Let $F: (Sch) \to (Sets)$ be a functor sends schemes to sets (for example, $F$ sends a scheme $S$ to families of K3 surfaces over $S$ with some fixed polarization). Then it is known that because of the existence of non trivial automorphism (in above example, it is because the non trivial automorphism of K3 surfaces), this functor cannot be represented by a scheme (i.e. we do not have $F \cong Hom(-,M)$).

My question is why the existence of automorphism makes a functor not being representable?

  • $\begingroup$ Read the discussion of the moduli space of elliptic curves in Chapter 1 of Moduli of Curves by Harris and Morrison. This will make everything clear. $\endgroup$
    – user64687
    Commented Apr 29, 2015 at 13:36
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    $\begingroup$ What you say is a helpful and good slogan, but actually the existence of automorphisms only becomes an issue when your moduli problem allows you to construct nontrivial isotrivial families (which should give you constant moduli maps). In fact there are examples of representable moduli functors where all objects involved do have automorphisms. $\endgroup$
    – Brenin
    Commented Apr 29, 2015 at 20:09
  • $\begingroup$ Related: math.stackexchange.com/questions/234213 $\endgroup$
    – Watson
    Commented Nov 23, 2018 at 20:17

1 Answer 1


Here's one concrete example that I like. Consider the functor $\mathcal{M}_{1,1} : (Sch)^{op} \to Set$. The $B$-valued points of this functor are elliptic curves over $B$. Now let us now consider two elliptic curves

$$E_1: y^2 = x^3 + x + 1, \hspace{5mm} E_2: 3y^2 = x^3 + x + 1.$$

It is not hard to show these are non-isomorphic over the rationals, but become isomorphic when we pass to the extension $\Bbb{Q}(\sqrt{3})$.

Conclusion: The presence of automorphisms implies that the functor $\mathcal{M}_{1,1}$ is not a sheaf in the fpqc topology, and hence cannot be representable by a scheme.

  • $\begingroup$ Sorry I confused: What is wrong with the restriction $\mathcal{M}_{1,1}(Spec(\mathbb{Q})) \to \mathcal{M}_{1,1}(Spec(\mathbb{Q(\sqrt{3})}))$ which sends $E_1, E_2$ to isomorphic curves. Why this indicates it is not a sheaf? $\endgroup$
    – Li Yutong
    Commented May 5, 2015 at 1:16
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    $\begingroup$ I may be wrong: you have a cover $U_i$ of $X$ (in some topology), and you show the functor $M$ is not injective on the restriction $F(X) \to F(U_i)$. Then why $F$ cannot be a sheaf? $\endgroup$
    – Li Yutong
    Commented May 5, 2015 at 13:16
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    $\begingroup$ @LiYutong Let $F$ be a sheaf in the fpqc topology say. This means for an fpqc cover $U_i$ of $X$, the following sequence is exact in the category of sets. Namely, $F(X) \to \prod_i F(U_i) \to \prod_{i,j} F(U_i \times U_j)$. In particular, this means $F(X) \to \prod_i F(U_i)$ is injective. In particular for $i = 1$, we have the following. Given any fpqc morphism $Y \to X$, we need that $F(X) \to F(Y)$ be injective. So if the map is not injective for some fpqc map, $F$ cannot be a sheaf in the fpqc topology. $\endgroup$ Commented May 5, 2015 at 15:20
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    $\begingroup$ Thank you very much! I understand it now - this seems really strange! $\endgroup$
    – Li Yutong
    Commented May 5, 2015 at 18:05
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    $\begingroup$ Really needed to read David's explanation to this $\endgroup$
    – kindasorta
    Commented Feb 8, 2023 at 20:37

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