bio | website | maths.ox.ac.uk/~greenbj |
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location | Auckland, New Zealand | |
age | 38 | |
visits | member for | 5 years, 3 months |
seen | 2 days ago | |
stats | profile views | 12,632 |
I'm a professor at Oxford University, on sabbatical in New Zealand until April 2014
Jun 1 |
awarded | Good Answer |
May 24 |
awarded | Nice Answer |
May 21 |
comment |
Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
I'll denote any points I get from this observation (which I stole from someone else anyway) to a good cause. An additional point to be made, now the preprint is available, is that the Bombieri-Fouvry-Friedlander-Iwaniec type results on level of distribution depend on an estimate for sums of Kloosterman sums that requires (via a lemma of Bombieri and Birch) Deligne's work on the Weil Conjectures. It seems to me (though I'm certainly not an expert) that these estimates are not accessible by the more elementary methods such as Dwork/Stepanov. |
May 21 |
answered | Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture |
May 20 |
comment |
Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
Mark - we added our comments at the same time. I was referring to the table on page 12 of GPY, whereas you were referring to the one on p9. I guess the one on page 9 tells you what you can get using just the basic GPY method, and the one on page 12 uses more complicated weights. I guess Zhang elaborates on the basic GPY method. Maybe his method can be combined with the more complicated GPY method which leads to the numerics on page 12; it seems likely that this will be one place any would-be 70000000-reducers will look first. |
May 20 |
comment |
Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
It's interesting to speculate on how much the 70000000 will be reduced. In this regard, the table on page 12 of Goldston-Pintz-Yildirim is relevant. If one had level of distribution 4/7 with no strings attached, it seems one might get gaps of size 500 or so. To get gaps under 100 without some completely new idea one would have to go out to level of distribution nearly 2/3, i.e. double the improvement of BFI. I'd wager that getting down to 10000 or so is going to prove pretty difficult. |
May 15 |
awarded | Nice Answer |
May 14 |
answered | Proof of the weak Goldbach Conjecture |
Apr 23 |
awarded | Yearling |
Apr 19 |
accepted | Additive Combinatorics - reference request |
Apr 19 |
comment |
Additive Combinatorics - reference request
I finally dug out the reference myself using a MathSciNet search. MR1931192 (2003f:11016) Reviewed Schoen, Tomasz(PL-POZN) The cardinality of restricted sumsets. (English summary) J. Number Theory 96 (2002), no. 1, 48–54. |
Apr 19 |
comment |
Additive Combinatorics - reference request
Thanks Seva - but I'm sure I've seen this exact argument somewhere. I guess not in one of your papers then? |
Apr 19 |
asked | Additive Combinatorics - reference request |
Apr 9 |
comment |
Minimal size of subsets $A,B$ in a finite group $G$ such that $AB=G$
Is this exactly "coupon collecting analysis"? For each choice of B = {b_1,..,b_k} you can try and estimate the probability that AB is all of G, if A is chosen randomly. But then you have to sum over a large selection of B's, and the triangle inequality may be too crude. In fact, I posed as an open problem (Barbados workshop, March 2012) the question of deciding whether the log N should really be there. Can Z/NZ be covered by O(N^{1/2}) translates of a random set A of size N^{1/2}? |
Apr 8 |
awarded | Good Answer |
Mar 26 |
comment |
Riemann Z function, bounds on number of non-trivial zeros along horizontal lines, rather than vertical ones
I think that from standard asymptotics on the number of zeros of height up to T you get a bound of O(log T). Apparently even on RH is is only known that the multiplicity of 1/2 + iT as a zero is O(log T/log log T). So I guess you shouldn't expect too much better than this O(log T) bound, |
Mar 11 |
comment |
On the $L^1$-norm of certain exponential sums.
This is nicely done. |
Mar 8 |
comment |
On the $L^1$-norm of certain exponential sums.
Yep, this issue (of whether a 1-dissociated set, or "subsum-distinct" set) has a positive density 2-dissociated subset seems a little unclear. It seems closely related to work of Pisier and Bourgain from the late 80's on Sidon sets. In particular I wonder whether Pisier's arithmetic characterisation of Sidon sets works with 2-dissociated in place of dissociated. Perhaps the proposer could comment on whether he needs the full generality? |
Mar 8 |
comment |
On the $L^1$-norm of certain exponential sums.
Noam, Very nice! To do the splitting, I was planning to use Holder: $\int |fg| \leq (\int |f|)^{1/3}(\int |g|)^{1/3}(\int |f g|^2)^{1/3}$. You have the trivial bound on each factor (note that fg is the FT of the set of subset sums of S), so you only need to win in one factor, say the first one. So actually, I think all you need is that any subsum-distinct set has a 2-dissociated subset of positive density. That's clear in the case of powers of 2 (just thin out every other power of two) but actually not so obvious in general. I'm looking at some literature on that right now. |
Mar 8 |
comment |
The polynomial Freiman-Ruzsa conjecture for the special case when $f$ is a bijection
It seems unlikely to me that the bijection assumption would help. If f is an arbitrary function then, provided f is not "really far" from a bijection I'd expect that some modification f'(x) = f(x) + eps(x) with eps varying in a small set would make f' pretty close to a bijection (perhaps use Hall's marriage theorem or something). PFR for f and for f' are basically the same problem. Being a bijection is not a property that is useful in connection with several of the existing techniques in the area, particularly Fourier analysis (cf. the work of Sanders). |