Rank 2 versus rank 3

One interesting feature of the heuristics of Garton, Park, Poonen, Wood, Voight, discussed here previously: they predict there are fewer elliptic curves of rank 3 than there are of rank 2.  Is this what we believe?  On one hand, you might believe that having three independent points should be “harder” than having only two.  But there’s the parity issue.  All right-thinking people believe that there are equally many rank 0 and rank 1 elliptic curves, because 100% of curves with even parity have rank 0, and 100% of curves with odd parity have rank 1.  If a curve has even parity, all that has to happen to force it to have rank 2 is to have a non-torsion point.  And if a curve has odd parity, all that has to happen to force it to have rank 3 is to have one more non-torsion point you don’t know about it.  So in that sense, it seems “equally hard” to have rank 2 or rank 3, given that parity should be even half the time and odd half the time.

So my intuition about this question is very weak.  What’s yours?  Should rank 3 be less common than rank 2?  The same?  More common?

3 thoughts on “Rank 2 versus rank 3

  1. Dick Gross says:

    Your guess is as good as mine. We could use some progress on the conjecture of Birch and Swinnerton-Dyer when L(E,1) = L'(E,1) = 0.

  2. Matt Young says:

    I would predict that there are more curves of rank 3 than rank 2. The numerical evidence (just looking at Cremona’s tables) supports this, from what I remember.

    It seems to me to be important to be clear what exactly is the model for higher rank curves. We can’t simply take a rank 0 or 1 curve, then “add a non-torsion point”, continue to enforce that the parity stays the same, and then get a rank 2 or 3 curve (even if we could, should we think of these as independent processes? and should it be equally easy to add points to a rank 1 curve as a rank 0 curve?).

    The approach via L-functions and random matrix theory seems to give reliable conjectures for quadratic twists having rank 2. There the idea is that random matrix theory gives the distribution of central values. There is an additional discretization constraint coming from BSD that forces the L-value to vanish provided it is sufficiently small. Now you can try to do the same thing for quadratic twists of a rank 1 curve. So then you want to look at BSD in the rank 1 case and use a similar discretization argument. Unfortunately, this is hard because you’ve got the regulator now, and it can jump around in size a lot. This makes the comparison between rank 2 and rank 3 quite difficult from this type of approach.

  3. Jeremy Rouse says:

    Cremona’s database (conductor up to 350000) contains 189420 isogeny classes of curves of rank 2, and 5061 isogeny classes of curves of rank 3. There are some more oddities in the data too – there are 36% more rank 1 classes than rank 0 classes, and the proportion of classes with rank 2 and rank 3 appear to grow as the conductor increases. Of course, this is still a “small” amount of data.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 560 other followers

%d bloggers like this: