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Let $M_2(\mathbb{C})$ be the group of all Möbius transformations $z\mapsto \frac{az+b}{cz+d}$ from $\mathbb{C}\cup\{ \infty\}$ to itself. Let $PSU(2,\mathbb{C})$ be the group of all Möbius transformations of the form $z\mapsto \frac{az+b}{-\overline{b}z+\overline{a}}$, with $|a|^2+|b|^2=1$. I want to see the proof of the following theorem:

Every finite group of Möbius transformations is conjugate to a subgroup of $PSU(2,\mathbb{C})$.

In the book "Complex Functions: Jones, Singerman", there is a proof which involves the applications of trace of Möbius transformation and the classification of Möbius transformations using trace.

Are there other elementary proofs of this theorem? Can one suggest references for it?

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Recall that $\mathrm{M}_2(\mathbf C)$ is nothing but $\mathrm{PGL}_2(\mathbf C)$. Your question is about a proof of the statement that any finite subgroup of $\mathrm{PGL}_2(\mathbf C)$ is conjugate to a subgroup of $\mathrm{PSU}_2(\mathbf C)$. In fact, this holds more generally:

Proposition. Let $n$ be a nonzero natural number. Then any finite subgroup of $\mathrm{PGL}_n(\mathbf C)$ is conjugate to a subgroup of $\mathrm{PSU}_n(\mathbf C)$.

Proof. Recall that $\mathrm{PGL}_n(\mathbf C)$ is the quotient of the general linear group $\mathrm{GL}_n(\mathbf C)$ by its normal subgroup ${\mathbf C}^\star I$ of all nonzero multiples of the $n\times n$ identity matrix $I$. Let $$ \pi\colon \mathrm{GL}_n(\mathbf C)\rightarrow \mathrm{PGL}_n(\mathbf C) $$ be the quotient map. Recall also that the special linear group $\mathrm{SL}_n(\mathbf C)$ is the subgroup of $\mathrm{GL}_n(\mathbf C)$ of all matrices of determinant equal to $1$. Since any nonzero complex number has a nonzero $n$-th root, $\pi$ maps this subgroup onto $\mathrm{PGL}_n(\mathbf C)$, i.e., $$ \pi(\mathrm{SL}_n(\mathbf C))=\mathrm{PGL}_n(\mathbf C). $$ Let $$ \rho\colon \mathrm{SL}_n(\mathbf C)\rightarrow \mathrm{PGL}_n(\mathbf C) $$ be the restriction of $\pi$ to $\mathrm{SL}_n(\mathbf C)$. By what has been said above, $\rho$ is surjective. Its kernel is $$ \ker(\rho)=\ker(\pi)\cap\mathrm{SL}_n(\mathbf C)=\mu_n I, $$ where $\mu_n$ is the group of $n$-th roots of unity in $\mathbf C$. In particular, the kernel of $\rho$ is finite.

Now, let $G$ be a finite group in $\mathrm{PGL}_n(\mathbf C)$. Since $\rho$ has finite kernel, the inverse image $H=\rho^{-1}(G)$ is a finite subgroup of $\mathrm{SL}_n(\mathbf C)$. In particular, $H$ is a finite subgroup of $\mathrm{GL}_n(\mathbf C)$ such that $\pi(H)=\rho(H)=G$ since $\rho$ is surjective.

Let $(\cdot|\cdot)$ denote the standard Hermitian inner-product on $\mathbf C^n$. Since $H$ is a finite subgroup of $\mathrm{GL}_n(\mathbf C)$, the form $(\cdot|\cdot)'$ on $\mathbf C^n$ defined by $$ (v|w)'=\sum_{h\in H}(hv|hw) $$ is a well-defined positive-definite Hermitian form on $\mathbf C^n$. Moreover, it is invariant under $H$, i.e., for all $h\in H$ and all $v,w\in\mathbf C^n$ one has $$ (hv|hw)'=(v|w)'. $$ This means that $H$ is contained in the unitary group of the Hermitian form $(\cdot|\cdot)'$. Since any positive-definite Hermitian form on $\mathbf C^n$ is equivalent to the standard Hermitian form, there is an element $a\in\mathrm{GL}_n(\mathbf C)$ such that $$ (av|aw)=(v|w)' $$ for all $v,w\in\mathbf C^n$. It follows that $aHa^{-1}$ is contained in the ordinary unitary group $\mathrm{U}_n(\mathbf C)$. Since all elements of $H$ have determinant equal to $1$, the conjugate $aHa^{-1}$ of $H$ is contained in the special unitary group $\mathrm{SU}_n(\mathbf C)$. It follows that $$ \pi(a)G\pi(a)^{-1}\subseteq\mathrm{PSU}_n(\mathbf C) $$ since $\mathrm{PSU}_n(\mathbf C)=\pi(\mathrm{SU}_n(\mathbf C))$. $\square$

Remark. Note that the above argument applies more generally to a compact subgroup $G$ of $\mathrm{PGL}_n(\mathbf C)$, replacing the sum over $H$ by a suitable integral.

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