Mixing
Are $Bbb R^2backslash{(x,y) : xle 0,y=0}$ and the unit disc homeomorphic? Difference between conformal map an...
$begingroup$
The Riemann mapping theorem says that there exsists a (bijective) conformal map $f$ between $Omega =Bbb Cbackslash {zinBbb C: Im(z)=0, Re(z)le0}$ and the unit disc $D_1$.
$f$ is the composition of $sqrt{z}$ and some Möbius transformation (I think $frac{-iz-i}{iz-i}$) and $sqrt z$ is not continuous.
But $f$ must be "biholomorphic" so holomorphic $implies$ continuous
How can $f$ be continuous?
general-topology mobius-transformation
asked Jan 23 at 8:18
John CataldoJohn Cataldo
1,1881316
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add a comment |
$begingroup$
The Riemann mapping theorem says that there exsists a (bijective) conformal map $f$ between $Omega =Bbb Cbackslash {zinBbb C: Im(z)=0, Re(z)le0}$ and the unit disc $D_1$.
$f$ is the composition of $sqrt{z}$ and some Möbius transformation (I think $frac{-iz-i}{iz-i}$) and $sqrt z$ is not continuous.
But $f$ must be "biholomorphic" so holomorphic $implies$ continuous
How can $f$ be continuous?
general-topology mobius-transformation
asked Jan 23 at 8:18
John CataldoJohn Cataldo
1,1881316
$endgroup$
$begingroup$
Why should it be homeomorphic to the unit disc ? One is compact, the other one is open and unbounded... Difference between conformal mapping and homeomorphism ? It's quite huge : one is analytic, the other one is only continuous.
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– Dylan
Jan 23 at 8:22
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You are quoting Riemann Mapping Theorem wrongly. $Omega$ is conformally equivalent to the open unit disk, not the closed unit disk.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:28
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If I understood correctly the slightly (or pretty...) confusing definition of that set, $,Omega,$ is the plane without the whole $;x,-$ axis ($,y=0,$) and without all the points with non-positive $;x,-$ entry ($,xle0,$). So it basically is the right half plane without the $;x,-$ axis...It isn't even path connected as there isn't path within $,Omega,$ between $;(1,-1);$ and $;(1,1);$ , for example...what am I missing here?
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– DonAntonio
Jan 23 at 8:40
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$Omega$ is simply connected, it's the whole plane minus the half line ${xle 0,y=0}$
$endgroup$
– John Cataldo
Jan 23 at 8:44
1
$begingroup$
"and $sqrt z$ is not continuous" On $Omega$, the usual square root is very much continuous.
$endgroup$
– Did
Jan 23 at 9:16
add a comment |
$begingroup$
The Riemann mapping theorem says that there exsists a (bijective) conformal map $f$ between $Omega =Bbb Cbackslash {zinBbb C: Im(z)=0, Re(z)le0}$ and the unit disc $D_1$.
$f$ is the composition of $sqrt{z}$ and some Möbius transformation (I think $frac{-iz-i}{iz-i}$) and $sqrt z$ is not continuous.
But $f$ must be "biholomorphic" so holomorphic $implies$ continuous
How can $f$ be continuous?
general-topology mobius-transformation
asked Jan 23 at 8:18
John CataldoJohn Cataldo
1,1881316
$endgroup$
The Riemann mapping theorem says that there exsists a (bijective) conformal map $f$ between $Omega =Bbb Cbackslash {zinBbb C: Im(z)=0, Re(z)le0}$ and the unit disc $D_1$.
$f$ is the composition of $sqrt{z}$ and some Möbius transformation (I think $frac{-iz-i}{iz-i}$) and $sqrt z$ is not continuous.
But $f$ must be "biholomorphic" so holomorphic $implies$ continuous
How can $f$ be continuous?
general-topology mobius-transformation
general-topology mobius-transformation
asked Jan 23 at 8:18
John CataldoJohn Cataldo
1,1881316
asked Jan 23 at 8:18
John CataldoJohn Cataldo
1,1881316
edited Jan 23 at 8:42
John Cataldo
asked Jan 23 at 8:18
John CataldoJohn Cataldo
1,1881316
asked Jan 23 at 8:18
John CataldoJohn Cataldo
1,1881316
asked Jan 23 at 8:18
John CataldoJohn Cataldo
1,1881316
1,1881316
$begingroup$
Why should it be homeomorphic to the unit disc ? One is compact, the other one is open and unbounded... Difference between conformal mapping and homeomorphism ? It's quite huge : one is analytic, the other one is only continuous.
$endgroup$
– Dylan
Jan 23 at 8:22
$begingroup$
You are quoting Riemann Mapping Theorem wrongly. $Omega$ is conformally equivalent to the open unit disk, not the closed unit disk.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:28
$begingroup$
If I understood correctly the slightly (or pretty...) confusing definition of that set, $,Omega,$ is the plane without the whole $;x,-$ axis ($,y=0,$) and without all the points with non-positive $;x,-$ entry ($,xle0,$). So it basically is the right half plane without the $;x,-$ axis...It isn't even path connected as there isn't path within $,Omega,$ between $;(1,-1);$ and $;(1,1);$ , for example...what am I missing here?
$endgroup$
– DonAntonio
Jan 23 at 8:40
$begingroup$
$Omega$ is simply connected, it's the whole plane minus the half line ${xle 0,y=0}$
$endgroup$
– John Cataldo
Jan 23 at 8:44
1
$begingroup$
"and $sqrt z$ is not continuous" On $Omega$, the usual square root is very much continuous.
$endgroup$
– Did
Jan 23 at 9:16
add a comment |
$begingroup$
Why should it be homeomorphic to the unit disc ? One is compact, the other one is open and unbounded... Difference between conformal mapping and homeomorphism ? It's quite huge : one is analytic, the other one is only continuous.
$endgroup$
– Dylan
Jan 23 at 8:22
$begingroup$
You are quoting Riemann Mapping Theorem wrongly. $Omega$ is conformally equivalent to the open unit disk, not the closed unit disk.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:28
$begingroup$
If I understood correctly the slightly (or pretty...) confusing definition of that set, $,Omega,$ is the plane without the whole $;x,-$ axis ($,y=0,$) and without all the points with non-positive $;x,-$ entry ($,xle0,$). So it basically is the right half plane without the $;x,-$ axis...It isn't even path connected as there isn't path within $,Omega,$ between $;(1,-1);$ and $;(1,1);$ , for example...what am I missing here?
$endgroup$
– DonAntonio
Jan 23 at 8:40
$begingroup$
$Omega$ is simply connected, it's the whole plane minus the half line ${xle 0,y=0}$
$endgroup$
– John Cataldo
Jan 23 at 8:44
1
$begingroup$
"and $sqrt z$ is not continuous" On $Omega$, the usual square root is very much continuous.
$endgroup$
– Did
Jan 23 at 9:16
$begingroup$
Why should it be homeomorphic to the unit disc ? One is compact, the other one is open and unbounded... Difference between conformal mapping and homeomorphism ? It's quite huge : one is analytic, the other one is only continuous.
$endgroup$
– Dylan
Jan 23 at 8:22
$begingroup$
Why should it be homeomorphic to the unit disc ? One is compact, the other one is open and unbounded... Difference between conformal mapping and homeomorphism ? It's quite huge : one is analytic, the other one is only continuous.
$endgroup$
– Dylan
Jan 23 at 8:22
$begingroup$
You are quoting Riemann Mapping Theorem wrongly. $Omega$ is conformally equivalent to the open unit disk, not the closed unit disk.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:28
$begingroup$
You are quoting Riemann Mapping Theorem wrongly. $Omega$ is conformally equivalent to the open unit disk, not the closed unit disk.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:28
$begingroup$
If I understood correctly the slightly (or pretty...) confusing definition of that set, $,Omega,$ is the plane without the whole $;x,-$ axis ($,y=0,$) and without all the points with non-positive $;x,-$ entry ($,xle0,$). So it basically is the right half plane without the $;x,-$ axis...It isn't even path connected as there isn't path within $,Omega,$ between $;(1,-1);$ and $;(1,1);$ , for example...what am I missing here?
$endgroup$
– DonAntonio
Jan 23 at 8:40
$begingroup$
If I understood correctly the slightly (or pretty...) confusing definition of that set, $,Omega,$ is the plane without the whole $;x,-$ axis ($,y=0,$) and without all the points with non-positive $;x,-$ entry ($,xle0,$). So it basically is the right half plane without the $;x,-$ axis...It isn't even path connected as there isn't path within $,Omega,$ between $;(1,-1);$ and $;(1,1);$ , for example...what am I missing here?
$endgroup$
– DonAntonio
Jan 23 at 8:40
$begingroup$
$Omega$ is simply connected, it's the whole plane minus the half line ${xle 0,y=0}$
$endgroup$
– John Cataldo
Jan 23 at 8:44
$begingroup$
$Omega$ is simply connected, it's the whole plane minus the half line ${xle 0,y=0}$
$endgroup$
– John Cataldo
Jan 23 at 8:44
1
1
$begingroup$
"and $sqrt z$ is not continuous" On $Omega$, the usual square root is very much continuous.
$endgroup$
– Did
Jan 23 at 9:16
$begingroup$
"and $sqrt z$ is not continuous" On $Omega$, the usual square root is very much continuous.
$endgroup$
– Did
Jan 23 at 9:16
add a comment |
1 Answer
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The set $Omega$ is open and simply connected. It does not contain $0$. The principal value $sqrt{cdot}$ of the square root function, given by
$$zmapsto w:=expbigl({1over2}{rm Log}(z)bigr),$$
is analytic, hence continuous on $Omega$. As is well known this function maps $Omega$ bijectively onto the open half plane $H: >{rm Re}(w)>0$, and a suitable Moebius transformation will then map $H$ onto $D$.
answered Jan 23 at 9:05
Christian BlatterChristian Blatter
175k8115327
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add a comment |
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$begingroup$
The set $Omega$ is open and simply connected. It does not contain $0$. The principal value $sqrt{cdot}$ of the square root function, given by
$$zmapsto w:=expbigl({1over2}{rm Log}(z)bigr),$$
is analytic, hence continuous on $Omega$. As is well known this function maps $Omega$ bijectively onto the open half plane $H: >{rm Re}(w)>0$, and a suitable Moebius transformation will then map $H$ onto $D$.
answered Jan 23 at 9:05
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
The set $Omega$ is open and simply connected. It does not contain $0$. The principal value $sqrt{cdot}$ of the square root function, given by
$$zmapsto w:=expbigl({1over2}{rm Log}(z)bigr),$$
is analytic, hence continuous on $Omega$. As is well known this function maps $Omega$ bijectively onto the open half plane $H: >{rm Re}(w)>0$, and a suitable Moebius transformation will then map $H$ onto $D$.
answered Jan 23 at 9:05
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
The set $Omega$ is open and simply connected. It does not contain $0$. The principal value $sqrt{cdot}$ of the square root function, given by
$$zmapsto w:=expbigl({1over2}{rm Log}(z)bigr),$$
is analytic, hence continuous on $Omega$. As is well known this function maps $Omega$ bijectively onto the open half plane $H: >{rm Re}(w)>0$, and a suitable Moebius transformation will then map $H$ onto $D$.
answered Jan 23 at 9:05
Christian BlatterChristian Blatter
175k8115327
$endgroup$
The set $Omega$ is open and simply connected. It does not contain $0$. The principal value $sqrt{cdot}$ of the square root function, given by
$$zmapsto w:=expbigl({1over2}{rm Log}(z)bigr),$$
is analytic, hence continuous on $Omega$. As is well known this function maps $Omega$ bijectively onto the open half plane $H: >{rm Re}(w)>0$, and a suitable Moebius transformation will then map $H$ onto $D$.
answered Jan 23 at 9:05
Christian BlatterChristian Blatter
175k8115327
answered Jan 23 at 9:05
Christian BlatterChristian Blatter
175k8115327
answered Jan 23 at 9:05
Christian BlatterChristian Blatter
175k8115327
answered Jan 23 at 9:05
Christian BlatterChristian Blatter
175k8115327
175k8115327
add a comment |
add a comment |
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$begingroup$
Why should it be homeomorphic to the unit disc ? One is compact, the other one is open and unbounded... Difference between conformal mapping and homeomorphism ? It's quite huge : one is analytic, the other one is only continuous.
$endgroup$
– Dylan
Jan 23 at 8:22
$begingroup$
You are quoting Riemann Mapping Theorem wrongly. $Omega$ is conformally equivalent to the open unit disk, not the closed unit disk.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 8:28
$begingroup$
If I understood correctly the slightly (or pretty...) confusing definition of that set, $,Omega,$ is the plane without the whole $;x,-$ axis ($,y=0,$) and without all the points with non-positive $;x,-$ entry ($,xle0,$). So it basically is the right half plane without the $;x,-$ axis...It isn't even path connected as there isn't path within $,Omega,$ between $;(1,-1);$ and $;(1,1);$ , for example...what am I missing here?
$endgroup$
– DonAntonio
Jan 23 at 8:40
$begingroup$
$Omega$ is simply connected, it's the whole plane minus the half line ${xle 0,y=0}$
$endgroup$
– John Cataldo
Jan 23 at 8:44
1
$begingroup$
"and $sqrt z$ is not continuous" On $Omega$, the usual square root is very much continuous.
$endgroup$
– Did
Jan 23 at 9:16