Mixing
Holomorphic Polynomial from the Riemann Sphere to Itself
$begingroup$
Helllo friends!
I want to prove that polynomial function $p(z)=z^2-4$ extends to a holomorphic mapping from the Riemann Sphere to itself. I know that four possible combinations of holomorphic charts in domain and codomain must be holomorphic but I do not know how to define these charts to show that the following 4 possible compositions are hol.:
$psi_1 o f o psi_1^{-1}$,
$psi_2 o f o psi_1^{-1}$,
$psi_1 o f o psi_2^{-1}$,
$psi_2 o f o psi_2^{-1}$
I appreciate any help in this problem.
complex-analysis complex-numbers
asked Jan 31 at 15:44
DavidDavid
976
$endgroup$
add a comment |
$begingroup$
Helllo friends!
I want to prove that polynomial function $p(z)=z^2-4$ extends to a holomorphic mapping from the Riemann Sphere to itself. I know that four possible combinations of holomorphic charts in domain and codomain must be holomorphic but I do not know how to define these charts to show that the following 4 possible compositions are hol.:
$psi_1 o f o psi_1^{-1}$,
$psi_2 o f o psi_1^{-1}$,
$psi_1 o f o psi_2^{-1}$,
$psi_2 o f o psi_2^{-1}$
I appreciate any help in this problem.
complex-analysis complex-numbers
asked Jan 31 at 15:44
DavidDavid
976
$endgroup$
add a comment |
$begingroup$
Helllo friends!
I want to prove that polynomial function $p(z)=z^2-4$ extends to a holomorphic mapping from the Riemann Sphere to itself. I know that four possible combinations of holomorphic charts in domain and codomain must be holomorphic but I do not know how to define these charts to show that the following 4 possible compositions are hol.:
$psi_1 o f o psi_1^{-1}$,
$psi_2 o f o psi_1^{-1}$,
$psi_1 o f o psi_2^{-1}$,
$psi_2 o f o psi_2^{-1}$
I appreciate any help in this problem.
complex-analysis complex-numbers
asked Jan 31 at 15:44
DavidDavid
976
$endgroup$
Helllo friends!
I want to prove that polynomial function $p(z)=z^2-4$ extends to a holomorphic mapping from the Riemann Sphere to itself. I know that four possible combinations of holomorphic charts in domain and codomain must be holomorphic but I do not know how to define these charts to show that the following 4 possible compositions are hol.:
$psi_1 o f o psi_1^{-1}$,
$psi_2 o f o psi_1^{-1}$,
$psi_1 o f o psi_2^{-1}$,
$psi_2 o f o psi_2^{-1}$
I appreciate any help in this problem.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Jan 31 at 15:44
DavidDavid
976
asked Jan 31 at 15:44
DavidDavid
976
edited Jan 31 at 16:24
David
asked Jan 31 at 15:44
DavidDavid
976
asked Jan 31 at 15:44
DavidDavid
976
asked Jan 31 at 15:44
DavidDavid
976
976
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, consider the extension defined at infinity as $p (infty) = infty$. By definition, $p$ is holomorphic at a point $x $ in the sphere iff there exists charts $phi$ and $psi$ on neighbourhoods of $x $ and $p (x) $ respectively, such that the function $f = psi circ p circ phi^{-1}$ is holomorphic (defined in an open set the complex plane) at the point $phi(x)$. If $z neq infty$, the charts are the identity charts (you are in the complex plane). Then, you simply have to check for the infinity point. Taking the chart of infinity at domain and codomain: $phi = psi = 1/z$, you can calculate:
$$
f (z) = psi circ p circ phi^{-1} = frac{1}{(frac{1}{z})^2 - 4} = frac{z}{1 - 4z^2}
$$
This new function is clearly holomorphic at $phi(infty) = 0$.
Edit: Denoting the sphere with $S$, open sets with $U = S setminus { infty }$ and $V = S setminus { 0}$ and define the charts $f_1(z) = z$ in $U $ and $f_2 (z) = 1/z$ (this one is the chart of infinity) in V. In the intersection of those you can use either of them.
answered Jan 31 at 16:30
Felipe MonteiroFelipe Monteiro
586
$endgroup$
$begingroup$
No. If you prove it in one, then they all are holomorphic. This is true because all the "coordinate changes" (that is, $phi circ psi^{-1}$) are holomorphic functions in the plane. Then if you know that this holds for one choice of coordinates, you can use this coordinate change to prove this holds for every choice.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:45
$begingroup$
Thanks friend. So, in the proof I can replace $phi$ and $psi$ with $f_1$ and $f_2$ respectively right? In case I want to show that the combination $f_2 circ p circ f_1^{-1}$ is hol, I should show that it is hol. at $f_1(0)=0$?
$endgroup$
– David
Jan 31 at 16:56
$begingroup$
You need to be careful with the neighbourhoods. Only the open set $V$ (with the function $f_2$) can be used in this case, because the open set $U$ does not contain the infinity.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:58
$begingroup$
Since we have to check for infinity point I have to use $V$. How about combination $f_1circ pcirc f_1^{-1}$ since $U $ does not contain infinity? How can I verify on appropriate open set?
$endgroup$
– David
Jan 31 at 17:04
$begingroup$
This is defined only for $z neq infty$, and will give you p in the plane
$endgroup$
– Felipe Monteiro
Jan 31 at 17:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First, consider the extension defined at infinity as $p (infty) = infty$. By definition, $p$ is holomorphic at a point $x $ in the sphere iff there exists charts $phi$ and $psi$ on neighbourhoods of $x $ and $p (x) $ respectively, such that the function $f = psi circ p circ phi^{-1}$ is holomorphic (defined in an open set the complex plane) at the point $phi(x)$. If $z neq infty$, the charts are the identity charts (you are in the complex plane). Then, you simply have to check for the infinity point. Taking the chart of infinity at domain and codomain: $phi = psi = 1/z$, you can calculate:
$$
f (z) = psi circ p circ phi^{-1} = frac{1}{(frac{1}{z})^2 - 4} = frac{z}{1 - 4z^2}
$$
This new function is clearly holomorphic at $phi(infty) = 0$.
Edit: Denoting the sphere with $S$, open sets with $U = S setminus { infty }$ and $V = S setminus { 0}$ and define the charts $f_1(z) = z$ in $U $ and $f_2 (z) = 1/z$ (this one is the chart of infinity) in V. In the intersection of those you can use either of them.
answered Jan 31 at 16:30
Felipe MonteiroFelipe Monteiro
586
$endgroup$
$begingroup$
No. If you prove it in one, then they all are holomorphic. This is true because all the "coordinate changes" (that is, $phi circ psi^{-1}$) are holomorphic functions in the plane. Then if you know that this holds for one choice of coordinates, you can use this coordinate change to prove this holds for every choice.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:45
$begingroup$
Thanks friend. So, in the proof I can replace $phi$ and $psi$ with $f_1$ and $f_2$ respectively right? In case I want to show that the combination $f_2 circ p circ f_1^{-1}$ is hol, I should show that it is hol. at $f_1(0)=0$?
$endgroup$
– David
Jan 31 at 16:56
$begingroup$
You need to be careful with the neighbourhoods. Only the open set $V$ (with the function $f_2$) can be used in this case, because the open set $U$ does not contain the infinity.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:58
$begingroup$
Since we have to check for infinity point I have to use $V$. How about combination $f_1circ pcirc f_1^{-1}$ since $U $ does not contain infinity? How can I verify on appropriate open set?
$endgroup$
– David
Jan 31 at 17:04
$begingroup$
This is defined only for $z neq infty$, and will give you p in the plane
$endgroup$
– Felipe Monteiro
Jan 31 at 17:07
add a comment |
$begingroup$
First, consider the extension defined at infinity as $p (infty) = infty$. By definition, $p$ is holomorphic at a point $x $ in the sphere iff there exists charts $phi$ and $psi$ on neighbourhoods of $x $ and $p (x) $ respectively, such that the function $f = psi circ p circ phi^{-1}$ is holomorphic (defined in an open set the complex plane) at the point $phi(x)$. If $z neq infty$, the charts are the identity charts (you are in the complex plane). Then, you simply have to check for the infinity point. Taking the chart of infinity at domain and codomain: $phi = psi = 1/z$, you can calculate:
$$
f (z) = psi circ p circ phi^{-1} = frac{1}{(frac{1}{z})^2 - 4} = frac{z}{1 - 4z^2}
$$
This new function is clearly holomorphic at $phi(infty) = 0$.
Edit: Denoting the sphere with $S$, open sets with $U = S setminus { infty }$ and $V = S setminus { 0}$ and define the charts $f_1(z) = z$ in $U $ and $f_2 (z) = 1/z$ (this one is the chart of infinity) in V. In the intersection of those you can use either of them.
answered Jan 31 at 16:30
Felipe MonteiroFelipe Monteiro
586
$endgroup$
$begingroup$
No. If you prove it in one, then they all are holomorphic. This is true because all the "coordinate changes" (that is, $phi circ psi^{-1}$) are holomorphic functions in the plane. Then if you know that this holds for one choice of coordinates, you can use this coordinate change to prove this holds for every choice.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:45
$begingroup$
Thanks friend. So, in the proof I can replace $phi$ and $psi$ with $f_1$ and $f_2$ respectively right? In case I want to show that the combination $f_2 circ p circ f_1^{-1}$ is hol, I should show that it is hol. at $f_1(0)=0$?
$endgroup$
– David
Jan 31 at 16:56
$begingroup$
You need to be careful with the neighbourhoods. Only the open set $V$ (with the function $f_2$) can be used in this case, because the open set $U$ does not contain the infinity.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:58
$begingroup$
Since we have to check for infinity point I have to use $V$. How about combination $f_1circ pcirc f_1^{-1}$ since $U $ does not contain infinity? How can I verify on appropriate open set?
$endgroup$
– David
Jan 31 at 17:04
$begingroup$
This is defined only for $z neq infty$, and will give you p in the plane
$endgroup$
– Felipe Monteiro
Jan 31 at 17:07
add a comment |
$begingroup$
First, consider the extension defined at infinity as $p (infty) = infty$. By definition, $p$ is holomorphic at a point $x $ in the sphere iff there exists charts $phi$ and $psi$ on neighbourhoods of $x $ and $p (x) $ respectively, such that the function $f = psi circ p circ phi^{-1}$ is holomorphic (defined in an open set the complex plane) at the point $phi(x)$. If $z neq infty$, the charts are the identity charts (you are in the complex plane). Then, you simply have to check for the infinity point. Taking the chart of infinity at domain and codomain: $phi = psi = 1/z$, you can calculate:
$$
f (z) = psi circ p circ phi^{-1} = frac{1}{(frac{1}{z})^2 - 4} = frac{z}{1 - 4z^2}
$$
This new function is clearly holomorphic at $phi(infty) = 0$.
Edit: Denoting the sphere with $S$, open sets with $U = S setminus { infty }$ and $V = S setminus { 0}$ and define the charts $f_1(z) = z$ in $U $ and $f_2 (z) = 1/z$ (this one is the chart of infinity) in V. In the intersection of those you can use either of them.
answered Jan 31 at 16:30
Felipe MonteiroFelipe Monteiro
586
$endgroup$
First, consider the extension defined at infinity as $p (infty) = infty$. By definition, $p$ is holomorphic at a point $x $ in the sphere iff there exists charts $phi$ and $psi$ on neighbourhoods of $x $ and $p (x) $ respectively, such that the function $f = psi circ p circ phi^{-1}$ is holomorphic (defined in an open set the complex plane) at the point $phi(x)$. If $z neq infty$, the charts are the identity charts (you are in the complex plane). Then, you simply have to check for the infinity point. Taking the chart of infinity at domain and codomain: $phi = psi = 1/z$, you can calculate:
$$
f (z) = psi circ p circ phi^{-1} = frac{1}{(frac{1}{z})^2 - 4} = frac{z}{1 - 4z^2}
$$
This new function is clearly holomorphic at $phi(infty) = 0$.
Edit: Denoting the sphere with $S$, open sets with $U = S setminus { infty }$ and $V = S setminus { 0}$ and define the charts $f_1(z) = z$ in $U $ and $f_2 (z) = 1/z$ (this one is the chart of infinity) in V. In the intersection of those you can use either of them.
answered Jan 31 at 16:30
Felipe MonteiroFelipe Monteiro
586
edited Jan 31 at 16:40
answered Jan 31 at 16:30
Felipe MonteiroFelipe Monteiro
586
answered Jan 31 at 16:30
Felipe MonteiroFelipe Monteiro
586
answered Jan 31 at 16:30
Felipe MonteiroFelipe Monteiro
586
586
$begingroup$
No. If you prove it in one, then they all are holomorphic. This is true because all the "coordinate changes" (that is, $phi circ psi^{-1}$) are holomorphic functions in the plane. Then if you know that this holds for one choice of coordinates, you can use this coordinate change to prove this holds for every choice.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:45
$begingroup$
Thanks friend. So, in the proof I can replace $phi$ and $psi$ with $f_1$ and $f_2$ respectively right? In case I want to show that the combination $f_2 circ p circ f_1^{-1}$ is hol, I should show that it is hol. at $f_1(0)=0$?
$endgroup$
– David
Jan 31 at 16:56
$begingroup$
You need to be careful with the neighbourhoods. Only the open set $V$ (with the function $f_2$) can be used in this case, because the open set $U$ does not contain the infinity.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:58
$begingroup$
Since we have to check for infinity point I have to use $V$. How about combination $f_1circ pcirc f_1^{-1}$ since $U $ does not contain infinity? How can I verify on appropriate open set?
$endgroup$
– David
Jan 31 at 17:04
$begingroup$
This is defined only for $z neq infty$, and will give you p in the plane
$endgroup$
– Felipe Monteiro
Jan 31 at 17:07
add a comment |
$begingroup$
No. If you prove it in one, then they all are holomorphic. This is true because all the "coordinate changes" (that is, $phi circ psi^{-1}$) are holomorphic functions in the plane. Then if you know that this holds for one choice of coordinates, you can use this coordinate change to prove this holds for every choice.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:45
$begingroup$
Thanks friend. So, in the proof I can replace $phi$ and $psi$ with $f_1$ and $f_2$ respectively right? In case I want to show that the combination $f_2 circ p circ f_1^{-1}$ is hol, I should show that it is hol. at $f_1(0)=0$?
$endgroup$
– David
Jan 31 at 16:56
$begingroup$
You need to be careful with the neighbourhoods. Only the open set $V$ (with the function $f_2$) can be used in this case, because the open set $U$ does not contain the infinity.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:58
$begingroup$
Since we have to check for infinity point I have to use $V$. How about combination $f_1circ pcirc f_1^{-1}$ since $U $ does not contain infinity? How can I verify on appropriate open set?
$endgroup$
– David
Jan 31 at 17:04
$begingroup$
This is defined only for $z neq infty$, and will give you p in the plane
$endgroup$
– Felipe Monteiro
Jan 31 at 17:07
$begingroup$
No. If you prove it in one, then they all are holomorphic. This is true because all the "coordinate changes" (that is, $phi circ psi^{-1}$) are holomorphic functions in the plane. Then if you know that this holds for one choice of coordinates, you can use this coordinate change to prove this holds for every choice.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:45
$begingroup$
No. If you prove it in one, then they all are holomorphic. This is true because all the "coordinate changes" (that is, $phi circ psi^{-1}$) are holomorphic functions in the plane. Then if you know that this holds for one choice of coordinates, you can use this coordinate change to prove this holds for every choice.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:45
$begingroup$
Thanks friend. So, in the proof I can replace $phi$ and $psi$ with $f_1$ and $f_2$ respectively right? In case I want to show that the combination $f_2 circ p circ f_1^{-1}$ is hol, I should show that it is hol. at $f_1(0)=0$?
$endgroup$
– David
Jan 31 at 16:56
$begingroup$
Thanks friend. So, in the proof I can replace $phi$ and $psi$ with $f_1$ and $f_2$ respectively right? In case I want to show that the combination $f_2 circ p circ f_1^{-1}$ is hol, I should show that it is hol. at $f_1(0)=0$?
$endgroup$
– David
Jan 31 at 16:56
$begingroup$
You need to be careful with the neighbourhoods. Only the open set $V$ (with the function $f_2$) can be used in this case, because the open set $U$ does not contain the infinity.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:58
$begingroup$
You need to be careful with the neighbourhoods. Only the open set $V$ (with the function $f_2$) can be used in this case, because the open set $U$ does not contain the infinity.
$endgroup$
– Felipe Monteiro
Jan 31 at 16:58
$begingroup$
Since we have to check for infinity point I have to use $V$. How about combination $f_1circ pcirc f_1^{-1}$ since $U $ does not contain infinity? How can I verify on appropriate open set?
$endgroup$
– David
Jan 31 at 17:04
$begingroup$
Since we have to check for infinity point I have to use $V$. How about combination $f_1circ pcirc f_1^{-1}$ since $U $ does not contain infinity? How can I verify on appropriate open set?
$endgroup$
– David
Jan 31 at 17:04
$begingroup$
This is defined only for $z neq infty$, and will give you p in the plane
$endgroup$
– Felipe Monteiro
Jan 31 at 17:07
$begingroup$
This is defined only for $z neq infty$, and will give you p in the plane
$endgroup$
– Felipe Monteiro
Jan 31 at 17:07
add a comment |
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