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Dec 12, 2019 at 20:58 comment added prochet The point is precisely that I don't want to modify the $\mathbb{A}^{1}$-factor.
Dec 12, 2019 at 20:12 comment added Piotr Achinger Re: @prochet's follow-up question. I doubt it, but an argument would have to be of global nature. There are moduli spaces of semistable reflexive sheaves on $\mathbb{P}^n$, and not all of these sheaves are locally free for $n>1$; if you take a curve $C$ in the moduli space whose intersection with the non-locally free locus is a finite set $S$, then your question with $C\setminus S$ instead of $\mathbb{A}^1$ will likely have a negative answer. But it is not obvious how to ensure $C\setminus S \simeq \mathbb{A}^1$.
Dec 12, 2019 at 14:43 comment added Minseon Shin I don't know, and I would also like to know.
Dec 12, 2019 at 13:44 comment added prochet and does it also hold for vector bundles over $\mathbb{P}^{n}\times\mathbb{A}^{1}$?
Dec 12, 2019 at 11:46 history answered Minseon Shin CC BY-SA 4.0