Weierstrass $wp$-function defines a map from the torus to an elliptic curve. Why is it injective?
$begingroup$
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
asked Dec 17 '18 at 1:45
D_SD_S
13.9k61553
$endgroup$
add a comment |
$begingroup$
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
asked Dec 17 '18 at 1:45
D_SD_S
13.9k61553
$endgroup$
add a comment |
$begingroup$
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
asked Dec 17 '18 at 1:45
D_SD_S
13.9k61553
$endgroup$
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
asked Dec 17 '18 at 1:45
D_SD_S
13.9k61553
asked Dec 17 '18 at 1:45
D_SD_S
13.9k61553
asked Dec 17 '18 at 1:45
D_SD_S
13.9k61553
asked Dec 17 '18 at 1:45
D_SD_S
13.9k61553
asked Dec 17 '18 at 1:45
D_SD_S
13.9k61553
13.9k61553
add a comment |
add a comment |
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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.
answered Dec 17 '18 at 2:14
Eric WofseyEric Wofsey
189k14216347
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1 Answer
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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.
answered Dec 17 '18 at 2:14
Eric WofseyEric Wofsey
189k14216347
$endgroup$
add a comment |
$begingroup$
[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.
answered Dec 17 '18 at 2:14
Eric WofseyEric Wofsey
189k14216347
$endgroup$
add a comment |
$begingroup$
[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.
answered Dec 17 '18 at 2:14
Eric WofseyEric Wofsey
189k14216347
$endgroup$
[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.
answered Dec 17 '18 at 2:14
Eric WofseyEric Wofsey
189k14216347
answered Dec 17 '18 at 2:14
Eric WofseyEric Wofsey
189k14216347
answered Dec 17 '18 at 2:14
Eric WofseyEric Wofsey
189k14216347
answered Dec 17 '18 at 2:14
Eric WofseyEric Wofsey
189k14216347
189k14216347
add a comment |
add a comment |
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