Weierstrass $wp$-function defines a map from the torus to an elliptic curve. Why is it injective?












2












$begingroup$


For $L$ a lattice in $mathbb C$, the Weierstrass $wp$-function is the meromorphic function



$$wp(z) = frac{1}{z^2} + sumlimits_{0 neq lambda in L}frac{1}{(z-lambda)^2} - frac{1}{lambda^2}$$
It can be shown to satisfy the differential equation $wp'(z) = 4wp(z)^3 - g_2wp(z) - g_3$, where



$$g_2 = 60 sumlimits_{0 neq lambda in L} frac{1}{lambda^4}$$



$$g_3 = 120 sumlimits_{0 neq lambda in L} frac{1}{lambda^6}$$
If $E$ is the elliptic curve in $mathbb P^2$ defined by the homogeneous polynomial $y^2z = 4x^3 - g_2xz^2-g_3z^3$, then



$$F(z) = begin{cases} (wp(z);wp'(z);1) & textrm{if }znotin L \ (0;1;0) & textrm{if } z in L end{cases}$$



can be shown to define a holomorphic function $mathbb C rightarrow E$. Since $mathscr P$ and $mathscr P'$ are well defined on $mathbb C/L$, so is $F$, and $F$ induces a holomorphic function



$$bar{F}: mathbb C /L rightarrow E$$
which is automatically surjective, because $F$ is an open map (being holomorphic and nonconstant), and $mathbb C/L$ and $E$ are compact. I want to say that $bar{F}$ is a biholomorphism, which is equivalent to saying $bar{F}$ is injective.



How do we know that $bar{F}$ is injective?






complex-analysis elliptic-curves riemann-surfaces complex-manifolds







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    For $L$ a lattice in $mathbb C$, the Weierstrass $wp$-function is the meromorphic function



    $$wp(z) = frac{1}{z^2} + sumlimits_{0 neq lambda in L}frac{1}{(z-lambda)^2} - frac{1}{lambda^2}$$
    It can be shown to satisfy the differential equation $wp'(z) = 4wp(z)^3 - g_2wp(z) - g_3$, where



    $$g_2 = 60 sumlimits_{0 neq lambda in L} frac{1}{lambda^4}$$



    $$g_3 = 120 sumlimits_{0 neq lambda in L} frac{1}{lambda^6}$$
    If $E$ is the elliptic curve in $mathbb P^2$ defined by the homogeneous polynomial $y^2z = 4x^3 - g_2xz^2-g_3z^3$, then



    $$F(z) = begin{cases} (wp(z);wp'(z);1) & textrm{if }znotin L \ (0;1;0) & textrm{if } z in L end{cases}$$



    can be shown to define a holomorphic function $mathbb C rightarrow E$. Since $mathscr P$ and $mathscr P'$ are well defined on $mathbb C/L$, so is $F$, and $F$ induces a holomorphic function



    $$bar{F}: mathbb C /L rightarrow E$$
    which is automatically surjective, because $F$ is an open map (being holomorphic and nonconstant), and $mathbb C/L$ and $E$ are compact. I want to say that $bar{F}$ is a biholomorphism, which is equivalent to saying $bar{F}$ is injective.



    How do we know that $bar{F}$ is injective?






    complex-analysis elliptic-curves riemann-surfaces complex-manifolds







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      For $L$ a lattice in $mathbb C$, the Weierstrass $wp$-function is the meromorphic function



      $$wp(z) = frac{1}{z^2} + sumlimits_{0 neq lambda in L}frac{1}{(z-lambda)^2} - frac{1}{lambda^2}$$
      It can be shown to satisfy the differential equation $wp'(z) = 4wp(z)^3 - g_2wp(z) - g_3$, where



      $$g_2 = 60 sumlimits_{0 neq lambda in L} frac{1}{lambda^4}$$



      $$g_3 = 120 sumlimits_{0 neq lambda in L} frac{1}{lambda^6}$$
      If $E$ is the elliptic curve in $mathbb P^2$ defined by the homogeneous polynomial $y^2z = 4x^3 - g_2xz^2-g_3z^3$, then



      $$F(z) = begin{cases} (wp(z);wp'(z);1) & textrm{if }znotin L \ (0;1;0) & textrm{if } z in L end{cases}$$



      can be shown to define a holomorphic function $mathbb C rightarrow E$. Since $mathscr P$ and $mathscr P'$ are well defined on $mathbb C/L$, so is $F$, and $F$ induces a holomorphic function



      $$bar{F}: mathbb C /L rightarrow E$$
      which is automatically surjective, because $F$ is an open map (being holomorphic and nonconstant), and $mathbb C/L$ and $E$ are compact. I want to say that $bar{F}$ is a biholomorphism, which is equivalent to saying $bar{F}$ is injective.



      How do we know that $bar{F}$ is injective?






      complex-analysis elliptic-curves riemann-surfaces complex-manifolds







      share|cite|improve this question









      $endgroup$




      For $L$ a lattice in $mathbb C$, the Weierstrass $wp$-function is the meromorphic function



      $$wp(z) = frac{1}{z^2} + sumlimits_{0 neq lambda in L}frac{1}{(z-lambda)^2} - frac{1}{lambda^2}$$
      It can be shown to satisfy the differential equation $wp'(z) = 4wp(z)^3 - g_2wp(z) - g_3$, where



      $$g_2 = 60 sumlimits_{0 neq lambda in L} frac{1}{lambda^4}$$



      $$g_3 = 120 sumlimits_{0 neq lambda in L} frac{1}{lambda^6}$$
      If $E$ is the elliptic curve in $mathbb P^2$ defined by the homogeneous polynomial $y^2z = 4x^3 - g_2xz^2-g_3z^3$, then



      $$F(z) = begin{cases} (wp(z);wp'(z);1) & textrm{if }znotin L \ (0;1;0) & textrm{if } z in L end{cases}$$



      can be shown to define a holomorphic function $mathbb C rightarrow E$. Since $mathscr P$ and $mathscr P'$ are well defined on $mathbb C/L$, so is $F$, and $F$ induces a holomorphic function



      $$bar{F}: mathbb C /L rightarrow E$$
      which is automatically surjective, because $F$ is an open map (being holomorphic and nonconstant), and $mathbb C/L$ and $E$ are compact. I want to say that $bar{F}$ is a biholomorphism, which is equivalent to saying $bar{F}$ is injective.



      How do we know that $bar{F}$ is injective?






      complex-analysis elliptic-curves riemann-surfaces complex-manifolds




      complex-analysis elliptic-curves riemann-surfaces complex-manifolds






      share|cite|improve this question













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      asked Dec 17 '18 at 1:45









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          $begingroup$

          [I will treat $wp$ and related functions always as functions with domain $mathbb{C}/L$, rather than $mathbb{C}$.]



          First, note that $wp$ has degree $2$: its only pole is a double pole at $0$, so it takes every value in $mathbb{C}cup{infty}$ twice, with multiplicity. Also, $wp$ is even. So, if $zneq -z$, then $z$ and $-z$ are two preimages of $wp(z)$ and thus must be the only such preimages (and both must have multiplicity $1$). If $z=-z$, we now see that $wp$ is a $2$-to-$1$ mapping in a deleted neighborhood of $z$, so $z$ is a preimage of $wp(z)$ of multiplicity $2$ and is the only such preimage. In particular, we see that for any $z,winmathbb{C}/L$, $wp(z)=wp(w)$ iff $z=pm w$



          So, it suffices to show that $wp'(z)neq wp'(-z)$ for any $z$ such that $zneq -z$. Now $wp'$ is odd, so $wp'(z)=wp'(-z)$ implies $wp'(z)=0$. But $wp'(z)=0$ means that $z$ is a preimage of $wp(z)$ of multiplicity $2$. By the previous paragraph, this implies $z=-z$, as desired.






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            1 Answer
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            1 Answer
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            active

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            $begingroup$

            [I will treat $wp$ and related functions always as functions with domain $mathbb{C}/L$, rather than $mathbb{C}$.]



            First, note that $wp$ has degree $2$: its only pole is a double pole at $0$, so it takes every value in $mathbb{C}cup{infty}$ twice, with multiplicity. Also, $wp$ is even. So, if $zneq -z$, then $z$ and $-z$ are two preimages of $wp(z)$ and thus must be the only such preimages (and both must have multiplicity $1$). If $z=-z$, we now see that $wp$ is a $2$-to-$1$ mapping in a deleted neighborhood of $z$, so $z$ is a preimage of $wp(z)$ of multiplicity $2$ and is the only such preimage. In particular, we see that for any $z,winmathbb{C}/L$, $wp(z)=wp(w)$ iff $z=pm w$



            So, it suffices to show that $wp'(z)neq wp'(-z)$ for any $z$ such that $zneq -z$. Now $wp'$ is odd, so $wp'(z)=wp'(-z)$ implies $wp'(z)=0$. But $wp'(z)=0$ means that $z$ is a preimage of $wp(z)$ of multiplicity $2$. By the previous paragraph, this implies $z=-z$, as desired.






            share|cite|improve this answer









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              2












              $begingroup$

              [I will treat $wp$ and related functions always as functions with domain $mathbb{C}/L$, rather than $mathbb{C}$.]



              First, note that $wp$ has degree $2$: its only pole is a double pole at $0$, so it takes every value in $mathbb{C}cup{infty}$ twice, with multiplicity. Also, $wp$ is even. So, if $zneq -z$, then $z$ and $-z$ are two preimages of $wp(z)$ and thus must be the only such preimages (and both must have multiplicity $1$). If $z=-z$, we now see that $wp$ is a $2$-to-$1$ mapping in a deleted neighborhood of $z$, so $z$ is a preimage of $wp(z)$ of multiplicity $2$ and is the only such preimage. In particular, we see that for any $z,winmathbb{C}/L$, $wp(z)=wp(w)$ iff $z=pm w$



              So, it suffices to show that $wp'(z)neq wp'(-z)$ for any $z$ such that $zneq -z$. Now $wp'$ is odd, so $wp'(z)=wp'(-z)$ implies $wp'(z)=0$. But $wp'(z)=0$ means that $z$ is a preimage of $wp(z)$ of multiplicity $2$. By the previous paragraph, this implies $z=-z$, as desired.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                [I will treat $wp$ and related functions always as functions with domain $mathbb{C}/L$, rather than $mathbb{C}$.]



                First, note that $wp$ has degree $2$: its only pole is a double pole at $0$, so it takes every value in $mathbb{C}cup{infty}$ twice, with multiplicity. Also, $wp$ is even. So, if $zneq -z$, then $z$ and $-z$ are two preimages of $wp(z)$ and thus must be the only such preimages (and both must have multiplicity $1$). If $z=-z$, we now see that $wp$ is a $2$-to-$1$ mapping in a deleted neighborhood of $z$, so $z$ is a preimage of $wp(z)$ of multiplicity $2$ and is the only such preimage. In particular, we see that for any $z,winmathbb{C}/L$, $wp(z)=wp(w)$ iff $z=pm w$



                So, it suffices to show that $wp'(z)neq wp'(-z)$ for any $z$ such that $zneq -z$. Now $wp'$ is odd, so $wp'(z)=wp'(-z)$ implies $wp'(z)=0$. But $wp'(z)=0$ means that $z$ is a preimage of $wp(z)$ of multiplicity $2$. By the previous paragraph, this implies $z=-z$, as desired.






                share|cite|improve this answer









                $endgroup$



                [I will treat $wp$ and related functions always as functions with domain $mathbb{C}/L$, rather than $mathbb{C}$.]



                First, note that $wp$ has degree $2$: its only pole is a double pole at $0$, so it takes every value in $mathbb{C}cup{infty}$ twice, with multiplicity. Also, $wp$ is even. So, if $zneq -z$, then $z$ and $-z$ are two preimages of $wp(z)$ and thus must be the only such preimages (and both must have multiplicity $1$). If $z=-z$, we now see that $wp$ is a $2$-to-$1$ mapping in a deleted neighborhood of $z$, so $z$ is a preimage of $wp(z)$ of multiplicity $2$ and is the only such preimage. In particular, we see that for any $z,winmathbb{C}/L$, $wp(z)=wp(w)$ iff $z=pm w$



                So, it suffices to show that $wp'(z)neq wp'(-z)$ for any $z$ such that $zneq -z$. Now $wp'$ is odd, so $wp'(z)=wp'(-z)$ implies $wp'(z)=0$. But $wp'(z)=0$ means that $z$ is a preimage of $wp(z)$ of multiplicity $2$. By the previous paragraph, this implies $z=-z$, as desired.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Dec 17 '18 at 2:14









                Eric WofseyEric Wofsey

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