# Automorphism group of Annulus

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

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Automorphism in the sense of "continuous bijection with continuous inverse"? Or something more strict? –  rschwieb Mar 8 '13 at 18:09
In what category. Do you want the annulus as a topological space. In this case you have all of the homeomorpisms. As a differentiable manifold, you have all of the diffeomorphisms, as a metric space, you have all of the distance preserving functions, as a conformal space you have all of the conformal equivalences ect. I would guess that the complex analysis tag indcates that you want the conformal equivalences. Also why the lie groups tag? –  Baby Dragon Mar 8 '13 at 18:09
yes you are right, as bcoz it is a lie group. –  La Belle Noiseuse Mar 8 '13 at 18:38

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.
what is $\mathbb{T}$? –  La Belle Noiseuse Mar 9 '13 at 6:24
@CityOfGod The unit circle, also known as multiplicative group of unimodular complex numbers. Also known as the $n$-dimensional torus $\mathbb T^n$ with $n=1$. –  user53153 Mar 9 '13 at 6:27
I am sorry to ask you or request to explain you that how to show that automorphisms of annulas are of the form $az$ or $\frac{ar}{z}$? –  La Belle Noiseuse Mar 9 '13 at 6:30
@CityOfGod I was referring to the cyclic group of order $2$, which I denoted by $\mathbb Z_2$. It appears as a factor in the direct product/sum, so I called it the cyclic factor. // Also, I recommend that you carefully read the post before firing questions one after another. I did not claim that the automorphisms are of that kind. –  user53153 Mar 9 '13 at 6:30