Re: New GP function ellrank (2-descent)

[John Cremona <john.cremona@gmail.com>]

Thanks for the additional information.  I think it would be a good
idea to make it clear when the results depend (in part) on
conjectures.

On Sat, 6 Feb 2021 at 01:41, Bill Allombert
<Bill.Allombert@math.u-bordeaux.fr> wrote:
>
> On Fri, Feb 05, 2021 at 04:49:28PM +0000, John Cremona wrote:
> > Bill,
> >
> > This is great news!  When mwrank is completely redundant then I can
> > really retire.
> >
> > Some questions:
> >
> > 1. When you show that the output is [1,1,[]] you are asserting that
> > the rank is exactly 1 even though you do not (yet) have any points.
> > Are you using the theorem that (analytic rank=1) => (rank=1)?
> > 2. Do you also use (analytic rank=0) => (rank=0)? In both these cases,
> > is the determination of the analytic rank rigorous?
>
> ellrank does not use analytic method, because the conductor is usually
> much too large.
>
> I am really not an expert on 2-descent but how I understand it is:
>
> - the rank of the 2-selmer is equal to the sum of the 2-rank of the torsion group,
> the rank of the curve, and the 2-rank of the Tate Shafarevich group.
>
> - the 2-rank of the Tate Shafarevich group is even.

This is not proved in general.  It follows from finiteness of Sha (or
of its 2-primary component).  It is not known (proved), for example,
that Sha could not be Q_2/Z_2 in which the 2-rank would be 1.

So if 2-descent gives an upper bound of 1, you know from this
conjecture that the rank must be 1, but to prove it you can either
find a point of infinite order, or prove that the analytic rank is 1
(and then use the theorem I mentioned).  If it is not possible to do
either of those (for example, if the conductor is huge, and the
generator is huge,  then as you say it may be impossible to find a
Heegner point, and also impossible to prove that the analytic rank is
1), then you have only proved that the rank is 0 or 1.  Knowing (from
the parity, i.e. the global root number) that the analytic rank is
odd, will only help if you know the "parity conjecture" that the
analytic and Mordell-Weil ranks have the same parity; and that is only
proved to follow from certain conditions, such as that the p-part of
Sha is finite for some p; and 2-descent cannot show that *except* by
showing that Sha has no 2-torsion.

>
> Assuming the above, then the parity of the rank is known.
> In the example, we conclude that the rank is odd and <= 2 so it must be
> 1.
>
> (Another way is to assume that the root number is equal to (-1)^rank, which
> as I understand was proved by Elkies at least in some case).

See above.

>
> > 3. Is the list of points always independent (modulo torsion)?  Denis's
> > output sometimes included torsion and/or included dependent points (if
> > I recall correctly).
>
> Denis script always include the 2-torsions points.
> For simplicty I removed them, because it was confusing.
> The remaining points are always independent.

Good!

>
> > 4. For analytic rank 1 curves do you use ellheegner() instead of
> > descent?  I hope so since ellheegner is totally and utterly brilliant!
>
> Unfortunately, ellheegner only works for curve with small conductor, so
> it is not used.

You could use it only when the conductor is less than some bound, I suppose.

>
> Note that Mark Watkins has published an algorithm which use 4-descent
> to speed up the Heegner point method. Unfortunately I do not know
> how to do 4-descent.

It is rather complicated to implement.  But you can also use 2-descent
for Heegner points:  the Heegner point construction of a rational
point (x,y) gives you real approximations to x and y.  2-descent gives
you an explicit map of degree 4 from a curve C of the form y^2=quartic
to the elliptic curve E.  In principle you can pull back (x,y) to give
4 points on C, at least one of which will be a rational point (and
exactly one if E has no 2-torsion).  The idea is that the height of
the rational point on C is less than that of ts image on E, so it
should be easier to recognise from a floating-point approximation.  In
Mark Watkins' examples (we both remember talking about all this in
Paris in 2004) he does some 2-descent examples, and also one with
4-descent (where C is an intersection of 2 quadrics in P^2 and the map
has degree 16 instead of 4).

>
> - - - -
>
> I implemented saturation but I do not know how to interface it with
> ellrank, in particular how to chose the bound.
> It seems the bound computed by mwrank is often impractical, so mwrank
> use 1000.

That's true.  I once tried to saturate a curve of rank 24 (possibly --
I think this was before Elkies had larger examples) and found that the
saturation bound was 10^100 or something like that;  I went up to a
million.  One problem is that the lattice constants are rather bad
except in low dimension.

It is possible to get better bounds by restricting to the subgroup (of
finite index) of points which have good reduction at all primes.
mwrank does this, and it helps a little.  To implement that you need
to be able to work rather explicitly in the local component groups,
and I wrote a short paper about that (ANTS 2008 I think).

>
> Did you see the paper https://arxiv.org/pdf/2007.09514.pdf ?
> Can it be used to improve the bound ?

I saw the paper but did not look at it in detail.  I suspect that
general formulas will not be as good as the algorithm Samir and I
developed to find a lower bound for the heights of non-torsion points.
(Given e>0 we give a method to try to prove that no points have height
<e (unless 0). If that succeeds, replace e by e/2 and repeat.  If it
fails, replace e by 2e and repeat).  That has been implemented too.

John

>
> Cheers,
> Bill
>