We are taught that, in general:

A type of objects that has nontrivial automorphisms cannot have a fine moduli space.

The proof generally goes along the lines of:

Take an object $X$ with a non-trivial automorphism $\phi$. Form a twisted family on the circle $S^1$ by taking $[0,1]\times X$ and identifying $(0,x)$ with $(1,\phi(x))$ (in a category other than $\mathbf{Top}$ this might need some work, or be impossible). Then the resulting family is non-trivial (but see below).

Suppose this family corresponded to some morphism $f\colon S^1\to M$, where $M$ is a fine moduli space for the problem. But the fibres are all isomorphic, so the map $f$ must be constant. But that means that the pullback along $f$ is trivial, and so cannot be equal to our family, which is a contradiction.

There are a few problems which might occur, outlined at this section of nLab. But in many cases, the existence of automorphisms does prevent the existence of a moduli space.

But take this example: a family of oriented topological circles over a space $X$ is a fibre bundle $E\xrightarrow{p}X$ with fibre $S^1$, and a continuous choice of orientation for the fibres; i.e., a group action of $S^1$ on $E$ which acts fibre-wise. In other words, it is a principal $S^1$-bundle.

But it is known that the automorphism classes of $S^1$-bundles over a space correspond to its second homology group. So any principal $S^1$-bundle over the circle (say) is trivial, and there are no examples arising in the manner outlined above. If we only looked at spaces with zero second homology, we might conclude that a moduli space for oriented topological circles is given by the point.

But there certainly are examples of non-trivial principal $S^1$-bundles (since there are spaces with non-zero second homology). The simplest example is the Hopf bundle over $S^2$. So there cannot be a fine moduli space for oriented topological circles. But this example doesn't arise in any way from the existence of non-trivial automorphisms of the oriented circle; of course, the circle has got non-trivial automorphisms, but they vary continuously among all circles; there are certainly no special circles which have extra automorphisms, as there are in the elliptic curve case.

Is there a more general obstruction to the existence of a moduli space for a moduli problem? You can restrict attention to moduli problems over $\mathbf{Top}$. It looks as if the original non-trivial-automorphism examples correspond to the first homology/fundamental group, but the oriented circle example corresponds to second homology, and it's not hard to construct similar examples for $n$-th homology using Eilenberg-Maclane spaces.


1 Answer 1


Your question, really, is about necessary conditions for a ($\mathbf{Set}$-valued) presheaf to be representable. I think that business with automorphisms is actually a red herring – let me illustrate.

Let $A$ be a set and let $\mathcal{F} : \mathbf{Set}^\mathrm{op} \to \mathbf{Grpd}$ be defined by $\mathcal{F} (X) = (\mathbb{B} \mathrm{Aut} (A))^X$, where by $\mathbb{B} G$ I mean a group $G$ considered as a one-object groupoid. As usual, we apply $\pi_0 : \mathbf{Grpd} \to \mathbf{Set}$ to get a presheaf $\pi_0 \mathcal{F} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$. Now, if $A$ has at least two elements, then $\mathcal{F}$ is a presheaf of non-discrete groupoids. Nonetheless $\pi_0 \mathcal{F}$ is representable – indeed, it is represented by $1$. In fact, even $\mathcal{F}$ is representable in the sense that it is the restriction of a representable $\mathbf{Grpd}$-valued presheaf on $\mathbf{Grpd}$. Thus, my conclusion is that automorphisms have nothing to do with representability.

So what's really going on in the classical argument? Well, I'd say that what the classical argument is really doing is showing that the presheaf in question fails to send epimorphisms to monomorphisms. I will illustrate by generalising.

Lemma. Let $\mathcal{C}$ be a category with a terminal object. A jointly epimorphic family in $\mathcal{C}$ is a sink $\{ U_i \to X : i \in I \}$ such that, for every object $Y$ in $\mathcal{C}$, the induced map $$\mathcal{C} (X, Y) \to \prod_{i \in I} \mathcal{C} (U_i, Y)$$ is injective. Let $F : \mathcal{C}^\mathrm{op} \to \mathbf{Set}$ be a presheaf that satisfies the following conditions:

  • For each object $X$ in $\mathcal{C}$, there is at least one element of $F (X)$.
  • For each object $X$ in $\mathcal{C}$, there is a jointly epimorphic family $\{ U_i \to X : i \in I \}$ in $\mathcal{C}$ such that each $F (U_i)$ is a singleton.

Then $F$ is representable if and only if $F (X)$ is a singleton for all objects $X$ in $\mathcal{C}$. (Exercise.)

As a special case, we note that $F : \mathbf{Top}^\mathrm{op} \to \mathbf{Set}$ satisfies the hypotheses of the lemma as soon as it satisfies these (stronger) conditions:

  • For every topological space $X$, there is at least one element of $F (X)$.
  • $F (1) \cong 1$.

This is because $\{ \{ x \} \hookrightarrow X : x \in X \}$ is a jointly epimorphic family for all topological spaces $X$. There are lots of examples of presheaves that satisfy this condition: for instance, for any abelian group $A$ and any positive integer $n$, $H^n (-, A) : \mathbf{Top}^\mathrm{op} \to \mathbf{Set}$ is such a presheaf. The conclusion, therefore, is that $H^n (-, A)$ is a representable presheaf if and only if $A$ is the trivial group $0$. (By e.g. considering cellular cohomology, $H^n (S^n, A) \cong A$.)

  • $\begingroup$ Thanks. What does the notation ${}^X$ mean in $(\mathbb B \Aut(A))^X$? $\endgroup$ Apr 15, 2015 at 16:25
  • 1
    $\begingroup$ The $X$-th power. Or the category of functors $X \to \mathbb{B} \mathrm{Aut} (A)$. $\endgroup$
    – Zhen Lin
    Apr 15, 2015 at 18:37

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