Timeline for Periodic functions over different lattices in $\mathbb R^d$ are linearly independent [closed]
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| when toggle format | what | by | license | comment | |
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| May 5, 2014 at 3:50 | review | Reopen votes | |||
| May 5, 2014 at 10:46 | |||||
| May 5, 2014 at 3:34 | history | edited | sweehong | CC BY-SA 3.0 |
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| May 5, 2014 at 3:33 | vote | accept | sweehong | ||
| May 4, 2014 at 22:28 | history | closed |
Qiaochu Yuan Ryan Budney S. Carnahan♦ |
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| May 4, 2014 at 18:11 | answer | added | paul garrett | timeline score: 4 | |
| May 4, 2014 at 17:40 | comment | added | Asaf | It was discussed recently here - mathoverflow.net/questions/162875/… Loosely speaking, this is some multi-dimensional generalization of the fact that irrational multiples are equidistributed in the torus, this appears in every basic book about ergodic theory, for example in Furstenberg's book or the recent book by Einsiedler-Ward book (volume I). I believe that the Fourier approach (in general LCA groups) appears in Katznelson book. | |
| May 4, 2014 at 17:36 | review | Close votes | |||
| May 4, 2014 at 23:22 | |||||
| May 4, 2014 at 17:28 | comment | added | sweehong | Thanks for your comment. Do you have reference for the Kronecker argument that you mention? | |
| May 4, 2014 at 17:19 | comment | added | Asaf | foliation of the torus by sub-tori (notice that Kronecker system is semi-simple, meaning it decomposes to a disjoint union of minimal ones, it is not necessarily minimal itself). The exact condition needed here is that any two sub-groups (not necessarily lattices (i.e. finite co-volume)) are not commensurable. | |
| May 4, 2014 at 17:17 | comment | added | Asaf | Well Fourier transform won't help you here, because periodic continuous functions are not $L^{1}$, if you want to go along this lines, you would need to go to Fourier series along the related tori. Anyways, an elementary solution would be to project the linear combination into say a fundemental domain for ome $L_i$ and then use Kronecker lemma/Weyl's theorem to ensure that the $L_i$ periodic function must be constant. But notice that in the settings you've described, the thm does not hold. From the Kronecker argument I've described, you get that the function must be constant along leafs of | |
| May 4, 2014 at 16:43 | history | asked | sweehong | CC BY-SA 3.0 |