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

Question

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?

Answer 1

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").