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Section 19.1 Discovery guide

In this set of discovery activities, we will attempt to use our knowledge of group actions and orbits to determine the possibilities for the structure of a finite subgroup of \(\SO_3(\R)\text{.}\)

Subsection Preliminary analysis

Recall that elements of \(\SO_3(\R)\) act on \(\R_3\) by matrix multiplication. Geometrically, the action of each individual element of \(\SO_3(\R)\) is a rotation about some axis through the origin. The vector space \(\R_3\) is a very large set; if we are only dealing with rotations then we can restrict to considering just the unit sphere \(\mathcal{S}\) in \(\R^3\text{.}\) (Here \(\mathcal{S}\) is just the surface of the sphere; points interior to the sphere are not included.)

Discovery 19.1.

How many fixed points on \(\mathcal{S}\) does a non-identity element of \(\SO_3(\R)\) have? How are these fixed points arranged on the surface of \(\mathcal{S}\text{?}\)

For reasons that should be obvious from the first activity above, a fixed point on \(\mathcal{S}\) for a particular element of \(\SO_3(\R)\) will be called a pole for the rotation.

Discovery 19.2.

Suppose \(g\) and \(h\) are elements of \(\SO_3(\R)\text{,}\) and \(x\) is a pole of \(g\text{.}\) That is, \(x\) is on \(\mathcal{S}\) and \(\grpact{g}{x} = x\text{.}\)

Demonstrate that the result \(\grpact{h}{x}\) of rotating \(x\) by \(h\) is a pole for some combination of \(g\) and \(h\text{.}\)

Now let \(G\) represent a finite subgroup of \(\SO_3(\R)\text{.}\) The unit sphere \(\mathcal{S}\) is still too large a set — to use our orbit-counting knowledge we need to have a finite group acting on a finite set. So let \(X\) represent the set of only those points on \(\mathcal{S}\) that are a pole for some element of \(G\) — then Discovery 19.2, combined with the closure properties of \(G\text{,}\) verifies that \(X\) is closed under the action of \(G\text{.}\)

The Counting Theorem says that the number of orbits is the average size of the fixed sets:

\begin{gather} N = \frac{1}{\grporder{G}} \sum_{G} \cardinality{X^g}\text{.}\tag{✶} \end{gather}

Discovery 19.3.

Keeping in mind that the elements of \(G\) are rotations of three-dimensional space, what are the possibilities for the value of \(\cardinality{X^g}\text{?}\)

Hint.

Careful: There is one exceptional case of group element \(g\text{.}\)

Based on Discovery 19.3, we can rewrite (✶) as

\begin{gather} N = \frac{\cardinality{X}}{\grporder{G}} + 2 \left( \frac{\grporder{G} - 1}{\grporder{G}} \right)\text{.}\tag{✶✶} \end{gather}

Now, the orbits partition \(X\text{,}\) so we have

\begin{gather} \cardinality{X} = \cardinality{\orbit{1}} + \cardinality{\orbit{2}} + \dotsb + \cardinality{\orbit{N}}\text{,}\tag{†} \end{gather}

where \(\orbit{1},\orbit{2},\dotsc,\orbit{N}\) is some enumeration of the \(N\) orbits. Furthermore, as \(N\) is a whole number we can think of it as the sum of \(N\) ones, and so we can rearrange our formula for \(N\) above to

\begin{gather} \sum_{k=1}^N \left( 1 - \frac{\cardinality{\orbit{k}}}{\grporder{G}} \right) = 2 \left( \frac{\grporder{G} - 1}{\grporder{G}} \right)\text{.}\tag{††} \end{gather}

Discovery 19.4.

(a)

Argue that each term in the sum on the left in (††) satisfies

\begin{equation*} \frac{1}{2} \le \left( 1 - \frac{\cardinality{\orbit{k}}}{\grporder{G}} \right) \lt 1 \text{.} \end{equation*}
Hint.

  • Consider the Orbit-Stabilizer Theorem.

  • Can an element of \(X\) have a trivial stabilizer under the action of \(G\text{?}\)

(b)

Assuming \(G\) is not just the trivial subgroup of \(\SO_3(\R)\text{,}\) how large can the right-hand side of (††) be? How small could it be?

(c)

Compare the left- and right-hand sides of (††) using Task a and Task b to come up with possibilities of \(N\text{.}\)

Hint.

Keep in mind that \(N\) is an integer, and is equal to the number of terms in the sum on the left-hand side of (††).

Subsection Case of Two Orbits

In this subsection, assume \(N = 2\text{.}\)

Discovery 19.5.

(a)

Using \(N = 2\text{,}\) solve for \(\cardinality{X}\) in (✶✶).

(b)

With \(N = 2\text{,}\) decomposition (†) becomes

\begin{equation*} \cardinality{X} = \cardinality{\orbit{1}} + \cardinality{\orbit{2}} \text{.} \end{equation*}

Given the value for \(\cardinality{X}\) you calculated in Task a, what must the size of each orbit be?

Note: An orbit cannot be empty.

(c)

Remember that elements of \(X\) are poles of the rotations in \(G\text{.}\) Given the size of \(X\) you calculated in Task a, how many different axes of rotation can there be for elements of \(G\text{?}\)

Note: Many different elements of \(G\) can share the same axis of rotation, just with different angles of rotation around that axis.

(d)

Based on your answer to Task c, argue that \(G\) must actual be isomorphic to a subgroup of \(\SO_2(\R)\text{.}\)

Hint.

Consider a plane through the origin, perpendicular to the axis of rotation.

Remark 19.1.1.

When you do the textbook reading for this chapter you will discover that all finite subgroups of \(\SO_2(\R)\) are cyclic, so the conclusion of Discovery 19.5 implies that in the current case of \(N = 2\) our subgroup \(G\) of \(\SO_3(\R)\) must be cyclic.

Subsection Case of Three Orbits

Using \(N = 3\) in (✶), substituting (†), and rearranging, we obtain

\begin{equation*} 1 + \frac{2}{\grporder{G}} = \frac{\cardinality{\orbit{1}}}{\grporder{G}} + \frac{\cardinality{\orbit{2}}}{\grporder{G}} + \frac{\cardinality{\orbit{3}}}{\grporder{G}} \text{.} \end{equation*}

Let \(x_1,x_2,x_3\) represent some arbitrary choice of orbit representatives, \(x_j \in \orbit{j}\text{.}\) Using the Orbit-Stabilizer Theorem, from the above equality we obtain

\begin{gather} 1 + \frac{2}{\grporder{G}} = \frac{1}{\grporder{G_{x_1}}} + \frac{1}{\grporder{G_{x_2}}} + \frac{1}{\grporder{G_{x_3}}}\text{.}\tag{✶✶✶} \end{gather}

Discovery 19.6.

Clearly the left-hand side of (✶✶✶) must be greater than \(1\text{.}\) But the three terms on the right-hand side must each be less than \(1\text{!}\) (Remember that each element of \(X\) is a pole of at least one rotation from \(G\text{,}\) so its stabilizer will not be trivial.)

Assuming that representative poles \(x_1,x_2,x_3\) have been listed in order the size of their stabilizers, what are the possible combinations of stabilizer orders to meet the restriction that the sum on the right-hand side of (✶✶✶) must be greater than \(1\text{?}\)

You should have come up with a number of possibilities in Discovery 19.6. We will proceed with the analysis of only one of those possibilities, as an example.

Assume

\begin{equation*} \grporder{G_{x_1}} = 2, \quad \grporder{G_{x_2}} = 3, \quad \grporder{G_{x_3}} = 3 \text{.} \end{equation*}

Discovery 19.7.

(a)

Using our assumptions on the stabilizer sizes, solve (✶✶✶) for \(\grporder{G}\text{,}\) and then use the Orbit-Stabilizer Theorem to determine the size of each of the three orbits.

(b)

Given their orders, what form of group must each of these three stabilizers be?

Now we will focus on the third representative pole \(x_3\text{,}\) its orbit \(\orbit{3}\text{,}\) and its stabilizer \(G_{x_3}\text{.}\) For notational simplicity, let us just write \(x = x_3\text{,}\) and also write \(\orbit{x}\) and \(G_x\) for the orbit and stabilizer, respectively, of this particular pole.

Let \(x'\) represent the pole in \(X\) that is diametrically opposed to \(x\text{.}\) Now, \(x'\) may or may not be in the same orbit as \(x\text{,}\) but given the size of \(\orbit{x}\) that you calculated in Discovery 19.7 (remember \(\orbit{x} = \orbit{3}\) now), \(\orbit{x}\) must contain a pole that is not diametrically opposed to \(x\text{.}\) Write \(a\) for one choice of such pole in \(\orbit{x}\text{.}\)

Two poles in the same orbit and a stabilizing group element.
Figure 19.1.2. Pole \(x\) and second point \(a\) from orbit \(\orbit{x}\text{,}\) along with the diametrically opposed pole \(x'\) and the stabilizing element \(g_x\text{.}\)

In Figure 19.1.2, group element \(g_x\) represents a generator for the cyclic stabilizer \(G_x\text{.}\)

Discovery 19.8.

(a)

Based on your calculation of \(\grporder{G_x}\text{,}\) what is the angle of rotation of \(g_x\text{?}\)

(b)

The image points \(b = \grpact{g_x}{a}\) and \(c = \grpact{g_x^2}{a}\) are also in the orbit \(\orbit{x}\text{.}\) Could either of these points be \(x\text{?}\) Where on the sphere must \(b\) and \(c\) be located? (Refer to Figure 19.1.2.)

Our analysis started with the arbitrary choice of point \(x\) from orbit \(\orbit{3} = \orbit{x}\text{.}\) So if we repeated the analysis with \(a\) in place of \(x\text{,}\) we would obtain similar results — the rotation axis in Figure 19.1.2 would pass through \(a\) and the stabilizing element \(g_a\) from cyclic \(G_a\) would fix \(a\) and rotate \(x, b, c\text{,}\) one to the next. Or similarly we could have started with \(b\) or with \(c\) in place of \(x\text{.}\)

Discovery 19.9.

(a)

Based on your results from Discovery 19.8 and the discussion following that activity, what geometric shape do the points \(x, a, b, c\) form, inscribed with the sphere?

(b)

In Discovery 19.7 you determined \(\grporder{G}\text{.}\) Combine this with the your answer to Task a to conjecture a familiar group from the very beginning of our study of group theory to which \(G\) must be isomorphic.

(c)

In Discovery 19.7 you also determined the size of all three orbits. Since the orbits partition \(X\text{,}\) this means you also know \(\cardinality{X}\text{.}\) Recall that \(X\) contains the poles of the non-identity elements of \(G\text{,}\) and that poles come in diametrically-opposed pairs. Based on your recollection of that familiar group identified in Task b, count up the number of poles created by those familiar axes of rotation, and make sure the number agrees with your previous number for \(\cardinality{X}\text{.}\)