bio | website | |
---|---|---|
location | Nottingham, UK | |
age | ||
visits | member for | 4 years, 5 months |
seen | 2 days ago | |
stats | profile views | 1,319 |
Aug 10 |
comment |
padic BSD vs. BSD for algorithm to compute rank
For BSD, we only know $r_{an}=0\Rightarrow r=0$ and $r_{an}=1\Rightarrow r=1$. For $p$-adic BSD we know this plus $r_{an} \geq r$. |
Aug 7 |
comment |
$p$-adic Regulators
What $p$ does not matter at all. The quotient is independent of $p$. In fact any quadratic form $h$ on $E(K)$ with $h(\varphi(P)) = \deg(\varphi) \cdot h(P)$ will give the same result. |
Aug 7 |
comment |
$p$-adic Regulators
... and as a consequence: I don't think there is a simple relation to the quotient of the minimal modular degree. Take non-semistable examples and examples with non-trivial Sha to find counter-examples. |
Aug 7 |
comment |
$p$-adic Regulators
Maybe I just should have said that both the p-adic and the real BSD formula are invariant under isogenies. |
Aug 7 |
answered | $p$-adic Regulators |
Aug 5 |
comment |
Rank of Elliptic Curves
Probably you mean "size" of the 2-Selmer group or "dimension" as $\mathbb{F}_2$-vector space, not "rank" since it is finite. The corank of the 2-primary Selmer group is equal to the rank of the curve unless Sha is infinite. |
Aug 5 |
answered | Rank of Elliptic Curves |
Jul 30 |
comment |
BSD leading-term coefficient in terms of places without distinction
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 |
comment |
BSD leading-term coefficient in terms of places without distinction
? 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 |
comment |
BSD leading-term coefficient in terms of places without distinction
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 |
comment |
BSD leading-term coefficient in terms of places without distinction
Another small correction: Sha also compares with the infintite place, not just finite primes. |
Jul 30 |
comment |
BSD leading-term coefficient in terms of places without distinction
What would be your uniform interpretation for the class number formula ? |
Jul 30 |
comment |
BSD leading-term coefficient in terms of places without distinction
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 |
comment |
BSD leading-term coefficient in terms of places without distinction
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 29 |
comment |
If the $L$-series does not vanish
Probably, that sort of question is best asked to the authors directly :) |
Jul 29 |
comment |
If the $L$-series does not vanish
"good" implies "semistable", so we are using a stronger hypothesis than the theorem we use, no ? |
Jul 28 |
comment |
Why is the norm map dual to restriction under Tate local duality?
Maybe you want to edit the question. It seems to me that the question you wanted to ask could be made more precise. |
Jul 28 |
comment |
Why is the norm map dual to restriction under Tate local duality?
I wouldn't know a good reference for the comparison with the bi-extension pairing. Maybe Milne's Airhtmetic Dualities ? |
Jul 28 |
comment |
Why is the norm map dual to restriction under Tate local duality?
It was not clear to me what definition of the pairing you are using. I thought you refer to the original definition by Tate (which is away from the characteristic). The compatibility between the two pairings is given in Proposition 2.1 in dpmms.cam.ac.uk/~taf1000/papers/ctpair.html by Tom Fisher (and probably in many other places). |
Jul 28 |
answered | Why is the norm map dual to restriction under Tate local duality? |