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Let $G$ be a bounded region in $\mathbb{C}$ (i.e. we have $G ≠ \emptyset$, and $G$ is open and connected), and let $G$ have a transitive automorphism group (that is, for each two points $z_1, z_2 \in G$, there exists a $\phi \in Aut(G)$ with $\phi(z_1) = z_2$, where $Aut(G)$ denotes the group of biholomorphic mappings from $G$ to $G$). In addition, let $G$ have "sufficiently smooth" edges: there exists an $a \in \partial G$ and an (open) neighborhood $U \subseteq G$ of $a$, so that $U \cap G$ is simply connected.

I now want to show that $G$ is biholomorphic to the open unit ball $B_1(0) = \{z \in \mathbb{C}: |z| < 1\}$.

I believe what I need to do in this exercise is to show that not only $U \cap G$ is simply connected, but also $G$ itself: the rest then should follow more or less from the Riemann mapping theorem. However, I don't know how I can show that. I think I must somehow "move" or "stretch" the simply connected set $U \cap G \subseteq G$ onto the other parts of $G$, to show that the rest of $G$ is simply connected too. But I'm not sure how to do that, seeing as we don't know too much about how the automorphisms of $G$ concretely look like, other than that there is always one that maps any point onto any other.

moran
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  • The information that G has a transitive automorphism group is redundant, since G is a manifold (You can look up Isotopy Lemma). What do you mean when you say there exists an open nbd. $U \subseteq G$ of $a$ ? Note that $a \notin G$. – Hmm. May 26 '16 at 20:55
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    I think by "automorphism" you mean a conformal self map. – Moishe Kohan May 26 '16 at 23:04
  • @studiosus Yes, you're right. I've specified it in the question, sorry if it was too ambiguous. – moran May 27 '16 at 09:53
  • @SoumyaSinha I mean a neighborhood of $a$ within $\mathbb{C}$. (Each open subset of $\mathbb{C}$ that contains $a$ also contains points from $G$, since $a \in \partial G$.) Also, thanks for the information about the Isotopy Lemma, I'll read into it. I just rephrased it like it was given to me, and the fact that this statement about $Aut(G)$ was explicitly stated makes me assume that I'll need it at some point in the proof. – moran May 27 '16 at 09:57
  • @moran, what if you let G be an annulus ? – Hmm. May 27 '16 at 10:08
  • @SoumyaSinha Is that possible, with all the given preassumptions? I assumed that the goal of the exercise was to show that it isn't possible that $G$ "has holes", i.e. that I need to show that $G$ also must be simply-connected. Otherwise, I couldn't apply the Riemann mapping theorem at the end. – moran May 27 '16 at 10:12
  • Yes, I think an annulus does constitute a counterexample. – Hmm. May 27 '16 at 10:21
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    @SoumyaSinha Ok, thanks. I must ask again though; are you sure that the Isotopy Lemma gives us that all (bounded) regions/domains in $\mathbb{C}$ have a transitive automorphism group when asking for a holomorphic structure (not just differentiable in the $\mathbb{C} \cong \mathbb{R}^2$-sense)? According to this thread: http://math.stackexchange.com/questions/16685/exist-domains-in-complex-plane-with-only-trivial-automorphisms there exist open (and also bounded) domains in $\mathbb{C}$ with a finite automorphism group (so it especially can't be transitive). Or am I overlooking something here? – moran May 27 '16 at 10:53
  • @moran, Yes in the holomorphic sense the isotopy lemma is useless. – Hmm. May 27 '16 at 11:03
  • @SoumyaSinha ok, that's what I originally meant in the question; sorry if it was too ambiguous at first. I added it to the question. – moran May 27 '16 at 11:11
  • @moran, rather surprisingly ( to me at least ), this indeed is true. Krantz gives quite a nice proof in the following - http://www.jstor.org/stable/2322791 – Hmm. May 27 '16 at 12:28
  • @SoumyaSinha Thanks, that's a very nice proof indeed. – moran May 27 '16 at 18:49
  • One more thing: The result that Krantz is proving does not need smooth boundary. The theorem is that if $U\subset {\mathbb C}$ is an open connected subset with transitive group of conformal automorphisms then either $U$ is simply connected or $U$ is the complement to a point in ${\mathbb C}$. However, I do not know any elementary proofs: The proof I have in mind uses the uniformization theorem, which is a harder form of the Riemann mapping theorem. – Moishe Kohan May 29 '16 at 23:43

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