It is well known that any smooth curve 
$C\in |\mathcal{O}\_{\mathbf{P}^1\times\mathbf{P}^1}(2,2)| $ has geometric genus equal to 1, so its isomorphism class is determined by its $j$-invariant. Nevertheless we have that $\dim(\mathbf P(H^0(\mathcal{O}\_{\mathbf{P}^1\times\mathbf{P}^1}(2,2))))=8$ and $\dim(\operatorname{Aut}(\mathbf P^1\times\mathbf P^1))=6$. So the dimension of the GIT quotient
$$
\mathbf P(H^0(\mathcal{O}\_{\mathbf{P}^1\times\mathbf{P}^1}(2,2)))//\operatorname{Aut}(\mathbf P^1\times\mathbf P^1)
$$
is 2. If we assume that the base field has characteristic 0, then by Castelnuovo theorem the quotient above is a rational surface. So, at least for a dense subset, there should be two numbers $I\_1,I\_2$ parametrizing the orbits of the action of $\operatorname{Aut}(\mathbf P^1\times\mathbf P^1)$.


My questions are:


1) Are these numbers known (in terms of the coefficients of the defining polynomial of $C$)?


2) How are they related to the $j$-invariant?


Thanks in advance.



Moduli of smooth curves in $|\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(2,2)| $ and their invariants
[**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.


