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Let $D\subset CP^1$ be a domain (a nonempty open connected subset) and let $S_D$ denote the space of conformal embeddings $D\to CP^1$ equipped with topology of uniform convergence on compacts. Is it true that this space is connected? Has homotopy type of $PSL(2,C)$?

The answer is easy and positive if $D$ is simply-connected, but I already do not know what happens if $D$ is an annulus.

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  • $\begingroup$ Sorry if this is a silly question, but who du you conclude that the answer to you question is easy and positive in the simply-connected case? Thanks in advance! $\endgroup$
    – ComplexF
    Commented Oct 9, 2018 at 6:42

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Here is a proof of connectivity for an annulus, but it generalizes to any domain of finite connectivity. WLOG, $D$ is an annulus $a<|z|<1$. Let $f:D\to P^1$ be the embedding. First we approximate it by an embedding which sends the boundary to quasicircles. For this we first restrict $f$ on $D'=\{ z: r_1<|z|<r_2\}$, where $a<r_1<r_2<1$. Let $\phi$ be a quasiconformal map $D\to D'$. Then there exists a quasiconformal map $\psi:P^1\to P^1$ such that $g=\psi\circ f\circ\phi$ is conformal on $D$. If $r_1$ and $r_2$ are close to $a$ and $1$, then $\phi$ can be chosen uniformly close to the identity, and $\psi$ can be chosen uniformly close to the identity.

On the second stage, we extend $g$ to the homeomorphism of the Riemann shpere which is quasiconformal on $|z|<a$ and $|z|>1$. This is possible to do because the images of these circles are quasicircles. Let $\mu$ be the Beltrami coefficient of $g$. Then we can deform $g$ to identity by constructing a family of normalized quasiconformal homeomorphisms $g_t$ with Beltrami coefficients $\mu_t=t\mu$, where $0\leq t\leq 1.$

This gives a deformation of your homeomorphism to the identity.

Reference on quasiconformal maps, quasicircles and Beltrami equation: L. Ahlfors, Lectures on quasiconformal mappings.

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  • $\begingroup$ I see: This is a nice argument for connectivity (I think, even path connectivity). Can it be promoted to contractibility in the case of domains of finite topological type? The key issue, as I see it, is lack of a canonical quasiconformal extension to the entire sphere. Maybe one can use extension by a harmonic quasiconformal map (if boundary of the image is nice enough) or Ahlfors-Beurling. $\endgroup$ Commented Apr 27, 2014 at 14:59
  • $\begingroup$ What you are saying is plausible, but the details should be checked. $\endgroup$ Commented Apr 27, 2014 at 18:37

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