bio | website | maths.ox.ac.uk/~greenbj |
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location | Auckland, New Zealand | |
age | 38 | |
visits | member for | 5 years, 1 month |
seen | 6 hours ago | |
stats | profile views | 12,506 |
I'm a professor at Oxford University, on sabbatical in New Zealand until April 2014
May 3 |
comment |
Historical question concerning Jordan's theorem
Jim: I agree completely. I'd still like an actual reference to the argument I sketched above though, if there is one earlier than Tao's blog. |
May 3 |
comment |
Historical question concerning Jordan's theorem
Yes indeed - and this paper has proved very useful indeed to many of us working in additive combinatorics recently. The paper also contains a proof in characteristic zero, but it is certainly a bit more complicated than the one I described above. |
May 3 |
comment |
Translation of Goldbach's 1742 letter to Euler
Brilliant - now we just need someone who reads German to take a look at the footnote at the bottom of page 127. |
May 3 |
asked | Translation of Goldbach's 1742 letter to Euler |
May 2 |
comment |
Historical question concerning Jordan's theorem
I should also point out that by $j(n)$ I think you mean the index of the biggest normal abelian subgroup of $A$; so your $j(n)$ is at most $F(n)!$, but they need not be the same. |
May 2 |
comment |
Historical question concerning Jordan's theorem
Igor, I believe this issue is comprehensively despatched in this paper of M. J. Collins: On Jordan's theorem for complex linear groups. J. Group Theory 10 (2007), no. 4, 411--423. He evaluates $j(n)$ for all $n$ and shows that $(n+1)!$ is the truth for $n \geq 71$ (and not for $n = 70$). But my interest is more in finding the simplest argument that gives some bound. |
May 2 |
awarded | Student |
May 2 |
asked | Historical question concerning Jordan's theorem |
May 2 |
awarded | Critic |
May 1 |
comment |
explicit big linearly independent sets
I think I agree! |
May 1 |
answered | explicit big linearly independent sets |
Apr 29 |
answered | Wanted: A constructive version of a theorem of Furstenberg and Weiss |
Apr 27 |
comment |
Are There Primes of Every Hamming Weight?
Qiaochu: no, but I believe it would give one in principle, in the sense that there is no ineffectivity arising from a potential Siegel Zero. People I know who have done the kind of explicit calculations necessary to extract actual bounds from arguments like this attest that it is very painful, and the bounds are often pretty awful to boot. |
Apr 27 |
awarded | Nice Answer |
Apr 27 |
awarded | Supporter |
Apr 27 |
awarded | Editor |
Apr 27 |
revised |
Are There Primes of Every Hamming Weight?
added 248 characters in body |
Apr 27 |
revised |
Are There Primes of Every Hamming Weight?
corrected a couple of typos; added 1 characters in body; added 14 characters in body |
Apr 27 |
answered | Are There Primes of Every Hamming Weight? |
Apr 26 |
awarded | Good Answer |