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I have found similar questions littered throughout this site and math.SE (for example [1], [2], [3],…), but I feel like like most of them usually just say that non-trivial automorphisms prevent the formation of fine moduli spaces, but actually mean something else. I wanted to see if my understanding is right or wrong. Firstly, let's define what a moduli problem on some category $\mathcal{C}$ is in two different ways:

Old Style: A moduli problem is just a functor: $$F:\mathcal{C}^\text{op}\rightarrow \mathcal{Sets}.$$

Fibered Style: A moduli problem is a fibered category $\phi: F\rightarrow \mathcal{C}$ along with a (level structure) functor $\mathfrak{L}:F^\text{op}\rightarrow \mathcal{Sets}$: $\require{AMScd}$ \begin{CD} F @>\displaystyle \mathfrak L^\text{op}>> \mathcal Sets^\text{op}\\ @V \displaystyle \phi VV\\ \mathcal C \end{CD}

We say a moduli problem as defined in the old style is representable, if it is naturally isomorphic to the "contravariant Yoneda functor" of some object in $\mathcal{C}$.

In the Fibered Style, one says that a moduli problem is representable when there is an object $\mathcal{E}\in \operatorname{Ob}(F)$, such that for any $E\in \operatorname{Ob}(F)$, $$\mathfrak{L}(E)=\operatorname{Hom}_F(E,\mathcal{E}).$$ (Recall that any morphism in $F$ gives a cartesion diagram.)

Now, if the moduli problem is representable in the fibered style, then one can show that it is representable by $(\phi(\mathcal{E}),\mathcal{E})$ in the old style. If one has a representability in the old style, and additionally for any object $E\in \operatorname{Ob}(F)$, the automorphism group $\operatorname{Aut}_{\phi(E)}(E)$ acts freely on $\mathfrak{L}(E)$, then we have representability in fibered style (this is often called the rigidity conditon, and it is a necessary condition for a moduli problem in the fibered style to be representable).

Now, let us come to the concrete example of the moduli problem of elliptic curves with no additional level structure (i.e. $\mathfrak{L}(E)$ is always a singleton set in the fibered definition). Because there exist elliptic curves with nontrivial automorphisms, we can show that it will not be representable in the fibered style. But, this tells you nothing about the old style (because I don't think these two styles are equivalent in general). Yet, often one hears that the moduli problem is not representable in the old style because of the existence of non-trivial automorphism — but when you search for a concrete proof of such a statement, you usually just see proofs which are constructed using quadratic twists of elliptic curves without mentioning the word automorphism anywhere. So, a few things I would like to know are:

  1. When people say that it is not representable because of non-trivial automorphisms, are they cheekily talking about the fibered style, or can one show that if $f:E\rightarrow E$ is a non-trivial automorphism of some elliptic scheme (nice family of elliptic curves) over some base scheme $S$, then this automorphism prevents representability in the old style (like one can prove in the fibered style)?

  2. Are these two styles somehow equivalent when talking about moduli problems of elliptic curves, and so one would be representable if and only if the other is?

  3. This question I care about the least: are there any 'simple' examples, where the old style is representable, but the fibered is not?

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    $\begingroup$ Let $E$ be an elliptic curve over $\mathbb{Q}$ with $Aut(E)={\pm 1}$. Then the quadratic twists of $E$ are classified by $H^1(\mathbb{Q},Aut(E))$. $\endgroup$
    – anon
    Commented Jun 20 at 2:06
  • $\begingroup$ @anon This comment is definitely helpful, but saying something as general as "automorphisms prevent fine moduli" would require more. For instance, if I restrict the problem to scheme over some $S$, where $\mathbb{Q}$-elliptic curves do not live, will automorphisms still create a problem (in the old style definition of moduli spaces)? All of this works out nicely in Fibered language. $\endgroup$
    – Coherent Sheaf
    Commented Jun 20 at 7:09
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    $\begingroup$ What prevents the existence of fine moduli is having forms of $X$ over $k$, say, that are nonisomorphic over $k$ but become isomorphic over the algebraic closure of $k$. Such forms are classified by $H^1(k,Aut(X))$, which is generally nontrivial if $Aut(X)$ is nontrivial. $\endgroup$
    – anon
    Commented Jun 20 at 11:13
  • $\begingroup$ @anon Appologies for the delayed reply, but I wanted to sleep on your comment once and be sure I understood it, and could generalise this to other moduli problems as well. Thank a lot for your comment. It makes complete sense and was really helpful! $\endgroup$
    – Coherent Sheaf
    Commented Jun 21 at 9:01

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