$\DeclareMathOperator\SL{\mathsf{SL}}$ Let $k$ be a field, let's suppose it's algebraically closed for convenience (take $\operatorname{char}k=0$ if you wish). Consider the (right) group action of $\SL_2(k)$ on $k[x,y]$ defined by $$g:P\begin{pmatrix}x\\y\end{pmatrix}\mapsto P\left(g\cdot \begin{pmatrix}x\\y\end{pmatrix}\right)$$ Restrict it to homogeneous polynomials of fixed degree $d$ and take the $4$th tensor power to obtain an action on polynomials in $x=(x_i)_{i\in\{1,2,3,4\}},y=(y_i)_{i\in\{1,2,3,4\}}$ that are homogeneous of degree $d$ in each pair $(x_i,y_i)$ (so of total degree $4d$).
Question: What are the invariants under this action? Is there any literature on this?
I conjecture that they are polynomials (homogeneous, of degree $d$) in the $\begin{vmatrix}x_i&x_j\\y_i&y_j\end{vmatrix}\cdot\begin{vmatrix}x_k&x_l\\y_k&y_l\end{vmatrix}$ where $(i,j,k,l)$ can be any permutation of $(1,2,3,4)$. Actually the polynomials $s_0,s_1,s_2$ for $(1,2,3,4),(1,3,4,2),(1,4,2,3)$ sum to zero, so it is equivalent to show that all invariants are polynomials in e.g. $s_0,s_1$. These aren't just any polynomials, in fact $-s_0/s_1$ descends to $(\mathbb P^1)^4\to\mathbb P^1$ as the cross ratio. More context below.
Ideas: Write an invariant polynomial using multi-indices as $$P=\sum_{a\in\mathbb N^4}c_ax^ay^{d-a}$$ where I denote $d=(d,d,d,d)$ for brevity. Applying $\begin{pmatrix}t&0\\0&t^{-1}\end{pmatrix}\in\SL_2$ with $t$ not an $(|a|-|d-a|)$th root of unity gives $c_{a}=0$ for $|a|\neq 2d$. So the only nonzero coefficients are for $|a|=|d-a|=2d$.
This shows that $P$ is a polynomial in the $x_ix_jy_ky_l$ with $i,j,k,l$ pairwise distinct (but the expression is not unique). This comes closer to the $s_0,s_1$ above.
Applying $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ gives $c_{a}=c_{d-a}$.
Applying $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ gives many more interesting relations: $$\sum_{a\in\mathbb N^4}c_{a}x^ay^{d-a}=\sum_{a\in\mathbb N^4}c_{a}(x+y)^ay^{d-a}$$ and the fact that $c_a=0$ for $|a|\neq2d$ e.g. gives $\sum c_a=0$ (if $d>0$). There is more, if we take $a\in\mathbb N^4$ with nonzero $c_a$ and $a_1$ minimal, the fact that $c_a$ is the only one who contributes to the coefficient of $x^{a-(1,0,0,0)}y^{d-a+(1,0,0,0)}$ in the RHS shows that $a_1=0$ (in $k$, or in $\mathbb Q$ if $\operatorname{char} k=0$). If $\operatorname{char} k=0$ and among the nonzero $c_a$ with $a_1=0$ we choose the $\alpha$ with $a_2$ minimal, we deduce $a_2c_\alpha=c_{\alpha+(1,-1,0,0)}$ because only two terms contribute to the coefficient of $x^{\alpha-(0,1,0,0)}y^{d-\alpha+(0,1,0,0)}$ in the RHS.
We can do much more with the above identity. I was thinking to do something with extremal values as above to hopefully split of a factor like $s_0$, and use the following
Conjectured lemma. If $s_0P+Q$ is $\SL_2$-invariant and $s_0$ does not divide $Q$, then $Q$ (and $P$) are invariant.
to proceed with induction.
Context: I wanted to characterize them in order to show that the cross-ratio $(\mathbb P^1)^4\to\mathbb P^1$ is the categorical quotient (in algebraic varieties over $k$) of $(\mathbb P^1)^4$ with the (diagonal) action of $\SL_2$ ("homographies"). After some work, the conjecture would imply this.