I was told (quickly) by my professor about the following yoga:

Say we want to define a functor $F:\operatorname{Sch}\to \operatorname{Set}$ which parametrizes some class of objects. If those objects have nontrivial automorphisms, then we cannot expect $F$ to be representable. There are two solutions to this issues:

  1. Sheafify the functor $F$ in order to obtain an fppf sheaf
  2. Rigidify the objects we want to parametrize, in order to kill their automorphisms

Now, at the present moment I don't have the time to go through the underlying theory. So


Could you give me some intuition on the above yoga?

Why are the automorphisms an obstruction to the representability of the functor?

What does it mean, in simple terms, for a functor to be a fppf sheaf?


I have a motivating example for this question:

Let $X$ and $T$ be schemes over $S$; the relative Picard functor $Pic_{X/S}:Sch_S \to Ab$ is defined by $$ T \mapsto \left\{ \text{Rigidified line bundles}\text{ on } X_T/T \right\} $$ and not as $$ T \mapsto \left\{ \text{Line bundles}\text{ on } X_T/T \right\} $$ The main difference between line bundles and rigidified ones is that the latter don't have any nontrivial automorphisms.

Under reasonable hypothesis the first functor turns out to be representable, while the first one doesn't.


1 Answer 1


Regarding #2, here is a related question. The idea is rather simple: the existence of nontrivial automorphisms implies existence of nontrivial isotrivial families (i.e. with all fibres isomorphic/equivalent). To see why this messes up representability, suppose we are given a nontrivial isotrivial family $\mathcal X\to B$. There exists a unique map $B\to P$, where $P$ supposedly represents whatever moduli functor we're interested in, such that $\mathcal X\to B$ is the pullback over this map of the universal family $\mathcal U\to P$. The map $B\to P$ is clearly constant, by the fact that the fibres are isomorphic. But this is impossible, since the pullback of $\mathcal U\to P$ over a constant map is the trivial family.

Hopefully someone else can say something about #1.

  • 2
    $\begingroup$ Your first statement, "the existence of nontrivial automorphisms implies existence of nontrivial isotrivial families", is that a mathematical fact, or something that "happens in practice"? For example, with topological spaces one can easily see this by taking a locally constant bundle over $S^1$ by gluing with the nontrivial automorphism. But what if I restrict my category to other objects so that, for example, $S^1$ is not one? Can you guarantee the existence of such a family? $\endgroup$
    – RghtHndSd
    Dec 17, 2013 at 15:52
  • $\begingroup$ Dear @rghthndsd, it seems every time this general statement is made, it is followed by some specific example and no proof. I suspect that this has to do with the difficulty of taking quotients by group actions in general for algebraic schemes. So I can only say it is a principal, that happens in practice, since I don't know in what generality it does hold. However, this phenomenon does already pop up in the the first, and one of the most interesting examples -- the moduli of curves. $\endgroup$
    – Andrew
    Dec 17, 2013 at 17:42
  • $\begingroup$ Thank you for the response. I have found exactly the same. It seems as if a category with no nontrivial isotrivial families would be very weird. $\endgroup$
    – RghtHndSd
    Dec 17, 2013 at 17:51
  • 1
    $\begingroup$ Thanks Andrew, this is exactly what I was looking for. Hope somebody say something about #2 as well! $\endgroup$
    – Abramo
    Dec 17, 2013 at 17:58
  • $\begingroup$ Whoops: principle, not principal :-P $\endgroup$
    – Andrew
    Dec 17, 2013 at 19:26

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