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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
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