Elliptic curve over projective line with four points of multiplicative reduction
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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 5 hours ago
Jared Weinstein
2,9631529
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up vote
5
down vote
favorite
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 5 hours ago
Jared Weinstein
2,9631529
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
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 5 hours ago
Jared Weinstein
2,9631529
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 5 hours ago
Jared Weinstein
2,9631529
asked 5 hours ago
Jared Weinstein
2,9631529
asked 5 hours ago
Jared Weinstein
2,9631529
asked 5 hours ago
Jared Weinstein
2,9631529
asked 5 hours ago
Jared Weinstein
2,9631529
2,9631529
add a comment |
add a comment |
1 Answer
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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 4 hours ago
Will Sawin
66.7k6134277
1
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
3 hours ago
add a comment |
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1 Answer
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1 Answer
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active
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active
oldest
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votes
up vote
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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 4 hours ago
Will Sawin
66.7k6134277
1
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
3 hours ago
add a comment |
up vote
4
down vote
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 4 hours ago
Will Sawin
66.7k6134277
1
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
3 hours ago
add a comment |
up vote
4
down vote
up vote
4
down vote
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 4 hours ago
Will Sawin
66.7k6134277
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 4 hours ago
Will Sawin
66.7k6134277
answered 4 hours ago
Will Sawin
66.7k6134277
answered 4 hours ago
Will Sawin
66.7k6134277
answered 4 hours ago
Will Sawin
66.7k6134277
66.7k6134277
1
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
3 hours ago
add a comment |
1
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
3 hours ago
1
1
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
3 hours ago
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
3 hours ago
add a comment |
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