Timeline for How small can a sum of a few roots of unity be?
Current License: CC BY-SA 2.5
28 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2021 at 9:53 | comment | added | Ben Barber | I turned this comment into a question. mathoverflow.net/questions/395281/… | |
| Jun 2, 2021 at 18:10 | comment | added | Ben Barber | What is the probabilistic intuition suggesting a polynomial lower bound? | |
| May 3, 2021 at 8:24 | answer | added | Ben Barber | timeline score: 21 | |
| Jun 19, 2019 at 15:38 | answer | added | Geoff Robinson | timeline score: 4 | |
| Jan 9, 2017 at 17:43 | comment | added | kodlu | I've asked a related question it this link mathoverflow.net/questions/259177/… | |
| Nov 15, 2010 at 20:03 | comment | added | Seva | A symbolic improvement over the trivial exponential lower bound can be found here: emis.de/journals/INTEGERS/papers/a1/a1.pdf. | |
| Nov 15, 2010 at 17:36 | comment | added | Warren Schudy | Are the $z_i$ complex numbers or is your question for roots of unity in a more general field? | |
| Nov 15, 2010 at 16:24 | comment | added | Terry Tao | Tim: The Galois group of the cyclotomic field has order phi(n), so unless one is somehow extremely careful not to lose a multiplicative factor for each Galois conjugate, attempting to control the algebraic integer S by analysing all the phi(n) Galois conjugates together would lead to inefficiencies that are exponential in phi(n). Perhaps there is a more "additive" way to exploit the Galois conjugates that would only lose polynomial factors rather than exponential ones, but it's hard to see how additive methods (e.g. moments) can lead to lower bounds on magnitudes, rather than upper bounds. | |
| Nov 15, 2010 at 15:55 | answer | added | Aaron Meyerowitz | timeline score: 6 | |
| Nov 15, 2010 at 15:26 | comment | added | JSE | That's already a good answer! | |
| Nov 15, 2010 at 15:25 | comment | added | Felipe Voloch | @Jordan: The set of (all) roots of unity is discrete in the p-adic topology but not in the complex topology, this already shows things will be different. I am not quite sure how to answer your "Why?". | |
| Nov 15, 2010 at 12:04 | comment | added | gowers | Having only a very poor grasp of Galois theory (in fact, even that is a rather flattering way of putting it), I'd be interested by an elaboration of your remark that going through Galois theory automatically leads to exponentially poor bounds. | |
| Nov 15, 2010 at 11:52 | answer | added | Roland Bacher | timeline score: 12 | |
| Nov 15, 2010 at 10:19 | answer | added | Denis Serre | timeline score: 5 | |
| Nov 15, 2010 at 5:05 | vote | accept | Terry Tao | ||
| Nov 15, 2010 at 1:58 | comment | added | JSE | So my guess that the archimedean story would be well-modeled by the non-archimedean story was totally wrong! Why? | |
| Nov 15, 2010 at 1:54 | comment | added | Felipe Voloch | My paper with Tate proves that the p-adic absolute value is bounded below by a constant independent of n. Of course, one can't expect that in the archimedian case. | |
| Nov 15, 2010 at 1:44 | comment | added | Gerry Myerson | Concerning the $p$-adic question, there may be something in a paper of Tate and Voloch, Linear forms in $p$-adic roots of unity, Internat. Math. Res. Notices 1996, no. 12, 589–601, MR 97h:11065. | |
| Nov 14, 2010 at 23:43 | comment | added | Gerry Myerson | The correct URL for the Calegari-Morrison-Snyder paper is math.northwestern.edu/~fcale/papers/cyclotomic.pdf | |
| Nov 14, 2010 at 23:05 | comment | added | JSE | (In case it's not obvious, I did not actually have any substantive ideas about the question at hand while walking with the kids, though I did buy a cute little drum shaped like a frog.) | |
| Nov 14, 2010 at 23:04 | comment | added | JSE | If we get a good answer to this one, I'm going to ask: if G is a finite group and rho: G -> M_n(C) an irreducible representation, how close can a nonzero sum rho(g_1) + ... rho(g_10) be to the zero matrix? Terry's question is the case G = Z/nZ. | |
| Nov 14, 2010 at 21:34 | comment | added | JSE | If I were trying to get a handle on what was true, I might try to prove a lower bound on a p-adic valuation of the sum of 10 nth roots -- why should the archimedean absolute value be any different? Off to take kids for a walk, will try to think about what I mean by this as I go. | |
| Nov 14, 2010 at 21:31 | comment | added | Terry Tao | By "degenerate" I mean "becomes much easier to solve". The first really difficult case is for 5 roots, which is confirmed by the reference given by noobcake below. | |
| Nov 14, 2010 at 21:30 | comment | added | Emerton | This joint paper of Frank Calegari, Scott Morrison, and Noah Snyder ( math.northwestern.edu/~fcale/cyclotomic.pdf ) might have some relevant techniques . | |
| Nov 14, 2010 at 21:23 | comment | added | JSE | Why is the problem degenerate for two roots? The difference between two distinct nth roots of unity is at most n^-1 or so, so if the sum is nonzero, it's at least n^-1, right? | |
| Nov 14, 2010 at 21:03 | answer | added | noobcake | timeline score: 62 | |
| Nov 14, 2010 at 21:00 | history | edited | Terry Tao | CC BY-SA 2.5 |
added 100 characters in body
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| Nov 14, 2010 at 20:54 | history | asked | Terry Tao | CC BY-SA 2.5 |