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Consider the algebraic group $G:=\operatorname{SL}_{2}\times\operatorname{SL}_{2}$ acting on $V:=\operatorname{Mat}_{2\times 2}\oplus\operatorname{Mat}_{2\times 2}$ via the action $(A,B)\,\cdot\,(X,Y)=(BXA^{-1},BYA^{-1})$. I'm interested in computing the invariant ring $\mathbb{K}[V]^{G}$. For this purpose I'm using the "InvariantRing" package in Macaulay2 (see here)and I did the following Below is the code But for some reason it is giving the wrong output. It returns the invariant ring to be trivial, which I know for sure is not the case. I repeated all of these over $\mathbb{C}$ and still got the same output.

I'm confused where is the mistake. I would highly appreciate your time and effort!!

For people who are familiar with quiver representations, I'm looking at Kronecker quiver with dimension vector (2,2).

Edit: I have done the above computation without taking $t$, i.e., by taking the ideal to be $(ps-rq-1, ad-bc-1)$ and without $t$ in the matrix, I still got the same output.

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  • $\begingroup$ What's that ideal? Modulo $I$, you have $t=1$. Is that what you intended? $\endgroup$
    – Zach Teitler
    Commented Aug 20, 2023 at 1:04
  • $\begingroup$ @ZachTeitler For defining the action, I need $1/det$, which is not in the polynomial ring, that's why I defined it like that. Just so you know I have done this computation without $t$, and I got the same output. $\endgroup$
    – It'sMe
    Commented Aug 20, 2023 at 1:09
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    $\begingroup$ That makes sense. I don't know how to answer your question, but can I suggest (1) the Macaulay2 discussion group might be more helpful, (2) have you tried a simpler example (simpler group action), (3) can you give an explicit invariant of this action and try the package command isInvariant to see what it says? (If it says that a known invariant isn't invariant, then either the package has a bug, or you might have to check the matrix $M$, or who knows what.) $\endgroup$
    – Zach Teitler
    Commented Aug 20, 2023 at 4:52
  • $\begingroup$ Why do you believe the invariant ring is larger than the constants? Certainly the invariant ring after inverting the determinants of $X$ and $Y$ is larger: the ring generated by the trace and determinant of $X^{-1}Y$ together with the inverse of the determinant. However, it seems to me that the only polynomials in this trace and determinant that come from invariants of the original ring are constants. $\endgroup$
    – Jason Starr
    Commented Aug 21, 2023 at 11:37
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    $\begingroup$ I missed that your group is a product of copies of $\textbf{SL}_2$, not $\textbf{GL}_2$. I agree, for the action of the subgroup $\textbf{SL}_2\times \textbf{SL}_2$, there are nonconstant invariants. Perhaps you should contact the maintainers of the "InvariantRing" package. $\endgroup$
    – Jason Starr
    Commented Aug 21, 2023 at 20:33

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