MIT 18.701 — Lecture 12

Orbit-Stabilizer, Symmetries of $SO_3(\mathbb{R})$

§ Orbit and Stabilizer Definitions

Definition (Orbit and Stabilizer). Let $G$ be a group acting on a set $S$ . Then for any $s \in S$ :

The orbit of $s$ is $Gs := \{gs : g \in G\} \subseteq S$ . (the set of reachable destinations)
The stabilizer of $s$ is $\Stab_G(s) := \{g \in G: gs = s\} \subseteq G$ . (the subgroup of do-nothing actions)

Example. Consider the group $D_4$ acting on the set $\mathbb{R}^2$ .

Considering the point $(0, 0)$ , we have $D_4(0, 0) = \{(0, 0)\}$ and $\Stab_{D_4}((0, 0)) = D_4$ .
Considering the point $(1, 2)$ , we have $D_4(1, 2) = \{(\pm 1, \pm 2), (\pm 2, \pm 1)\}$ and $\Stab_{D_4}((1, 2)) = \{\mathrm{id}\}$ .
Considering the point $(1, 1)$ , we have $D_4(1, 1) = \{(\pm 1, \pm 1)\}$ and $\Stab_{D_4}((1, 1)) = \{\mathrm{id}, \text{reflect about $y = x$}\}$ .

Intuitively, any group action $G$ on a set $S$ yields a partitioning of $S$ into orbits.

Definition (Transitive). A group $G$ acting on $S$ is transitive if $Gs = S$ for all $s \in S$ ; that is, there is only one orbit.

Remark. Any group action $G$ on a set $S$ is the disjoint union of transitive group actions $G$ on each of its orbits.

§ Same Orbit $\implies$ Conjugate Stabilizers

Theorem. (Same Orbit $\implies$ Conjugate Stabilizers) Let $G$ be a group acting on a set $S$ . Then for any $s_1$ and $s_2$ in the same orbit, the subgroups $\Stab_G(s_1)$ and $\Stab_G(s_2)$ are conjugates of each other.

In particular, for every $s \in S$ and $g \in G$ , we have $\Stab_G(gs) = g \Stab_G(s) g^{-1}$ .

Proof: Notice $g' \in \Stab_G(gs)$ iff $g'(gs) = gs$ , meaning $g^{-1}g'g \in \Stab_G(s)$ , which says $g' \in g\Stab_G(s)g^{-1}$ . $\blacksquare$

The above theorem is very powerful! Some implications:

Fact 1. Any two elements of $S$ in the same orbit have stabilizers of the same size.
- In particular, if $G$ acts transitively on $S$ , then the stabilizer of every $s \in S$ has the same size.
Fact 2. If the stabilizer $\Stab_G(s) = H$ satisfies $H \unlhd G$ , then $\Stab_G(t) = H$ for all $t \in Gs$ .

§ Group Actions are Coset Actions

Here's a natural question:

Question: For any group $G$ and subgroup $H \subseteq G$ , how can we designate $G$ as a transitive group action on some set $S$ such that $H = \Stab_G(s)$ for some $s \in S$ ?

Here's a natural answer:

Answer: Designate $G$ as a group action on $G / H$ by left multiplication, where $G / H$ is the set of left cosets of $H$ in $G$ . Then $G$ acts transitively on $G / H$ .

Note that $\Stab_G(gH) = gHg^{-1}$ for all $g \in G$ ; in particular, $\Stab_G(H) = H$ . If $H \unlhd G$ , then in fact $\Stab_G(gH) = H$ for every $g$ .

It turns out that the above is the only right answer!

Theorem. (Transitive Actions are Coset Actions) Suppose $G$ is a transitive group action on $S$ , and for some $s \in S$ , we have $\Stab_G(s) = H$ . Then the group action of $G$ on $S$ is isomorphic to the group action of $G$ on $G / H$ .

Proof: Equivalently, we seek a bijection $f: S \to G/H$ such that $f(g \cdot s') = g \cdot f(s')$ for all $g \in G$ and $s' \in S$ .

Since $G$ is transitive, we can write every $s' \in S$ as $s' = g' s$ for some $g' \in G$ . So equivalently, we seek an $f: S \to G / H$ such that $f(g \cdot g's) = g \cdot f(g's)$ for all $g, g' \in G$ .

Turns out picking $f$ is really easy: choosing $f(gs) = gH$ for all $g \in G$ just works. But we still need to make sure this choice of $f$ is well-defined; that is, we must check:

Claim. If we have $g_1s = g_2s$ for some $g_1, g_2 \in G$ , then we must also have $g_1 H = g_2 H$ .

Proof: Since $g_1s = g_2s$ , we have $s = g_1^{-1}g_2s$ , meaning $g_1^{-1}g_2 \in \Stab_G(s) = H$ . This means $g_2 \in g_1H$ , so $g_1$ and $g_2$ correspond to the same left coset of $H$ , done. $\blacksquare$

The above theorem is very strong! It has the following corollary:

Theorem. (Group Actions are Coset Actions) If $G$ is a group action on $S$ , and $\Stab_G(s) = H$ , then the group action of $G$ on the orbit $Gs$ of $s$ is isomorphic to the group action of $G$ on $G/H$ .

§ The Orbit-Stabilizer Theorem

The previous theorem can be interpreted combinatorially, too.

Theorem. (Orbit-Stabilizer) For any group $G$ acting on a set $S$ , and any $s \in S$ , we have $|G| = |Gs| \cdot |\Stab_G(s)|$ .

Proof: Recall that actions of $G$ on $Gs$ are isomorphic to actions of $G$ on $G / \Stab_G(s)$ . Thus, $Gs \cong G/\Stab_G(s)$ .

Therefore, $|Gs| = [G : \Stab_G(s)] = |G| / |\Stab_G(s)|$ , which implies $|G| = |Gs| \cdot |\Stab_G(s)|$ , as desired. $\blacksquare$

Example (Cube Symmetries). Let $G \subseteq SO_3(\mathbb{R})$ be the symmetry group of the cube.

Since $G$ acts transitively on $V = \{\text{all $8$ vertices}\}$ , and $|\Stab_G(v)| = 3$ for any vertex $v$ , we have $|G| = 8 \times 3 = 24$ .
Since $G$ acts transitively on $E = \{\text{all $12$ edges}\}$ , and $|\Stab_G(e)| = 2$ for any edge $e$ , we have $|G| = 12 \times 2 = 24$ .
Since $G$ acts transitively on $F = \{\text{all $6$ faces}\}$ , and $|\Stab_G(f)| = 4$ for any face $f$ , we have $|G| = 6 \times 4 = 24$ .

Of course, all three choices of orbit $Gs$ yield the same result $|Gs| \times |\Stab_G(s)| = 24$ .

§ Classifying Symmetries of $SO_3(\mathbb{R})$

We've been classifying geometric groups in $\mathbb{R}^2$ so far. Now it's time to handle $\mathbb{R}^3$ .

Question. What do all of the finite subgroups of $SO_3(\mathbb{R})$ look like?

Let $G \subseteq SO_3(\mathbb{R})$ be a finite subgroup of $SO_3(\mathbb{R})$ with size $N = |G|$ . The nice thing about $SO_3(\mathbb{R})$ is that—as shown at the end of Lecture 9—it consists entirely of pure rotations about an axis through the origin.

Definition (Pole). Let $S^2 := \{x \in \mathbb{R}^3: |x| = 1\}$ be the unit sphere. Then every $g \in G \setminus \{\mathrm{id}\}$ has two poles, those being the two points of $S^2$ fixed by $g$ . Furthermore, we define:

P := \{(g, p) \mid g \in G \setminus \{\mathrm{id}\} \text{ and } p \text{ is a pole of } g\}.

Here comes the clever observation; the proof of the following theorem is plain computation.

Theorem. (Subgroup Acts on Its Poles) We can view $G$ as a group action on $P$ defined by:

(g_1, p_1) \in P ~ \xrightarrow{\text{multiplication by } g_2 \in G} ~ (g_2g_1g_2^{-1}, g_2p_1) \in P.

Remark. Where does this group action come from? Well, to construct a group action of $G$ on $P$ , we would first like to show that if $p_1$ is a pole of $g_1$ , then $g_2p_1$ is a pole of some element of $G$ as well.

To prove this, the key is to notice that $p$ is a pole if and only if $\Stab_G(p)$ is nontrivial. But recall that $\Stab_G(g_2p_1)$ is the conjugation of $\Stab_G(p_1)$ by $g_2$ , so $\Stab_G(g_2p_1)$ is nontrivial if and only if $\Stab_G(p_1)$ is nontrivial.

Thus, $g_2p_1$ is a pole if and only if $p_1$ is a pole! It then remains to answer the question: which element of $G$ is $g_2p_1$ a pole of? The answer is $g_2g_1g_2^{-1}$ , again because $\Stab_G(g_2p_1)$ is the conjugation of $\Stab_G(p_1)$ by $g_2$ .

Say the induced group action on the set of poles partitions it into $k$ orbits of sizes $n_1, \dots, n_k$ . We now count $|P|$ in two ways:

For each of the $n_i$ poles $p$ in the $i^{\text{th}}$ orbit, there are $\frac{N}{n_i} - 1$ nontrivial group actions that fix that pole; this is because of the Orbit-Stabilizer theorem.

Thus, the $i^{\text{th}}$ orbit contributes $n_i \cdot \left(\frac{N}{n_i} - 1\right)$ ordered pairs $(g, p)$ to $P$ . Therefore, $|P| = \sum_{i = 1}^{k} n_i\left(\frac{N}{n_i} - 1\right)$ .
For each of the $N - 1$ group actions $g \in G$ , there are $2$ poles $p$ that it fixes. Therefore, $|P| = 2(N - 1)$ .

Setting these two quantities equal to each other yields the following Diophantine Equation:

2(N - 1) = \sum_{i = 1}^k n_i\left(\dfrac{N}{n_i} - 1\right) ~ \implies ~ 2 - \dfrac{2}{N} = \sum_{i = 1}^k \left(1 - \dfrac{1}{r_i}\right) \text{ for integers } r_i := \dfrac{N}{n_i}.

The rest of the argument is solely number-theoretic; the first step is bounding.

Claim. If $k = 1$ , then $N = 1$ , meaning $G$ is trivial.

Proof: This is because if $k = 1$ , then the RHS is less than $1$ , whereas the LHS is at least $1$ for $N > 1$ .
Claim. It is impossible to have $k \geq 4$ .

Proof: Recall that there are $\frac{N}{n_i} - 1 = r_i - 1$ nontrivial group actions that fix any pole in the $i^{\text{th}}$ orbit. But every pole—by nature of being a pole—is fixed by at least $1$ nontrivial group action. Thus, $r_i \geq 2$ for all $i$ .

This means that for $k \geq 4$ , the RHS is at least $4 - \frac{4}{2} = 2$ , whereas the LHS is less than $2$ , contradiction.

Thus, either $k = 2$ or $k = 3$ . Now this is just a pure number theory problem; it turns out the solutions are:

For $k = 2$ , we have $\frac{1}{r_1} + \frac{1}{r_2} = \frac{2}{N}$ , which forces $(r_1, r_2) = (N, N)$ .
For $k = 3$ , we have $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = 1 + \frac{2}{N}$ . It turns out this has solutions:
$(r_1, r_2, r_3, N) \in \left \{ \left(2, 2, \frac{N}{2}, N\right), \ (2, 3, 3, 12), \ (2, 3, 4, 24), \ (2, 3, 5, 60) \right \}.$

That's just five cases in total! It still remains to consider what each of these five cases means geometrically for the entire group, though—but the details are messy, so we omit them here.

It turns out that these five solutions correspond to the following five symmetry groups.

Rotation Symmetry Groups in $\mathbb{R}^3$ .

$(N, N, N)$ : the symmetry group $G \cong C_N$ of a pyramid with a regular $N$ -gon as a base.
$(2, 2, \tfrac{N}{2}, N)$ : the symmetry group $G \cong D_n$ of a prism with a regular $n$ -gon as a base.
$(2, 3, 3, 12)$ : the symmetry group $G$ of the tetrahedron.
$(2, 3, 4, 24)$ : the symmetry group $G$ of the cube (or its dual, the octahedron).
$(2, 3, 5, 60)$ : the symmetry group $G$ of the icosahedron (or its dual, the dodecahedron).

And that's all the finite rotation symmetry groups in $\mathbb{R}^3$ .

Remark. Recall the following step in the above argument.

Say the induced group action on the set of poles partitions it into $k$ orbits of sizes $n_1, \dots, n_k$ . We now count $|P|$ in two ways:

For each of the $n_i$ poles $p$ in the $i^{\text{th}}$ orbit, there are $\frac{N}{n_i} - 1$ nontrivial group actions that fix that pole; this is because of the Orbit-Stabilizer theorem.

Thus, the $i^{\text{th}}$ orbit contributes $n_i \cdot \left(\frac{N}{n_i} - 1\right)$ ordered pairs $(g, p)$ to $P$ , so $|P| = \sum_{i = 1}^{k} n_i\left(\frac{N}{n_i} - 1\right)$ .
For each of the $N - 1$ group actions $g \in G$ , there are $2$ poles $p$ that it fixes, so $|P| = 2(N - 1)$ .

Is the truth of these statements unique to the setting of $SO_3(\mathbb{R})$ ?

§ Orbit and Stabilizer Definitions

§ Same Orbit ⟹ \implies⟹ Conjugate Stabilizers

§ Group Actions are Coset Actions

§ The Orbit-Stabilizer Theorem

§ Classifying Symmetries of SO3(R)SO_3(\mathbb{R})SO3​(R)

§ Same Orbit $\implies$ Conjugate Stabilizers

§ Classifying Symmetries of $SO_3(\mathbb{R})$