Timeline for Pull-back of factor of automorphy
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 2, 2022 at 14:12 | vote | accept | Mjr | ||
| May 2, 2022 at 12:04 | history | edited | DamienC | CC BY-SA 4.0 |
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| May 2, 2022 at 11:54 | comment | added | DamienC | Hence the pull-back of $J$ shall probably be $J(\varphi(\gamma),\tilde{f}(x))$. | |
| May 2, 2022 at 11:47 | comment | added | DamienC | As for equivariant I should have been a bit more precise. $\tilde{f}$ is the composition of a homomorphism with a translation. A translation is equivariant in an obvious sense; $T(\gamma\cdot x)=\gamma\cdot T(x)$. A homomorphism $\varphi$ sends $\Gamma$ to $\Gamma$, and thus is equivariant in the following sense: $\varphi(\gamma\cdot x)=\varphi(\gamma)\cdot\varphi(x)$. | |
| May 2, 2022 at 11:44 | comment | added | DamienC | @Mjr: You should use $\tilde{f}$ rather than $f$. | |
| May 2, 2022 at 11:29 | comment | added | Mjr | If I get your means correctly, by the $\Gamma$-equivariance the pull-back of $J$ shoud be $J(\gamma,f(x))$. But I am not clear about why $f$ is equivariant, could you please give me some more explanations? | |
| May 2, 2022 at 9:22 | comment | added | Mjr | I am not sure that whether the pull back of $J$ is given by $J(\gamma, f(x))$ or $J(f(\gamma),f(x))$. | |
| May 2, 2022 at 9:15 | history | edited | DamienC | CC BY-SA 4.0 |
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| May 2, 2022 at 9:06 | history | answered | DamienC | CC BY-SA 4.0 |