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

Question

The reference I am using is Hoffmann - 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 $$\operatorname{Bun}_{r,d}(S) = \{ \mathcal{E} \text{ vector bundles on } C\times_k S \text{ of rank $r$ and degree $d$ } \}/{\backsim}$$ 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 $$ f^* : \operatorname{Bun}_{r,d}(S) \to \operatorname{Bun}_{r,d}(T) $$ with $$ [\mathcal{E}] \mapsto [f^* \mathcal{E}]. $$ Question 1: Should it not be $$ f^* : \operatorname{Bun}_{r,d}(T) \to \operatorname{Bun}_{r,d}(S) \,?$$ 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) $$ \operatorname{Bun}_{r,d}(-) : \text{Schemes over $k$} \to \text{Sets} $$ 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 $\operatorname{Bun}_{r,d}(-)$.

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 $\operatorname{Bun}_{r,d}(-)$ 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 by 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 these two objects behave completely differently under gluing but since I do not see their functorial isomorphism I do not see the author's 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.

Answer 1

This is more of a long comment:

Question 1: Shouldn't it be

$$f^*:\operatorname{Bun}_{r,d}(T)\to\operatorname{Bun}_{r,d}(S)?$$

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

The functor is contravariant, as you point out, so we actually want $\operatorname{Bun}_{r,d}(-) : \mathsf{Sch}_k^\text{op} \rightarrow \mathsf{Set}$ 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 $\operatorname{Bun}_{r,d}(-)$. 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:

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

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

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

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 $^2$.

Answer 2

Somehow this question came back up after many years. I didn't think the answers really got to the essence of the situation.

Asking for a fine moduli space is asking for a several different things at once that are related but not quite the same. So not only are the nontrivial automorphisms a problem but also that isomorphisms between bundles can degenerate. I'll leave it to a licensed Algebraic Geometer to fill out the requirements for being a fine moduli space. I think it is instructive to explain that even in the simplest nontrivial example, rank 2 bundles on $\mathbb{P}^1$ the situation is already subtle. A key phenomenon is that a naive moduli space (isomorphism classes of bundles with a topology that forces holomolorphic (or algebraic) families of bundles is to give rise to a continuous maps to iso classes) typically is not Hausdorff. To see this consider rank two bundles on $\mathbb{P}^1$ (working over $\mathbb{C}$). Suppose further that the degree of the bundle is zero. By the Dedekind-Weber, del Pezzo-Bertini, Birkhoff, Grothendieck theorem such bundle is isomorphic to a sum $\cal{O}(d)\oplus \cal{O}(-d)$ (say $d\ge 0$ to rid ourselves of some ambiguity). So you great the set of isomorphism classes is parametrized by $\mathbb{Z}_{\ge 0}$. Viewing $\mathbb{P}^1$ as $\mathbb{C}\cup \infty$ with coordinate $z$ around $0$ and $1/z$ around infinity you can write the transition function for such bundle as

\begin{pmatrix} z^{-d} & 0 \\ 0 & z^{d} \end{pmatrix}

which admits the deformation \begin{equation}g_{d,t}=\begin{pmatrix} z^{-d} & 0 \\ t & z^{d} \end{pmatrix}.\end{equation}

Define a new trivialization (when $t\ne 0$) away from $\infty$ by \begin{equation} \{\begin{pmatrix} z^d\\-t \end{pmatrix}, \begin{pmatrix} 1 \\0\end{pmatrix}\}. \end{equation} This gets mapped under $g_{d,t}$ to \begin{equation} \{\begin{pmatrix}1 \\ 0 \end{pmatrix},\begin{pmatrix} z^{-d} \\ t \end{pmatrix}\}. \end{equation} This is trivialization that extends over $\infty$ (when $t\ne 0$). Thus the family of bundles with transition function $g_{d,t}$ gives the trivial bundle for $t\ne 0$ and $\cal{O}(d)\oplus \cal{O}(-d)$ for $t=0$. This shows that the closure of the trivial bundle contains every bundle. More generally the closure of $\cal{O}(d)\oplus \cal{O}(-d)$ contains $\cal{O}(e)\oplus \cal{O}(-e)$ for all $e \ge d$.

Answer 3

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 $S$.