16
$\begingroup$

It's conjectured that, asymptotically, half of elliptic curves have rank 0, half have rank 1, and elliptic curves of rank $\geq 2$ have density 0. But what if we disregard elliptic curves of rank 0 or 1: Are there any conjectures about the average rank of elliptic curves of rank $\geq 2$?

More generally, for any integer $n \geq 2$, what value should one expect for the average rank of elliptic curves of rank $\geq n$?

Improve this question
$\endgroup$
2
  • $\begingroup$ I think the parity conjecture forces 2 and 3 to have about the same (once excluding rank 1), and we expect that rank higher than 3 is much more rare. So, one, if one were me, would expect half of elliptic curves of rank at least two to actually be of rank two, another half to be of rank 3, and the rest to have density zero. $\endgroup$
    – Dror Speiser
    Commented Nov 1, 2013 at 19:47
  • 3
    $\begingroup$ The parity conjecture definitely does not force this. We could take the actual set of all elliptic curves and turn a density-$0$ subset of the rank $0$ curves into rank $2$ curves. This would preserve the parity and average rank conjectures, but could change the average rank among curves of rank $\geq 2$, if the set we changed no longer had density $0$ when restricted to that subset of curves. $\endgroup$
    – Will Sawin
    Commented Nov 1, 2013 at 21:13

3 Answers 3

Reset to default
11
$\begingroup$

A very simple random matrix heuristic says, based on the function field model, that the rank of a random elliptic curve is the dimension of the invariant subspace of a random element of $O(n)$ for large $n$.

We can easily compute the dimension of the space of matrices that preserve exactly $k$-dimensional subspace. We can write each such matrix as a nondegenerate invariant subspace of dimension $k$ plus an orthogonal matrix with no fixed vectors of dimension $n-k$. The dimension of the space of $k$-dimensional subspaces is $k(n-k)$, and nondegenerate ones are generic. The dimension of the space of matrices in $O(n-k)$ that with no fixed points is $(n-k)(n-k-1)/2$, because that's a dense set of one of the connected components. So the total number is $k(n-k)+(n-k)(n-k-1)/2 = (n-k)(n+k-1)/2$.

This dimension takes the same value at $k=0$ and $k=1$ but takes rapidly decreasing values after that. So this heuristic suggests that most elliptic curves of rank at least $n$ have rank exactly $n$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks for the answer. You refer to this as a "very simple" heuristic. Does that mean that it's known to be a significant oversimplification? Also, by "most elliptic curves", do you mean density $1$ or something weaker? $\endgroup$
    – Daniel Hast
    Commented Nov 2, 2013 at 2:17
  • 1
    $\begingroup$ As far as I know, it is not known to be an oversimplification, except that heuristics like this can trip up when you change the ordering that you use to define the density. I mean density 1. $\endgroup$
    – Will Sawin
    Commented Nov 2, 2013 at 2:53
  • $\begingroup$ I am not sure how this squares with the "linear decay" in density suggested by the other 2 answers. It seems that to make this analogue, you think of the $O(n)$ over some large field of size $q$, and the number of rank $k$ curves is about $q^{(n-k)(n+k-1)/2}$. The heuristic mentioned by Matt Young OTOH says at "size $X$" the chance of a random curve being rank $r\ge1$ is $\sim1/X^{(r-1)/24}$ with $X^{5/6}$ total curves. Balancing $q$ and $X$ as you like, your calculation suggests a $1/q$ chance of $k=2$, then a $1/q^3$ of rank 3, then $1/q^6$ of rank 4, etc., or faster than linear decay. $\endgroup$
    – ThisNameForSale
    Commented Nov 2, 2013 at 3:01
  • 1
    $\begingroup$ It's not clear if one should do this, since the $O(n)$ I'm thinking of is over $\mathbb Q_l$, being related to the $l$-adic cohomology. So I don't think one can directly conclude that the average rank among $\geq n$s is $n$ in the $l$-adic case, either. But intuition suggests there will be a lot more $n$s than $n+1$s. But I don't think this heuristic is precise enough to predict the rate of decay for finite $D$. $\endgroup$
    – Will Sawin
    Commented Nov 2, 2013 at 3:14
  • $\begingroup$ I would not say it is much of an oversimplification either. A similar heuristic can be seen in Poonen-Rains (or earlier work of de Jong) for Selmer groups. The idea at the bottom can be said: first to note the dimension of the invariant subspace is the same as the number of eigenvalues equal to 1, and then that RMT speculates such eigenvalues should account for the rank (exactly). In the function field analogue, much more is known about the validity of this second step. $\endgroup$
    – ThisNameForSale
    Commented Nov 2, 2013 at 4:07
8
$\begingroup$

These types of questions are pretty speculative. One should be aware firstly that there is no reason for there to be roughly equal numbers of rank 2 curves and rank 3 curves.

Mark Watkins has a paper where he comes up with a conjecture, using random matrix theory, that the number of rank 2 elliptic curves with conductor up to $X$ is asymptotically $c X^{19/24} (\log X)^{3/8}$. The paper is: Mark Watkins, Some heuristics about elliptic curves. Experiment. Math. 17 (2008), no. 1, 105–125. At the end of section 4 of the paper, he remarks that possibly there are around $X^{\frac{21-r}{24}}$ elliptic curves of rank $r$, for each $r \geq 2$, compared to around $X^{5/6}$ total elliptic curves.

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ So this would imply that the average value is $n$ for $n\leq 22$ and undetermined for $n >22$? $\endgroup$
    – Will Sawin
    Commented Nov 1, 2013 at 20:23
  • $\begingroup$ Thanks for the answer; I figured it would be pretty speculative. Is the conjecture you mention based in a more refined version of the heuristic described in Will Sawin's answer, or is it a different approach? $\endgroup$
    – Daniel Hast
    Commented Nov 2, 2013 at 2:21
  • $\begingroup$ I think it is a different approach. The Watkins paper only references a to-come work with Granville, which seems to be the one cited in my answer. They work in the twist case there, but would basically heuristically bound the number of (integral?) points $(A,B,X,Y,Z)$ on $Y^2Z=X^3+AXZ^2+BZ^3$ in some ranges, and compare this to the number of such points (via ellipsoids) that a curve of rank $r$ with parameters of size $(A,B)$ would generate in said ranges. I think Lang-Vojta guessing is similar. Maybe I will try to write this as an answer, if I can work out how the non-twist case differs. $\endgroup$
    – ThisNameForSale
    Commented Nov 2, 2013 at 3:43
  • $\begingroup$ The only place where Watkins uses RMT is the 3/8 exponent for log; he says the 19/24 can be derived more crudely. For RMT, model $L$-values via charpolys and apply a discretization process in an arithmetic analog. One has $Prob(P(1)\le t)\sim M^{3/8}\sqrt{t}$ as $t\sim 0$ for the charpoly $P$ of an orthog matrix, where $M$ is the matrix size. For ellcurv we guess the same for $Prob(L(E,1)\le t)$ with an arith factor that averages out. He matches $M\sim\log N$ as ellcurv conductor, mucks cond vs disc vs $1/\Omega^{1/12}$, discretizes as Sha is integral, and ends with the 3/8 he started with. $\endgroup$
    – ThisNameForSale
    Commented Nov 2, 2013 at 4:01
8
$\begingroup$

Regarding the work of Watkins as mentioned by Matt Young, he has a different paper where the question of rank distribution in quadratic twist families is considered. There the RMT prediction for rank 2 is well shown by the Rubinstein data to demonstrate about $D^{3/4}$ twists of rank 2 (or more) in the even parity subclass, while Watkins suggests that 3/4 is too high for rank 3. He does not posit anything exactly, but the data for the congruent number curve, divided into 2 natural classes (Section 3.3) gives best-fit exponents of 0.44 and 0.55, so about $D^{1/2}$ is probably a best current guess for rank 3, though in initial regions of data collection the logarithmic factors can be difficult.

http://archive.numdam.org/article/JTNB_2008__20_3_829_0.pdf‎

The natural linear extrapolation for these types of problems might be semi-valid at least initially, so $D^{1/4}$ for rank 4 and then some power of logarithm for rank 5 are as knowledgeable as guesses as any. But the 2-torsion plays a role here, and it is not clear whether it affects the exponent on the $D$-power. Watkins has a recent preprint (joint with 5 others) where data for the congruent number curve is given, finding "lots" of rank 6 examples, but "few" of rank 7 (and none of rank 8). Granville has a heuristic (see Section 4 loc. cit.) that might suggest ranks 5 and 7 are the correct cutoffs in the generic and 2-torsion cases.

http://magma.maths.usyd.edu.au/~watkins/papers/RANK7.pdf

But really no one has any factual idea, and a number of caveats can be listed, concerning specially parametrised (sparse) families. Indeed, the rank 28 example of Elkies starts from a rank 17 special family and he gains 11 from searching on specializations, and the same was approximately true for the NSA curve, they had rank 24 starting from I think a rank 13 family of Mestre or Nagao. So again 11 more than the family rank. However, generically it seems one should not expect more than 10 "small" points (polynomial height) on an elliptic curve except in such special families, and one actually reachieves the bound of 21 when appending 11 as above.

Edit: Elkies says that the NSA searched in a rank 11 family, so they beat the family rank by 13 in fact, see page 5 of his arxiv.org/pdf/0709.2908v1

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .