Timeline for Finitely many arithmetic progressions
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16 events
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| Sep 13, 2010 at 16:06 | history | edited | Charles |
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| Jun 3, 2010 at 12:42 | comment | added | Hugh Thomas | There are a couple of papers by David Feldman, Jim Propp, and Sinai Robins (arXiv:0905.0441 (one page long) and arXiv:1006.0472) which treat a generalization to $\mathbb Z^d$. In the first paper, they give an example showing that the most obvious generalization to sublattices fails for dimension greater than or equal to three, and they prove a generalization where there is an additional assumption that the lattices are "straight", i.e. there is some basis with respect to which each of the lattices is generated by integer multiples of that same basis. | |
| May 20, 2010 at 23:28 | comment | added | Gerry Myerson | @fedja I think the other approach, the one that forms the main topic of the Zeilberger piece, does apply to the group setting, but I don't know the exact statement. Follow the references in the Zeilberger essay. | |
| May 20, 2010 at 15:19 | comment | added | fedja | @Gerry Of course, but I didn't really mean as strong concluson as that 2 subgroups must coinside. Just that two subgroups must have the same index. | |
| May 20, 2010 at 12:29 | comment | added | Gerry Myerson | @Michael, Zeilberger wrote up a history of the two proofs of the result. It's at emis.de/journals/EJC/Volume_8/PDF/v8i2a1.pdf | |
| May 20, 2010 at 11:59 | comment | added | Gerry Myerson | @fedja, no. You can partition $S_3$ into cosets of the 2-element subgroups. That is, if $A$, $B$, and $C$ are the 2-element subgroups, then there are group elements $g$, $h$, and $k$ such that $S_3$ is the disjoint union of $gA$, $hB$, and $kC$. Also, if $G$ is any finitely generated abelian group of rank at least 3, finite or infinite, then you can partition $G$ into cosets of distinct subgroups. So far as I know, the question is wide open for abelian groups of rank 2. @Michael, yes, there is another proof, but I don't have access to my references. I'll try to look it up soon. | |
| May 20, 2010 at 11:44 | answer | added | Jorge Vitório Pereira | timeline score: 10 | |
| May 20, 2010 at 10:21 | answer | added | Tamas Hausel | timeline score: 18 | |
| May 20, 2010 at 5:18 | comment | added | fedja |
What I wonder is if you can prove the same for any partition of a finite (not necessarily abelian) group into left cosets of some subgroups (the partition of $\mathbb Z$ is actually a partition of $\mathbb Z_N$ where $N$ is any common multiple of all differences). Here the generating function approach will meet some troubles.
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| May 20, 2010 at 4:13 | answer | added | Hugh Thomas | timeline score: 7 | |
| May 20, 2010 at 4:07 | comment | added | Qiaochu Yuan | I find the generating function proof quite illuminating. It can be summarized as follows: arithmetic progressions have Fourier coefficients, and the nonzero Fourier coefficients of the natural numbers are all zero. But arithmetic progressions with pairwise distinct common differences have Fourier coefficients which are too different to sum to zero. | |
| May 20, 2010 at 3:28 | comment | added | Michael Lugo | Is there a proof not using generating functions? (I'm surprised to be asking this, since I'm usually a fan of generating functions, but they seem pretty unilluminating here.) | |
| May 20, 2010 at 2:55 | vote | accept | Andrés E. Caicedo | ||
| May 20, 2010 at 2:54 | answer | added | Andrés E. Caicedo | timeline score: 17 | |
| May 20, 2010 at 2:52 | answer | added | Gerry Myerson | timeline score: 14 | |
| May 20, 2010 at 2:46 | history | asked | Andrés E. Caicedo | CC BY-SA 2.5 |