and D2 are not symmetry groups of regular polygons. They are not really dihedral groups at all. Nevertheless
1. we take D1 to be the symmetry group of an isosceles triangle,
2. and we take D2 to be the symmetry group of a rectangle.
The isosceles triangle only has two symme- tries: a rotation of zero, aka the identity transformation, and a vertical reflection in the triangle’s single axis of symmetry.
D1 ≈ Z2.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups
A rectangle has two rotational symmetries, the identity and a rotation of 180 degrees. It has two axes of symmetry, a vertical and a horizontal axis. Its symmetry group is
D2 ≈ Z2 ⊕ Z2.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups
What Is a Rotation in R3?
3 A rotation of θ in R is a rotation about a line so that every point in a plane normal to the line is rotated in the plane by θ.
In terms of linear algebra, a matrix A represents a rotation E1(A) 3 in R if it is orthogonal and det(A) = 1. The axis of rota- tion is the eigenspace of A cor- responding to the eigenvalue λ = 1. For every point ~x in ¨¨¨H ¨¨ ¨ ¨ H ¨ a plane normal to the axis of ¨ ¨ θ H A~x ¨ ¨ ¨ ~x r Hj ¨ rotation, A~x is the vector ob- ¨ ¨ tained by rotating ~x by θ.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups
Finite Groups of Rotations in R3 Theorem 27.2: up to isomorphism, the finite groups of rotations 3 in R are Zn, Dn, A4, S4, and A5. Proof: see for instance, Algebra, 2/ed by Michael Artin, Sec. 6.12. Comments: 3 1. Figures in a plane can be rotated by rotations in R if the rotation is around an axis perpendicular to the plane. Thus figures in the plane that give rise to symmetry groups Zn or Dn will give rise to groups of rotations isomorphic to Zn or Dn 3 3 in R . In the latter case, a rotation in R of order 2 can have the same effect as a reflection in the plane. See Example 3. 2. Recall: the rotational symmetries of a tetrahedron form a group isomorphic to A4, and the rotational symmetries of a cube (or an octahedron) form a group isomorphic to S4.
3. The group A5 is isomorphic to the group of rotational symmetries of a dodecahedron or icosahedron.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups Example 3
Consider again the square of Example 1, but now consider it as a 3 square in the xy-plane in R , with equation z = 0. The square has four rotational symmetries, with axis of rotation the z-axis, represented by the matrices, of orders 1, 4, 2 and 4, respectively,
1 0 0 0 −1 0 −1 0 0 0 1 0 0 1 0 , 1 0 0 , 0 −1 0 , −1 0 0 . 0 0 1 0 0 1 0 0 1 0 0 1
Also the square has four rotational symmetries of order 2 given by
1 0 0 0 1 0 −1 0 0 0 −1 0 0 −1 0 , 1 0 0 , 0 1 0 , −1 0 0 , 0 0 −1 0 0 −1 0 0 −1 0 0 −1
with axes of rotation the x-axis; the line x = y, z = 0; the y-axis; and the line x = −y, z = 0, respectively. Compare with Example 1.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Example 3, Continued
Let 0 −1 0 1 0 0 R = 1 0 0 , F = 0 −1 0 . 0 0 1 0 0 −1 Since R4 = I , F 2 = I , and FRF = R3, as you can check, the above eight rotations do in fact form a group isomorphic to D4. A word about calculations: rotational symmetries are always represented by orthogonal matrices A with det(A) = 1. In particular, A−1 = AT . So the above calculation could be rewritten as FRF = RT .
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups
Example 4: The Symmetry Group of a 3-Prism in R3
A 3-prism is shown in the figure below. What is its symmetry group, G? The first thing you can do is use the Orbit- Stabilizer Theorem (Theorem 7.4) to find |G|. The two ends of the 3-prism consist of equilateral triangles, each with a sym- metry group of D3, which has order 6. So |G| = 2×6 = 12. Or you can look at the 3-prism as consisting of three congruent rectangles, each of which has symmetry group Z2 ⊕ Z2, which has order 4. Thus |G| = 3 × 4 = 12. Either way, G contains a normal subgroup of order 6 isomorphic to D3 and so G ≈ D3 ⊕ Z2 ≈ D6.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Example 5: Symmetry Groups of the Platonic Solids
Figure: Tetrahedron, Octahedron, Cube, Icosahedron, Dodecahedron. What are the symmetry groups of the Platonic solids?
1. |S(T )| = 4 × |D3| = 4 × 6 = 24; S(T ) ≈ S4. 2. |S(O)| = 8 × |D3| = 8 × 6 = 48; S(O) ≈ S4 ⊕ Z2. 3. |S(C)| = 6 × |D4| = 6 × 8 = 48; S(C) ≈ S4 ⊕ Z2. 4. |S(I )| = 20 × |D3| = 20 × 6 = 120; S(I ) ≈ A5 ⊕ Z2. 5. |S(D)| = 12 × |D5| = 12 × 10 = 120; S(D) ≈ A5 ⊕ Z2.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups
Classification of Rotations in R3 if det(A) = 1 tr (A) action of A |A| The axis of rotation is the eigenspace corresponding to 3 identity, A = I 1 the eigenvalue λ = 1 of A. If the 3 × 3 matrix represents −1 rotation 2 a rotation of θ then its eigen- values are
0 rotation 3 iθ −iθ λ1 = 1, λ2 = e , λ3 = e
1 rotation 4 and so tr (A) = √ 1− 5 λ + λ + λ = 1 + 2 cos θ, ψ = 2 rotation 5 1 2 3 √ 1+ 5 by Euler’s formula: φ = 2 rotation 5 ei θ = cos θ + i sin θ. 2 rotation 6
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups
Classification of Non-Rotations in R3 if det(A) = −1 tr (A) action of A |A| −I represents inversion, a ‘re- −3 inversion, A = −I 2 flection’ in the origin. (It is the composition of three separate 1 reflection 2 reflections.) If det(A) = 1, then det(−A) = −1, and 0 improper rotation 6 tr (−A) = −tr (A).
−1 improper rotation 4 Then A and −A have the same order if |A| is even, and −ψ improper rotation 10 | − A| = 2|A|, −φ improper rotation 10 if |A| is odd. −2 improper rotation 6
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups One way to establish these results without using complex numbers is by using the Cayley-Hamilton Theorem. If A 6= I is the matrix of a rotation then the characteristic polynomial of A is (as you can check) 3 2 CA(x) = x − tr (A) x + tr (A) x − 1, whence A3 − tr (A) A2 + tr (A) A − I = O. In particular A3 = tr (A) A2 − tr (A) A + I . If A2 = I then the equation reduces to A = tr (A) I − tr (A) A + I , and we can conclude tr (A) = −1. If A3 = I then tr (A) = 0. Successive powers of A can be calculated in terms of I , A and A2 recursively: A3 = tr (A) A2 − tr (A) A + I ⇒ A4 = tr (A) A3 − tr (A) A2 + A ⇒ A4 = tr (A)(tr (A) A2 − tr (A) A + I ) − tr (A)A2 + A ⇒ A4 = ((tr (A))2 − tr (A))A2 + (1 − (tr (A))2)A + tr (A) I and you can see that if A4 = I and A2 6= I then tr (A) = 1.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups
Next: A5 = ((tr (A))2 − tr (A))A3 + (1 − (tr (A))2)A2 + tr (A) A
(after much algebra) = [(tr (A))3 − 2(tr (A)2 + 1]A2+ [(tr (A))3 − (tr (A))2 − tr (A)]A + [(tr (A))2 − tr (A)]I
If the order of A is 5, then A5 = I , and (in particular) √ 1 ± 5 (tr (A))2 − tr (A) = 1 ⇔ tr (A) = . 2 Let √ √ 1 + 5 1 − 5 φ = and ψ = . 2 2 Thus A5 = I implies tr (A) = φ or tr (A) = ψ. Similarly, you can check that if A6 = I , but A2 6= I and A3 6= I , then tr (A) = 2.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups
What Is a Reflection in R3? The 3 × 3 orthogonal matrix A with A−1 = A, represents a reflection if its eigenvalues are λ1 = 1, λ2 = 1 and λ3 = −1. Then: I The eigenspace corresponding to the eigenvalue λ = 1 is 3 two-dimensional; it is a plane in R which is the mirror of reflection. I The eigenspace corresponding to the eigenvalue λ = −1 is the line normal to the plane. If the unit eigevector is n~, then A n~ = −n~. If A represents a reflection in the plane passing through the origin normal to the unit vector n~, then A = I − 2n~ n~T . Correspondingly, −A represents a rotation of order 2 around the axis parallel to the unit vector n~, and −A = 2n~ n~T − I .
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Example 6
If 2/7 n~ = 3/7 6/7 then the matrix of reflection in the plane with equation 2x + 3y + 6z = 0 is A = I − 2n~ n~T =
1 0 0 2/7 0 1 0 − 2 3/7 2/7 3/7 6/7 0 0 1 6/7
41 −12 −24 1 = −12 31 −36 . 49 −24 −36 −23 Check that: A−1 = AT = A, tr (A) = 1, det(A) = −1, A n~ = −n~.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups
What Is an Improper Rotation in R3?
In terms of linear algebra, a E−1(A) matrix A represents an im- 3 proper rotation in R if it is ~z n~ orthogonal, det(A) = −1, and 6@I 6 ~y @ the only real eigenvalue of A @ is −1. If n~ is an eigenvector of ¨¨¨@H ¨¨ ¨ ¨ A H A~x ¨ A corresponding to the eigen- ¨ ¨ θr HHj ¨ ¨¨ ¨ ~x A ¨¨ value λ = −1 then A n~ = −n~. A For every point ~x in the plane A A −~y normal to the vector n~ and A passing through the origin, A~x AU?A ~z is the vector obtained by rotating ~x by θ. The action of A on all 3 other vectors ~z ∈ R is a combination:
A~z = A(~x + ~y) = A ~x − ~y.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Example 7
Let 0 0 −1 1 A = −1 0 0 , n~ = 1 . 0 −1 0 1 Check that the only real eigenvalue of A is −1, det(A) = −1, and A n~ = −n~. Thus A represents an improper rotation, of order 6. Restricted to the plane with equation x + y + z = 0, A represents a rotation of order 6, as the following calculations show: x x −z ~v = −x − z ⇒ A ~v = A −x − z = −x z z x + z and ~v · A~v x2 + xz + z2 1 π cos θ = = = ⇒ θ = . k~vk kA ~vk 2x2 + 2xz + 2z2 2 3
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups The Symmetry Group of a Cube Consider the cube in the figure to the right, with vertices (±1, ±1, ±1), with respect to the usual x, y and z axes, and centre at the origin (0, 0, 0). Let S(C) be the symmetry group of this cube. Recall from Chapter 7 that S(C) consists of the following 48 matrices, with usual matrix multiplication:
±1 0 0 0 ±1 0 0 0 ±1 0 ±1 0 , 0 0 ±1 , ±1 0 0 , 0 0 ±1 ±1 0 0 0 ±1 0
±1 0 0 0 0 ±1 0 ±1 0 0 0 ±1 , 0 ±1 0 , ±1 0 0 . 0 ±1 0 ±1 0 0 0 0 ±1
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Rotational Symmetries of a Cube
The 24 matrices A ∈ S(C) with det(A) = 1 represent the rotational symmetries of the cube. One of these is I , the identity. What is the geometry of the other 23? The possible axes of rotation of a cube are illustrated in the figure below:
They are, respectively: the three lines joining the centres of opposite faces; the four diagonals of the cube, joining opposite corners; the six lines joining midpoints of opposite edges.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups Rotations Around the Coordinate Axes about x-axis about y-axis about z-axis order
1 0 0 0 0 1 0 −1 0 A 0 0 1 0 1 0 1 0 0 4 0 −1 0 −1 0 0 0 0 1
1 0 0 −1 0 0 −1 0 0 A2 0 −1 0 0 1 0 0 −1 0 2 0 0 −1 0 0 −1 0 0 1
1 0 0 0 0 −1 0 1 0 A3 0 0 −1 0 1 0 −1 0 0 4 0 1 0 1 0 0 0 0 1
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Rotations of Order 3 Around the Diagonals of a Cube
0 1 0 0 1 0 0 −1 0 0 −1 0 0 0 1 0 0 −1 0 0 1 0 0 −1 1 0 0 −1 0 0 −1 0 0 1 0 0 and and and and 0 0 1 0 0 −1 0 0 −1 0 0 1 1 0 0 1 0 0 −1 0 0 −1 0 0 0 1 0 0 −1 0 0 1 0 0 −1 0 about about about about diagonal diagonal diagonal diagonal parallel to parallel to parallel to parallel to the vector the vector the vector the vector 1 1 −1 1 1 1 1 −1 1 −1 1 1
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups The Remaining Rotations of Order 2
The remaining six rotation matrices in S(C) all represent rotations of order 2 around lines that join the midpoints of opposite edges of the cube. They are, with the direction vectors of their axes: −1 0 0 0 −1 0 0 0 0 0 −1 about −1 ; 0 0 1 about 1 ; 0 −1 0 1 0 1 0 1
0 0 −1 −1 0 0 1 1 0 −1 0 about 0 ; 0 −1 0 about 0 ; −1 0 0 1 1 0 0 1
0 −1 0 −1 0 1 0 1 −1 0 0 about 1 ; 1 0 0 about 1 . 0 0 −1 0 0 0 −1 0
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Reflections in S(C) With Their Planes of Reflection
−1 0 0 1 0 0 1 0 0 reflection 0 1 0 0 −1 0 0 1 0 0 0 1 0 0 1 0 0 −1 in plane x = 0 y = 0 z = 0
1 0 0 0 0 1 0 1 0 reflection 0 0 1 0 1 0 1 0 0 0 1 0 1 0 0 0 0 1 in plane y = z x = z x = y
1 0 0 0 0 −1 0 −1 0 reflection 0 0 −1 0 1 0 −1 0 0 0 −1 0 −1 0 0 0 0 1 in plane y = −z x = −z x = −y
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups The Six Improper Rotations of Order 4 in S(C)
about x-axis about y-axis about z-axis
−1 0 0 0 0 −1 0 1 0 0 0 −1 0 −1 0 −1 0 0 0 1 0 1 0 0 0 0 −1
−1 0 0 0 0 1 0 −1 0 0 0 1 0 −1 0 1 0 0 0 −1 0 −1 0 0 0 0 −1
Geometrically, an improper rotation of order 4 about an axis, rotates each point on a plane perpendicular to the axis by ±90◦ while simultaneously reflecting the point to the opposite plane.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Example 8
Let −1 0 0 A = 0 0 −1 , 0 1 0 and consider its action on the face of the cube in the plane x = 1.
1 1 0 −1 0 −1 A y = A 0 +A y = 0 + −z = −z , z 0 z 0 y y
which is in the plane x = −1. Observe also that
0 0 y · −z = 0, z y
so these two vectors are perpendicular.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups The Eight Improper Rotations of Order 6 in S(C)
0 −1 0 0 −1 0 0 1 0 0 1 0 0 0 −1 0 0 1 0 0 −1 0 0 1 −1 0 0 1 0 0 1 0 0 −1 0 0 and and and and 0 0 −1 0 0 1 0 0 1 0 0 −1 −1 0 0 −1 0 0 1 0 0 1 0 0 0 −1 0 0 1 0 0 −1 0 0 1 0 about about about about diagonal diagonal diagonal diagonal parallel parallel parallel parallel to vector to vector to vector to vector 1 1 −1 1 1 1 1 −1 1 −1 1 1
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups Example 9
By Example 7,
0 0 −1 A = −1 0 0 0 −1 0
acts as a rotation of 60◦ on the plane with with equation x + y + x = 0. What is its action on the cube? In the figures to the left, the green triangle joins the vertices of the cube in the plane x + y + z = 1 and the blue triangle joins the vertices of the cube in the plane x + y + z = −1. The action of A on these triangles is to rotate a vertex on one triangle by 60◦, and then to reflect it onto the other triangle.
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla Chapter 27: Symmetry Groups The Symmetry Group of a Hypercube
If we move a one-dimensional line perpendicular to itself through a 2-nd dimension, we obtain a square. If we now take the square and move it perpendicular to itself through a 3-rd dimension, we obtain a cube. If we take the cube and move it perpendicular to itself through a 4-th dimension we obtain a hypercube. Even though we cannot physically do this in our 3-dimensional world, we can do it mathematically. How many symmetries does the hypercube have?
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla
Chapter 27: Symmetry Groups
1. The line with vertices ±1 has 2 symmetries. 2. The square with vertices (±1, ±1) has four sides and has 4 × 2 = 8 = 22 × 2! symmetries. 3. The cube with vertices (±1, ±1, ±1) has six square faces and has 6 × 8 = 48 = 23 × 3! symmetries. 4. The hypercube with vertices (±1, ±1, ±1, ±1) has eight cubic ‘faces’ and has 8 × 48 = 384 = 24 × 4! symmetries. (The pattern is that the number of symmetries of an n-dimensional ‘cube’ is 2n × n! .) The symmetries of the hypercube can all be represented by the 24 possible permutations of the sixteen 4 × 4 matrices ±1 0 0 0 0 ±1 0 0 . 0 0 ±1 0 0 0 0 ±1
Week 11 Lecture Notes MAT301H1S Lec5101 Burbulla