Let $\mathscr{M}_g$ be the moduli stack of genus $g$ curves ($g \geq 2$). That is, $\mathscr{M}_g$ is the category whose objects are proper smooth morphisms $f: C \to S$ whose geometric fibers are connected, genus $g$ curves. A morphism of curves $(f: C \to S)$ to $(f' : C' \to S')$ is an appropriate cartesian diagram.

Where can I find a reference containing a proof of the fact that the diagonal $\Delta : \mathscr{M}_g \to \mathscr{M}_g \times \mathscr{M}_g$ is representable?

In most places I've seen (for example in Edidin's notes), they prove representability of the diagonal by showing that $\mathscr{M}_g$ is a quotient stack . However, can one prove representability by hand? I know for instance that it boils down to the fact that if $C_1 \to S$ and $C_2 \to S$ are two curves, then we want that the functor $\text{Isom}_S(C_1,C_2)$ be representable by a scheme.


The Isom scheme will be part of Hilbert scheme.

Suppose given $T \to S,$ and consider the pull-backs $C_{1/T}$ and $C_{2/T}$. If these are isomorphic, then the graph of the isomorphism will be a certain closed subscheme in $C_{1/T} \times_T C_{2/T}$ (which is the pull-back under $T \to S$ of the fibre product $C_1 \times_S C_2$) which is flat over $T$.

So Isom_S(C_1,C_2) is a certain locally closed subscheme of the Hilbert scheme of $C_1\times_S C_2$ over $S$. (To verify this, you have to think about what the conditions are on a closed subscheme of $C{1/T}\times_T C_{2/T}$ which is flat over $T$ for it to actually be the graph of an isomorphism.)

The construction of Isom schemes in this way is due to Grothendieck (perhaps in the same Bourbaki seminar where he describes Hilbert schemes themselves?).

  • $\begingroup$ Thanks for your answer. Do you have any references I can look at? $\endgroup$ – Ben Lim May 6 '15 at 7:00
  • $\begingroup$ I think Mumford's original paper on Picard groups of moduli problems at least mentions Isom schemes (my vague memory is that this is the first place that I saw them); so you could try there. $\endgroup$ – Just passing by May 9 '15 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.