The reference I am using is Norbert Hoffmann's The moduli stack of vector bundles on a curve . The question is about the moduli space of vector bundles. I am trying to understand why the fine moduli scheme does not exist. Let $C$ a projective curve. Let $S$ be a $k$-scheme then we denote by $$\text{Bun}_{r,d}(S) = \{ \mathcal{E} \text{ vector bundles on } C\times_k S \text{ rank $r$ and degree $d$ } \}/\backsim$$ the set of isomorphism classes of vector bundles $\mathcal{E}$ on $C \times_k S$. Now, every morphism of $k$-schemes $f:T\to S$ induces a pullback map $$ f^* : \text{Bun}_{r,d}(S) \to \text{Bun}_{r,d}(T) $$ with $$ [\mathcal{E}] \mapsto [f^* \mathcal{E}]. $$ Question 1: Should it not be $$ f^* : \text{Bun}_{r,d}(T) \to \text{Bun}_{r,d}(S) \,?$$ Thus we get the contravariant functor $$ \text{Bund}_{r,d}(-) : \text{Schemes over $k$} \to \text{Sets} $$ from the category of schemes to the category of sets. Then, we have the definition of the fine moduli scheme.

Definition A scheme $M$ over $k$ is a fine moduli scheme for vector bundles (of rank $r$ and degree $d$) on $C$ if $M$ represents the functor Bun$_{r,d}(-)$.

Question 2: What does this requirement actually mean? I.e. that a scheme represents a functor as above?

More explicitly, the author continues, and this is where I get confused mostly, $M$ is a fine moduli scheme of vector bundles if there exists the following functorial bijection: $$ \{ \phi : S \to M \text{ a $k$-morphism} \} = \{ \mathcal{E} \text{ vect. bundle of rank $r$ and degree $d$ } \} /\backsim $$

Question 3 How exactly can I understand this equality? It is not quite clear what the objects a morphism in both sides are and why there is some isomorphism between them.

Finally, the whole point is to show that $M$ does not represent the functor Bund$_{r,d}(-)$ which actually is not representable (thus the need for the moduli stack). To show this the author uses the gluing example. In specific

  • for any $k$-scheme $M$ a $k$-morphism $\phi : S \to M$ is given b a $k$-morphism $\phi_i:U_i \to M$ such that in intersection $U_{ij}=U_i \cap U_j$ we have $\phi_i=\phi_j$.
  • a vector bundle $\mathcal{E}$ over $C \times_k S$ is given by a vector bundle $\mathcal{E}_i :C \times_k U_i $, $U_i \subset \mathcal{E}$,for each $i$, an isomorphism $a_{il} = \mathcal{E}_i \to \mathcal{E}_j$ in the intersection, and the cocycle condition $a_{il} = a_{jl} \circ a_{ij}$ on triple intersections.

The author says that the these two objects behave completely differently under gluing but since I do not see their functorial isomorphism I do not see the authors point.

Question 4 Would you be able to clear this point out and explain it?

  • $\begingroup$ The first three are more elementary. What's confusing about the first? Do you understand how pullback of sheaves works? For the second two you should read about and prove the basic forms of the Yoneda lemma. I'm sure searching will turn up something helpful. You can also read the start of the second edition of EGA I. $\endgroup$
    – Hoot
    Jun 14, 2016 at 17:30
  • $\begingroup$ For the last one I think it's a heuristic rather than a complete argument. I think it would be best for your purposes to actually get a concrete example where this fails. This isn't my area but I'll try to think of something. There might be something in Le Potier. $\endgroup$
    – Hoot
    Jun 14, 2016 at 18:13
  • $\begingroup$ We should link the MO question too. mathoverflow.net/questions/242191/… $\endgroup$
    – Hoot
    Jun 14, 2016 at 18:14
  • $\begingroup$ Aha, I think I found something. You never said anything about stability so the news is even worse: there is not even a coarse moduli space. See 11.32 of Mukai, Introduction to Invariants and Moduli. Hopefully I'm interpreting his words correctly. $\endgroup$
    – Hoot
    Jun 14, 2016 at 18:21
  • $\begingroup$ @Hoot in the reference I provide Bun$_{r,d}$ works, apparently, for vector bundles independently of wether we impose stability or not. Also, I am a physicist, so no, the first three are not elementary at all. I understand that if the functor is contravariant it works as it is written but I do not know why it is a contravariant functor. The basic problem is, though, I do not see how exactly we can compare the equality of the main question and in turn make conclusion about the gluing. They seem very different objects. $\endgroup$
    – Marion
    Jun 14, 2016 at 22:18

1 Answer 1


I'm worried that I'm going to rewrite a Wikipedia article but let's try this. The general setup is this: one has a category $\mathscr{C}$ and a contravariant functor $F$ from $\mathscr{C}$ to the category of sets. $F$ is representable if there is an object $M$ of $\mathscr{C}$ and an isomorphism of functors $\Theta\colon h_M \to F$, where $h_M = \operatorname{Hom}_{\mathscr{C}}(-, M)$. Great.

Now, Yoneda tells you something interesting. The morphism $\Theta$ is determined by the element $\xi = \Theta_M(1_M)$ of $F(M)$. Why? One checks that for any object $X$ we're forced to have the bijection $\Theta_X\colon \operatorname{Hom}(X,M) \to F(X)$ given by $\Theta_X(f) = F(f)(\xi)$.

In your case, the requirement is that there is some scheme $M$ and a vector bundle $\mathscr{U}$ of rank $r$, degree $d$ on $M \times C$. So you also have to construct this bundle. From any morphism $f\colon X\to M$ we get a bundle $(f \times 1_C)^*\mathscr{U}$ on $X \times C$, and we have to get each bundle on $X \times C$ from a unique $f$.

  • $\begingroup$ thanks for your answer. I am still confused on why functorially a morphism of schemes $\phi: M \to S$ is bijective to the isomorphism of vector bundles $\mathcal{E}$ on a curve $C$ times $S$. I am not sure I can phrase my confusion better since it seems I am quite confused indeed. $\endgroup$
    – Marion
    Jun 15, 2016 at 11:50
  • $\begingroup$ @Marion I wrote down what the bijection is. To prove that it really is a bijection you need to construct $M$. That's less formal. $\endgroup$
    – Hoot
    Jun 15, 2016 at 15:54

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