Timeline for Bounds on Fourier coefficients for $GL(3)$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2020 at 6:51 | comment | added | Subhajit Jana | If you are asking for Ramanujan on average uniformly in $n$, to prove that it will be as hard as proving Ramanujan: Let $n$ be any large prime bigger than $x$ so that $(m,n)=1$ for all $m<x$ so that $$\lambda(n,m)=\lambda(n,1)\lambda(1.m).$$ If we have $$\sum_{m<x} |\lambda(n,m)|^2\ll x^{1+\epsilon}$$ uniformly in $n$ then using the fact that $$\sum_{m<x}|\lambda(1,m)|^2\gg x^{1-\epsilon}$$ we obtain $$\lambda(n,1)\ll x^\epsilon < n^\epsilon.,$$ i.e. Ramanujan. | |
| Mar 28, 2020 at 4:55 | comment | added | Amomentum | But this is wasteful since we have a rough $n^2$ bound on the right then, which is a loss compared to Ramanujan conjecture. Is there any way to do better? | |
| Mar 26, 2020 at 8:22 | comment | added | Subhajit Jana | I thought that $n$ in the second sum is fixed, say $n_0$. Then the second sum is bounded by $\sum_{n^2m<n_0^2x} |\lambda(n,m)|^2\ll_{n_0} x^{1+\epsilon}$ from the first estimate in your question. | |
| Mar 26, 2020 at 1:48 | comment | added | Amomentum | But how do you get the bound on the second sum? (for now it seems a bit circular) | |
| Mar 25, 2020 at 9:30 | vote | accept | Amomentum | ||
| Mar 26, 2020 at 1:46 | |||||
| Mar 24, 2020 at 8:29 | history | answered | Subhajit Jana | CC BY-SA 4.0 |