# Moduli functor $\mathcal{M}_g$ of smooth curves of genus $g$ not representable

Let $$S$$ be a scheme. By a smooth curve of genus $$g$$ over $$S$$ we mean a proper, flat, family $$C \to S$$ whose geometric fibers are smooth, connected $$1$$-dimensional schemes of genus $$g$$. The moduli functor $$\mathcal{M}_g$$ of smooth curves of genus $$g$$ over a noetherian base $$S$$ is the functor that sends each $$S$$-scheme $$B$$ to the set $$\mathcal{M}_g(B)$$ of isomorphism classes of smooth and proper morphisms $$C \to B$$(where $$C$$ is also an $$S$$-scheme) whose fibers are geometrically connected curves of genus $$g$$.

I have a question about the argument that this moduli functor $$\mathcal{M}_g$$ of smooth curves of genus $$g$$ over a noetherian base $$S$$ is not representable (in spirit of Yoneda-Lemma this means that $$\mathcal{M}_g$$ isn't a sheaf). I found it in Pedro Castillejo's paper Introduction to stacks and he refered for a "detailed" proof to Dan Edidin's "Notes on the construction of the moduli space of curves". Now one argument from Edidin's paper I not understand:

The key point is that some curves have non trivial isomorphisms, i.e. not identity maps, and it makes it possible to construct non-trivial families $$C \to B$$ where each fiber has the same isomorphism class. The construction in Edindin's paper on page 3 I understand.

What I not understand is why the existence of such non-trivial family $$C \to B$$ of isomorphic fiber imply that the functor $$\mathcal{M}_g$$ is not representable. In the paper on page 2 the argument is

...As a result, it is possible to construct non-trivial families $$C \to B$$ where each fiber has the same isomorphism class. Since the image of $$B$$ under the corresponding map to the moduli space is a point (???), if the moduli space represented the functor $$\mathcal{M}_g$$ then $$C \to B$$ would be isomorphic to the trivial product family (???)

Assume $$\mathcal{M}_g$$ is representable by an $$S$$-scheme $$M$$, thus we have natural equivalence $$\mathcal{M}_g \cong Hom_S(-, M)$$.

Q_1: Why this imply that then the image of $$B$$ is a point of $$M$$?

Q_2: I not see how the sheaf axiom would give a contradiction to representability.

Suppose $$M$$ represents your functor $$\mathcal{M}_g$$ and you have an isotrivial family $$C\rightarrow B$$ which is globally non-trivial. Then by definition, there exists a universal family $$\mathcal{C}_g \rightarrow \mathcal{M}_g$$ map $$B\rightarrow M_g$$ such that the pullback (fiber product) gives you $$C\rightarrow B$$.
Now if you take any $$b\in B$$, then under the composition $$b\hookrightarrow B \rightarrow M_g$$ corresponds to the family $$C_b\rightarrow b$$, where $$C_b$$ is the fiber of $$C\rightarrow B$$ over $$b\in B$$.
Since $$C\rightarrow B$$ is assumed to be isotrivial, $$C_b \equiv C_{b'}$$ for any $$b,b' \in B$$. Therefore $$b,b'$$ will map to the same point in $$\mathcal{M}_g$$. In particular we deduce that $$B\rightarrow \mathcal{M}_g$$ is the constant map, where every point maps to the point corresponding to the curve $$C_b$$.
However since $$C\rightarrow B$$ is the fiber product, it follows that $$C\rightarrow B$$ is the trivial family. This contradicts the assumption that $$C\rightarrow B$$ is non-trivial.
• $2\frac{1}{2}$ remarks on your answer: the "universal family" $\mathcal{C}_g \rightarrow \mathcal{M}_g$ is exactly that one that determines the natural isomorphism $\mathcal{M}_g \cong Hom(-,M)$ in spirit of Yoneda-Lemma as explaned here ncatlab.org/nlab/show/representable+functor,right? And by $M_g$ you mean $M$? – user526728 Dec 24 '19 at 6:24
• Secondly, I understand how you conclude that $B\rightarrow \mathcal{M}_g$ must be a constant map with image $c \in M$ corresponding to curve $C_g$. What I not understand is why does this imply that $C \to B$ most be the trivial family? – user526728 Dec 24 '19 at 6:24
• @TimGrosskreutz 1. Yes and yes -- oops! 2. So if you know that it's the constant map, then you know that the map $B\rightarrow M$ factors through some point $B\rightarrow b = \operatorname{Spec} k \hookrightarrow M$. The map $b\rightarrow M$ also corresponds to a family, and it's precisely the corresponding curve over a point. i.e. it's $C_b\rightarrow \operatorname{Spec} k$. Now you take the fibered product of $B\rightarrow b = \operatorname{Spec} k$ and $C_b \rightarrow \operatorname{Spec} k$ and you'll get a trivial family. – loch Dec 24 '19 at 7:26