Showing algebraic dependence of meromorphic functions on a compact Riemann surface
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I have been given the following question to do: Let $f,g$ be meromorphic functions on a compact Riemann Surface $R$. Show that there is some polynomial such that $P(f,g) = 0$ (i.e. show that any two meromorphic functions
on $R$ are algebraically dependent). I have seen this result over the torus which follows from looking at the Weierstrass $wp$ function, however I have no idea how to generalise that to every compact Riemann Surface.
There is a hint which says I should let $d = m+n$ where $m,n$ are the valencies of $f,g$ respectively and consider $P(f,g) = sumlimits_{j = 0}^dsumlimits_{k = 0}^d a_{jk}f(z)^jg(z)^k$ and show that this has at most $d^2$ poles and that I can choose the $a_{jk}$ so that $P(f,g)$ has at least $d^2+2d$ roots and so is constant by the valency theorem.
Showing that there are at most $d^2$ poles is easy but I don't know how to select the $a_{jk}$ to get $d^2+2d$ roots. I don't see whether I should try and find them explicitly (seems hard) or use some indirect argument (but I can't see where to start). Any help is much appreciated.
complex-analysis polynomials riemann-surfaces compact-manifolds meromorphic-functions
asked Dec 17 '18 at 18:44
Abdul Hadi KhanAbdul Hadi Khan
518419
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add a comment |
$begingroup$
I have been given the following question to do: Let $f,g$ be meromorphic functions on a compact Riemann Surface $R$. Show that there is some polynomial such that $P(f,g) = 0$ (i.e. show that any two meromorphic functions
on $R$ are algebraically dependent). I have seen this result over the torus which follows from looking at the Weierstrass $wp$ function, however I have no idea how to generalise that to every compact Riemann Surface.
There is a hint which says I should let $d = m+n$ where $m,n$ are the valencies of $f,g$ respectively and consider $P(f,g) = sumlimits_{j = 0}^dsumlimits_{k = 0}^d a_{jk}f(z)^jg(z)^k$ and show that this has at most $d^2$ poles and that I can choose the $a_{jk}$ so that $P(f,g)$ has at least $d^2+2d$ roots and so is constant by the valency theorem.
Showing that there are at most $d^2$ poles is easy but I don't know how to select the $a_{jk}$ to get $d^2+2d$ roots. I don't see whether I should try and find them explicitly (seems hard) or use some indirect argument (but I can't see where to start). Any help is much appreciated.
complex-analysis polynomials riemann-surfaces compact-manifolds meromorphic-functions
asked Dec 17 '18 at 18:44
Abdul Hadi KhanAbdul Hadi Khan
518419
$endgroup$
add a comment |
$begingroup$
I have been given the following question to do: Let $f,g$ be meromorphic functions on a compact Riemann Surface $R$. Show that there is some polynomial such that $P(f,g) = 0$ (i.e. show that any two meromorphic functions
on $R$ are algebraically dependent). I have seen this result over the torus which follows from looking at the Weierstrass $wp$ function, however I have no idea how to generalise that to every compact Riemann Surface.
There is a hint which says I should let $d = m+n$ where $m,n$ are the valencies of $f,g$ respectively and consider $P(f,g) = sumlimits_{j = 0}^dsumlimits_{k = 0}^d a_{jk}f(z)^jg(z)^k$ and show that this has at most $d^2$ poles and that I can choose the $a_{jk}$ so that $P(f,g)$ has at least $d^2+2d$ roots and so is constant by the valency theorem.
Showing that there are at most $d^2$ poles is easy but I don't know how to select the $a_{jk}$ to get $d^2+2d$ roots. I don't see whether I should try and find them explicitly (seems hard) or use some indirect argument (but I can't see where to start). Any help is much appreciated.
complex-analysis polynomials riemann-surfaces compact-manifolds meromorphic-functions
asked Dec 17 '18 at 18:44
Abdul Hadi KhanAbdul Hadi Khan
518419
$endgroup$
I have been given the following question to do: Let $f,g$ be meromorphic functions on a compact Riemann Surface $R$. Show that there is some polynomial such that $P(f,g) = 0$ (i.e. show that any two meromorphic functions
on $R$ are algebraically dependent). I have seen this result over the torus which follows from looking at the Weierstrass $wp$ function, however I have no idea how to generalise that to every compact Riemann Surface.
There is a hint which says I should let $d = m+n$ where $m,n$ are the valencies of $f,g$ respectively and consider $P(f,g) = sumlimits_{j = 0}^dsumlimits_{k = 0}^d a_{jk}f(z)^jg(z)^k$ and show that this has at most $d^2$ poles and that I can choose the $a_{jk}$ so that $P(f,g)$ has at least $d^2+2d$ roots and so is constant by the valency theorem.
Showing that there are at most $d^2$ poles is easy but I don't know how to select the $a_{jk}$ to get $d^2+2d$ roots. I don't see whether I should try and find them explicitly (seems hard) or use some indirect argument (but I can't see where to start). Any help is much appreciated.
complex-analysis polynomials riemann-surfaces compact-manifolds meromorphic-functions
complex-analysis polynomials riemann-surfaces compact-manifolds meromorphic-functions
asked Dec 17 '18 at 18:44
Abdul Hadi KhanAbdul Hadi Khan
518419
asked Dec 17 '18 at 18:44
Abdul Hadi KhanAbdul Hadi Khan
518419
edited Dec 18 '18 at 8:08
Abdul Hadi Khan
asked Dec 17 '18 at 18:44
Abdul Hadi KhanAbdul Hadi Khan
518419
asked Dec 17 '18 at 18:44
Abdul Hadi KhanAbdul Hadi Khan
518419
asked Dec 17 '18 at 18:44
Abdul Hadi KhanAbdul Hadi Khan
518419
518419
add a comment |
add a comment |
1 Answer
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You can prove that there exists two polynomial $p,qin mathbb{C}[x,y]$ such that
$ord_a(frac{p(f,g)}{q(f,g)})geq 0$ for every $ain X$
so it is an olomorphic function on $X$ that is a Compact R.S. so there exist a costant $c$ such that
$frac{p(f,g)}{q(f,g)})=c$
Then
$p(f,g)-cq(f,g)=0$
But I don’t know how built these two polynomials.
We can indicate ${a_1,dots , a_n}$ the set of the poles of $f$ and ${b_1,dots b_m}$ the set of the poles of $g$.
If $l:=ord_b(g)<0$ and $ord_b(f)geq 0$ then you have that
$ord_b((f-f(b))^l g)geq 0$
So you can resolve the problem in this particular case.
answered Dec 17 '18 at 19:19
Federico FalluccaFederico Fallucca
2,280210
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$begingroup$
Why I’m wrong? I don’t understand
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– Federico Fallucca
Jan 11 at 12:39
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can prove that there exists two polynomial $p,qin mathbb{C}[x,y]$ such that
$ord_a(frac{p(f,g)}{q(f,g)})geq 0$ for every $ain X$
so it is an olomorphic function on $X$ that is a Compact R.S. so there exist a costant $c$ such that
$frac{p(f,g)}{q(f,g)})=c$
Then
$p(f,g)-cq(f,g)=0$
But I don’t know how built these two polynomials.
We can indicate ${a_1,dots , a_n}$ the set of the poles of $f$ and ${b_1,dots b_m}$ the set of the poles of $g$.
If $l:=ord_b(g)<0$ and $ord_b(f)geq 0$ then you have that
$ord_b((f-f(b))^l g)geq 0$
So you can resolve the problem in this particular case.
answered Dec 17 '18 at 19:19
Federico FalluccaFederico Fallucca
2,280210
$endgroup$
$begingroup$
Why I’m wrong? I don’t understand
$endgroup$
– Federico Fallucca
Jan 11 at 12:39
add a comment |
$begingroup$
You can prove that there exists two polynomial $p,qin mathbb{C}[x,y]$ such that
$ord_a(frac{p(f,g)}{q(f,g)})geq 0$ for every $ain X$
so it is an olomorphic function on $X$ that is a Compact R.S. so there exist a costant $c$ such that
$frac{p(f,g)}{q(f,g)})=c$
Then
$p(f,g)-cq(f,g)=0$
But I don’t know how built these two polynomials.
We can indicate ${a_1,dots , a_n}$ the set of the poles of $f$ and ${b_1,dots b_m}$ the set of the poles of $g$.
If $l:=ord_b(g)<0$ and $ord_b(f)geq 0$ then you have that
$ord_b((f-f(b))^l g)geq 0$
So you can resolve the problem in this particular case.
answered Dec 17 '18 at 19:19
Federico FalluccaFederico Fallucca
2,280210
$endgroup$
$begingroup$
Why I’m wrong? I don’t understand
$endgroup$
– Federico Fallucca
Jan 11 at 12:39
add a comment |
$begingroup$
You can prove that there exists two polynomial $p,qin mathbb{C}[x,y]$ such that
$ord_a(frac{p(f,g)}{q(f,g)})geq 0$ for every $ain X$
so it is an olomorphic function on $X$ that is a Compact R.S. so there exist a costant $c$ such that
$frac{p(f,g)}{q(f,g)})=c$
Then
$p(f,g)-cq(f,g)=0$
But I don’t know how built these two polynomials.
We can indicate ${a_1,dots , a_n}$ the set of the poles of $f$ and ${b_1,dots b_m}$ the set of the poles of $g$.
If $l:=ord_b(g)<0$ and $ord_b(f)geq 0$ then you have that
$ord_b((f-f(b))^l g)geq 0$
So you can resolve the problem in this particular case.
answered Dec 17 '18 at 19:19
Federico FalluccaFederico Fallucca
2,280210
$endgroup$
You can prove that there exists two polynomial $p,qin mathbb{C}[x,y]$ such that
$ord_a(frac{p(f,g)}{q(f,g)})geq 0$ for every $ain X$
so it is an olomorphic function on $X$ that is a Compact R.S. so there exist a costant $c$ such that
$frac{p(f,g)}{q(f,g)})=c$
Then
$p(f,g)-cq(f,g)=0$
But I don’t know how built these two polynomials.
We can indicate ${a_1,dots , a_n}$ the set of the poles of $f$ and ${b_1,dots b_m}$ the set of the poles of $g$.
If $l:=ord_b(g)<0$ and $ord_b(f)geq 0$ then you have that
$ord_b((f-f(b))^l g)geq 0$
So you can resolve the problem in this particular case.
answered Dec 17 '18 at 19:19
Federico FalluccaFederico Fallucca
2,280210
edited Dec 18 '18 at 14:49
answered Dec 17 '18 at 19:19
Federico FalluccaFederico Fallucca
2,280210
answered Dec 17 '18 at 19:19
Federico FalluccaFederico Fallucca
2,280210
answered Dec 17 '18 at 19:19
Federico FalluccaFederico Fallucca
2,280210
2,280210
$begingroup$
Why I’m wrong? I don’t understand
$endgroup$
– Federico Fallucca
Jan 11 at 12:39
add a comment |
$begingroup$
Why I’m wrong? I don’t understand
$endgroup$
– Federico Fallucca
Jan 11 at 12:39
$begingroup$
Why I’m wrong? I don’t understand
$endgroup$
– Federico Fallucca
Jan 11 at 12:39
$begingroup$
Why I’m wrong? I don’t understand
$endgroup$
– Federico Fallucca
Jan 11 at 12:39
add a comment |
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