Timeline for On holomorphic branched coverings of a domain in the plane to the unit disk
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Aug 20, 2013 at 12:51 | vote | accept | Malik Younsi | ||
| Aug 20, 2013 at 12:50 | answer | added | Malik Younsi | timeline score: 2 | |
| Aug 20, 2013 at 12:48 | history | edited | Malik Younsi | CC BY-SA 3.0 |
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| Jun 22, 2012 at 21:22 | history | bounty ended | Malik Younsi | ||
| Jun 20, 2012 at 16:21 | comment | added | Ian Agol | @unknown(google): you're right, I don't know where I got 2g-3 from! One can obtain the n-punctured disk branched over the n-disk with n branch points, e.g. a generic polynomial of degree n sends the complement of the n roots to the complement of zero (in the Riemann sphere) has n-1 branched points of degree 2 (at the roots of the derivative) and one branched point of degree n at infinity. But I guess given the correct dimension computation, this will not be onto (except possibly for n=3). One could also consider maps of higher degree. | |
| Jun 20, 2012 at 3:09 | comment | added | John Pardon | @Agol: I certainly believe this would work, I think though that the Teichmuller spaces are not of the same dimension. The Teichmuller space of a sphere minus n disks has dimension 3n-6. An euler characteristic argument shows that the map to the disk is branched over 2n-2 points, so that Teichmuller space has dimension 4n-7. So the map will be surjective with a large dimensional fiber. | |
| Jun 19, 2012 at 15:07 | comment | added | Malik Younsi | @Agol : Your heuristic remark about the corresponding Teichmuller spaces is very interesting, thank you. | |
| Jun 19, 2012 at 15:06 | comment | added | Malik Younsi | @Geoge : I thought about doing this, but it's not clear to me how you can choose $z_1, \dots, z_n$ so that the resulting function will be single valued... Note that this requirement puts constraints on $z_1, \dots, z_n$, so there are choices of the $z_j$'s that won't work. | |
| Jun 18, 2012 at 21:44 | comment | added | George Lowther | Or, which amounts to the same thing, choose n points $P=\lbrace z_1,\ldots,z_n\rbrace$ in the domain and define a harmonic function on the domain minus P by $f(z)=\log\vert z-z_1\vert+\cdots+\log\vert z-z_n\vert+u(z)$ where $u$ is harmonic and chosen such that $f=0$ on the boundary. You can extend $f$ to a (multiple-valued) harmonic function $g(z)=f(z)+iv(z)$. Choosing $z_1,\ldots,z_n$ carefully, you can make this a singly valued function, and the mapping is given by $h(z)=\exp(g(z)$. | |
| Jun 18, 2012 at 20:38 | comment | added | George Lowther | I think you can do this by multiplying together functions coming from the Riemann mapping, one for each Jordan curve, then multiply by a term of the form $\exp(u(z)+iv(z))$ where $u$ is harmonic and chosen to make the Jordan curves map to the unit circle and $v$ is chosen to make $u+iv$ analytic. It could be that $v$ is multiple-valued (so, only defined on the covering surface) but, by combining the Riemann map terms with Mobius transformations on the unit disc, you can make sure that $v$ is singly valued. That's a rough sketch I have in my head anyway. | |
| Jun 18, 2012 at 17:23 | comment | added | Ian Agol | I don't know how to answer your question, but heuristically, this should hold true by consider branched covers over the disk with appropriate branching data. One may obtain an n-punctured sphere as a branched cover over a disk with n branched points. The dimension of Teichmuller space of an n-punctured sphere is 2n-3, which is the same as the dimension of the space of n points in a disk, modulo the mobius group. So this gives a map between Teichmuller spaces, which one would like to prove is onto. I'm not sure how to give an elementary proof of this though. | |
| Jun 18, 2012 at 16:40 | history | edited | Malik Younsi | CC BY-SA 3.0 |
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| Jun 18, 2012 at 16:26 | history | edited | Malik Younsi | CC BY-SA 3.0 |
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| Jun 15, 2012 at 20:59 | history | bounty started | Malik Younsi | ||
| Jun 13, 2012 at 17:38 | history | edited | Malik Younsi | CC BY-SA 3.0 |
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| Jun 13, 2012 at 14:39 | history | asked | Malik Younsi | CC BY-SA 3.0 |