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Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.
2
votes
1
answer
171
views
How can one test whether a given analytic curve in the plane is algebraic or not?
Suppose $\Gamma$ is an analytic Jordan curve in the complex plane $\mathbb{C}$. What are ways to test whether $\Gamma$ is (contained in) an algebraic curve, i.e., whether there exists a real polynomia …
7
votes
Non-locally connected polynomial Julia sets
There is a necessary and sufficient condition due to Yoccoz in terms of continued fraction expansion. The condition is that $\lambda=e^{2\pi i \theta}$ should be such that $\theta$ is not a Brjuno num …
7
votes
Accepted
Riemann mapping theorem with smooth boundary
The main reference on this topic is the book "Boundary behavior of conformal maps" by Pommerenke.
If the curve is $C^\infty$, then the biholomorphic mapping extends to a smooth map on the closure of t …
9
votes
1
answer
937
views
Removable sets for harmonic functions and Hardy spaces of general domains
Let $\Omega$ be a domain of the complex plane. The Hardy space $H^p(\Omega)$ is defined, for $1 \leq p<\infty$, as the class of functions $f$ that are holomorphic on $\Omega$ such that $|f|^p$ has a h …
2
votes
0
answers
79
views
On the (Brouwer-Koebe) Continuity Method
The so-called Continuity Method is a simple yet powerful method to show that a given continuous injective map is surjective. Namely, suppose that $f:X \to Y$ is a map between two connected manifolds $ …
1
vote
Accepted
Conformal mappings for domains whose complement is totally disconnected
The only example I know of a conformal map that "stretches" a point boundary component to a nondegenerate continuum was constructed by Gehring and Martio in Quasiextremal distance domains and extensio …
2
votes
Is there an explicit formula for the modulus of an annulus given a parameterization of the i...
I also think there is no "explicit formula" for the conformal modulus, even in simple cases. However, as mentioned in Igor Rivin's answer, there are methods for approximating the conformal map and the …
2
votes
2
answers
367
views
On the set of zero radial limits of bounded analytic functions
Hi,
Let $f$ be a non-identically zero bounded analytic function in the open unit disk $\mathbb{D}$. It is well-known that $f$ has radial limits almost everywhere on the unit circle $\mathbb{T}$. Let …
9
votes
3
answers
717
views
Can the limit set of an infinitely generated Schottky group have positive area?
Dear Mathoverflow Community,
Suppose that $\Omega$ is a domain in the Riemann Sphere $\widehat{\mathbb{C}}$ with $\infty \in \Omega$, and assume that every connected component of $\partial \Omega$ is …
2
votes
$\mathcal P(K)=\mathcal R(K)$ iff $\Bbb C\backslash K$ is connected
Perhaps a more elementary proof is the following :
Assume $\mathbb{C} \setminus K$ is disconnected. Let $V$ be a bounded component of $\mathbb{C} \setminus K$, and let $w \in V$. Let us show that the …
14
votes
4
answers
3k
views
A learning roadmap to the Schramm-Loewner evolution (SLE) for the complex analyst
I would like some good references to learn about the Schramm-Loewner evolution (SLE), for a complex analyst with no background in probability.
A quick google search gave a lot of references on SLE th …
11
votes
Harmonic map proof of Riemann mapping theorem
Yes, there is a classical proof of the Riemann mapping theorem using harmonic maps and the Dirichlet problem. Riemann's original assumption of boundary smoothness can be removed using Perron's method …
4
votes
A question on Ahlfors covering surface
Yes it can, see the paper A new proof of the Ahlfors five islands theorem by Walter Bergweiler. The proof is based on a result on Nevanlinna concerning perfectly branched values, Zalcman's rescaling l …
13
votes
3
answers
713
views
How bad can a circle domain get?
Let $X$ be a domain in the Riemann sphere $\widehat{\mathbb{C}}$. We say that $X$ is a circle domain if every connected component of its boundary is either a circle or a point.
It was conjectured by …
4
votes
Accepted
A Generalization of the Ahlfors function to have varying degrees?
The answer to your question is yes. Indeed, one can replace $\mathbb{D}$ with the right half plane, applying a Mobius transformation. Then the argument is quite simple if you are familiar with the fol …