Timeline for On V. Arnold's trinities regarding PSL(2,5), PSL(2,7), and PSL(2,11)?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 12, 2023 at 14:35 | comment | added | Somos | Let us continue this discussion in chat. | |
| Jan 12, 2023 at 13:17 | comment | added | Tito Piezas III | @Somos I've looked at level 55. In comparison, for level 25 with 24 functions, what Ramanujan did was eliminate 4 (multiples of 5), so $24-4 = 20$, grouped them by 4, and ended up with a quintic with coefficients determined by $\eta(\tau)/\eta(25\tau)$. (He re-discovered the Emma Lehmer quintic but didn't know that.) For level 55 with with 54 functions, we can eliminate 14 (multiples of 5 and 11), so $54-14 = 40$. We can then group those by 4 to end up with a 10-deg, or even by 8 to end up with a quintic again. But to know which group of functions go together is beyond my powers, I'm afraid. | |
| Jan 12, 2023 at 5:00 | comment | added | Tito Piezas III | @Somos I just finished my answer. It seems $a^2b+b^2c+c^2a$ is a property of a certain generic cubic. Kindly see my answer here. P.S. I will explore the level $n=55$ soon. | |
| Jan 12, 2023 at 2:49 | comment | added | Somos | I did a search and 55 was the first with 660 denominator. | |
| Jan 12, 2023 at 2:33 | comment | added | Tito Piezas III | @Somos I must have been sleepy when I tried it because the denominator $660$ reduces only if $k=5m$. And believe I know why you chose $55$ since $24/(p-1)$ for $p=11$ is $12/\color{blue}5$. The $33$ version is not pretty? | |
| Jan 11, 2023 at 18:51 | comment | added | Somos | Have you tried $q^{(4565-330k+6k^2)/660}f(-q^k,-q^{55-k})f(-q)$, although it is not a quotient? | |
| Jan 11, 2023 at 18:27 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Corrected numerators
|
| Jan 11, 2023 at 16:29 | answer | added | Somos | timeline score: 2 | |
| Jan 11, 2023 at 14:57 | comment | added | Tito Piezas III | @Somos I've finished my addendum. Kindly check pls. Also, can you write a tentative answer and show what is the product of your six level 13 functions? | |
| Jan 11, 2023 at 14:55 | answer | added | Tito Piezas III | timeline score: 1 | |
| Jan 11, 2023 at 13:37 | comment | added | Tito Piezas III | @Somos. Hmm, your formula gives the same denominators as mine, but the numerator for $p=11,13$ is different. Yours give $q^{199/264}$ and $q^{149/156}$. Give me time to write an addendum. | |
| Jan 11, 2023 at 13:20 | comment | added | Somos | My calculations show that for odd level $p>1$ that the function is $q^{(24-21p+5p^2)/(48p)}f(-q,-q^{p-1})/f(-q^{(p-3)/2}).$ Your power of $q$ factor is wrong for $p=11,13$. | |
| Jan 11, 2023 at 3:45 | history | asked | Tito Piezas III | CC BY-SA 4.0 |