Let $k$ be a field and let $\operatorname{SL}\_2(k)$ act on $k[x\_1,x\_2]$ and $k[y\_1,y\_2]$ in the usual ways. These actions induce an action on the tensor product $k[x\_1,x\_2,y\_1,y\_2]$ that preserves the subspace $k[x\_1,x\_2,y\_1,y\_2]\_{s,k}$ of polynomials that are homogeneous of degree $s+k$ with total $x\_i$ degree $s$ and total $y\_i$ degree $k$. I think these are sometimes said to be of bidegree $(s,k)$, but I'm not entirely sure that's standard terminology.


A computation I've performed in a seemingly unrelated mathematical field has led me to believe that for all $d \geq 0$, there should be a nonzero $\operatorname{SL}\_2(k)$-invariant polynomial in $k[x\_1,x\_2,y\_1,y\_2]\_{d,d}$ that is unique up to scaling.


**Question**: Assuming I'm right, how can I go about writing this polynomial down explicitly?



$\operatorname{SL}_2(k)$ invariant polynomials in $k[x_1,x_2,y_1,y_2]$
The polynomial you gave in the [comments](https://mathoverflow.net/questions/372617/operatornamesl-2k-invariant-polynomials-in-kx-1-x-2-y-1-y-2#comment942654_372617), $x\_1y\_2 - y\_2 x\_1$, after correcting the typo to $x\_1 y\_2 - x\_2 y\_1$, is invariant under $\operatorname{SL}\_2$.


Proof: It's the determinant of
$$ \begin{pmatrix} x\_1 & y\_1 \\ x\_2 & y\_2 \end{pmatrix}$$ and determinants are invariant under left multiplication by matrices of determinant $1$.


It indeed generates the ring of invariants. You can check this using representation theory (bidegree $s, k$ polynomials form the representation $\operatorname{Sym}^s \otimes \operatorname{Sym}^k$ of $\operatorname{SL}\_2$, and because $\operatorname{Sym}^j$ is irreducible this has one invariant if $s=k$ and $0$ otherwise) or by observing that any two nonzero matrices with the same determinant are equal up to the action of $\operatorname{SL}\_2$.


The same idea can be used to find the $\operatorname{SL}\_n$-invariants in the tensor product of $n$ copies of $k[x\_1,\dotsc,x\_n]$.


