Timeline for Cosets representatives of congruence subgroups
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2022 at 11:42 | comment | added | Tireless and hardworking | Thanks for the time you've spent to me. I appreciate that. Finally I found the "correspondence" between your answer and his answer, and that was just a very simple congruence excercise. I think I understood everything. Thanks for everything, specially for your kindness and your generousity dear Francois Brunault. | |
| Mar 28, 2022 at 11:13 | comment | added | François Brunault | @Tirelessandhardworking I think I came up with the formula starting with the description of $\Gamma_1(N) \backslash \mathrm{SL}_2(\mathbf{Z})$ as the set of pairs $(u,v) \in (\mathbf{Z}/N\mathbf{Z})^2$ such that $u,v$ generate $\mathbf{Z}/N\mathbf{Z}$, which is easier. Then one can work out the orbit a given pair $(u,v)$ under $(\mathbf{Z}/N\mathbf{Z})^\times$. But this doesn't really give an intuition. Anyway, thank you very much for pointing the error. | |
| Mar 27, 2022 at 19:33 | comment | added | Tireless and hardworking | Thank you. The map is impressive and the excercise was interesting. I verified that the map is well-defined, injective and surjective. I read also the answer by user "Max Horn", and now I can see some relations between his answer and your answer (the relation about "projective line", and the role of $u$ and $(c, d)$), but yet I can not understand where the "congruence mod $\frac{N}{c}$ over $d$" emerged form? | |
| Mar 27, 2022 at 17:59 | history | edited | François Brunault | CC BY-SA 4.0 |
Specified we take right cosets
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| Mar 27, 2022 at 17:17 | comment | added | François Brunault | @Tirelessandhardworking I see, the definition of the map $\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \mapsto (c,d)$ is incorrect. One should first find $u \in (\mathbb{Z}/N\mathbb{Z})^\times$ such that $\gamma u \equiv \mathrm{gcd}(\gamma,N) \bmod{N}$, and multiply $\delta$ by $u$ accordingly. | |
| Mar 27, 2022 at 13:45 | comment | added | Tireless and hardworking | Let $N=9$, and let $M_1=\begin{pmatrix} 1 & 1 \\ 3 & 4 \end{pmatrix}$, and let $M_2=\begin{pmatrix} A & B \\ 9C & D \end{pmatrix} \in \Gamma_0(9)$, where $D \equiv 2 \mod{3}$. No matter if we consider left cosets or right cosets, then the map is not well defined, because $M_1M_2=\begin{pmatrix} A+9c & B+D \\ 3A+36C & 3B+4D \end{pmatrix}$, and $M_2M_1=\begin{pmatrix} A+3B & A+4B \\ 9C+3D & 9C+4D \end{pmatrix}$, and we have: $$M_1 \mapsto (3, 1),$$ $$M_1M_2 \mapsto (3, 2),$$ $$M_2M_1 \mapsto (3, 2),$$ so two different represntatives do not send to a unique element. | |
| Dec 20, 2011 at 23:08 | history | answered | François Brunault | CC BY-SA 3.0 |