Automorphism group of Annulus

Question

Could any one tell me precisely how to compute the automorphism group of an annulus say $r<|z|<R$? Thank you!

Answer 1

The group of conformal automorphisms of an annulus is $\mathbb T\times \mathbb Z_2$. The cyclic factor comes from the inversion $z\mapsto Rr/z$ which exchanges the order of boundary components. The first factor comes from rotations.

Preparation. Any homeomorphism (in particular any conformal map) between finitely connected domains induces a bijection of their boundary components as sets. That is, if $f$ is a homeomorphism between $\Omega$ and $\Omega'$, then for every boundary component $C$ of $\Omega$ there is a boundary component $C'$ of $\Omega'$ such that $\operatorname{dist} (f(z),C')\to 0$ as $\operatorname{dist} (z,C)\to 0$. (Idea of proof: suppose some points approach $C'$ while others approach $C''$. Draw a curve in $\Omega'$ separating $C'$ from $C''$. The preimage of this curve under $f$ is compact and therefore does not meet sufficiently small neighborhood of $C$. Conclude.)

Returning to annulus, we observe that there are only two possible bijections on the set of boundary components. If we classify conformal automorphisms that preserve the order of components, then their compositions with inversion will give the rest of the group. Suppose $f$ is such an automorphism. Then $g(z)=f(z)/z$ is a holomorphic function such that $|g(z)|$ tends to $1$ when $z$ approaches either boundary component. By the maximum principle, $|g|\equiv 1$. Thus, $g(z)\equiv \zeta\in\mathbb T$ and we are done.