Timeline for Confusion regarding the invariant rational functions
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 4, 2023 at 14:11 | comment | added | Friedrich Knop | It should still be $n\chi$ because there might not be enough semiinvariants of weight $\chi$. Possibly the simplest example is $X=\mathbf A^2$ and $G=\mathbf G_m$ acting by $t\cdot(x,y)=(tx,t^2y)$. Take $\chi(t)=t$. Then all points are $\chi$-stable except for the origin and $f=x^2/y$ is a counterexample. So there is a glitch in Mukai's book. The last sentence of the Proposition is not affected, though. | |
| Dec 1, 2023 at 19:02 | comment | added | It'sMe | @FriedrichKnop But say there exists semiinvariants of weight $\chi$, then is the statement of the proposition true? Or should it still be $n\chi$? And, in this case is the $n$ fixed or does $n$ depend on the invariant rational function that I choose? | |
| Dec 1, 2023 at 9:07 | comment | added | Friedrich Knop | I think you are right. There might not even exist semiinvariants of weight $\chi$. | |
| Dec 1, 2023 at 3:12 | history | edited | It'sMe | CC BY-SA 4.0 |
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| Nov 28, 2023 at 23:41 | history | edited | It'sMe | CC BY-SA 4.0 |
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| Nov 28, 2023 at 22:50 | history | asked | It'sMe | CC BY-SA 4.0 |