Timeline for Help with Macaulay2 computation of invariant ring
Current License: CC BY-SA 4.0
8 events
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| Aug 21, 2023 at 20:33 | comment | added | Jason Starr | 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. | |
| Aug 21, 2023 at 20:15 | comment | added | It'sMe | @JasonStarr The invariant ring in this case is a polynomial ring in three variables. You calculate $det(X+sY)$, where $s$ is just some variable. You'll get a quadratic polynomial in $s$. The coefficients of $1,s,s^2$ are the three generators of this invariant ring. Kindly have a look at Example 10.5.2, pg-200, in "An Introduction to Quiver Representations" by Harm Derksen and Jerzy Weyman. I'm looking at dimension vector (2,2). That's why I'm confused about what am I doing wrong in the code. | |
| Aug 21, 2023 at 11:37 | comment | added | Jason Starr | 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. | |
| Aug 20, 2023 at 4:52 | comment | added | Zach Teitler |
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.)
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| Aug 20, 2023 at 1:12 | history | edited | It'sMe | CC BY-SA 4.0 |
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| Aug 20, 2023 at 1:09 | comment | added | It'sMe | @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. | |
| Aug 20, 2023 at 1:04 | comment | added | Zach Teitler | What's that ideal? Modulo $I$, you have $t=1$. Is that what you intended? | |
| Aug 20, 2023 at 0:52 | history | asked | It'sMe | CC BY-SA 4.0 |