# The non-existence of the fine moduli scheme of vector bundles. Why?

The reference I am using is this one. 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 (only because we have arrow reversal, but how do we know this? I.e. this is the same as Question 1) $$\text{Bund}_{r,d}(-) : \text{Schemes over k} \to \text{Sets}$$ 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?

I asked this question on Math.StackExchange but nobody answered.

## 2 Answers

This is more of a long comment:

Question 1: Shouldn't it be

$f^∗:Bun_{r,d}(T)→Bun{r,d}(S)$

No. You can't push vector bundles forward along general morphisms.

The functor is contravariant, as you point out, so we actually want $Bun_{r,d}(-) : \mathsf{Sch}_k^{op} \rightarrow \mathsf{Set}$ which agrees with the direction the pull-back functor goes.

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

Yes it means exactly that, e.g. there is a scheme $M/\operatorname{spec} k$ such that the Yoneda functor $h_M = \mathrm{Hom}_{\mathsf{Sch}_k}(-,M)$ is naturally isomorphic to the functor $Bun_{r,d}(-)$. That means that the set of maps of varieties from $X$ to $M$ exactly 'classifies' bundles of rank $r$ and degree $d$ on $X$. This idea, to my knowledge, originally comes from topology.

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.

The map comes from the Yoneda embedding and is essentially a very abstract tautology. We have a bijection which I'll lazily write as an equality:

$Bun_{r,d}(M) = \mathrm{Hom}_{\mathsf{Sch}_k}(M,M)$

But the $\mathrm{Hom}$ set has a distinguished element, the identity $\mathrm{id}_M$! That means that -- up to the ambiguity which might be involved in choosing a natural isomorphism -- we have a canonical bundle $E_{r,d}/M$ of rank $r$ and degree $d$. Furthermore if you go back and carefully look through a proof of the Yoneda embedding and apply it in this situation, you'll find out that the above actually describes the function

$\{\varphi:S\rightarrow M$ a $k$-morphism $\}=\{E$ vect. bundle of rank $r$ and degree $d\} / \sim$

exactly as the operation $\phi \mapsto \phi^* E_{r,d}$

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

I'm not sure if I can do this one! This is quite a subtle point, and my own point of view is poisoned$^1$ by the higher categorical language.

The point is that the Yoneda functor obtained from a scheme is a contravariant functor to the category of sets, i.e. a pre-sheaf. When we want to classify objects with automorphisms, the naive way to do it will often fail to be a sheaf when one appropriately generalizes the notion of a sheaf to include such things. Instead what one can do is consider this as a sheaf of 'groupoids', that is a category whose morphisms are all isomorphisms.

A set naturally gives a groupoid by considering each element as an object with no morphisms besides the identity morphisms.

Classically, a sheaf is a presheaf $\mathcal{F}$ which satisfies for every open cover $\{U_i \}$ of an open set $U$ we have that the sequence

$\mathcal{F}(U) \rightarrow \prod_{i} \mathcal{F}(U_i) \rightrightarrows \prod_{j,k} \mathcal{F}(U_j \cap U_k)$

is an equalizer diagram (the analogue of a kernel for $\mathsf{Set}$).

The analogous notion for groupoids is not so simple. It involves not only the pairwise intersections $U_j \cap U_k$, but the triple-wise intersections $U_j \cap U_k \cap U_l$ and a cocycle condition on them.

Now for the poison: groupoids are equivalently$^2$ described as topological spaces who have a potentially non-trivial $\pi_0$ and $\pi_1$ but whose higher homotopy groups are trivial, i.e. $\pi_i=0$ for $i>1$.

A classical sheaf is then$^3$ a presheaf $\mathcal{G}$ of spaces which have possibly nontrivial $\pi_0$ but no higher homotopy groups, and we have that

$\pi_0 \mathcal{G}(U) = \lim\Big\{ \prod_i \pi_0 \mathcal{G}(U_i) \rightrightarrows \prod_{j,k} \pi_0 \mathcal{G}(U_j \cap U_k) \Big\}$

To talk about sheaves of groupoids from a topological point of view we want that same above condition, but we want a condition relating the fundamental groups too! It is a general fact in topology that homotopy groups beyond $\pi_0$ do not behave well under limits in the category $\mathsf{Top}$, even fibered products can show this behavior.

$\pi_1 \mathcal{G}(U) = \lim\Big\{ \prod_i \pi_1 \mathcal{G}(U_i) \rightrightarrows \prod_{j,k} \pi_1 \mathcal{G}(U_j \cap U_k) \overset{\rightarrow}{\underset{\rightarrow}{\rightarrow}} \prod_{s,r,t} \pi_1 \mathcal{G}(U_s \cap U_r \cap U_t)\Big\}$

This triple intersection cocycle condition is what guarantees this property of the $\pi_1$'s.

$^1$: I'm being facetious

$^2$: This is more subtle than I'm letting on

$^3$: See $^2$

• Thanks a lot, I need to take some time to read this off. I will comment later. – Marion Jun 16 '16 at 10:42

The idea is that there is no fine moduli space if the objects in question have nontrivial automorphisms.

On representability: $M$ is a fine moduli space if there is a "universal" vector bundle $\mathcal{E}$ on $M$ such that morphisms $\phi: S \to M$ to $M$ and pulling back $\mathcal{E}$ along them gives all vector bundles $\phi^*\mathcal{E}$ on $S$.

• More precisely, $\mathcal E$ is a vector bundle on $C \times_k M$, namely corresponding to the identity morphism $\phi = \text{id}\!: M \to M$. – Thomas Poguntke Jun 15 '16 at 13:33
• If rank and degree are coprime, there exists a fine moduli space of stable vector bundles, even though the bundles all have non-trivial automorphisms. – euklid345 Jun 15 '16 at 16:13