Timeline for Moduli of smooth curves in $|\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(2,2)| $ and their invariants
Current License: CC BY-SA 4.0
10 events
when toggle format | what | by | license | comment | |
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May 12, 2020 at 9:01 | vote | accept | DDT | ||
May 8, 2020 at 1:22 | answer | added | Noam D. Elkies | timeline score: 11 | |
May 7, 2020 at 22:40 | comment | added | Abdelmalek Abdesselam | @NoamD.Elkies: I guess it makes more sense to index the $J$'s by their degree...:) | |
May 7, 2020 at 22:21 | comment | added | Noam D. Elkies | The ring of invariants is ${\bf C}[J_2,J_3,J_4]$ where $J_k$ (each $k=2,3,4$) is a polynomial invariant of degree $k$. The Jacobian of $C$ is $y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2$, with a visible rational point $(x,y) = (0,J_3)$. This must be classical but it's more fun to work it out than to try to hunt it down in the literature. I'll post one approach soon. | |
May 7, 2020 at 21:27 | comment | added | Abdelmalek Abdesselam | If $G(x)$ is a binary quartic in $x=(x_1,x_2)$, one can define a biform $F$ by polarizing twice, i.e., applying $(y\cdot\partial_x)^2$ to $G$. Pulling back a $SL_2\times SL_2$ invariant of the resulting biform $F(x,y)$ gives an $SL_2$ invariant of $G$. This could help in finding the connection to $j$ invariants of elliptic curves. | |
May 7, 2020 at 21:11 | comment | added | Abdelmalek Abdesselam | There is a single (up to scale) invariant $J_1$ in degree two, and $J_2$ in degree three. I still have to think about $J_3$. | |
May 7, 2020 at 21:02 | comment | added | Abdelmalek Abdesselam | I added the tag "classical invariant theory" which is the one most relevant to the question, I think. That's because what is asked is to figure out some explicit generators for the ring of invariants of bihomogeneous forms $F(x,y)$ of bidegree $(2,2)$ in $x=(x_1,x_2)$ and $y=(y_1,y_2)$ under the group $SL_2\times SL_2$. I suspect one should be able to construct three explicit invariants $J_1,J_2,J_3$ and take for $I_1$ and $I_2$ suitable ratios of powers of these invariants to make them absolute. | |
May 7, 2020 at 20:58 | history | edited | Abdelmalek Abdesselam |
edited tags
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May 7, 2020 at 16:03 | comment | added | Noam D. Elkies | Geometrically we can see not just a $j$-invariant but also a divisor class on Jac$(C)$, coming from the difference of the degree-$2$ classes ${\mathcal O}(1,0)$ and ${\mathcal O}(0,1)$. Conversely, given a genus-$1$ curve $C$ and two degree-$2$ divisors $D_1,D_2$ you can recover the embedding $\iota: C \to {\bf P}^1 \times {\bf P}^1$ from the two degree-$2$ maps $f_i : C \to {\bf P}^1$ obtained from sections of $D_i$. $$ $$ Somebody else will probably post a description or reference to the invariant theory of $(2,2)$ forms before I have the chance to work it out or look it up. | |
May 7, 2020 at 15:29 | history | asked | DDT | CC BY-SA 4.0 |