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S Apr 26, 2021 at 21:08 history suggested tinlyx CC BY-SA 4.0
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Apr 26, 2021 at 18:25 review Suggested edits
S Apr 26, 2021 at 21:08
Nov 7, 2020 at 12:54 comment added Mark Wildon I’m voting to close this question because it appears to have been answered to the poster's satisfaction in a comment.
Nov 5, 2020 at 6:30 comment added Kim @abx Now I got it. Thank you!!
Nov 4, 2020 at 14:44 comment added abx Use the surjective homomorphism $V\otimes _{\mathbb{C}}\mathscr{O}_{\mathbb{P}}\rightarrow \mathscr{O}_{\mathbb{P}}(1)$. The natural $\operatorname{GL}(V) $-linearization of $V\otimes _{\mathbb{C}}\mathscr{O}_{\mathbb{P}}$ induces one on $\mathscr{O}_{\mathbb{P}}(1)$.
Nov 4, 2020 at 14:02 comment added Kim @abx Can you give any more details about the "Conversely" part? Thanks!
Nov 4, 2020 at 12:58 comment added abx For any line bundle $L$ on a variety $X$, a $G$-linearization induces a linear action of $G$ on $H^0(X,L)$. Conversely, a linear action of $G$ on $V$ induces an action on $\mathbb{P}(V)$ with a $G$-linearization of $\mathscr{O}_{\mathbb{P}}(1)$.
Nov 4, 2020 at 9:23 review Suggested edits
Nov 4, 2020 at 19:34
Nov 4, 2020 at 9:02 comment added Kim @abx It seems that $H^0(\mathbb{P}(V), \mathcal{O}(1))=V$ in this book... I add the picture of the proof of Proposition 1.7 in question.
Nov 4, 2020 at 9:01 history edited Kim CC BY-SA 4.0
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Nov 4, 2020 at 7:48 review Close votes
Nov 7, 2020 at 19:33
Nov 4, 2020 at 7:30 comment added abx Are you aware that $H^0(\mathbb{P}(V),\mathscr{O}_{\mathbb{P}}(1))=V^*$? That helps...
Nov 4, 2020 at 2:04 history edited Kim CC BY-SA 4.0
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Nov 4, 2020 at 2:02 comment added Kim @abx Sorry, I mean the dual action.
Nov 3, 2020 at 9:15 comment added abx What is a coaction? The term does not appear in Mumford's book.
Nov 3, 2020 at 7:56 review First posts
Nov 3, 2020 at 8:32
Nov 3, 2020 at 7:48 history asked Kim CC BY-SA 4.0