PS3 | Updated to version 4.85 issue

[EDITED to exhibit $j$ as a rational function of $J_2,J_3,J_4$, and to fix various local errors etc.]

The action of ${\rm SL_2} \times {\rm SL_2}$ on the $9$-dimensional space of $(2,2)$ forms has a polynomial ring of invariants, with generators in degrees $2,3,4$. If we write a general $(2,2)$ form $P(x_1,x_2;y_1,y_2)$ as $(x_1^2, x_1 x_2, x_2^2) M_3 (y_1^2, y_1 y_2, y_2^2)^{\sf T}$ where $M_3$ is the $3 \times 3$ matrix $$ M_3 = \left( \begin{array}{ccc} a_{00} & a_{01} & a_{02} \cr a_{10} & a_{11} & a_{12} \cr a_{20} & a_{21} & a_{22} \end{array} \right) \; , $$ then $J_k$ ($k=2,3,4$) can be taken to be the $x^{4-k}$ coefficient of the characteristic polynomial $\chi^{\phantom.}_{M_4}$ of the $4 \times 4$ matrix $$ M_4 = \left( \begin{array}{cccc} \frac12 a_{11} & -a_{10} & -a_{01} & 2 a_{00} \cr a_{12} & -\frac12 a_{11} & -2 a_{02} & a_{01} \cr a_{21} & -2 a_{20} & -\frac12 a_{11} & a_{10} \cr 2 a_{22} & -a_{21} & -a_{12} & \frac12 a_{11} \end{array} \right) . $$ This matrix is characterized by the identity $$ P(x_1,x_2;y_1,y_2) = (z_{11},z_{12},z_{21},z_{22}) M_4 (z_{22},-z_{21},-z_{12},z_{11})^{\sf T} $$ where each $z_{ij} = x_i y_j$, together with the requirement that $M_4$ has trace zero and becomes symmetric when its columns are listed in reverse order and columns $2,3$ are multiplied by $-1$. The invariants of degree $2$ and $3$ can also be written as $$ J_2 = -\frac12 a_{11}^2 + 2(a_{01} a_{21} + a_{10} a_{12}) - 4 (a_{00} a_{22} + a_{20} a_{02}), \quad J_3 = -4 \det M_3; $$ of course $J_4 = \det M_4$. The (Jacobian of the) genus-$1$ curve $P=0$ is isomorphic with the elliptic curve $$ y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2. $$ In particular this lets us compute the $j$-invariant of this curve as a rational function of $J_2,J_3,J_4$: $$ j = \frac{256 (J_2^2 + 12 J_4)^3}{16 J_2^4 J_4 - 4 J_2^3 J_3^2 - 128 J_2^2 J_4^2 + 144 J_2 J_3^2 J_4 + 256 J_4^3 - 27 J_3^4} \, . $$

One way to obtain these results is as follows. First compute the Hilbert series of the invariant ring. We find that it is $1 / \bigl( (1-t^2) (1-t^3) (1-t^4) \bigr)$; this suggests a polynomial ring of invariants with generators of degrees $2,3,4$, and shows that if we find independent invariants $J_2,J_3,J_4$ of those degrees then ${\bf C}[J_2,J_3,J_4]$ is the full invariant ring.

Now use the basis $\{z_{ij}\}$ of the four-dimensional space, call it $Z$, of sections of ${\cal O}(1,1)$; it is well-known that $\{z_{ij}\}$ embeds ${\bf P}^1 \times {\bf P}^1$ into ${\bf P}^3$ as the quadric $z_{11} z_{22} = z_{12} z_{21}$, identifying ${\rm SL_2} \times {\rm SL_2}$ with the special orthogonal group ${\rm SO}(Q)$ where $Q$ is the quadratic form $z_{11} z_{22} - z_{12} z_{21}$. This identifies $P$ with some other quadratic form $\tilde P$ in the $z_{ij}$, determined uniquely modulo ${\bf C} Q$.

Now $Q$ is nondegenerate, so it identifies $Z$ with its dual $Z^*$, and thus identifies quadratic forms on $Z$ with self-adjoint maps $T: Z \to Z$, with $Q$ itself mapping to the identity map. It is known that generically ${\rm SO}(q)$ orbits of such $T$ are determined by their spectrum, and thus by the characteristic polynomial $\chi^{\phantom.}_T$. There is a unique translate $\tilde P + cQ$ of trace zero, represented by the above matrix $M_4$. Hence the coefficients $J_2,J_3,J_4$ of $\chi^{\phantom.}_{M_4}$ are invariant and independent, as claimed.

To identify the elliptic curve, write $C$ as a double cover of one of the ${\bf P}^1$'s by taking the discriminant of $P$ with respect to the other ${\bf P}^1$, and then use classical formulas for the Jacobian of a genus-$1$ curve $y^2 = {\rm quartic}$. The formulas, though not pretty to look at, are short enough to let us identify the coefficients with polynomials in $J_2,J_3,J_4$. The resulting curve has a rational point whose $x$-coordinate is a multiple of $J_2$; translating $x$ to put this point at $x=0$ yields the model $y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2$ exhibited above. The visible rational point $(x,y) = (0,J_3)$ ought to correspond to the difference between the divisors ${\mathcal O(0,1)}_C$ and ${\mathcal O(1,0)}_C$, but I haven't checked this.