Riemann mapping theorem with pathological boundary
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From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $mathbb{C}$ which is not all of $mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.
Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?
complex-analysis riemann-surfaces
asked Nov 30 '18 at 23:26
CO2CO2
1529
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add a comment |
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From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $mathbb{C}$ which is not all of $mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.
Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?
complex-analysis riemann-surfaces
asked Nov 30 '18 at 23:26
CO2CO2
1529
$endgroup$
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The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
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– zhw.
Nov 30 '18 at 23:54
add a comment |
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From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $mathbb{C}$ which is not all of $mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.
Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?
complex-analysis riemann-surfaces
asked Nov 30 '18 at 23:26
CO2CO2
1529
$endgroup$
From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $mathbb{C}$ which is not all of $mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.
Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?
complex-analysis riemann-surfaces
complex-analysis riemann-surfaces
asked Nov 30 '18 at 23:26
CO2CO2
1529
asked Nov 30 '18 at 23:26
CO2CO2
1529
asked Nov 30 '18 at 23:26
CO2CO2
1529
asked Nov 30 '18 at 23:26
CO2CO2
1529
asked Nov 30 '18 at 23:26
CO2CO2
1529
1529
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The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
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– zhw.
Nov 30 '18 at 23:54
add a comment |
$begingroup$
The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
$endgroup$
– zhw.
Nov 30 '18 at 23:54
$begingroup$
The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
$endgroup$
– zhw.
Nov 30 '18 at 23:54
$begingroup$
The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
$endgroup$
– zhw.
Nov 30 '18 at 23:54
add a comment |
1 Answer
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It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.
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1 Answer
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1 Answer
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active
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It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.
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It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.
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add a comment |
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It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.
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It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.
answered Nov 30 '18 at 23:59
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Paul Frost
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The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
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– zhw.
Nov 30 '18 at 23:54