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?


1 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").

  • $\begingroup$ 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 $H^1(F)=0$? $\endgroup$
    – user267839
    Nov 26, 2020 at 2:12
  • $\begingroup$ 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? $\endgroup$
    – Bernie
    Nov 26, 2020 at 12:19
  • $\begingroup$ 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? $\endgroup$
    – user267839
    Nov 27, 2020 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.