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I saw this question, which brought up the question: can we classify all the Möbius transformations (with complex coefficients) of finite order? In particular, do these only consist of the rotations? I sense symbolic manipulation isn't the way to go about this—and indeed didn't get anywhere with this—and would appreciate some insight.

bookworm
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    Depending on what you mean by “rotations,” yes, they only consist of rotations. Are you familiar with how to relate Mobius transformations to $2 \times 2$ matrices? – Qiaochu Yuan Sep 02 '20 at 06:53
  • Yes, I am familiar with the matrix thing. By rotations I'm thinking of those around the origin (as in the linked answer) when we consider the Möbius transformations as maps on the complex plane/Riemann sphere? – bookworm Sep 02 '20 at 07:03

2 Answers2

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Every Möbius transformation which is not the identity is conjugate to either a translation $$ z \to z + a \, \quad (a \in \Bbb C, a \ne 0) $$ or a complex-linear map (rotation/dilation) $$ z \to \lambda z \, \quad (\lambda \in \Bbb C, \lambda \ne 0, 1) $$ depending on whether it has one or two fixed points. Translations do not have finite order, and rotations if and only if $\lambda$ is some root of unity.

It follows that all Möbius transformation of finite order are of the form $$ T(z) = S^{-1}(\lambda S(z)) $$ with some Möbius transformation $S$ and $\lambda = e^{2 \pi i k/n} \ne 1$ for some integers $k, n$.

Martin R
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  • How would you prove that first statement? I've not seen this result before. – bookworm Sep 02 '20 at 07:08
  • @bookworm: If $T$ has exactly one fixed point $a$ then $S \circ T \circ S^{-1}$ with $S(z) = 1/(z-a)$ has exactly one fixed point $z = \infty$, and that can only be a translation. – If $T$ has two fixed points then you can conjugate them to $0$ and $\infty$ and conclude that $S \circ T \circ S^{-1}$ is a rotation. – Martin R Sep 02 '20 at 07:14
  • @bookworm: You can also find it in https://en.wikipedia.org/wiki/Möbius_transformation: Every non-parabolic transformation is conjugate to a dilation/rotation, and parabolic transforms are conjugate to a translation. – Martin R Sep 02 '20 at 07:17
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    Right, but you should not call the second type a rotation: Some are dilations, etc.; you can call it complex-linear. – Moishe Kohan Sep 02 '20 at 10:59
  • @MoisheKohan: Agreed, thanks! – Martin R Sep 02 '20 at 11:13
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Thinking of the group of Mobius transformations as the projective linear group $PGL_2(\mathbb{C})$, we can argue as follows. A matrix $X \in GL_2(\mathbb{C})$ has the property that its image in $PGL_2(\mathbb{C})$ has finite order iff $X^n = \lambda I$ for some scalar $\lambda$. Because $\mathbb{C}$ is algebraically closed (I mention this explicitly to show that it doesn't generalize to more general fields) some $n^{th}$ root $\sqrt[n]{\lambda}$ exists and we can rescale $X$ by it, so we can assume WLOG that $X^n = I$; that is, that $X$ itself has finite order. So it suffices to classify such matrices.

Exercise: A matrix of finite order is diagonalizable. More generally, a matrix satisfying a polynomial $f(X) = 0$ with no repeated roots is diagonalizable.

Now things are very easy: a diagonalizable matrix satisfies $X^n = I$ iff its eigenvalues are $n^{th}$ roots of unity, and this generalizes to all $GL_d(\mathbb{C})$. So $X$ must be conjugate to a diagonal matrix with diagonal entries $\zeta_n^i, \zeta_n^j$ ($\zeta_n$ a primitive $n^{th}$ root of unity), which means the corresponding Mobius transformation is conjugate to

$$z \mapsto \frac{\zeta_n^i z}{\zeta_n^j} = \zeta_n^{i-j} z.$$

In the special case of $PGL_2(\mathbb{C})$ more can be said: it is actually possible to classify completely the finite subgroups of Mobius transformations (here we classify the finite cyclic groups). This classification turns out to be identical to the classification of finite subgroups of $SO(3)$: other than the cyclic and dihedral groups, there are three "exceptional" groups corresponding to symmetries of the Platonic solids, namely the tetrahedral group $A_4$, the octahedral group $S_4$, and the icosahedral group $A_5$.

Qiaochu Yuan
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