Timeline for Is $U\subseteq X^{s}(L)$?
Current License: CC BY-SA 4.0
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| Nov 10, 2022 at 14:33 | comment | added | It'sMe | Thanks for your answer! | |
| Nov 10, 2022 at 10:30 | comment | added | Jason Starr | No that is not true. Let $k$ be a field. Let $n\geq 2$ be an integer. Let $G$ be $\mathbb{G}_m = \text{Spec}\ k[t,t^{-1}]$. Let $X$ be affine space $\mathbb{A}^n_k=\text{Spec}\ k[x_1,\dots,x_n]$. Let the action of $G$ on $X$ be the one pulling back each $x_i$ to $tx_i$ on $G\times X$. Let $L$ be the structure sheaf with the linearization for which the global section $1$ is $G$-invariant. Then $X^{ss}(L)$ equals all of $X$, i.e., the distinguished open affine of the invariant section $1$. The GIT quotient is $\text{Spec}(k)$. So $U=X\setminus\{0\}$ is a counterexample. | |
| Nov 10, 2022 at 5:23 | history | edited | It'sMe | CC BY-SA 4.0 |
added 11 characters in body
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| Nov 9, 2022 at 8:30 | history | asked | It'sMe | CC BY-SA 4.0 |