bio | website | maths.ox.ac.uk/~greenbj |
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location | Auckland, New Zealand | |

age | 38 | |

visits | member for | 4 years, 11 months |

seen | Mar 23 at 10:48 | |

stats | profile views | 12,378 |

I'm a professor at Oxford University, on sabbatical in New Zealand until April 2014

Mar 19 |
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Cricket and the Hardy-Littlewod maximal function
From what I understand the theorem is rather trivial when applied to Bollobas's own cricketing career - run out for 0 off the second ball of his only innings. |

Aug 26 |
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Maximum sets of lattice points such that only a few points collinear
It is, in fact, a very unsolved problem to decide whether you can put $2n$ points in $[n] \times [n]$ with no three collinear, if $n$ is large. I suspect the answer is no, and that in fact one can only put $(c + o(1))n$ such points for some $c < 2$; possibly $c = 3/2$. There is a construction which achieves this. |

Jun 12 |
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What are some examples of mathematicians who had an unconventional education?
The windmill in Nottingham operated by Green (no relation) can still be visited: en.wikipedia.org/wiki/Green's_Mill,_Sneinton though it is something of a detour even if one happens to be in Nottingham. Isaac Newton's home, 28 miles away, could be ticked off too for a mathematical tour of the East Midlands of England. |

Jun 10 |
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Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
I may try and read BFI. If you do it just in some critical range, rather than in the extreme generality they do with unspecified parameters $C, D, H, R, N,\dots$, my guess is it will boil down to the same basic ingredients in a different order, once one has applied a suitable decomposition into bilinear forms: Cauchy, Weyl shift, completion of sums/Fourier expansion. What else is there? |

Jun 9 |
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Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
And by the way, I think he should have called his paper "On a new bound for an incomplete Kloosterman sum to composite moduli, with applications". |

Jun 9 |
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Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
I know very little of Kloostermania, but my impression is the arguments in the BFI papers you speak of are basically the same, only the inputs are bounds for averages of Kloosterman sums over different moduli (?) coming from automorphic form theory, rather than bounds for products of Kloosterman sums to a distinct modulus. So it's just a different black box you have to believe, right? |

Jun 9 |
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Philosophy behind Yitang Zhang's work on the Twin Primes Conjecture
I have to say that, now that I have studied the paper in some detail, this seems to me to be an extraordinarily accurate (if perhaps difficult to parse on a first reading) answer to the original question. |

May 21 |
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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 20 |
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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 |
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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. |

Apr 19 |
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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 |
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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 9 |
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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}? |

Mar 26 |
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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 |
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On the $L^1$-norm of certain exponential sums.
This is nicely done. |

Mar 8 |
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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 |
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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 |
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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). |

Mar 7 |
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On the $L^1$-norm of certain exponential sums.
I should add, though, that I have to prepare an undergraduate lecture now - so I'll try and write down the details tomorrow. |

Mar 7 |
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On the $L^1$-norm of certain exponential sums.
I chatted with Tom Sanders about this today in Oxford, and I think we can more-or-less solve it, at least the second case. The key idea is to write |1 + e(t)| as 2^{1/2} (1 + cos(2 pi t))^{1/2}, then use the inequality (1 + x)^{1/2} \leq 1 + x/2 - cx^2, valid for some $c > 0 (in fact for c = 3/2 - 2^{1/2}). Now expand everything out, and you get a bound for your integral of 2^{n/2}(1 - c/4)^n if S is "good": has no relations with coefficients <= 2. A bit of fiddling should give exactly what you want, with your weaker assumption on S; in the powers of two case you can split S into two good sets |