Timeline for Reference for an algebraic group preserving a cubic form
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2014 at 12:51 | answer | added | Jérémy Blanc | timeline score: 3 | |
| May 13, 2014 at 20:11 | comment | added | Steven Sam | @abx: Thanks, I've edited my answer. | |
| May 13, 2014 at 17:28 | comment | added | abx | Yes, by choosing an appropriate origin. For instance if your curve is given in Weierstrass form $Y^2T=X^3+pXT^2+qT^3$, the symmetry is just $(X,Y,T)\mapsto (X,-Y,T)$. | |
| May 13, 2014 at 17:16 | comment | added | Steven Sam | Can those additional automorphisms be realized as linear transformations of the projective plane? | |
| May 13, 2014 at 6:55 | comment | added | Kenneth | @abx Right, so this $\mathbb{Z}/2\mathbb{Z}$ becomes $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/6\mathbb{Z}$ for $j(E) = 1728, 0$ (at least for $\mathrm{char}(k)=0$). | |
| May 13, 2014 at 4:45 | comment | added | abx | Except that Steven forgot that there are other automorphisms than translations... You have to add a term $\mathbb{Z}/2$ in any case (so that the quotient is $(\mathbb{Z}/3)^2\rtimes \mathbb{Z}/2$), and in some particular cases a slightly larger cyclic group. | |
| May 12, 2014 at 22:29 | comment | added | Kenneth | @abx Thanks for the clarification. So by Steven's comment below this quotient is actually $\mathbb{Z}/3 \times \mathbb{Z}/3$ (in the case $E$ is smooth and $k=\bar{k}$). | |
| May 12, 2014 at 22:23 | vote | accept | Kenneth | ||
| May 12, 2014 at 18:31 | comment | added | abx | I was talking about the group of projective automorphisms of a plane cubic. | |
| May 12, 2014 at 18:18 | comment | added | Kenneth | @abx I think the group of automorphisms of a smooth elliptic curve $E$ as an algebraic variety is infinite (at least when $k$ is infinite), it has the group automorphisms as well as translations. | |
| May 12, 2014 at 15:05 | answer | added | Steven Sam | timeline score: 4 | |
| May 12, 2014 at 13:02 | answer | added | Jim Humphreys | timeline score: 3 | |
| May 12, 2014 at 12:29 | comment | added | abx | I assume that your form $p$ is irreducible. Your group $G$ contains the central subgroup $\mathbb{G}_m\subset \mathrm{GL}_3$ of homotheties; the quotient is the group of projective automorphisms of the curve $E$ defined by $p=0$, and this is well-known to be finite. | |
| May 12, 2014 at 9:23 | answer | added | Alexander Premet | timeline score: 3 | |
| May 12, 2014 at 9:08 | review | First posts | |||
| May 12, 2014 at 9:21 | |||||
| May 12, 2014 at 8:51 | history | asked | Kenneth | CC BY-SA 3.0 |