Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am currently writing my master's thesis and I was wondering if the rank distribution conjecture was ever formally written down. Recall that it says that:

Half of all elliptic curves have rank $0$, half have rank $1$, and the rest have rank $\geq 2$.

I am looking for a reference for where the conjecture was made. Right now all I have for a reference is the linked talk and I'm not very comfortable with using that.

Thanks.

share|improve this question
    
Did you write to the person who wrote the slides at your link, or the person to whom he attributes the conjecture? –  KConrad May 25 '12 at 0:02
1  
In the Bhargava–Shankar article (arxiv.org/abs/1006.1002), they say it "originates" in the work of Goldfeld and Katz–Sarnak and provide a reference for each. –  Rob Harron May 25 '12 at 1:44
    
Perfect! Thank you! –  Eugene May 25 '12 at 3:40
    
Do you want to include your comment as an answer so I can close up this question? –  Eugene May 25 '12 at 3:42

1 Answer 1

up vote 2 down vote accepted

Goldfeld made the conjecture for the family of quadratic twists of a given curve; it would be interesting to check the paper to see if he wrote anything about the more general case (I don't have it at hand). For the latter, the possibility that the average rank might be 1/2 was certainly folklore quite a while before Katz-Sarnak, but the numerical evidence did not (still doesn't, I think) really support this, so it was not usually stated as a formal conjecture. But see, for instance, the introduction to "The average rank of elliptic curves, I" by Brumer (Invent. math. 109, 1992), or even the last lines of the first page of "The rank of elliptic curves", by Brumer-Kramer (Duke Math. J. 44, 1977):

Our bounds suggest that, in general, curves of prime conductor have the smallest rank compatible with the parity predictions of Birch and Swinnerton-Dyer.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.