Why do Bhargava-Skinner-Zhang consider the ordering by height?Can you show rank E(Q) = 1 exactly for infinitely many elliptic curves E over Q without using BSD?Is there a “Basic Number Theory” for elliptic curves?Deducing BSD from Gross-Zagier and KolyvaginAverage rank of elliptic curves over function fieldsIs there an analog of the Birch/Swinnerton-Dyer conjecture for abelian varieties in higher dimensions?BSD and congruent numbersAverage size of $p$-part of the Tate-Shafarevich group for elliptic curvesBSD and generalisation of Gross-Zagier formulaWhat is the smallest positive integer for which the congruent number problem is unsolved?Height pairings of Heegner points of nontrivial conductor
Multi tool use
Why do Bhargava-Skinner-Zhang consider the ordering by height?
Can you show rank E(Q) = 1 exactly for infinitely many elliptic curves E over Q without using BSD?Is there a “Basic Number Theory” for elliptic curves?Deducing BSD from Gross-Zagier and KolyvaginAverage rank of elliptic curves over function fieldsIs there an analog of the Birch/Swinnerton-Dyer conjecture for abelian varieties in higher dimensions?BSD and congruent numbersAverage size of $p$-part of the Tate-Shafarevich group for elliptic curvesBSD and generalisation of Gross-Zagier formulaWhat is the smallest positive integer for which the congruent number problem is unsolved?Height pairings of Heegner points of nontrivial conductor
$begingroup$
A result of Bhargava-Skinner-Zhang says that a majority of elliptic curves over $mathbbQ$ satisfy the BSD rank conjecture.
The are infinitely many isomorphism classes of $mathbbQ$ so one naturally has to be more precise about what "a majority" means; in their case, the elliptic curves are done by height.
The question is what are the arguments for the ordering by height being the correct ordering to consider (in the sense that Bhargava-style results actually constitute non-zero progress to the BSD rank conjecture)?
The appearance of percentages in these theorems for some reason reminds me of traders on a bazaar advertising their products ("Ours can deal with 50% of the elliptic curves", "Forget him, mine can do 60%"). I just want to understand whether there is in fact a deep connection between the results in question and BSD or whether it is extraordinarily good sales pitch of the authors (this question is not meant to cast a shadow of doubt on anything other than the OP's knowledge of the subject).
nt.number-theory elliptic-curves
$endgroup$
|
show 2 more comments
$begingroup$
A result of Bhargava-Skinner-Zhang says that a majority of elliptic curves over $mathbbQ$ satisfy the BSD rank conjecture.
The are infinitely many isomorphism classes of $mathbbQ$ so one naturally has to be more precise about what "a majority" means; in their case, the elliptic curves are done by height.
The question is what are the arguments for the ordering by height being the correct ordering to consider (in the sense that Bhargava-style results actually constitute non-zero progress to the BSD rank conjecture)?
The appearance of percentages in these theorems for some reason reminds me of traders on a bazaar advertising their products ("Ours can deal with 50% of the elliptic curves", "Forget him, mine can do 60%"). I just want to understand whether there is in fact a deep connection between the results in question and BSD or whether it is extraordinarily good sales pitch of the authors (this question is not meant to cast a shadow of doubt on anything other than the OP's knowledge of the subject).
nt.number-theory elliptic-curves
$endgroup$
4
$begingroup$
The work only uses what is known about BSD for ranks 0 and 1. It gives no insight on cases of rank 2 and more, and in that sense it is not a fundamental breakthrough on our understanding of BSD.
$endgroup$
– user1728
Jun 9 at 23:47
1
$begingroup$
We can't prove BSD in full generality. So counting and showing that it holds a lot of the time is the next best thing.
$endgroup$
– Daniel Loughran
Jun 10 at 7:09
$begingroup$
@DanielLoughran in that sense, yes, I understand. But could we reorder the elliptic curves in some cunning way and get higher percentages than BSZ? I was trying to understand whether their ordering was arbitrarily chosen to get them more impressive results (it is not of course, but I wanted to see some reasons why). Silverman gave some reasons I think.
$endgroup$
– user141498
Jun 10 at 7:13
$begingroup$
Once we know the result is true for infinitely many elliptic curves, we can of course (like a dishonest salesman) reorder and get 100% by just putting scarcely the elliptic curves of analytic rank $geq 2$...
$endgroup$
– François Brunault
Jun 10 at 10:55
1
$begingroup$
meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
$endgroup$
– Steven Landsburg
Jun 10 at 13:46
|
show 2 more comments
$begingroup$
A result of Bhargava-Skinner-Zhang says that a majority of elliptic curves over $mathbbQ$ satisfy the BSD rank conjecture.
The are infinitely many isomorphism classes of $mathbbQ$ so one naturally has to be more precise about what "a majority" means; in their case, the elliptic curves are done by height.
The question is what are the arguments for the ordering by height being the correct ordering to consider (in the sense that Bhargava-style results actually constitute non-zero progress to the BSD rank conjecture)?
The appearance of percentages in these theorems for some reason reminds me of traders on a bazaar advertising their products ("Ours can deal with 50% of the elliptic curves", "Forget him, mine can do 60%"). I just want to understand whether there is in fact a deep connection between the results in question and BSD or whether it is extraordinarily good sales pitch of the authors (this question is not meant to cast a shadow of doubt on anything other than the OP's knowledge of the subject).
nt.number-theory elliptic-curves
$endgroup$
A result of Bhargava-Skinner-Zhang says that a majority of elliptic curves over $mathbbQ$ satisfy the BSD rank conjecture.
The are infinitely many isomorphism classes of $mathbbQ$ so one naturally has to be more precise about what "a majority" means; in their case, the elliptic curves are done by height.
The question is what are the arguments for the ordering by height being the correct ordering to consider (in the sense that Bhargava-style results actually constitute non-zero progress to the BSD rank conjecture)?
The appearance of percentages in these theorems for some reason reminds me of traders on a bazaar advertising their products ("Ours can deal with 50% of the elliptic curves", "Forget him, mine can do 60%"). I just want to understand whether there is in fact a deep connection between the results in question and BSD or whether it is extraordinarily good sales pitch of the authors (this question is not meant to cast a shadow of doubt on anything other than the OP's knowledge of the subject).
nt.number-theory elliptic-curves
nt.number-theory elliptic-curves
edited Jun 9 at 22:33
asked Jun 9 at 22:14
user141498
4
$begingroup$
The work only uses what is known about BSD for ranks 0 and 1. It gives no insight on cases of rank 2 and more, and in that sense it is not a fundamental breakthrough on our understanding of BSD.
$endgroup$
– user1728
Jun 9 at 23:47
1
$begingroup$
We can't prove BSD in full generality. So counting and showing that it holds a lot of the time is the next best thing.
$endgroup$
– Daniel Loughran
Jun 10 at 7:09
$begingroup$
@DanielLoughran in that sense, yes, I understand. But could we reorder the elliptic curves in some cunning way and get higher percentages than BSZ? I was trying to understand whether their ordering was arbitrarily chosen to get them more impressive results (it is not of course, but I wanted to see some reasons why). Silverman gave some reasons I think.
$endgroup$
– user141498
Jun 10 at 7:13
$begingroup$
Once we know the result is true for infinitely many elliptic curves, we can of course (like a dishonest salesman) reorder and get 100% by just putting scarcely the elliptic curves of analytic rank $geq 2$...
$endgroup$
– François Brunault
Jun 10 at 10:55
1
$begingroup$
meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
$endgroup$
– Steven Landsburg
Jun 10 at 13:46
|
show 2 more comments
4
$begingroup$
The work only uses what is known about BSD for ranks 0 and 1. It gives no insight on cases of rank 2 and more, and in that sense it is not a fundamental breakthrough on our understanding of BSD.
$endgroup$
– user1728
Jun 9 at 23:47
1
$begingroup$
We can't prove BSD in full generality. So counting and showing that it holds a lot of the time is the next best thing.
$endgroup$
– Daniel Loughran
Jun 10 at 7:09
$begingroup$
@DanielLoughran in that sense, yes, I understand. But could we reorder the elliptic curves in some cunning way and get higher percentages than BSZ? I was trying to understand whether their ordering was arbitrarily chosen to get them more impressive results (it is not of course, but I wanted to see some reasons why). Silverman gave some reasons I think.
$endgroup$
– user141498
Jun 10 at 7:13
$begingroup$
Once we know the result is true for infinitely many elliptic curves, we can of course (like a dishonest salesman) reorder and get 100% by just putting scarcely the elliptic curves of analytic rank $geq 2$...
$endgroup$
– François Brunault
Jun 10 at 10:55
1
$begingroup$
meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
$endgroup$
– Steven Landsburg
Jun 10 at 13:46
4
4
$begingroup$
The work only uses what is known about BSD for ranks 0 and 1. It gives no insight on cases of rank 2 and more, and in that sense it is not a fundamental breakthrough on our understanding of BSD.
$endgroup$
– user1728
Jun 9 at 23:47
$begingroup$
The work only uses what is known about BSD for ranks 0 and 1. It gives no insight on cases of rank 2 and more, and in that sense it is not a fundamental breakthrough on our understanding of BSD.
$endgroup$
– user1728
Jun 9 at 23:47
1
1
$begingroup$
We can't prove BSD in full generality. So counting and showing that it holds a lot of the time is the next best thing.
$endgroup$
– Daniel Loughran
Jun 10 at 7:09
$begingroup$
We can't prove BSD in full generality. So counting and showing that it holds a lot of the time is the next best thing.
$endgroup$
– Daniel Loughran
Jun 10 at 7:09
$begingroup$
@DanielLoughran in that sense, yes, I understand. But could we reorder the elliptic curves in some cunning way and get higher percentages than BSZ? I was trying to understand whether their ordering was arbitrarily chosen to get them more impressive results (it is not of course, but I wanted to see some reasons why). Silverman gave some reasons I think.
$endgroup$
– user141498
Jun 10 at 7:13
$begingroup$
@DanielLoughran in that sense, yes, I understand. But could we reorder the elliptic curves in some cunning way and get higher percentages than BSZ? I was trying to understand whether their ordering was arbitrarily chosen to get them more impressive results (it is not of course, but I wanted to see some reasons why). Silverman gave some reasons I think.
$endgroup$
– user141498
Jun 10 at 7:13
$begingroup$
Once we know the result is true for infinitely many elliptic curves, we can of course (like a dishonest salesman) reorder and get 100% by just putting scarcely the elliptic curves of analytic rank $geq 2$...
$endgroup$
– François Brunault
Jun 10 at 10:55
$begingroup$
Once we know the result is true for infinitely many elliptic curves, we can of course (like a dishonest salesman) reorder and get 100% by just putting scarcely the elliptic curves of analytic rank $geq 2$...
$endgroup$
– François Brunault
Jun 10 at 10:55
1
1
$begingroup$
meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
$endgroup$
– Steven Landsburg
Jun 10 at 13:46
$begingroup$
meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
$endgroup$
– Steven Landsburg
Jun 10 at 13:46
|
show 2 more comments
1 Answer
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$begingroup$
Roughly speaking, the height of an arithmetic object (number, variety, ...) is a natural measure of its complexity, say in the sense of "how many bits does it take to describe the object." (This is not meant to be rigorous, but you seem to want to know why people use "heights".) One can then ask for heights (complexity measures) that have nice properties, for example, transform nicely (functorially) for maps, i.e., ht(f(object)) is related to some nice function of ht(object). Weil heights and morphisms constitute a nice example of this, and canonical heights on abelian varieties behave even more nicely. For [BSZ], counting elliptic curves by height is more-or-less counting by (1) # of bits to describe the $j$-invariant (i.e., the $barmathbb Q$ isomorphism class of $E$) plus (2) # of bits to describe how twisted the curve is.
answered Jun 9 at 23:17
Joe SilvermanJoe Silverman
32.2k1 gold badge94 silver badges164 bronze badges
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$begingroup$
Roughly speaking, the height of an arithmetic object (number, variety, ...) is a natural measure of its complexity, say in the sense of "how many bits does it take to describe the object." (This is not meant to be rigorous, but you seem to want to know why people use "heights".) One can then ask for heights (complexity measures) that have nice properties, for example, transform nicely (functorially) for maps, i.e., ht(f(object)) is related to some nice function of ht(object). Weil heights and morphisms constitute a nice example of this, and canonical heights on abelian varieties behave even more nicely. For [BSZ], counting elliptic curves by height is more-or-less counting by (1) # of bits to describe the $j$-invariant (i.e., the $barmathbb Q$ isomorphism class of $E$) plus (2) # of bits to describe how twisted the curve is.
answered Jun 9 at 23:17
Joe SilvermanJoe Silverman
32.2k1 gold badge94 silver badges164 bronze badges
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add a comment |
$begingroup$
Roughly speaking, the height of an arithmetic object (number, variety, ...) is a natural measure of its complexity, say in the sense of "how many bits does it take to describe the object." (This is not meant to be rigorous, but you seem to want to know why people use "heights".) One can then ask for heights (complexity measures) that have nice properties, for example, transform nicely (functorially) for maps, i.e., ht(f(object)) is related to some nice function of ht(object). Weil heights and morphisms constitute a nice example of this, and canonical heights on abelian varieties behave even more nicely. For [BSZ], counting elliptic curves by height is more-or-less counting by (1) # of bits to describe the $j$-invariant (i.e., the $barmathbb Q$ isomorphism class of $E$) plus (2) # of bits to describe how twisted the curve is.
answered Jun 9 at 23:17
Joe SilvermanJoe Silverman
32.2k1 gold badge94 silver badges164 bronze badges
$endgroup$
add a comment |
$begingroup$
Roughly speaking, the height of an arithmetic object (number, variety, ...) is a natural measure of its complexity, say in the sense of "how many bits does it take to describe the object." (This is not meant to be rigorous, but you seem to want to know why people use "heights".) One can then ask for heights (complexity measures) that have nice properties, for example, transform nicely (functorially) for maps, i.e., ht(f(object)) is related to some nice function of ht(object). Weil heights and morphisms constitute a nice example of this, and canonical heights on abelian varieties behave even more nicely. For [BSZ], counting elliptic curves by height is more-or-less counting by (1) # of bits to describe the $j$-invariant (i.e., the $barmathbb Q$ isomorphism class of $E$) plus (2) # of bits to describe how twisted the curve is.
answered Jun 9 at 23:17
Joe SilvermanJoe Silverman
32.2k1 gold badge94 silver badges164 bronze badges
$endgroup$
Roughly speaking, the height of an arithmetic object (number, variety, ...) is a natural measure of its complexity, say in the sense of "how many bits does it take to describe the object." (This is not meant to be rigorous, but you seem to want to know why people use "heights".) One can then ask for heights (complexity measures) that have nice properties, for example, transform nicely (functorially) for maps, i.e., ht(f(object)) is related to some nice function of ht(object). Weil heights and morphisms constitute a nice example of this, and canonical heights on abelian varieties behave even more nicely. For [BSZ], counting elliptic curves by height is more-or-less counting by (1) # of bits to describe the $j$-invariant (i.e., the $barmathbb Q$ isomorphism class of $E$) plus (2) # of bits to describe how twisted the curve is.
answered Jun 9 at 23:17
Joe SilvermanJoe Silverman
32.2k1 gold badge94 silver badges164 bronze badges
edited Jun 10 at 2:59
answered Jun 9 at 23:17
Joe SilvermanJoe Silverman
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answered Jun 9 at 23:17
Joe SilvermanJoe Silverman
32.2k1 gold badge94 silver badges164 bronze badges
answered Jun 9 at 23:17
Joe SilvermanJoe Silverman
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vxGiEIbPHLw7,lnqX,mwpk6 nx3KTxE0umqF
4
$begingroup$
The work only uses what is known about BSD for ranks 0 and 1. It gives no insight on cases of rank 2 and more, and in that sense it is not a fundamental breakthrough on our understanding of BSD.
$endgroup$
– user1728
Jun 9 at 23:47
1
$begingroup$
We can't prove BSD in full generality. So counting and showing that it holds a lot of the time is the next best thing.
$endgroup$
– Daniel Loughran
Jun 10 at 7:09
$begingroup$
@DanielLoughran in that sense, yes, I understand. But could we reorder the elliptic curves in some cunning way and get higher percentages than BSZ? I was trying to understand whether their ordering was arbitrarily chosen to get them more impressive results (it is not of course, but I wanted to see some reasons why). Silverman gave some reasons I think.
$endgroup$
– user141498
Jun 10 at 7:13
$begingroup$
Once we know the result is true for infinitely many elliptic curves, we can of course (like a dishonest salesman) reorder and get 100% by just putting scarcely the elliptic curves of analytic rank $geq 2$...
$endgroup$
– François Brunault
Jun 10 at 10:55
1
$begingroup$
meta.mathoverflow.net/questions/4200/flood-of-similar-new-users
$endgroup$
– Steven Landsburg
Jun 10 at 13:46