The valuation of j-functions vs number of isomorphisms for an elliptic curve

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Gross and Zagier prove the following fantastic result in their paper "Singular Moduli":



Let $R$ be a discrete valuation ring over $mathbb Z_p$ with uniformizer $pi$ such that $k = R/pi$ is algebraically closed and normalize the valuation so that $v(pi) = 1$.



Now, let $E_1,E_2$ be two distinct (ie, non isomorphic) elliptic curves over $R$ with complex multiplication with $j$-invariants $j_i = j(E_i)$. These are algebraic integers in $R$ and we can talk about their reduction mod $pi$. Define $$i(n) = frac2.$$



Then Gross-Zagier show that:
$$v(j_1-j_2) = sum_ngeq 1i(n).$$



Unfortunately, their proof is a very explicit case by case analysis (depending on both $n,p$) of both sides of the equation.



This proof is very unsatisfying to me because of how uniform the statement of the result is (it doesn't depend on p or the Elliptic curves, it doesn't even distinguish between the curves with extra automorphisms and those without).



However, I suspect that there should be a very nice uniform proof using maybe a moduli space (stack..?) of Elliptic curves. For instance, we easily see that $v(j_1-j_2) geq 1 iff i(1) geq 1$ because over an algebraically closed field the $j$-invariant determines the isomorphism class.



Does anyone know such a uniform proof?



(The paper I am referring to is here, the theorem is 2.3 on page 196.)










share|cite|improve this question























  • By Aut you mean the number of isomorphisms between those two elliptic curves defined over that ring?
    – Will Sawin
    2 days ago










  • Yes, I am sorry that is bad notation.
    – Asvin
    2 days ago














up vote
19
down vote

favorite
3












Gross and Zagier prove the following fantastic result in their paper "Singular Moduli":



Let $R$ be a discrete valuation ring over $mathbb Z_p$ with uniformizer $pi$ such that $k = R/pi$ is algebraically closed and normalize the valuation so that $v(pi) = 1$.



Now, let $E_1,E_2$ be two distinct (ie, non isomorphic) elliptic curves over $R$ with complex multiplication with $j$-invariants $j_i = j(E_i)$. These are algebraic integers in $R$ and we can talk about their reduction mod $pi$. Define $$i(n) = frac2.$$



Then Gross-Zagier show that:
$$v(j_1-j_2) = sum_ngeq 1i(n).$$



Unfortunately, their proof is a very explicit case by case analysis (depending on both $n,p$) of both sides of the equation.



This proof is very unsatisfying to me because of how uniform the statement of the result is (it doesn't depend on p or the Elliptic curves, it doesn't even distinguish between the curves with extra automorphisms and those without).



However, I suspect that there should be a very nice uniform proof using maybe a moduli space (stack..?) of Elliptic curves. For instance, we easily see that $v(j_1-j_2) geq 1 iff i(1) geq 1$ because over an algebraically closed field the $j$-invariant determines the isomorphism class.



Does anyone know such a uniform proof?



(The paper I am referring to is here, the theorem is 2.3 on page 196.)










share|cite|improve this question























  • By Aut you mean the number of isomorphisms between those two elliptic curves defined over that ring?
    – Will Sawin
    2 days ago










  • Yes, I am sorry that is bad notation.
    – Asvin
    2 days ago












up vote
19
down vote

favorite
3









up vote
19
down vote

favorite
3






3





Gross and Zagier prove the following fantastic result in their paper "Singular Moduli":



Let $R$ be a discrete valuation ring over $mathbb Z_p$ with uniformizer $pi$ such that $k = R/pi$ is algebraically closed and normalize the valuation so that $v(pi) = 1$.



Now, let $E_1,E_2$ be two distinct (ie, non isomorphic) elliptic curves over $R$ with complex multiplication with $j$-invariants $j_i = j(E_i)$. These are algebraic integers in $R$ and we can talk about their reduction mod $pi$. Define $$i(n) = frac2.$$



Then Gross-Zagier show that:
$$v(j_1-j_2) = sum_ngeq 1i(n).$$



Unfortunately, their proof is a very explicit case by case analysis (depending on both $n,p$) of both sides of the equation.



This proof is very unsatisfying to me because of how uniform the statement of the result is (it doesn't depend on p or the Elliptic curves, it doesn't even distinguish between the curves with extra automorphisms and those without).



However, I suspect that there should be a very nice uniform proof using maybe a moduli space (stack..?) of Elliptic curves. For instance, we easily see that $v(j_1-j_2) geq 1 iff i(1) geq 1$ because over an algebraically closed field the $j$-invariant determines the isomorphism class.



Does anyone know such a uniform proof?



(The paper I am referring to is here, the theorem is 2.3 on page 196.)










share|cite|improve this question















Gross and Zagier prove the following fantastic result in their paper "Singular Moduli":



Let $R$ be a discrete valuation ring over $mathbb Z_p$ with uniformizer $pi$ such that $k = R/pi$ is algebraically closed and normalize the valuation so that $v(pi) = 1$.



Now, let $E_1,E_2$ be two distinct (ie, non isomorphic) elliptic curves over $R$ with complex multiplication with $j$-invariants $j_i = j(E_i)$. These are algebraic integers in $R$ and we can talk about their reduction mod $pi$. Define $$i(n) = frac2.$$



Then Gross-Zagier show that:
$$v(j_1-j_2) = sum_ngeq 1i(n).$$



Unfortunately, their proof is a very explicit case by case analysis (depending on both $n,p$) of both sides of the equation.



This proof is very unsatisfying to me because of how uniform the statement of the result is (it doesn't depend on p or the Elliptic curves, it doesn't even distinguish between the curves with extra automorphisms and those without).



However, I suspect that there should be a very nice uniform proof using maybe a moduli space (stack..?) of Elliptic curves. For instance, we easily see that $v(j_1-j_2) geq 1 iff i(1) geq 1$ because over an algebraically closed field the $j$-invariant determines the isomorphism class.



Does anyone know such a uniform proof?



(The paper I am referring to is here, the theorem is 2.3 on page 196.)







ag.algebraic-geometry algebraic-number-theory elliptic-curves complex-multiplication






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edited 2 days ago

























asked 2 days ago









Asvin

1,2301820




1,2301820











  • By Aut you mean the number of isomorphisms between those two elliptic curves defined over that ring?
    – Will Sawin
    2 days ago










  • Yes, I am sorry that is bad notation.
    – Asvin
    2 days ago
















  • By Aut you mean the number of isomorphisms between those two elliptic curves defined over that ring?
    – Will Sawin
    2 days ago










  • Yes, I am sorry that is bad notation.
    – Asvin
    2 days ago















By Aut you mean the number of isomorphisms between those two elliptic curves defined over that ring?
– Will Sawin
2 days ago




By Aut you mean the number of isomorphisms between those two elliptic curves defined over that ring?
– Will Sawin
2 days ago












Yes, I am sorry that is bad notation.
– Asvin
2 days ago




Yes, I am sorry that is bad notation.
– Asvin
2 days ago










1 Answer
1






active

oldest

votes

















up vote
14
down vote



accepted










Yes, let's use the fact that the moduli stack of elliptic curves is etale-locally a scheme. We could also use the formal deformation space of the elliptic curve mod $pi$ (of course we may assume $E_1 cong E_2 mod pi$). We pick an etale-local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an $R$-point. (This many involve henselizing or completing $R$, which obviously won't affect things.)



Over the moduli stack of pairs of elliptic curves, and thus over any etale-local model, the scheme of isomorphisms is a disjoint union of smooth closed curves (Proof is deformation theory - the deformation space of an isomorphism is one-dimensional and always maps injectively to the deformation space of a pair of curves, because it is just the deformation space of one curves). The sum $i(n)$ is the sum over isomorphisms mod $pi$ of half the power of $pi$ they lift too, which is a sum over these curves of half the intersection number with $operatornameSpec R$. In other words, each of these curves is locally defined by an equation in the moduli stack, and we restrict that equation to $R$ and look at its degree, then sum these degrees and divide by two.



On the other hand, the valuation of $j(E_1)-j(E_2)$ is the intersection number with the divisor of $j(E_1)- j(E_2)$. This divisor is supported at the union of all the curves in the Isom scheme. So it suffices to show that the divisor of $j(E_1)-j(E_2)$ is half the sum of all the curves in the Isom scheme. Because these are divisors supported at the same union of curves, it suffices to check at their mutual generic points. But these generic points are generic elliptic curves with an automorphism of order two, so the Isom divisor appears with multiplicity two while the $j$ divisor has multiplicity one.






share|cite|improve this answer
















  • 4




    Where is it used that the $E_i$ are CM? Or is doesn't matter?
    – Felipe Voloch
    yesterday










  • Thanks for the very interesting answer! However, I am afraid I don't quite know enough to follow it completely. In particular, is there any place where I can read enough deformation theory to understand how it gets used here?
    – Asvin
    yesterday










  • @FelipeVoloch I think it's used in the first paragraph when Will chooses an etale local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an R-point.
    – Ariyan Javanpeykar
    yesterday






  • 1




    @FelipeVoloch I don't think that it matters, unless $j(E_1)-j(E_2)$ has a pole, which can't happen in the CM case.
    – Will Sawin
    yesterday










  • @ArithmeticGeometer The main thing I'm using here is that the moduli stack is a smooth Deligne-Mumford staxk, i.e. etale-locally a smooth scheme. Are you comfortable with this? The fact that the diagonal of the etale-local model is a disjoint union of smooth curves follows from this also, because it's etale-locally the diagonal of the stack, which is just the stack again. To check that the diagonal is locally a closed immersion is maybe something more, but in any case I think this can be proved the same way as the construction of the space.
    – Will Sawin
    yesterday










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
14
down vote



accepted










Yes, let's use the fact that the moduli stack of elliptic curves is etale-locally a scheme. We could also use the formal deformation space of the elliptic curve mod $pi$ (of course we may assume $E_1 cong E_2 mod pi$). We pick an etale-local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an $R$-point. (This many involve henselizing or completing $R$, which obviously won't affect things.)



Over the moduli stack of pairs of elliptic curves, and thus over any etale-local model, the scheme of isomorphisms is a disjoint union of smooth closed curves (Proof is deformation theory - the deformation space of an isomorphism is one-dimensional and always maps injectively to the deformation space of a pair of curves, because it is just the deformation space of one curves). The sum $i(n)$ is the sum over isomorphisms mod $pi$ of half the power of $pi$ they lift too, which is a sum over these curves of half the intersection number with $operatornameSpec R$. In other words, each of these curves is locally defined by an equation in the moduli stack, and we restrict that equation to $R$ and look at its degree, then sum these degrees and divide by two.



On the other hand, the valuation of $j(E_1)-j(E_2)$ is the intersection number with the divisor of $j(E_1)- j(E_2)$. This divisor is supported at the union of all the curves in the Isom scheme. So it suffices to show that the divisor of $j(E_1)-j(E_2)$ is half the sum of all the curves in the Isom scheme. Because these are divisors supported at the same union of curves, it suffices to check at their mutual generic points. But these generic points are generic elliptic curves with an automorphism of order two, so the Isom divisor appears with multiplicity two while the $j$ divisor has multiplicity one.






share|cite|improve this answer
















  • 4




    Where is it used that the $E_i$ are CM? Or is doesn't matter?
    – Felipe Voloch
    yesterday










  • Thanks for the very interesting answer! However, I am afraid I don't quite know enough to follow it completely. In particular, is there any place where I can read enough deformation theory to understand how it gets used here?
    – Asvin
    yesterday










  • @FelipeVoloch I think it's used in the first paragraph when Will chooses an etale local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an R-point.
    – Ariyan Javanpeykar
    yesterday






  • 1




    @FelipeVoloch I don't think that it matters, unless $j(E_1)-j(E_2)$ has a pole, which can't happen in the CM case.
    – Will Sawin
    yesterday










  • @ArithmeticGeometer The main thing I'm using here is that the moduli stack is a smooth Deligne-Mumford staxk, i.e. etale-locally a smooth scheme. Are you comfortable with this? The fact that the diagonal of the etale-local model is a disjoint union of smooth curves follows from this also, because it's etale-locally the diagonal of the stack, which is just the stack again. To check that the diagonal is locally a closed immersion is maybe something more, but in any case I think this can be proved the same way as the construction of the space.
    – Will Sawin
    yesterday














up vote
14
down vote



accepted










Yes, let's use the fact that the moduli stack of elliptic curves is etale-locally a scheme. We could also use the formal deformation space of the elliptic curve mod $pi$ (of course we may assume $E_1 cong E_2 mod pi$). We pick an etale-local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an $R$-point. (This many involve henselizing or completing $R$, which obviously won't affect things.)



Over the moduli stack of pairs of elliptic curves, and thus over any etale-local model, the scheme of isomorphisms is a disjoint union of smooth closed curves (Proof is deformation theory - the deformation space of an isomorphism is one-dimensional and always maps injectively to the deformation space of a pair of curves, because it is just the deformation space of one curves). The sum $i(n)$ is the sum over isomorphisms mod $pi$ of half the power of $pi$ they lift too, which is a sum over these curves of half the intersection number with $operatornameSpec R$. In other words, each of these curves is locally defined by an equation in the moduli stack, and we restrict that equation to $R$ and look at its degree, then sum these degrees and divide by two.



On the other hand, the valuation of $j(E_1)-j(E_2)$ is the intersection number with the divisor of $j(E_1)- j(E_2)$. This divisor is supported at the union of all the curves in the Isom scheme. So it suffices to show that the divisor of $j(E_1)-j(E_2)$ is half the sum of all the curves in the Isom scheme. Because these are divisors supported at the same union of curves, it suffices to check at their mutual generic points. But these generic points are generic elliptic curves with an automorphism of order two, so the Isom divisor appears with multiplicity two while the $j$ divisor has multiplicity one.






share|cite|improve this answer
















  • 4




    Where is it used that the $E_i$ are CM? Or is doesn't matter?
    – Felipe Voloch
    yesterday










  • Thanks for the very interesting answer! However, I am afraid I don't quite know enough to follow it completely. In particular, is there any place where I can read enough deformation theory to understand how it gets used here?
    – Asvin
    yesterday










  • @FelipeVoloch I think it's used in the first paragraph when Will chooses an etale local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an R-point.
    – Ariyan Javanpeykar
    yesterday






  • 1




    @FelipeVoloch I don't think that it matters, unless $j(E_1)-j(E_2)$ has a pole, which can't happen in the CM case.
    – Will Sawin
    yesterday










  • @ArithmeticGeometer The main thing I'm using here is that the moduli stack is a smooth Deligne-Mumford staxk, i.e. etale-locally a smooth scheme. Are you comfortable with this? The fact that the diagonal of the etale-local model is a disjoint union of smooth curves follows from this also, because it's etale-locally the diagonal of the stack, which is just the stack again. To check that the diagonal is locally a closed immersion is maybe something more, but in any case I think this can be proved the same way as the construction of the space.
    – Will Sawin
    yesterday












up vote
14
down vote



accepted







up vote
14
down vote



accepted






Yes, let's use the fact that the moduli stack of elliptic curves is etale-locally a scheme. We could also use the formal deformation space of the elliptic curve mod $pi$ (of course we may assume $E_1 cong E_2 mod pi$). We pick an etale-local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an $R$-point. (This many involve henselizing or completing $R$, which obviously won't affect things.)



Over the moduli stack of pairs of elliptic curves, and thus over any etale-local model, the scheme of isomorphisms is a disjoint union of smooth closed curves (Proof is deformation theory - the deformation space of an isomorphism is one-dimensional and always maps injectively to the deformation space of a pair of curves, because it is just the deformation space of one curves). The sum $i(n)$ is the sum over isomorphisms mod $pi$ of half the power of $pi$ they lift too, which is a sum over these curves of half the intersection number with $operatornameSpec R$. In other words, each of these curves is locally defined by an equation in the moduli stack, and we restrict that equation to $R$ and look at its degree, then sum these degrees and divide by two.



On the other hand, the valuation of $j(E_1)-j(E_2)$ is the intersection number with the divisor of $j(E_1)- j(E_2)$. This divisor is supported at the union of all the curves in the Isom scheme. So it suffices to show that the divisor of $j(E_1)-j(E_2)$ is half the sum of all the curves in the Isom scheme. Because these are divisors supported at the same union of curves, it suffices to check at their mutual generic points. But these generic points are generic elliptic curves with an automorphism of order two, so the Isom divisor appears with multiplicity two while the $j$ divisor has multiplicity one.






share|cite|improve this answer












Yes, let's use the fact that the moduli stack of elliptic curves is etale-locally a scheme. We could also use the formal deformation space of the elliptic curve mod $pi$ (of course we may assume $E_1 cong E_2 mod pi$). We pick an etale-local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an $R$-point. (This many involve henselizing or completing $R$, which obviously won't affect things.)



Over the moduli stack of pairs of elliptic curves, and thus over any etale-local model, the scheme of isomorphisms is a disjoint union of smooth closed curves (Proof is deformation theory - the deformation space of an isomorphism is one-dimensional and always maps injectively to the deformation space of a pair of curves, because it is just the deformation space of one curves). The sum $i(n)$ is the sum over isomorphisms mod $pi$ of half the power of $pi$ they lift too, which is a sum over these curves of half the intersection number with $operatornameSpec R$. In other words, each of these curves is locally defined by an equation in the moduli stack, and we restrict that equation to $R$ and look at its degree, then sum these degrees and divide by two.



On the other hand, the valuation of $j(E_1)-j(E_2)$ is the intersection number with the divisor of $j(E_1)- j(E_2)$. This divisor is supported at the union of all the curves in the Isom scheme. So it suffices to show that the divisor of $j(E_1)-j(E_2)$ is half the sum of all the curves in the Isom scheme. Because these are divisors supported at the same union of curves, it suffices to check at their mutual generic points. But these generic points are generic elliptic curves with an automorphism of order two, so the Isom divisor appears with multiplicity two while the $j$ divisor has multiplicity one.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









Will Sawin

65.1k6131273




65.1k6131273







  • 4




    Where is it used that the $E_i$ are CM? Or is doesn't matter?
    – Felipe Voloch
    yesterday










  • Thanks for the very interesting answer! However, I am afraid I don't quite know enough to follow it completely. In particular, is there any place where I can read enough deformation theory to understand how it gets used here?
    – Asvin
    yesterday










  • @FelipeVoloch I think it's used in the first paragraph when Will chooses an etale local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an R-point.
    – Ariyan Javanpeykar
    yesterday






  • 1




    @FelipeVoloch I don't think that it matters, unless $j(E_1)-j(E_2)$ has a pole, which can't happen in the CM case.
    – Will Sawin
    yesterday










  • @ArithmeticGeometer The main thing I'm using here is that the moduli stack is a smooth Deligne-Mumford staxk, i.e. etale-locally a smooth scheme. Are you comfortable with this? The fact that the diagonal of the etale-local model is a disjoint union of smooth curves follows from this also, because it's etale-locally the diagonal of the stack, which is just the stack again. To check that the diagonal is locally a closed immersion is maybe something more, but in any case I think this can be proved the same way as the construction of the space.
    – Will Sawin
    yesterday












  • 4




    Where is it used that the $E_i$ are CM? Or is doesn't matter?
    – Felipe Voloch
    yesterday










  • Thanks for the very interesting answer! However, I am afraid I don't quite know enough to follow it completely. In particular, is there any place where I can read enough deformation theory to understand how it gets used here?
    – Asvin
    yesterday










  • @FelipeVoloch I think it's used in the first paragraph when Will chooses an etale local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an R-point.
    – Ariyan Javanpeykar
    yesterday






  • 1




    @FelipeVoloch I don't think that it matters, unless $j(E_1)-j(E_2)$ has a pole, which can't happen in the CM case.
    – Will Sawin
    yesterday










  • @ArithmeticGeometer The main thing I'm using here is that the moduli stack is a smooth Deligne-Mumford staxk, i.e. etale-locally a smooth scheme. Are you comfortable with this? The fact that the diagonal of the etale-local model is a disjoint union of smooth curves follows from this also, because it's etale-locally the diagonal of the stack, which is just the stack again. To check that the diagonal is locally a closed immersion is maybe something more, but in any case I think this can be proved the same way as the construction of the space.
    – Will Sawin
    yesterday







4




4




Where is it used that the $E_i$ are CM? Or is doesn't matter?
– Felipe Voloch
yesterday




Where is it used that the $E_i$ are CM? Or is doesn't matter?
– Felipe Voloch
yesterday












Thanks for the very interesting answer! However, I am afraid I don't quite know enough to follow it completely. In particular, is there any place where I can read enough deformation theory to understand how it gets used here?
– Asvin
yesterday




Thanks for the very interesting answer! However, I am afraid I don't quite know enough to follow it completely. In particular, is there any place where I can read enough deformation theory to understand how it gets used here?
– Asvin
yesterday












@FelipeVoloch I think it's used in the first paragraph when Will chooses an etale local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an R-point.
– Ariyan Javanpeykar
yesterday




@FelipeVoloch I think it's used in the first paragraph when Will chooses an etale local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an R-point.
– Ariyan Javanpeykar
yesterday




1




1




@FelipeVoloch I don't think that it matters, unless $j(E_1)-j(E_2)$ has a pole, which can't happen in the CM case.
– Will Sawin
yesterday




@FelipeVoloch I don't think that it matters, unless $j(E_1)-j(E_2)$ has a pole, which can't happen in the CM case.
– Will Sawin
yesterday












@ArithmeticGeometer The main thing I'm using here is that the moduli stack is a smooth Deligne-Mumford staxk, i.e. etale-locally a smooth scheme. Are you comfortable with this? The fact that the diagonal of the etale-local model is a disjoint union of smooth curves follows from this also, because it's etale-locally the diagonal of the stack, which is just the stack again. To check that the diagonal is locally a closed immersion is maybe something more, but in any case I think this can be proved the same way as the construction of the space.
– Will Sawin
yesterday




@ArithmeticGeometer The main thing I'm using here is that the moduli stack is a smooth Deligne-Mumford staxk, i.e. etale-locally a smooth scheme. Are you comfortable with this? The fact that the diagonal of the etale-local model is a disjoint union of smooth curves follows from this also, because it's etale-locally the diagonal of the stack, which is just the stack again. To check that the diagonal is locally a closed immersion is maybe something more, but in any case I think this can be proved the same way as the construction of the space.
– Will Sawin
yesterday

















 

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