Showing algebraic dependence of meromorphic functions on a compact Riemann surface












3












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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







share|cite|improve this question











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    3












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 18 '18 at 8:08







      Abdul Hadi Khan

















      asked Dec 17 '18 at 18:44









      Abdul Hadi KhanAbdul Hadi Khan

      518419




      518419






















          1 Answer
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          -1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why I’m wrong? I don’t understand
            $endgroup$
            – Federico Fallucca
            Jan 11 at 12:39











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          -1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why I’m wrong? I don’t understand
            $endgroup$
            – Federico Fallucca
            Jan 11 at 12:39
















          -1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why I’m wrong? I don’t understand
            $endgroup$
            – Federico Fallucca
            Jan 11 at 12:39














          -1












          -1








          -1





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 14:49

























          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


















          • $begingroup$
            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


















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