Timeline for Construction of an atlas for the moduli stack $\mathcal{Bun}_X^{n,d}$ in F. Neumann's 'Algebraic Stacks and Moduli of Vector Bundles'

5 events

when toggle format	what		by	license	comment
Nov 27, 2020 at 1:02	vote	accept	user267839
Nov 27, 2020 at 0:50	comment	added	user267839		Oh yes sorry, I think I understand your point now. Since $F$ is loc free (so flat) the canonical $R^1(pr_2)_*F \otimes k(u) \to H^1(F_u)$ is surjection and to show that $r_m$ is smooth with respect arbitrary point $U \to R_m$ we check it fiberwise for every $u \in U$ simply by definition of smoothness, that's it, right?
Nov 26, 2020 at 12:19	comment	added	Bernie		No in this case we have $R^1(pr_2)_{*}F=0$ which implies $H^1(F_u)=0$ for the sheaves on the fiber over a point $u\in U$. As I said, I don't really know what is meant here, only some ideas. But look at arXiv:1602.05267 on the beginning of p.17. There the authors do it exactly as I think. As one of the authors is Neumann, maybe you can just ask him, what is meant in the book?
Nov 26, 2020 at 2:12	comment	added	user267839		Now I'm a bit confused about your argument on $H^1(F)=R^1(pr_2)_{}F=0$. Clearly if $U$ is a spectrum of a field $k$, then $pr_2:= \pi: X_k \to Spec(k)$ is the canonical structure map and the global section functor coinsides with push-forward: ie $\Gamma((X,-) = \pi_$, so indeed $H^i(X,F)= R^i \pi_* F$ as you wrote. But what if $U$ is an arbitrary abstract point of $R_m$, ie a map $f: U \to R_m$ where $U$ is an arbitrary scheme. Then $\Gamma(X,-) = (pr_2)_*$ is no longer true, right? Why we nevertheless can conclude that $H^1(F)=0$?
Nov 25, 2020 at 11:02	history	answered	Bernie	CC BY-SA 4.0