11.3. FINITE ROTATION GROUPS
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contained in the interior of the unit sphere S . The projection x 7! x=jjxjj
onto the sphere maps the polyhedron bijectively onto the sphere. For example, Figure 11.2.3 on the facing page shows the projection of a dodecahedron onto the sphere. The image of the edges and vertices is a connected graph G on the sphere with e edges and v vertices, and each face of the polyhedron projects onto a connected component in S n G . Hence, Euler’s theorem for polyhedra follows from Theorem 11.2.6.
n
Exercises 11.2 11.2.1. Prove Lemma 11.2.2.
11.2.2. Explain how the data p 3, q D 2 can be sensibly interpreted, in the context of Lemma 11.2.2.
11.2.3. Determine by case-by-case inspection that the symmetry groups of the five regular polyhedra satisfy jGj D vq D fp:
Find an explanation for these formulas. Hint: Consider actions of G on vertices and on faces.
11.2.4. Show that a tree always has a vertex with valence 1, and that in a connected tree, there is exactly one path between any two vertices.
11.3. Finite Rotation Groups
In this section we will classify the finite subgroups of SO.3; R/ and O.3; R/, largely by means of exercises.
It’s easy to obtain the corresponding result for two dimensions: A fi nite subgroup of SO.2; R/ is cyclic. A finite subgroup of O.2; R/ is either cyclic or a dihedral group Dn for n 1 (Exercise 11.3.1).
The classification for SO.3; R/ proceeds by analyzing the action of the finite group on the set of points left fixed by some element of the group. Let G be a finite subgroup of the rotation group SO.3; R/. Each nonidentity element g 2 G is a rotation about some axis; thus, g has two fixed points on the unit sphere S , called the poles of g. The group G acts on the set P of poles, since if x is a pole of g and h 2 G, then hx is a pole of hgh 1.
You are asked to show in the Exercises that the stabilizer of a pole in G is a nontrivial cyclic group.
Let M denote the number of orbits of G acting on P . Applying Burn side’s Lemma 5.2.2 to this action gives
1
M D .jP j C 2.jGj 1//;
jGj
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11. ISOMETRY GROUPS because the identity of G fixes all elements of P , while each nonidentity element fixes exactly two elements. We can rewrite this equation as
.M
2/ jGj D jP j 2;
(11.3.1)
which shows us that M 2:
We have the following data for several known finite subgroups of the rotation group:
group
jGj orbits orbit sizes stabilizer sizes
Zn
n
2
1; 1 n; n
Dn
2n
3
2; n; n n; 2; 2
tetrahedron
12
3
6; 4; 4 2; 3; 3
cube/octahedron
24
3
12; 8; 6 2; 3; 4
dodec/icosahedron
60
3
30; 20; 12 2; 3; 5 Theorem 11.3.1. The finite subgroups of SO.3; R/ are the cyclic groups Zn, the dihedral groups Dn, and the rotation groups of the regular polyhedra.
Proof. Let x1; : : : ; xM be representatives of the M orbits. We rewrite
M
M
X jO.xi /j
X
1
jP j=jGj D
D
:
jGj
jStab.xi /j
i D1 i D1
Putting this into the orbit counting equation, writing M as PM i D1 1, and rearranging gives
M
X
1
1
.1
/ D 2.1 /:
(11.3.2)
jStab.xi /j
jGj
i D1
Now, the right-hand side is strictly less than 2, and each term on the left is at least 1=2, since the size of each stabilizer is at least 2, so we have 2 M 3: If M D 2, we obtain from Equation (11.3.1) or (11.3.2) that there are exactly two poles, and thus two orbits of size 1. Hence, there is only one rotation axis for the elements of G, and G must be a finite cyclic group.
If M D 3, write 2 a b c for the sizes of the stabilizers for the three orbits. Equation (11.3.2) rearranges to
1
1
1
2
C
C
D 1 C :
(11.3.3)
a
b
c
jGj
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Since the right side is greater than 1 and no more than 2, the only solutions are .2; 2; n/ for n 2, .2; 3; 3/, .2; 3; 4/, and .2; 3; 5/.
The rest of the proof consists of showing that the only groups that can realize these data are the dihedral group Dn acting as rotations of the regular n–gon, and the rotation groups of the tetrahedron, the octahedron, and the icosahedron. The idea is to construct the geometric figures from the data on the orbits of poles.
It’s a little tedious to read the details of the four cases but fun to work out the details for yourself. The Exercises provide a guide.
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Exercises 11.3 11.3.1. Show that the finite subgroup of SO.2; R/ is cyclic and, consequently, a finite subgroup of SO.3; R/ that consists of rotations about a single axis is cyclic. Show that a finite subgroup of O.2; R/ is either cyclic or a dihedral group Dn for n 1.
11.3.2. Show that the stabilizer of any pole is a cyclic group of order at least 2. Observe that the stabilizer of a pole x is the same as the stabilizer of the pole x, so the orbits of x and of x have the same size. Consider the example of the rotation group G of the tetrahedron. What are the orbits of G acting on the set of poles of G? For which poles x is it true that x and x belong to the same orbit?
11.3.3. Let G be a finite subgroup of SO.3; R/ with the data M D 3, .a; b; c/ D .2; 2; n/ with n 2. Show that jGj D 2n, and the sizes of the three orbits of G acting on P are n; n, and 2. The case n D 2 is a bit special, so consider first the case n 3. There is one orbit of size 2, which must consist of a pair of poles fx; xg; and the stabilizer of this pair is a cyclic group of rotations about the axis determined by fx; xg. Show that there is an n-gon in the plane through the origin perpendicular to x, whose n vertices consist of an orbit of poles of G, and show that G is the rotation group of this n-gon.
11.3.4. Extend the analysis of the last exercise to the case n D 2. Show that G must be D2 Š Z2 Z2, acting as symmetries of a rectangular card.
11.3.5. Let G be a finite subgroup of SO.3; R/ with the data M D 3, .a; b; c/ D .2; 3; 3/. Show that jGj D 12, and the size of the three orbits of G acting on P are 6; 4, and 4. Consider an orbit O P of size 4, and let
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11. ISOMETRY GROUPS u 2 O. Choose a vector v 2 O nfu; ug. Let g 2 G generate the stabilizer of u (so o.g/ D 3).
Conclude that fu; v; gv; g2vg D O. Deduce that fv; gv; g2vg are the three vertices of an equilateral triangle, which lies in a plane perpendicular to u and that jju vjj D jju gvjj D jju g2vjj.
Now, let v play the role of u, and conclude that the four points of O are equidistant and are, therefore, the four vertices of a regular tetrahedron
T . Hence, G acts as symmetries of T ; since jGj D 12, conclude that G is the rotation group of T .
11.3.6. Let G be a finite subgroup of SO.3; R/ with the data M D 3, .a; b; c/ D .2; 3; 4/. Show that jGj D 24, and the size of the three orbits of G acting on P are 12; 8, and 6.
Consider the orbit O P of size 6. Since there is only one such orbit, O must contain together with any of its elements x the opposite vector x.
Let u 2 O. Choose a vector v 2 O nfu; ug. The stabilizer of fu; ug is cyclic of order 4; let g denote a generator of this cyclic group.
Show that fv; gv; g2v; g3vg is the set of vertices of a square that lies in a plane perpendicular to u. Show that v D g2v and that the plane of the square bisects the segment Œu; u.
Using rotations about the axis through v; v, show that O consists of the 6 vertices of a regular octahedron. Show that G is the rotation group of this octahedron.
11.3.7. Let G be a finite subgroup of SO.3; R/ with the data M D 3, .a; b; c/ D .2; 3; 5/.Show that jGj D 60, and the size of the three orbits of G acting on P are 30; 20, and 12.
Consider the orbit O P of size 12. Since there is only one such orbit, O must contain together with any of its elements x the opposite vector x.
Let u 2 O and let v 2 O satisfy jju vjj jju yjj for all y 2 Onfug.
Let g be a generator of the stabilizer of fu; ug, o.g/ D 5. Show that the 5 points fgi v W 0 i 4g are the vertices of a regular pentagon that lies on a plane perpendicular to u.
Show that the plane of the pentagon cannot bisect the segment Œu; u, that u and the vertices of the pentagon lie all to one side of the bisector of Œu; u, and finally that the 12 points f˙u; ˙gi v W 0 i 4g comprise the orbit O. Show that these 12 points are the vertices of a regular icosahedron and that G is the rotation group of this icosahedron.
It is not very much work to extend our classification results to finite subgroups of O.3; R/. If G is a finite subgroup of O.3; R/, then H D
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