Holomorphic Automorphism Group

By a domain I mean an open connected subset of ${\mathbb C}$. If $D$ is a domain, let $\operatorname{Aut}(D)$ denote the collection of holomorphic bijections $f:D\to D$. It is well-known that if $f$ is holomorphic, so is its inverse, so $\operatorname{Aut}(D)$ is actually a group.

We can give a topology to $H(D)$, the set of holomorphic maps with domain $D$, by setting $f_n\to f$ iff $f_n$ converges uniformly to $f$ on any compact subset of $D$. This actually makes $H(D)$ a complete metric space (Section 2.2 of Berenstein-Gay "").

I have a few questions about $\operatorname{Aut}(D)$ with this topology. First,

Is $\operatorname{Aut}(D)$ a topological group?

It is easy to check that $f_n\circ g_n\to f\circ g$ if $f_n\to f$ and $g_n\to g$ in $H(D)$, so the question here is whether $f_n^{-1}\to f^{-1}$ if $f_n\to f$, and $f_n,f\in \operatorname{Aut}(D)$. (This may be trivial.) Second,

If $f_n\to f$ in $H(D)$ and $f_n\in \operatorname{Aut}(D)$ for all $n$, is $f\in \operatorname{Aut}(D)$ as well? Are there any reasonable assumptions on $D$ that make this true? (As pointed out in the comments below, if $D$ is simply connected, then $f$ needs not be injective.)

By Rouche, we have that $f$ is either constant or injective (Prop 2.6.19 in Berenstein-Gay "Complex Variables"). Perhaps, if it is injective, showing it is surjective should be easy?


I've heard that $\operatorname{Aut}(D)$ is actually a Lie group. I suppose the topology is the one I mentioned, but I do not see where the manifold structure will come from if that's the case. Could you sketch why this is so, or point me in the right direction if I'm completely off the mark here?

2022-07-25 20:40:42
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Answers: 1

To answer your first question:

Suppose $f_n\in \mathrm{Aut}(D)\to f\in\mathrm{Aut}(D)$. Fix $z_0\in D$, and let $w_0 = f(z_0)$. We need to show that $f_n^{-1}\to f^{-1}$ uniformly in a neighborhood of $w_0$.

Lemma: Let $C = |f'(z_0)|$. Then there is an $r>0$ such that $|f_n(z) - f_n(z_0)|\geq (C/2)|z - z_0|$ whenever $n\gg1$ and $|z - z_0|<r$.

To see this, first define $$g_n(z) = \begin{cases}\frac{f_n(z) - f_n(z_0)}{z - z_0} & z\neq z_0\\ f_n'(z_0) & z = z_0\end{cases}\,\mbox{ and }\,g(z) = \begin{cases}\frac{f(z) - f(z_0)}{z - z_0} & z\neq z_0\\ f'(z_0) & z = z_0\end{cases}.$$ One then has that $g_n\to g$ locally uniformly. Since $|g(z_0)| = C>0$, it follows that there is an $r>0$ such that $|g_n(z)|\geq C/2$ whenever $|z - z_0|<r$ and $n\gg1$.

From the lemma we can conclude that $f_n(\mathbb{D}(z_0,r))\supseteq \mathbb{D}(w_0,Cr/2)$ whenever $n\gg 1$. Suppose that $w\in \mathbb{D}(w_0,Cr/2)$. The lemma also gives that $$|f_n^{-1}(w) - f^{-1}(w)|\leq \frac{2}{C}|w -f_n(f^{-1}(w))|$$ when $n\gg 1$. Since $f_n\to f$ uniformly on $\mathbb{D}(z_0,r)$, the right hand side of the inequality $\to 0$ uniformly in $w\in \mathbb{D}(w_0,Cr/2)$ as $n\to \infty$. Therefore $f^{-1}_n\to f^{-1}$ uniformly on $\mathbb{D}(w_0,Cr/2)$.

About your third question: I can't say I know what the proof of this, though it's definitely true. My understanding is that this was first proved by H. Cartan in the paper Sur les groupes de transformations analytiques (1935). There are probably better references, but I don't know them.

2022-07-25 21:19:57