I'm reading Frank Neumann's ["Algebraic Stacks and Moduli of
Vector Bundles"](https://impa.br/wp-content/uploads/2017/04/PM_36.pdf) 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?



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 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").


