Permutation Groups vis-à-vis Conformal Maps of the Riemann Sphere

In this post, we discuss a few ways in which the symmetric and alternating groups can be realized as finite collections of self-maps on the Riemann sphere. Throughout, our group operation will be composition of functions: as such, the maps we choose will necessarily be homeomorphisms of $\mathbb{P}^1$ . Within this broad framework, two classes are of particular interest:

1. The group of biholomorphic maps $f: \mathbb{P}^1 \to \mathbb{P}^1$ (those that respect the structure of $\mathbb{P}^1$ as a Riemann surface). It is well-known that such maps are given by Möbius transformations, i.e. rational functions of the form

$\displaystyle f(z) = \frac{a z +b}{cz +d},$

satisfying $ad-bc \neq 0 \in \mathbb{C}$ . The group of Möbius transformations (also known as the Möbius Group and herein denoted $\mathcal{M}$ ) is naturally isomorphic to $\mathrm{PSL}_2(\mathbb{C})$ , the projective (special) linear group, via:

$\displaystyle\frac{az+b}{cz+d} \mapsto \left(\begin{matrix} a & b \\ c & d \end{matrix} \right).$

2. The group of conformal maps $f: \mathbb{P}^1 \to \mathbb{P}^1$ , denoted $\mathcal{C}$ for brevity. To be clear, here we refer to those maps $f$ which preserve unsigned angle measure. (In contrast, some authors require conformal maps to preserve orientation as well.) We recall the fundamental result that such maps contain the Möbius group as a subgroup of index two. To be specific, any conformal self-map on $\mathbb{P}^1$ is either biholomorphic (returning to case (1)), or bijective and anti-holomorphic: a biholomorphic function of the complex conjugate $\overline{z}$ .

In this post, we enact a two-part program to calculate the maximal $n$ such that the symmetric group $S_n$ injects into $\mathcal{M}$ (resp. $\mathcal{C}$ ). Along the way, we study injections of the alternating group into $\mathcal{M}$ , and highlight some exceptional cases in which our injections can be attached to group actions on a finite invariant set.

— PART I (Injections $\varphi: S_n \to \mathcal{M}$ ) —

Our first goal will be to verify the existence of injective maps $\varphi: S_4 \to \mathcal{M}$ . While it suffices here to provide a single example, it is infinitely more enlightening to show how such an example may be found. As a happy consequence, we’ll end up classifying all possible images $\varphi(S_4)$ up to inner automorphism. To be specific,

Theorem 1.1: There exists an injection $\varphi: S_4 \to \mathcal{M}$ . Moreover, this injection is unique (up to inner automorphism on $\mathcal{M}$ ).

Proof: Recall the group presentation

$S_4 \cong \langle \alpha,\beta \mid \alpha^2 =\beta^3 = (\beta\alpha)^4\rangle.$

Now, suppose that $\varphi: S_4 \to \mathcal{M}$ is given, and let $\sigma,\tau \in \mathcal{M}$ denote the images of $\alpha,\beta$ , respectively. Because $\tau(z) \in \mathrm{PSL}_2(\mathbb{C})$ is elliptic, we may assume $\tau(z)=\lambda z$ (up to inner automorphism on $\mathcal{M}$ ), in which $\lambda \in \mu_3$ is primitive. Since $f \in \mathcal{M}$ has order two if and only if $\mathrm{tr} f =0$ , $\sigma(z)$ takes the form

$\displaystyle\sigma(z) =\frac{az+b}{cz-a},$

Secondly, the relation $(\tau \sigma)^4(z)=z$ (with plenty of algebra!) forces

$a(\lambda-1)(a^2+2bc+a^2\lambda^2)(cz^2-(a+a\lambda)z-b\lambda)=0,$

as a polynomial in $\mathbb{C}[z]$ . Moreover, since $\tau\sigma$ has order exactly four, we find $a^2+2bc+a^2\lambda^2=0$ (else $\tau \sigma$ has order dividing two). Next, because $\lambda^2+\lambda+1=0$ , we can rewrite this as $\lambda(2bc-a^2)=0$ , hence $a^2=2bc$ . In particular, $a$ is non-zero, by a determinant calculation. We may thus projectivize $a=1$ , such that

(1) $\displaystyle\sigma(z)=\frac{z+1/2c}{cz-1},$

in which we have used that $2bc=1$ . Because the centralizer $C_\mathcal{M}(\tau)$ contains the subgroup of diagonal (projective) matrices, we may freely conjugate $\sigma$ by diagonal elements. In particular, we note that

$\displaystyle\left(\begin{matrix} c & 0 \\ 0 & 1 \end{matrix} \right)\left( \begin{matrix} 1 & 1/2c \\ c & -1 \end{matrix} \right)\left( \begin{matrix} c^{-1} & 0 \\ 0 & 1 \end{matrix} \right)=\left(\begin{matrix} 1 & c^2/2 \\ 1 & -1 \end{matrix} \right),$

hence $\sigma(z)$ is conjugate (via $C_\mathcal{M}(\tau)$ ) to a Möbius transformation mapping $\infty \mapsto 1$ . In other words, we may assume $c=1$ in line (1) – again, up to inner automorphism on $\mathcal{M}$ . In this sense, $\sigma$ is uniquely determined by $\tau$ , whereby there exist at most two injections $\varphi: S_4 \to \mathcal{M}$ modulo inner automorphism: one for each choice of $\lambda$ among the roots $\lambda_1,\lambda_2$ of $f(t)=t^2+t+1$ .

As it happens, each choice of $\lambda$ induces a valid injection $\varphi: S_4 \to \mathcal{M}$ (this is finite computation). As a final claim, we leave to the reader to show that conjugation by the Möbius transformation $f(z)=-1/(2z)$ fixes $\sigma(z)$ while interchanging the choices for $\tau(z)$ associated to $\lambda_1$ and $\lambda_2$ . That is, any two injections $\varphi,\varphi': S_4 \to \mathcal{M}$ differ by an inner automorphism of $\mathcal{M}$ , precisely as claimed. $\square$

Next, we claim that the constant $n=4$ taken in Theorem 1.1 is maximal. Here, two avenues of proof seem promising:

A direct proof (along the lines of Theorem 1.1), built from a manageable group presentation for $S_5$ .
A proof built upon Theorem 1.1, noting that any injection $\varphi: S_5 \to \mathcal{M}$ restricts to a map on $S_4$ of the type studied in Theorem 1.1.

Both work well. The first is given below, while the second is sketched in the Exercises:

Theorem 1.2: There exists no injection $\varphi: S_5 \to \mathcal{M}$ .

Proof: We begin with the group presentation

$S_5 \cong \langle \alpha,\beta \mid \alpha^2=\beta^5=(\beta \alpha)^4=[\beta,\alpha]^3 \rangle,$

known at the time of Burnside’s article Note on the Symmetric Group (1897). If $\varphi$ exists, let $\sigma,\tau \in \mathcal{M}$ denote the images of $\alpha,\beta$ , respectively. For definiteness, we may assume that $\tau(z)=\lambda z$ (up to inner automorphism). For $f \in \mathcal{M}$ , we recall that $\mathrm{tr} f =0$ if and only if $\mathrm{ord} f =2$ . Thus $\sigma(z)$ takes the form

$\displaystyle\sigma(z)=\frac{az+b}{cz-a},$

and some computer-assisted algebra quickly gives us the following relations:

$(\tau\sigma)^4: \quad a^2+2bc\lambda+a^2 \lambda^2=0;$

$[\tau,\sigma]^3: \quad (bc+bc\lambda^2+a^2 \lambda-bc\lambda)(bc+bc\lambda^2+3a\lambda^2+bc\lambda)=0.$

The first of these implies $a^2(1-\lambda)^2=2\lambda \det \sigma$ , whereas the second yields

$\lambda^2(\det \sigma)^2 = (2a^2\lambda + bc(1+\lambda^2))^2=((1+\lambda^2)(a^2+\det \sigma)-2a^2\lambda)^2$

$= (a^2(1-\lambda)^2+(1+\lambda^2)\det \sigma)^2=(2\lambda \det \sigma + (1+\lambda^2)\det \sigma)^2$

$= (1+\lambda)^4 (\det \sigma)^2.$

Since $\det \sigma \neq 0$ , we obtain $(1+\lambda)^4=\lambda^2$ , which contradicts that $\lambda^5=1$ . $\square$

— PART II (Injections $\varphi: S_n \to \mathcal{C}$ ) —

The questions answered in the previous section admit natural analogues with $\mathcal{C}$ in place of $\mathcal{M}$ . Answering them, however, will be a bit harder than before, owing to the greater complexity of the group structure on $\mathcal{C}$ . For this reason, we try when possible to reduce questions regarding $\mathcal{C}$ to questions of $\mathcal{M}$ alone.

For example, the identification $\mathcal{M} \cong \mathrm{PSL}_2(\mathbb{C})$ induces a semi-direct product structure on $\mathcal{C}$ ; namely,

$\mathcal{C} \cong \mathrm{PSL}_2(\mathbb{C}) \rtimes \mathbb{Z}/2\mathbb{Z}$ ,

in which the $\mathbb{Z}/2\mathbb{Z}$ -action is given by complex conjugation. It follows that any injective map $\varphi: S_n \to \mathcal{C}$ induces a restricted map $\varphi: A_n \to \mathcal{M}$ , i.e. an injection of the alternating group $A_n$ into the Möbius group. To see how this might benefit us, consider the following:

Proposition 2.1: There exists no injection $\varphi: A_6 \to \mathcal{M}$ . It follows that $S_6$ does not inject into $\mathcal{C}$ .

Proof: See the Exercises. $\square$

As it turns out, there does exists an injection of $A_5$ into $\mathcal{M}$ , which is maximal in the sense of the preceding Proposition. Unfortunately, this fact does not a priori imply that $S_5$ injects into $\mathcal{C}$ (so Proposition 2.1 won’t serve a major role in our classification of symmetric subgroups). In fact, the opposite is true:

Theorem 2.2: There exists no injection $\varphi: S_5 \to \mathcal{C}$ .

Proof: As in Theorem 1.2, we begin with the group presentation

$S_5 \cong \langle \alpha,\beta \mid \alpha^2=\beta^5=(\beta \alpha)^4=[\beta,\alpha]^3\rangle$ .

Suppose that $\varphi : S_5 \to \mathcal{C}$ exists. Since $\beta$ has cycle type $(5)$ , it follows that $\beta \in A_5$ , hence $\tau:=\varphi(\beta)$ lies in $\mathcal{M}$ . Regarded as an element of $\mathrm{PSL}_2(\mathbb{C})$ , $\tau(z)$ is torsion (hence elliptic), so we may assume that $\tau(z) =\lambda z$ , in which $\lambda^5=1$ (up to inner automorphism). As $\sigma(z) \notin \mathcal{M}$ by Theorem 1.2, we may write

$\displaystyle\sigma(z) = \overline{z} \circ \frac{az+b}{cz+d}$

(in general form). The relation $\sigma^2(z)=z$ holds if and only if

(2) $\overline{a}b+\overline{b}d=0$ , $\vert a \vert^2-\vert d \vert^2=b \overline{c}-\overline{b}c$ , and $a \overline{c}+c\overline{d}=0$ .

(Case 1): If $c=0$ , we may scale $a=1$ ; then $b+\overline{b}d=0$ and $\vert d \vert =1$ . It follows that $b^2=-d \vert b \vert^2$ , in which $\vert b \vert =0$ or $\vert b \vert =1$ . If $b=0$ , we then have $\sigma(z)=\overline{z/d}$ , which implies that $(\tau \sigma)^2(z)=z$ , a contradiction. Thus $\vert b \vert=1$ , and we may write

$\displaystyle\sigma(z) = \overline{z} \circ \frac{z+b}{-b^2}.$

On the other hand, that $\mathrm{ord}(\tau \sigma) =4$ forces $\vert b \vert^4+\vert \lambda \vert^2 =0$ , an impossibility. This case is therefore untenable, and we turn to:

(Case 2): If $c\neq 0$ , we may projectivize to assume $c=1$ ; then $a+\overline{d}=0$ and $\vert a \vert^2-\vert d \vert^2=b-\overline{b}$ . This last expression is both real and (purely) imaginary, hence $0$ . Thus $b \in \mathbb{R}$ , and $\sigma(z)$ takes the form

$\displaystyle\sigma(z)= \overline{z} \circ \frac{a z+b}{z-\overline{a}}.$

As for the relation $(\tau \sigma)^4(z)=z$ , it follows that $2\vert a \vert^2+\lambda \overline{b}+b \overline{\lambda}=0$ . Since $b \in \mathbb{R}$ , this implies $b=-\vert a \vert^2/\mathrm{Re}(\lambda)$ . After substitutiting this into our expression for $\sigma$ , we find that the relation $[\tau,\sigma]^3(z)=z$ forces

$(\lambda-1)^4(1+\lambda+\lambda^2)(1+3\lambda+\lambda^2)=0.$

Coupled with $\lambda^5=1$ , we find $\lambda=1$ , but this contradicts that $\tau(z)$ has order five. Thus there exists no injection $\varphi: S_5 \to \mathcal{C}$ . $\square$

We note that Theorem 2.2 recovers Theorem 1.2. Regardless, Theorem 2.2 can’t be used to prove Theorem 1.2, because we have invoked that first theorem in assuming $\sigma \notin \mathcal{M}$ . Avoiding such circularity would require — in essence — reproving Theorem 1.2 as an early claim.

For free, we obtain

Corollary 2.3: There exists an injection $\varphi: S_n \to \mathcal{C}$ if and only if $n \leq 4$ .

— PART III (Group Actions and Invariant Sets) —

We’ve seen that $\mathcal{C}$ contains no symmetric subgroups $S_n$ beyond those that inject into $\mathcal{M}$ , despite the fact that $\mathcal{C}$ contains $\mathcal{M}$ properly as an index two subgroup. Why, then, might we be interested in subgroups $\varphi(S_n) \subset \mathcal{C}$ not contained in $\mathcal{M}$ ?

For starters, let $\varphi : S_n \to \mathcal{C}$ be any injection, and consider the sequence of maps

$f : S_n \stackrel{\varphi}{\longrightarrow} \mathcal{C} \stackrel{\pi}{\longrightarrow} \mathcal{C}/\mathcal{M} \cong \mathbb{Z}/2\mathbb{Z},$

in which $\pi$ denotes the quotient of $\mathcal{C}$ by its normal subgroup $\mathcal{M}$ . As mentioned previously, we have $A_n \subset \ker f$ . (In particular, we cannot have $\ker f =V_4$ , Klein’s four group, despite the fact that $V_4$ is normal in $S_4$ .) Thus $A_n = \ker f$ if and only if $\varphi(S_n) \not\subset \mathcal{M}$ . When these equivalent conditions hold, membership in $\mathcal{M}$ is detected by the sign character on $S_n$ . In this one sense, the theory of symmetric subgroups in $\mathcal{C}$ is more robust than the analogue theory in $\mathcal{M}$ .

There is a second, much more compelling reason to study symmetric subgroups in $\mathcal{C}$ , which begins with the following construction:

Let $K \subset \mathbb{P}^1$ be any finite set, $\# K =n$ , and suppose that $H \leq \mathcal{C}$ is a subgroup of conformal mappings such that each $\sigma \in H$ permutes the points of $K$ . (In other words, $K$ is an invariant set for the action of $H$ .) This gives an induced map $\rho : H \to S_n$ , known as the permutation representation (associated to the group action $H \circlearrowright K$ ). Our interest in such maps is simple: if $\rho: H \to S_n$ bijects, then $\rho^{-1}: S_n \to \mathcal{C}$ is exactly the sort of map we’ve been looking for.

If this permutation representation bijects, then our general theory implies $n \leq 4$ . Of these $n$ , the extremal case $n=4$ naturally presents the greatest interest. Here, there are two cases to consider:

All points in $K$ lie on a circle in $\mathbb{P}^1$ . Up to inner automorphism, we may assume that $K=\{0,1,\infty,z_0\}$ , with $z_0 \in \mathbb{R}$ .
The points in $K$ lie on no common circle, but we may assume after conjugation that $K$ is given by $\{0,1,\infty,z_0\}$ , with $z_0 \notin \mathbb{R}$ .

If case (1) holds, note that we may assume $H \subset \mathcal{M}$ , by modding out by the trivial action of $\overline{z}$ . As it turns out, however, case (1) cannot actually occur:

Example 3.1: Suppose that $\rho: H \to S_4$ surjects, and that case (1) holds. We may assume $H \subset \mathcal{M}$ . If $\tau(z)$ corresponds to the 3-cycle $(0\,1\,\infty)$ , brief computation shows that

$\displaystyle\tau(z)=\frac{1}{1-z}.$

Since $\tau(z)$ fixes $z_0$ , we must have $z_0^2-z_0+1=0$ . But this polynomial has no real roots, which contradicts that $z_0 \in \mathbb{R}$ . It follows that case (1) cannot occur.

Note: If case (2) can be realized, this gives strong motive to consider injections of $S_n$ into $\mathcal{C}$ , as opposed to injections into the smaller group $\mathcal{M}$ . As it happens, case (2) is realized, in an essentially unique way.

To prove this result, it will be advantageous to present first a general Lemma. For the moment, let’s relax our assumptions and suppose that the permutation representation $\rho: H \to S_4$ merely surjects. By the previous Example, it follows that the invariant set $K= \{0,1,\infty,z_0\}$ has $z_0 \notin \mathbb{R}$ (after inner automorphism). As the following (general) Lemma shows, the map $\rho$ injects without further hypotheses:

Lemma 3.2: Suppose that $K$ contains four points not lying on a circle in $\mathbb{P}^1$ . If $\psi(z) \in \mathcal{C}$ fixes each point in $K$ , then $\psi(z)=z$ . In other words, the permutation representation $\rho : H \to S_n$ injects.

Proof: Suppose that a non-identity element $\psi \in \mathcal{C}$ fixes $K$ . Conjugating $\psi$ by an element of $\mathcal{M}$ , we may assume that $K$ contains $\{0,1,\infty,z_0\}$ , with $z_0 \in \mathbb{C} \smallsetminus \mathbb{R}$ . We note that $\psi \notin \mathcal{M}$ , as $\mathcal{M}$ acts sharply 3-transitively on $\mathbb{P}^1$ . Thus, writing $\psi(z) = \overline{z} \circ \sigma(z)$ with $\sigma \in \mathcal{M}$ , it follows that $\sigma$ fixes the three points $0,1$ , and $\infty$ . Because $\mathcal{M}$ acts sharply 3-transitively on $\mathbb{P}^1$ , we have $\sigma(z)=z$ . Thus $\psi(z)=\overline{z}$ , which fixes $\mathbb{R} \cup \{\infty\}$ (and nothing else). This contradicts that $\psi$ fixes $z_0$ , whence no such $\psi$ exists. $\square$

And now, our final Theorem:

Theorem 3.3: There exists a subgroup $H \leq \mathcal{C}$ and an invariant set $K$ , $\# K=4$ , such that the permutation representation $\rho : H \to S_4$ bijects. If $H'$ and $K'$ are as $H$ and $K$ above, there exists an element $\psi \in \mathcal{C}$ such that $H' = \psi H \psi^{-1}$ and $K'=\psi K$ . It follows that the associated injection $\rho^{-1} : S_4 \to \mathcal{C}$ is unique up to inner automorphism.

Proof: Suppose that $H$ and $K$ exist as above. Up to inner automorphism by $\mathcal{M}$ , we may assume that $K=\{0,1,\infty,z_0\}$ , with $z_0 \in \mathbb{C} \smallsetminus \mathbb{R}$ . Moreover, Example 3.1 implies that $z_0$ is a root of $t^2-t+1$ . In the group presentation

$S_4 \cong \langle \alpha,\beta \mid \alpha^2=\beta^3=(\beta \alpha)^4 \rangle,$

we may take $\beta \leftrightarrow \tau(z)$ , wherein $\tau(z)$ is as in Example 3.1. Since $H \not\subset \mathcal{M}$ the image of $\alpha$ ( $\sigma$ , say) is anti-holomorphic, hence a simple transposition (by order). Moreover, $\sigma$ can be chosen to fix $(0,\infty)$ while transposing $(1,z_0)$ , as $S_4$ is complete. Thus $\sigma(z)$ takes the form $\sigma(z)=z_0 \overline{z}$ , for which we note that $\rho: \langle \sigma,\tau \rangle \to S_4$ surjects.

By Lemma 3.2, we have $H= \langle \sigma,\tau \rangle$ . As such, $H$ is uniquely determined by $K$ , and $K$ is determined by choice of $z_0$ (among the roots of $t^2-t+1$ ). These two choices of $K$ are related by conjugation by $\overline{z} \in \mathcal{C}$ , which gives uniqueness of $H,K$ (in the sense of this Theorem). It follows that $\rho^{-1}:S_4 \to \mathcal{C}$ is unique up to inner automorphism, as claimed. $\square$

— EXERCISES —

Exercise: Herein, we outline a second proof of Theorem 1.2. Assuming that $\varphi : S_5 \to \mathcal{M}$ injects, let $H \leq S_5$ denote any subgroup isomorphic to $S_4$ . By Theorem 1.1, we may assume that $H$ is generated by

$\displaystyle\sigma(z)= \frac{z+1/2}{z-1}$ and $\tau(z) = \lambda z$ ,

in which $\lambda$ is some primitive cube root of unity. Show that $\varphi(S_5)$ contains an element $\nu \notin H$ that commutes with $\sigma(z)$ such that $\mathrm{ord}(\nu)=2$ and $\mathrm{ord}(\tau \nu)=4$ . Use these relations to find a contradiction.