Timeline for 2x2-determinantal representations of cubic curves
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2017 at 19:32 | comment | added | Zach Teitler | ... Explicitly, a general cubic curve can be written in Hesse normal form as $f = x^3+y^3+z^3+6hxyz$ for some $h$, after a linear change of coordinates. The point $[-1:1:0] = V(x+y,z) \in V(f)$. Then $f = \det \begin{pmatrix} z^2+6hxy & -(x^2-xy+y^2) \\ x-y & z \end{pmatrix}.$ This leaves some non-generic curves to be handled individually. | |
| Jun 21, 2017 at 19:29 | comment | added | Zach Teitler | Wow, I was super terse this morning. In case any students come across this I will add bit of detail: If $V(\ell_1,\ell_2) \subset V(f)$, then $f \in (\ell_1,\ell_2)$. So we can write $f = q_1 \ell_1 + q_2 \ell_2$ for some $q_1,q_2$. By homogeneity, $q_1,q_2$ are quadratic forms. Then $$f = \det \begin{pmatrix} q_2 & -q_1 \\ \ell_1 & \ell_2 \end{pmatrix}$$ (minor rearrangement of indices). ... | |
| Jun 21, 2017 at 14:54 | vote | accept | Dima Pasechnik | ||
| Jun 21, 2017 at 14:52 | comment | added | Dima Pasechnik | Thanks, this explains why the difficulty starts with symmetric representations, or ones of size bigger than 2x2... | |
| Jun 21, 2017 at 14:45 | history | answered | Zach Teitler | CC BY-SA 3.0 |