Proposition
Let \(F:(\text{Sch}/\mathbb C)\to (\text{Sets})\) be a moduli functor. If there is a family of objects \(\mathscr E\in F(S)\) over a variety \(S\) such that
- the fibers of \(\mathscr E_{s}\) are all isomorphic to each other for \(s \in S(\mathbb C)\); and
- the family \(\mathscr E\) is non-trivial, i.e. is not equal to the pullback of an object \(E\in F(\mathbb C)\) along the structure map \(S\to \Spec \mathbb C\),
then \(F\) is not representable.
Proof
Suppose we have a moduli functor \(F\) represented by some scheme \(X\). Let \(U\to X\) be the universal family in \(F(X)\) corresponding to the identity map \(X\to X\) in \(\Hom(X,X)\). Consider a family \(\mathscr E\) over a variety \(S\) together with a fiber \(\mathscr E_s\) over a closed point \(s\in S(\mathbb C)\). This fits into a diagram:
If \(\mathscr E\to S\) satisfies condition (1) above, then the fiber families \(\mathscr E_s\to \Spec \mathbb C\) are all mutually isomorphic for different choices of \(s\), and hence represent the same element in \(F(\Spec \mathbb C)\). Via representability, this means the map \(\Spec \mathbb C\to X \in X(\mathbb C)\) corresponding to \(\mathscr E_s\to \Spec \mathbb C\) is independent of \(s\). Said another way, the composition forming the bottom row of the diagram above picks out the same \(x\in X(\mathbb C)\) irrespective of the choice of \(s\in S(\mathbb C)\). This implies \(S\to X\) is a constant map to some \(x\in X(\mathbb C)\) and hence factors through the structure map \(S\to \Spec \mathbb C \xrightarrow{x} X\). Hence \(S\) is a trivial family and in particular cannot also satisfy condition (2).