Timeline for Serre's formula for $\Delta^{1/3}$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 28, 2019 at 7:09 | comment | added | Shimrod | @LSpice, The three cube roots correspond to the three partitions of the four element set (sorry for late answer). | |
| May 13, 2019 at 2:09 | comment | added | LSpice | Sorry for being obtuse. What I meant to ask about \eqref{1} was: is it that the equation holds for any of the $\binom 4 2$ possible arrangements, for a particular one, or for the sum over all of them? | |
| May 12, 2019 at 20:32 | comment | added | Shimrod | @LSpice, no, they just indicate the four different $x$-coordinates in some order. In the second relation, the $j$ is used for both the $j$ invariant and an index, $j$ is the $j$ invariant whenever it is not a subscript. | |
| May 12, 2019 at 20:23 | comment | added | LSpice | Do the free variables $i$, $j$, $k$, and $l$ in \eqref{1} (currently (1)) indicate summation? Is the $j$ on the left-hand side of \eqref{2} (currently (2)) the same as the index $j$ on the right-hand side? | |
| May 11, 2019 at 21:14 | answer | added | François Brunault | timeline score: 5 | |
| May 10, 2019 at 20:08 | answer | added | David E Speyer | timeline score: 11 | |
| May 10, 2019 at 18:24 | comment | added | Abdelmalek Abdesselam | I should also add, essentially you would be computing the ternary resultant of two cubics and a line which reduces to a binary resultant of two binary cubics and expressible by a $6\times 6$ Sylvester determinant. I think it should be doable by hand. | |
| May 10, 2019 at 18:15 | comment | added | Abdelmalek Abdesselam | Just an idea for 1): one could use a little bit of classical invariant theory of ternary cubics and compute the Hessian of the cubic plane curve $E$. The 8 points of order three together with the point at infinity are the intersections of the curve with its Hessian (i.e., the nine inflection points). | |
| May 10, 2019 at 15:48 | history | asked | Shimrod | CC BY-SA 4.0 |