Timeline for Level vs. conductor of a supercuspidal representation
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 19, 2020 at 19:21 | history | edited | LSpice | CC BY-SA 4.0 |
Reference to @Kimball's better answer
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| May 17, 2020 at 23:14 | vote | accept | user15243 | ||
| May 17, 2020 at 23:14 | comment | added | user15243 | Thanks a lot... | |
| May 17, 2020 at 23:02 | comment | added | LSpice | @Kiddo, as the statement of the theorem says, it is notation (6.2.1): $m/e = \max (m_1/e_1, m_2/e_2)$. | |
| May 17, 2020 at 22:58 | comment | added | user15243 | Theorem 6.5,(ii) says $f(\sigma_1' \otimes \sigma_2)= n_1n_2(1+m/e)$. What is this $m$? It is not mentioned in the theorem. Is it the level of $\sigma_1' \otimes \sigma_2$? | |
| May 17, 2020 at 22:55 | comment | added | user15243 | Thank you for the reply. | |
| May 17, 2020 at 22:48 | comment | added | LSpice | @Kiddo, there is no "the conductor of $\pi$"; it depends on a choice of $\psi$, so that (as the reference indicates) one should really write $f(\pi, \psi)$. We may certainly choose $\psi$ as you say, and it simplifies the formula, but of course doesn't affect $\pi$ at all. | |
| May 17, 2020 at 22:34 | comment | added | user15243 | If we choose $c(\psi)$ to be $0$, can we say something about $\pi$? | |
| May 17, 2020 at 21:28 | history | answered | LSpice | CC BY-SA 4.0 |