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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?

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  • $\begingroup$ By looking at the action of the diagonal torus, it seems to me that a fixed bidegree-$(1, 1)$ polynomial would have to be in the span of $x_1 y_2$ and $y_2 x_1$; but this span contains no non-$0$ $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$-fixed polynomial. What am I missing? $\endgroup$
    – LSpice
    Commented Sep 25, 2020 at 22:46
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    $\begingroup$ @LSpice: Doesn't $x_1 y_2 - y_2 x_1$ work, or is my calculation totally off? $\endgroup$
    – Helen
    Commented Sep 25, 2020 at 22:49
  • $\begingroup$ Of course you meant $x_1 y_2 - x_2 y_1$. Yes, it works. I'm not sure what I was thinking …. $\endgroup$
    – LSpice
    Commented Sep 25, 2020 at 23:19
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    $\begingroup$ @Lspice: Thanks for noticing that typo in my comment! I just can't seem to type without errors today... $\endgroup$
    – Helen
    Commented Sep 25, 2020 at 23:27

1 Answer 1

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The polynomial you gave in the comments, $x_1y_2 - y_2 x_1$, 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 $k[x_1,\dotsc,x_n]$.

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  • $\begingroup$ This is great, thanks! I understand the representation theory argument you give (or, rather, I understand what it is claiming and know where to look up the relevant facts). Can you say a little more about the alternate argument about matrices with the same determinant? $\endgroup$
    – Helen
    Commented Sep 25, 2020 at 23:00
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    $\begingroup$ @Helen We need to check that, if there's an invariant polynomial not in the ring generated by det, then it takes different values on two different matrices with the same determinant. For this you can use the fact that the ring of functions on matrices with determinant $1$ are the ring of functions modulo $x_1y_2-x_2y_1-1$. So if your function is constant on that space, it is equal to a constant times a multiple of $x_1y_2-x_2y_1-1$. Subtract the constant and divide by $x_1y_2-x_2y_1-1$, and it's still invariant, of lower degree. Now iterate / induct. $\endgroup$
    – Will Sawin
    Commented Sep 25, 2020 at 23:17
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    $\begingroup$ @LSpice Thanks! $\endgroup$
    – Will Sawin
    Commented Sep 25, 2020 at 23:29
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    $\begingroup$ @WillSawin: Are you assuming $k$ is infinite? Otherwise you can't check whether two polynomials are the same by evaluating them on matrices. Furthermore, is the modular representation theory of $SL_n(k)$ for finite $k$ that easy? $\endgroup$
    – user124543
    Commented Sep 26, 2020 at 3:49
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    $\begingroup$ @LSpice Over a finite field, I would consider invariants to be invariants of the algebraic group, and not just its group of rational points. On the algebraic group level this argument works fine, for instance, as LSpice suggested, because you can pass to points over a larger field. For the group of rational points, these invariants will remain invariants, but there will be further invariants (since the ring of invariants under a finite group action will have the same dimension as the original ring.) $\endgroup$
    – Will Sawin
    Commented Sep 26, 2020 at 12:20

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