Timeline for Average rank of elliptic curves, excluding those of low rank
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2013 at 19:41 | vote | accept | Daniel Hast | ||
| Nov 4, 2013 at 2:43 | comment | added | Qiaochu Yuan | Some details on this heuristic can be found in Ulmer's "Function fields and random matrices" (arxiv.org/abs/1002.3289), particularly the second example in section 3.3. | |
| Nov 2, 2013 at 4:07 | comment | added | ThisNameForSale | 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. | |
| Nov 2, 2013 at 3:14 | comment | added | Will Sawin | 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 $D$. | |
| Nov 2, 2013 at 3:01 | comment | added | ThisNameForSale | 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. | |
| Nov 2, 2013 at 2:53 | comment | added | Will Sawin | 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. | |
| Nov 2, 2013 at 2:17 | comment | added | Daniel Hast | 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? | |
| Nov 1, 2013 at 21:10 | history | answered | Will Sawin | CC BY-SA 3.0 |