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

I'm reading Frank Neumann's "Algebraic Stacks and Moduli of Vector Bundles" and have some problems to understand a construction from the proof of:

Theorem 2.67. (page 81) The moduli stack $$\mathcal{Bun}_X^{n,d}$$ of vector bundles of rank n and degree $$d$$ on a smooth projective irreducible algebraic curve $$X$$ of genus $$g \ge 2$$ is an Artin algebraic stack which is smooth and locally of finite type.

The proof is long therefore I will quote only the relevant parts containing the steps in not understand. The whole proof can be looked up since the source is free available.

Proof. [...] Let us now describe the construction of an atlas for the moduli stack $$\frak{Bun}$$ $$_X^{n,d}$$. Let $$P_{n,d}$$ be polynomial

$$P_{n,d}(x) := nx + d + n(1 - g)$$

For every integer $$m$$ let $$P(m) = P{n,d}(m)$$ and consider the Quot scheme $$\operatorname{Quot} (\mathcal{O}_X^{P(m)}, P(x+m))$$ parametrizing quotient sheaves of $$\mathcal{O}_X$$-modules $$\mathcal{O}_X^{P(m)}$$ with prescribed Hilbert polynomial $$P_{n,d}$$. In general, a Quot scheme $$\operatorname{Quot}(\mathcal{F}, P)$$ is a fine moduli space for the moduli functor $$\frak{Quot}$$ $$(Sch/S)^{op} \to (Sets)$$
of the moduli problem of classifying quotient sheaves of $$\mathcal{O}_X$$-modules $$\mathcal{F}$$ with prescribed Hilbert polynomial $$P$$ and there exists a universal family of such quotient sheaves over the Quot-scheme $$\operatorname{Quot}(\mathcal{F}, P)$$.

For every integer $$m$$ we define an open subscheme

$$R_m \hookrightarrow \operatorname{Quot} (\mathcal{O}_X^{P(m)}, P(x+m))$$

by requiring that

(i) the quotient sheaves $$\mathcal{O}^{P(m)}_X \to \mathcal{F} \to 0$$ parametrized by $$R_m$$ are vector bundles, i.e. $$\mathcal{F}$$ is a locally free $$\mathcal{O}_X$$-sheaf.

(ii) for every $$U$$-point of $$R_m$$ defined by the family $$\mathcal{O}^{P(m)}_{X \times U} \to \mathcal{F} \to 0$$ we have that derived image $$R^1(pr_2)_* \mathcal{F} =0$$ and $$(pr_2)_*: \mathcal{O}^{P(m)}_{X \times U} \cong (pr_2)_* \mathcal{F}$$ is an isomorphism.

Induced by the universal family over $$\operatorname{Quot} (\mathcal{O}_X^{P(m)}, P(x+m))$$ we get now a universal family $$\mathcal{E}_{univ}$$ of vector bundles over $$X$$ of rank $$n$$ and degree $$d$$ parametrized by the subscheme $$R_m$$. Therefore we get a morphism of stacks

$$r_m: R_m \to \mathcal{Bun}_X^{n,d}.$$

From (ii) it follows (?) that if a point of $$R_m$$ is represented by a quotient sheaf of the form

$$0\to \mathcal{G} \to \mathcal{O}^{P(m)}_{X \times U} \to \mathcal{F} \to 0$$

then $$H^1(\mathcal{F} \otimes \mathcal{G}^{\vee}) =0$$ (?), which implies that $$r_m$$ is a smooth morphism. [...]

Question 1: Why (ii) implies $$H^1(\mathcal{F} \otimes \mathcal{G}^{\vee}) =0$$?

Question 2: Why $$H^1(\mathcal{F} \otimes \mathcal{G}^{\vee}) =0$$ implies that $$r_m$$ is smooth? $$H^1(\mathcal{F} \otimes \mathcal{G}^{\vee})$$ classifies extensions of $$\mathcal{F}$$ by $$\mathcal{G}$$. Why the conclusion that all extension are equivalent to the trivial $$\mathcal{G} \oplus \mathcal{F}$$ gives smoothness for $$r_m$$?

a note on question 1: $$(pr_2)_*$$ is a functor from $$\mathcal{O}_{X \times U}$$- modules to $$\mathcal{O}_U$$-modules, so $$\mathcal{O}^{P(m)}_{X \times U}$$ and $$(pr_2)_* \mathcal{F}$$ live in different categories. How does it make sense to talk about "isomorphism" $$(pr_2)_*: \mathcal{O}^{P(m)}_{X \times U} \cong (pr_2)_* \mathcal{F}$$ in (ii)? Does anybody see what the author has here in mind?

I don't know what is going on exactly (misprints?), but here are some ideas:

If you take a point of $$q\in R_m$$ (i.e. $$U=Spec(k)$$) defined by a sequence

$$0 \rightarrow G \rightarrow \mathcal{O}_X^{P(m)}\rightarrow F \rightarrow 0$$

then by (ii) we have $$H^1(F)=R^1(pr_2)_{*}F=0$$.

If you apply $$Hom(-,F)$$ to the exact sequence gives at the end of the long exact sequence:

$$Ext^1(\mathcal{O}_X^{P(m)},F)\rightarrow Ext^1(G,F)\rightarrow 0$$.

But $$Ext^1(\mathcal{O}_X^{P(m)},F)=H^1(F)^{P(m)}=0$$ so we get $$Ext^1(G,F)=H^1(F\otimes G^{\vee})=0$$.

Now $$Ext^1(G,F)=0$$ implies that the Quot scheme ist smooth at $$q$$, see e.g. the book of Huybrechts-Lehn (the chapter "Grothendieck's Quot-Scheme").

• 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 $H^1(F)=0$? – Ghost in Grothendieck universe Nov 26 '20 at 2:12
• 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? – Bernie Nov 26 '20 at 12:19
• 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? – Ghost in Grothendieck universe Nov 27 '20 at 0:50