Timeline for BSD leading-term coefficient in terms of places without distinction
Current License: CC BY-SA 3.0
16 events
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| Jul 30, 2014 at 16:59 | answer | added | Ben Wieland | timeline score: 3 | |
| Jul 30, 2014 at 15:25 | answer | added | guest | timeline score: 3 | |
| Jul 30, 2014 at 14:33 | comment | added | Chris Wuthrich | Ok, I saw 3 on the linked post. I would disagree with that "canonical volume" interpretation to be very good for understanding the regulator. There is an symmetric bilinear form on $E(\mathbb{Q})$ with values in $\mathbb{R}$ defined for instance in Silverman's book VIII.9. The target here is the completion where the $L$-series takes values. In the $p$-adic BSD conjecture, the leading term of the $p$-adic $L$-series is linked to the canonical $p$-adic regulator coming from a bilinear form with values in $\mathbb{Q}_p$. | |
| Jul 30, 2014 at 14:17 | comment | added | Chris Wuthrich | ? Sorry, I don't understand this. If the rank is 1, you have a "stack" $\mathbb{R}//\mathbb{Z}$. What sort of measure to you put on this to get the canonical height of the generator ? | |
| Jul 30, 2014 at 13:39 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
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| Jul 30, 2014 at 13:35 | comment | added | David Roberts♦ | Well, for instance, $Reg_E/\#E_{tors}(\mathbb{Q})$ is the volume of the stack $E(\mathbb{Q})\otimes \mathbb{R}//E(\mathbb{Q})$, do the other terms assemble into a measure of such a thing? Namely, is there a geometric object, say over adeles, such that this is the measure of it? | |
| Jul 30, 2014 at 13:15 | comment | added | Chris Wuthrich | Even this formulation is uniform in the different completions. Sha is the kernel to all completions; the regulator is the determinant of the height, which is a sum over all completions; the Tamagawa numbers and the archemedian periods are gathered as an adelic measure over all places (in the number field case, the product does not split canonically). | |
| Jul 30, 2014 at 12:25 | comment | added | David Roberts♦ | I really don't know anything about this, I'm just asking if there's a more sophisticated formulation that doesn't make a distinction between different completions. Certainly I'm expecting a more fancy interpretation of the terms, cf the Tamagawa number conjecture. | |
| Jul 30, 2014 at 12:21 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
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| Jul 30, 2014 at 11:30 | comment | added | Chris Wuthrich | Another small correction: Sha also compares with the infintite place, not just finite primes. | |
| Jul 30, 2014 at 11:28 | comment | added | Chris Wuthrich | What would be your uniform interpretation for the class number formula ? | |
| Jul 30, 2014 at 11:26 | comment | added | Joe Silverman | First, you say denominator, but mean numerator. Second, I don't at all agree that the regulator compares the rational and the real points of $E$. But the canonical height (when written as a sum of local heights) is a sort of average measure of a rational point over how it sits in $E(\mathbb{Q}_v)$ for all places $v$, including the real place. But it's certainly not just the real place. I'm not sure whether this helps or hurts what you're looking for. | |
| Jul 30, 2014 at 11:22 | comment | added | David Roberts♦ | A single expression is perhaps too ambitious, but treating the different places in a uniform way is really what would be nice. | |
| Jul 30, 2014 at 9:28 | comment | added | Chris Wuthrich | Yet, in the end, whatever formulation you will see, there will always be the input of several sides. The Bloch Kato conjecture will also put together comparison isomorphisms etc to get to a formulation. So I don't think there is a single expression that covers the full quotient. | |
| Jul 30, 2014 at 9:26 | comment | added | Chris Wuthrich | Tate's original forumlation archive.numdam.org/ARCHIVE/SB/SB_1964-1966__9_/… using adelic integration uniformises $\Omega_E\cdot \prod c_p$ into one expression. Since the regulator measures something on the Mordell-Weil group modulo torsion, one could also consider the quotient of $\mathrm{Reg}_E/(\#E(\mathbb{Q})_{\mathrm{tors}}))^2$ as one thing. | |
| Jul 30, 2014 at 7:57 | history | asked | David Roberts♦ | CC BY-SA 3.0 |