Timeline for Discrete Fourier Transform of the Möbius Function
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 6, 2011 at 18:41 | vote | accept | Gil Kalai | ||
| Mar 4, 2011 at 3:26 | comment | added | Matt Young | If there is a Landau-Siegel zero of modulus $q$ with $q$ some power of $\log{n}$ then there could be many bad $k$'s. The issue is that our best bounds on $\hat{\mu}(k)$ come from rational approximations $|k/n-a/q|\leq q^{-2}$. I guess roughly $n/q$ of the $k$'s may have a rational approximation of this form. All of these $k$'s would then be affected by this bad $q$. Known methods do not directly give a bound on $\hat{\mu}(k)$ by using the rationals $k/n$ because the denominator $n$ is too big. | |
| Mar 3, 2011 at 20:00 | comment | added | Gil Kalai | Many thanks Ben and Matt. Ben you wrote: "one can show that there cannot be two bad real characters but there might still be one." This is good for us since we want to show that the entire "energy is not concentrated on a few characters". What can be said about the second worse character? can we show that (log n)^({log log n}^A}) characters cannot captire a substantial amount of L_2 norm for perhaps every positive A. | |
| Mar 3, 2011 at 18:46 | comment | added | Matt Young | Ick, I lost the comment I had typed up. Luckily, Ben covered the A-dependence. I would just add that the issue is if $k/n$ has a good rational approximation of the form $a/q$ with $q$ corresponding to a Landau-Siegel zero. If $k=1$ or is very small then this is not going to happen so perhaps for 2) one can take $A$ larger. I didn't work it out though. I would be careful on the size of $s$ since it's reminiscent of the Mertens conjecture. As for the "truth", it comes from the heuristic that the M\"{o}bius function is not correlated with additive characters. | |
| Mar 3, 2011 at 18:28 | comment | added | Ben Green | Yes, I believe what I just said is correct. It's a paper of Baker and Harman with MS reference MR1111578 (92d:11087). Apparently Montgomery and Vaughan had this independently but didn't publish. | |
| Mar 3, 2011 at 18:24 | comment | added | Ben Green | Gil, without any extra hypotheses $A$ cannot be taken to be a slowly growing function of $n$, since the constant $C_A$ is ineffective. This comes from a well-known source of ineffectivity related to possible Siegel zeros: one can show that there cannot be two bad real characters but there might still be one. On the GRH you can certainly prove $\hat{\mu}(k) \ll n^{-c}$ pointwise. I'd have to look up the best value of $c$ currently known, due to Harman I think. From memory it's about 1/4. | |
| Mar 3, 2011 at 17:18 | comment | added | Gil Kalai | That's great, Matt. Do you know what A stands really for? (usually when you can prove things for A arbitrary large the proof allows A to be a function of n) E.g. if A is (log log n)^B still ok? regarding edit 2 can we expect that s should be at least a constant times n? Is the guess about the truth follows from RH or some version? | |
| Mar 3, 2011 at 16:15 | history | edited | Matt Young | CC BY-SA 2.5 |
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| Mar 3, 2011 at 16:10 | history | edited | Matt Young | CC BY-SA 2.5 |
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| Mar 3, 2011 at 15:45 | history | answered | Matt Young | CC BY-SA 2.5 |