Timeline for $\operatorname{SL}_2(k)$ invariant polynomials in $k[x_1,x_2,y_1,y_2]$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2020 at 12:20 | comment | added | Will Sawin | @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.) | |
| Sep 26, 2020 at 4:08 | comment | added | LSpice | @user124543, an invariant polynomial remains such after extending scalars (since $\operatorname{SL}_2(k)$ is generated by its intersection with $1 + \mathfrak{sl}_2(k)$), so there is no issue with assuming $k$ is infinite. This exact argument breaks down for $\operatorname{SL}_n$ when $\operatorname{char}(k) < n$, but maybe there's another argument in that case. | |
| Sep 26, 2020 at 3:49 | comment | added | user124543 | @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? | |
| Sep 25, 2020 at 23:29 | comment | added | Will Sawin | @LSpice Thanks! | |
| Sep 25, 2020 at 23:29 | comment | added | Will Sawin | @Helen Yes, a constant plus a multiple of. | |
| Sep 25, 2020 at 23:28 | comment | added | Helen | @WillSawin: Awesome, thanks! (ps: I assume you mean a constant plus a multple of, not a constant times a multiple of, right?) | |
| Sep 25, 2020 at 23:26 | vote | accept | Helen | ||
| Sep 25, 2020 at 23:22 | comment | added | LSpice | I'm scarcely in a position to complain since I couldn't even be bothered to do my 2D linear algebra, but you and the comment both accidentally used $x_1 y_2 - y_2 x_1$ when you clearly meant $x_1 y_2 - x_2 y_1$. I took the liberty of correcting this when I edited in a link to the comment; I hope that is all right. | |
| Sep 25, 2020 at 23:21 | history | edited | LSpice | CC BY-SA 4.0 |
Link to comment; minor TeXing
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| Sep 25, 2020 at 23:17 | comment | added | Will Sawin | @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. | |
| Sep 25, 2020 at 23:00 | comment | added | Helen | 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? | |
| Sep 25, 2020 at 22:56 | history | answered | Will Sawin | CC BY-SA 4.0 |