Timeline for Potential modularity and the Ramanujan conjecture
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2011 at 22:33 | comment | added | GH from MO | OK, but then we need the relevant properties for $L(s,\text{sym}^n f\otimes\chi^m)$ for all $n$ and many $m$'s, right? I am not familiar with Clebsch-Gordon, and I don't know if the recent work on potential modularity covers these character twists as well. | |
| Jan 17, 2011 at 22:18 | comment | added | David Hansen | Sure, just argue with $L(s,\mathrm{sym}^n(f) \otimes \mathrm{sym}^n(\overline{f}))$ instead, and use the fact that $\overline{f}=f\otimes \overline{\chi}$ where $\chi$ is the nebentypus. | |
| Jan 2, 2011 at 23:30 | comment | added | GH from MO | That's a very good point, thanks! Does it also work when the nebentypus of $f$ is nontrivial? | |
| Jan 2, 2011 at 22:12 | comment | added | David Hansen | To deduce Ramanujan: You need to write $L(s,\mathrm{sym}^n(f)\otimes \mathrm{sym}^n(f))$ as a product of $L(s,\mathrm{sym}^j(f))$'s for various j's (use Clebsch-Gordon) and then appeal to the fact that a Dirichlet series with nonnegative coefficients converges absolutely up to its first pole. | |
| Jan 2, 2011 at 12:55 | comment | added | GH from MO | On the other hand, the following is clear: if in a fixed right half-plane each $L(s,sym^n f)$ is given by an absolutely convergent Euler-product, then Ramanujan holds. This is because the assumption implies that all the Euler factors are holomorphic in a fixed right half-plane which is equivalent to Ramanujan. It is also worth mentioning that an elaboration of this idea (pole of an Euler factor or gamma factor forces a zero of the remaining partial $L$-function), originally due to Serre, lead to nontrivial bounds for cusp forms on GL(n), see the work of Luo-Rudnick-Sarnak. | |
| Jan 2, 2011 at 10:34 | history | answered | GH from MO | CC BY-SA 2.5 |