Symmetry in R³

**Review**

**Reflection**

**The Dihedral Group**

**Glide Reflection**

**Proper Rotations**

**Polyhedra**

**Sources**

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REVIEW

Rigid motion of Rⁿ is m: Rⁿ to Rⁿ that preserves distance

Orthogonal group = O_n = all rigid motions that fix the origin = {A 0 GL(n, R) | A^T,A = I }

In other words, the rigid motion can be represented by an orthogonal matrix.

In R², the members of O_n are rotations around the origin and reflections about lines through the origin.

The reason the orthogonal group is so important is because all rigid motions are m(X) = AX + b where A 0 O_n.

For every A in O_n, det (A) = ± 1

det (A) = 1 then it is orientation-preserving (rotations)

det (A) = -1 then it is orientation-reversing (reflections)

Symmetry of a set F

m:F to F

m(F) = F

Symmetry group of F is a subgroup of M, the group of all rigid motions. The symmetry group is the set of all symmetries of F.

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THE CYCLIC GROUP

C_n = {e, a, a², ... , a^n-1} where aⁿ = e.

In R², C_n represents the group of rotations around point O.

In R³, think of C_n as rotations around an axis through O.

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REFLECTION

Umklappung is what we call “pancake”

Look at a reflection in R² around line l.

The same reflection in R³ is really a rotation around l.

The effect is a pancake rotation.

There are three types of reflections: point reflections, line reflections, and plane reflections. To reflect a point P, continue the line perpendicular to the point, line, or plane that also goes through P. The reflected point, P', occurs on that line when the distance from P to the point/line/plane is the same as the distance from P' to the point/line/plane.

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THE DIHEDRAL GROUP

D_n = {e, x, x², ... , x^n-1, y, xy, ... , x^n-1y} where xⁿ = e, y² = e, yx = x^-1y.

In R², D_n represents the group of reflections and rotations in the plane. (contains rotations of the cyclic group)

In R³, think of those reflections as pancakes around the old line of reflection. There are n pancakes (one for each old line of reflection).

The new group we’ll call D' _n = rotations in R³.

D₁ has only one rotation, which is the identity, and one line of reflection.

Therefore D'₁ has only one pancake rotation

D'₁ = {e, a} where a² = e. So D'₁ = C₂

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GLIDE REFLECTION

Ingredients:

one glide plane

one vector moving parallel to glide plane

Steps:

1. Translate by vector

2. Reflect in glide plane

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PROPER ROTATIONS

Special Orthogonal Group = SO_n= {A in O_n| det (A) = 1}

In R³, A in SO₃ iff A represents proper rotation of R³

In other words, proper rotations fix a line through the origin. Proper rotations go around an axis.

All proper rotations in R³:

C_n (n = 1, 2, 3, ... )

D_n (n = 2, 3, ... )

T = symmetries of tetrahedron

W = symmetries of cube/octahedron

P = symmetries of dodecahedron/icosahedron

Proof (from Weyl):

We’ll say we have a group of rotations and then show that it has to be one of those above.

Let G be a group of rotations = {e, g₁, g₂, ... , g_n-1}

|G| = n

We can think of the points on a unit sphere.

Take an arbitrary element g with order v. Then v is bigger than 1 if g is not e.

The rotation g fixes an axis.

There are exactly v rotations (including e) that fix that axis.

Let p be a point on the axis.

G_p = rotations fixing p = {e, a₁, a₂, ... , a_n-1} a^v = e

This is a cyclic group with order v.

Let q be a point not on the axis.

Then g takes q to q_i

q ~ q₁ ~ q₂ ~ ... ~q_v-1

“~” means you can get there by g

The equivalence class of q is all places q can end up.

There are n possible g’s (rotations) so n equivalence classes.|

A Useful Thing in Weyl: |C_p| = n/v (see Weyl for proof)

Thus, v divides n and (number of p’s)·v = n

The number of combos (g is not e, p) = 2(n - 1) because there are 2 p’s for each rotation and (n - 1) is the number of rotations possible excluding the identity.

For each p there are (v - 1) rotations that are not the identity

So the number of combos (g is not e, p) =SUM(v_p - 1) as we range over p.

Put our two formulae together to get 2(n - 1) =SUM(v_p - 1) =SUM(# of p’s)(# of rotations for each) as we range over C_p. Plug in to get

SUM(n/v)(v - 1) as we range over C_p

Divide by n on both sides. Now we have Formula $:

2 - (2/n) = SUM(1 - 1(1/v)) as we range over C_p

If n = 1 then G = {e} = C₁

Else n >= 2 (there are at least 2 rotations in G)

Then 2 - (2/n) is at least 1

So we get Formula @:

1 <= 2 - (2/n) < 2

n >= 2 means there are at least 2 C_p’s.

v >= 2 means that 1 - (1/v) >= ½

Use $ and @ to get 2 - (2/n) = SUM(1 - 1(1/v)) < 2

We see that the number of C_p’s must be less than 4.

So the number of C_p’s is either 2 or 3.

Case 1: There are 2 C_p’s

2 - (2/n) = 1 - (1/v₁) + 1 - (1/v₂)

2 - (2/n) = 2 - ((1/v₁)+(1/v₂))

2/n = (1/v₁) + (1/v₂)

2 = (n/v₁) + (n/v₂)

Since n, v₁, v₂are integers, (n/v₁) = 1 and (n/v₂) = 1

So n = v₁ = v₂

Size of C_p = (n/v) = 1 (ie there is only one pole in class or |g| = n)

Then G = C_n = {rotations around vertical axis}

Case 2: There are 3 C_p’s

2 - (2/n) = 1 - (1/v₁) + 1 - (1/v₂) + 1 - (1/v₃)

2 - (2/n) = 3 - ((1/v₁) + (1/v₂) + (1/v₃))

-(1 + (2/n) = - ((1/v₁) + (1/v₂) + (1/v₃))

1 + (2/n) = (1/v₁) + (1/v₂) + (1/v₃)

Arrange so that v₁ <= v₂ <= v₃

Suppose v₁ > 2

Then at the biggest, (1/v₁) + (1/v₂) + (1/v₃) = 1/3 + 1/3 + 1/3 = 1

So (1/v₁) + (1/v₂) + (1/v₃) = 1 + (2/n) <= 1

Subtract 1 from both sides. We get (2/n) <= 0, which contradicts n being a positive integer.

So v₁ = 2.

2 <= v₂<= v₃

Then v₂ = 2 or 3 because (1/v₂) + (1/v₃) = ½ + (2/n) so v₂ < 4.

Subcase a: v₂ = 2

Then 1 + 2/n = ½ + ½ + v₃

2/n = v₃

n = 2 v₃

2 = n/v₃ = # of p’s in class C_p = # of places g₃ takes p

g₃ = pancake so G = D' _n

Subcase b: v₂ = 3

Then 1 + 2/n = ½ + 1/3 + 1/v₃

1/6 + 2/n = 1/v₃

Since v₂ <= v₃, 3 <= v₃

If v₃= 3, n = 12 and we have G = T

If v₃= 4, n = 24 and we have G = W

If v₃= 5, n = 60 and we have G = P

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POLYHEDRA

There are five regular polyhedra in space: tetrahedron, cube, octahedron, dodecahedron, and icosahedron.

Symmetries of polyhedra are not cyclic or dihedral. The Platonic solids have rotations around axes and reflections about planes.

Rotations of Tetrahedron:

There are three axes of rotation for rotations by P (one for each pair of opposite edges)

There are four axes of rotation for rotations by P /3 (one for each face and its opposite vertex)

Reflections of Tetrahedron:

There are six planes of reflection (one for each edge)

Rotations of Cube:

There are three axes of rotation for rotations by P /2 (one for each pair of faces)

There are four axes of rotation for rotations by 2 P /3 (one for each pair of opposite vertices)

There are six axes of rotation for rotations by rotations by P (one for each pair of opposite edges)

Reflections of Cube:

There are four planes of reflection that cut the cube into two identical rectangular prisms

There are six planes of reflection that cut the cube diagonally across two opposite faces

The cube also has point reflection symmetry about its center

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SOURCES

For stuff we did last semester, see our class notes starting 10/17/2000 and handout for 10/19 and 10/20/2000.

Sources for new stuff:

Holden, Alan. Shapes, Space, and Symmetry. Dover Publications, Inc. (New York, 1971).

Lockwood, E.H. and R.H. MacMillan. Geometric Symmetry. Cambridge University Press (New York, 1978).

Rosen, Joe. Symmetry Discovered: Concepts and Applications in Nature and Science. Cambridge University Press (New York, 1975).

Weyl, Hermann. Symmetry. Princeton University Press (Princeton, 1952).

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