Elliptic curve over projective line with four points of multiplicative reduction
Consider the elliptic surface $E$ with affine equation
$$y^2 = x(x-1)(x-t^2)$$
over the base $mathbf{P}^1$ with parameter $t$ (with complex scalar field). Then $E$ has four points of bad reduction, namely $0$, $1$, $-1$, and $infty$. One can check that the reduction type is multiplicative at each bad place. Is this the only elliptic surface with multiplicative reduction at those four places? I understand that the condition of multiplicative reduction means that the corresponding local system has unipotent monodromy around each of the four bad points, but I don't know how to use this to classify such elliptic surfaces.
ag.algebraic-geometry
asked Dec 17 at 21:31
Jared Weinstein
2,9901530
add a comment |
Consider the elliptic surface $E$ with affine equation
$$y^2 = x(x-1)(x-t^2)$$
over the base $mathbf{P}^1$ with parameter $t$ (with complex scalar field). Then $E$ has four points of bad reduction, namely $0$, $1$, $-1$, and $infty$. One can check that the reduction type is multiplicative at each bad place. Is this the only elliptic surface with multiplicative reduction at those four places? I understand that the condition of multiplicative reduction means that the corresponding local system has unipotent monodromy around each of the four bad points, but I don't know how to use this to classify such elliptic surfaces.
ag.algebraic-geometry
asked Dec 17 at 21:31
Jared Weinstein
2,9901530
add a comment |
Consider the elliptic surface $E$ with affine equation
$$y^2 = x(x-1)(x-t^2)$$
over the base $mathbf{P}^1$ with parameter $t$ (with complex scalar field). Then $E$ has four points of bad reduction, namely $0$, $1$, $-1$, and $infty$. One can check that the reduction type is multiplicative at each bad place. Is this the only elliptic surface with multiplicative reduction at those four places? I understand that the condition of multiplicative reduction means that the corresponding local system has unipotent monodromy around each of the four bad points, but I don't know how to use this to classify such elliptic surfaces.
ag.algebraic-geometry
asked Dec 17 at 21:31
Jared Weinstein
2,9901530
Consider the elliptic surface $E$ with affine equation
$$y^2 = x(x-1)(x-t^2)$$
over the base $mathbf{P}^1$ with parameter $t$ (with complex scalar field). Then $E$ has four points of bad reduction, namely $0$, $1$, $-1$, and $infty$. One can check that the reduction type is multiplicative at each bad place. Is this the only elliptic surface with multiplicative reduction at those four places? I understand that the condition of multiplicative reduction means that the corresponding local system has unipotent monodromy around each of the four bad points, but I don't know how to use this to classify such elliptic surfaces.
ag.algebraic-geometry
ag.algebraic-geometry
asked Dec 17 at 21:31
Jared Weinstein
2,9901530
asked Dec 17 at 21:31
Jared Weinstein
2,9901530
asked Dec 17 at 21:31
Jared Weinstein
2,9901530
asked Dec 17 at 21:31
Jared Weinstein
2,9901530
asked Dec 17 at 21:31
Jared Weinstein
2,9901530
2,9901530
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The classification of semistable families of elliptic curves with 4 singular points is due to Beauville in . The trick for the classification is to study, not the monodromy of the family of elliptic curves, but the monodromy of the $j$ invariant map as a cover of the projective line.
It seems Beauville doesn't calculate in his paper the actual singular points of these families. This is easy to do from the formula, but this is also done in a paper of Katz. In particular, there are two whose singular locus look like ${0,1, -1, infty}$, and these are isogenous. The isogeny is produced by modding out by the $2$-torsion subgroup $(1,0)$, for instance.
But this is a classification up to automorphism, and one still has to check how the automorphisms of $mathbb P^1$ that permute these four points act on the set of elliptic curves. The group of automorphisms is $D_4$. For your elliptic curve, the stabilizer has order $4$, generated by $t to -t$ and $t to 1/t$, so there is one conjugate, which looks like $y^2 =x (x-1) (x - (t+1) / (t-1))$. For the isogenous elliptic curve, the stabilizer only has order $2$ (you can upper bound the stabilizer by looking at the description of the type of the singular fibers in Beauville's paper, and lower bound by finding explicit automorphisms) so there are actually $4$ isomorphism classes, for $6$ total.
However, I think these are all isogenous by some further isogenies (e.g. modding out by another $2$-torsion point).
answered Dec 17 at 22:07
Will Sawin
67k6135279
4
I checked that the quotient of $E$ by the $2$-torsion point $(t^2,0)$ is isomorphic to the family associated to $Gamma_1(4) cap Gamma_0(8)$ in Beauville's list. In fact this family also has the nice equation $x+1/x+y+1/y=4t$.
– François Brunault
Dec 17 at 22:53
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: "504"
};
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%2fmathoverflow.net%2fquestions%2f318920%2felliptic-curve-over-projective-line-with-four-points-of-multiplicative-reduction%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
The classification of semistable families of elliptic curves with 4 singular points is due to Beauville in . The trick for the classification is to study, not the monodromy of the family of elliptic curves, but the monodromy of the $j$ invariant map as a cover of the projective line.
It seems Beauville doesn't calculate in his paper the actual singular points of these families. This is easy to do from the formula, but this is also done in a paper of Katz. In particular, there are two whose singular locus look like ${0,1, -1, infty}$, and these are isogenous. The isogeny is produced by modding out by the $2$-torsion subgroup $(1,0)$, for instance.
But this is a classification up to automorphism, and one still has to check how the automorphisms of $mathbb P^1$ that permute these four points act on the set of elliptic curves. The group of automorphisms is $D_4$. For your elliptic curve, the stabilizer has order $4$, generated by $t to -t$ and $t to 1/t$, so there is one conjugate, which looks like $y^2 =x (x-1) (x - (t+1) / (t-1))$. For the isogenous elliptic curve, the stabilizer only has order $2$ (you can upper bound the stabilizer by looking at the description of the type of the singular fibers in Beauville's paper, and lower bound by finding explicit automorphisms) so there are actually $4$ isomorphism classes, for $6$ total.
However, I think these are all isogenous by some further isogenies (e.g. modding out by another $2$-torsion point).
answered Dec 17 at 22:07
Will Sawin
67k6135279
4
I checked that the quotient of $E$ by the $2$-torsion point $(t^2,0)$ is isomorphic to the family associated to $Gamma_1(4) cap Gamma_0(8)$ in Beauville's list. In fact this family also has the nice equation $x+1/x+y+1/y=4t$.
– François Brunault
Dec 17 at 22:53
add a comment |
The classification of semistable families of elliptic curves with 4 singular points is due to Beauville in . The trick for the classification is to study, not the monodromy of the family of elliptic curves, but the monodromy of the $j$ invariant map as a cover of the projective line.
It seems Beauville doesn't calculate in his paper the actual singular points of these families. This is easy to do from the formula, but this is also done in a paper of Katz. In particular, there are two whose singular locus look like ${0,1, -1, infty}$, and these are isogenous. The isogeny is produced by modding out by the $2$-torsion subgroup $(1,0)$, for instance.
But this is a classification up to automorphism, and one still has to check how the automorphisms of $mathbb P^1$ that permute these four points act on the set of elliptic curves. The group of automorphisms is $D_4$. For your elliptic curve, the stabilizer has order $4$, generated by $t to -t$ and $t to 1/t$, so there is one conjugate, which looks like $y^2 =x (x-1) (x - (t+1) / (t-1))$. For the isogenous elliptic curve, the stabilizer only has order $2$ (you can upper bound the stabilizer by looking at the description of the type of the singular fibers in Beauville's paper, and lower bound by finding explicit automorphisms) so there are actually $4$ isomorphism classes, for $6$ total.
However, I think these are all isogenous by some further isogenies (e.g. modding out by another $2$-torsion point).
answered Dec 17 at 22:07
Will Sawin
67k6135279
4
I checked that the quotient of $E$ by the $2$-torsion point $(t^2,0)$ is isomorphic to the family associated to $Gamma_1(4) cap Gamma_0(8)$ in Beauville's list. In fact this family also has the nice equation $x+1/x+y+1/y=4t$.
– François Brunault
Dec 17 at 22:53
add a comment |
The classification of semistable families of elliptic curves with 4 singular points is due to Beauville in . The trick for the classification is to study, not the monodromy of the family of elliptic curves, but the monodromy of the $j$ invariant map as a cover of the projective line.
It seems Beauville doesn't calculate in his paper the actual singular points of these families. This is easy to do from the formula, but this is also done in a paper of Katz. In particular, there are two whose singular locus look like ${0,1, -1, infty}$, and these are isogenous. The isogeny is produced by modding out by the $2$-torsion subgroup $(1,0)$, for instance.
But this is a classification up to automorphism, and one still has to check how the automorphisms of $mathbb P^1$ that permute these four points act on the set of elliptic curves. The group of automorphisms is $D_4$. For your elliptic curve, the stabilizer has order $4$, generated by $t to -t$ and $t to 1/t$, so there is one conjugate, which looks like $y^2 =x (x-1) (x - (t+1) / (t-1))$. For the isogenous elliptic curve, the stabilizer only has order $2$ (you can upper bound the stabilizer by looking at the description of the type of the singular fibers in Beauville's paper, and lower bound by finding explicit automorphisms) so there are actually $4$ isomorphism classes, for $6$ total.
However, I think these are all isogenous by some further isogenies (e.g. modding out by another $2$-torsion point).
answered Dec 17 at 22:07
Will Sawin
67k6135279
The classification of semistable families of elliptic curves with 4 singular points is due to Beauville in . The trick for the classification is to study, not the monodromy of the family of elliptic curves, but the monodromy of the $j$ invariant map as a cover of the projective line.
It seems Beauville doesn't calculate in his paper the actual singular points of these families. This is easy to do from the formula, but this is also done in a paper of Katz. In particular, there are two whose singular locus look like ${0,1, -1, infty}$, and these are isogenous. The isogeny is produced by modding out by the $2$-torsion subgroup $(1,0)$, for instance.
But this is a classification up to automorphism, and one still has to check how the automorphisms of $mathbb P^1$ that permute these four points act on the set of elliptic curves. The group of automorphisms is $D_4$. For your elliptic curve, the stabilizer has order $4$, generated by $t to -t$ and $t to 1/t$, so there is one conjugate, which looks like $y^2 =x (x-1) (x - (t+1) / (t-1))$. For the isogenous elliptic curve, the stabilizer only has order $2$ (you can upper bound the stabilizer by looking at the description of the type of the singular fibers in Beauville's paper, and lower bound by finding explicit automorphisms) so there are actually $4$ isomorphism classes, for $6$ total.
However, I think these are all isogenous by some further isogenies (e.g. modding out by another $2$-torsion point).
answered Dec 17 at 22:07
Will Sawin
67k6135279
answered Dec 17 at 22:07
Will Sawin
67k6135279
answered Dec 17 at 22:07
Will Sawin
67k6135279
answered Dec 17 at 22:07
Will Sawin
67k6135279
67k6135279
4
I checked that the quotient of $E$ by the $2$-torsion point $(t^2,0)$ is isomorphic to the family associated to $Gamma_1(4) cap Gamma_0(8)$ in Beauville's list. In fact this family also has the nice equation $x+1/x+y+1/y=4t$.
– François Brunault
Dec 17 at 22:53
add a comment |
4
I checked that the quotient of $E$ by the $2$-torsion point $(t^2,0)$ is isomorphic to the family associated to $Gamma_1(4) cap Gamma_0(8)$ in Beauville's list. In fact this family also has the nice equation $x+1/x+y+1/y=4t$.
– François Brunault
Dec 17 at 22:53
4
4
I checked that the quotient of $E$ by the $2$-torsion point $(t^2,0)$ is isomorphic to the family associated to $Gamma_1(4) cap Gamma_0(8)$ in Beauville's list. In fact this family also has the nice equation $x+1/x+y+1/y=4t$.
– François Brunault
Dec 17 at 22:53
I checked that the quotient of $E$ by the $2$-torsion point $(t^2,0)$ is isomorphic to the family associated to $Gamma_1(4) cap Gamma_0(8)$ in Beauville's list. In fact this family also has the nice equation $x+1/x+y+1/y=4t$.
– François Brunault
Dec 17 at 22:53
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2fmathoverflow.net%2fquestions%2f318920%2felliptic-curve-over-projective-line-with-four-points-of-multiplicative-reduction%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
