Invariants of tuples of matrices under $\mathrm{GL}(p)\otimes \mathrm{GL}(q) \subseteq \mathrm{GL}(n)$?

Question

$\DeclareMathOperator\GL{GL}$Consider the $\GL(n)$ over some field of characteristic zero (I'm thinking of either the rationals, reals or complex) and the subgroup $\GL(p)\otimes \GL(q)$ which embeds irreducibly into $\GL(n)$.

It is known that the diagonal action of $g\in \GL(n)$ on d-tuples of matrices $(X_1,\ldots,X_d)\mapsto (gX_1g^{-1},\ldots,gX_dg^{-1})$ results in a ring of invariants generated by $Tr(X_{i_1}\cdots X_{i_k})$ for $i_j\in \{1,\ldots,d\}$. Similarly, it is known when the group is the orthogonal or symplectic group.

Question: Is there some known result when the group is $\GL(p)\otimes \GL(q)$? Specifically, I'm interested in the lowest dimensional non-trivial case, i.e. when $p=q=2$ and $n=4$.

As far as I understand the question is unfeasible to answer (computationally) when $G$ is some generic reductive group. However I would expect that this can be answered thanks to the tensor product structure and also assuming that both $p$ and $q$ are small.

I'm quite new to invariant theory so I'm not fully familiar with all classical results. Also, I have found some possible relevant question on MO: Invariant polynomials for a product of algebraic groups.

Answer 1

For $GL(p)\times GL(q)$ there are no nonconstant invariants as that group contains the scalars and no nonconstant polynomial is invariant under scaling. I guess you mean $SL(p)\times SL(q)$.

$SL(p) \times SL(q)$ for $p=q=2$ is the same as $SO(4)$ (well technically its spin double cover) so it reduces to the invariants of $SO(n)$ on tuples of vectors. For $d=1$ the invariants are generated by the quadratic form, and for higher $d$ I believe the invariants are generated by the pullback of that invariant together with, for $d \geq 4$, the determinants of $4$-tuples of vectors.