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
$endgroup$
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
$endgroup$
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
$endgroup$
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
13.9k61553
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
[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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043451%2fweierstrass-wp-function-defines-a-map-from-the-torus-to-an-elliptic-curve-wh%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
[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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
[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
189k14216347
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043451%2fweierstrass-wp-function-defines-a-map-from-the-torus-to-an-elliptic-curve-wh%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown