Timeline for The Wilton-type bounds involving half-integral weight cusp forms
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2023 at 15:41 | comment | added | Pratim Mitra | I'm not sure but I think we can find this in Iwaniec's Topics in Classical Automorphic Forms book. | |
| Jun 16, 2023 at 14:36 | comment | added | hofnumber | Sorry, .....we see that $a_f(n)$... , $a_f(n)$ should be $a_f(n)=\lambda_f(n)n^{1/4}$. | |
| Jun 16, 2023 at 14:28 | comment | added | hofnumber | Great answer! According to the normalization, we see that $a_f(n)=\lambda_f(n)n^{-1/4}$. in other words, one has the analog of square-root cancellation. That is, $$\sum_{n\le X}a_f(n)e(n)\alpha \ll X^{1/2}\log (2X).$$ However, could you please show me how $f(z)\ll y^{-(k+1/2)/2}$ if $f$ is a cusp form of half-weight $k+1/2$?? Could you show me certain reference? If $f$ is a cusp form of integral weight $k$, we know $f(z)\ll y^{-k/2}$ . | |
| S Jun 16, 2023 at 12:01 | review | First answers | |||
| Jun 16, 2023 at 12:16 | |||||
| S Jun 16, 2023 at 12:01 | history | answered | Pratim Mitra | CC BY-SA 4.0 |