Timeline for Fricke involution and Atkin operator
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 20, 2022 at 7:45 | history | edited | François Brunault | CC BY-SA 4.0 |
Clarified the action of $U_q$
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| Jun 20, 2022 at 7:18 | comment | added | Akash Yadav | Yes I agree with your comment. Thanks for all the help. | |
| Jun 20, 2022 at 7:18 | vote | accept | Akash Yadav | ||
| Jun 19, 2022 at 23:25 | comment | added | François Brunault | @AkashYadav If I understood your definitions correctly, your operator $U_q$ doesn't preserve the space of forms of level $N$. For example let $f=\sum_{n \geq 1} a_n x^n$ be a newform of level $N$, then $U_q f = q \sum_{n \geq 1} a_{qn} x^{qn} = q a_q f(qz)$ since the Fourier coefficients of $f$ are completely multiplicative. So $U_q f$ has level $Nq$, and applying $W_N$ doesn't make sense. | |
| Jun 19, 2022 at 10:18 | comment | added | Akash Yadav | Just a final question, the operators still commute for a newform right? | |
| Jun 18, 2022 at 13:21 | comment | added | François Brunault | I see, your notation is not standard (usually $U_p(\sum a_n q^n) = \sum a_{pn} q^n$). I think Henri Cohen's example works. | |
| Jun 18, 2022 at 12:35 | comment | added | Akash Yadav | I am not sure but have you observed that the stated operator U_q is different from the usual Hecke operator for prime q dividing N. | |
| Jun 18, 2022 at 12:14 | comment | added | Henri Cohen | Simply take $q=N=2$ and $f(\tau)=\Delta(2\tau)$ with $\Delta$ the usual weight $12$ cusp form of level $1$. | |
| Jun 18, 2022 at 12:04 | history | answered | François Brunault | CC BY-SA 4.0 |