So I've been trying to understand why in general automorphisms stop a moduli problem being representable. I've seen the answer to Killing the automorphisms to make a functor representable

To see why this messes up representability, suppose we are given a nontrivial isotrivial family X→B. There exists a unique map B→P, where P supposedly represents whatever moduli functor we're interested in, such that X→B is the pullback over this map of the universal family U→P. The map B→P is clearly constant, by the fact that the fibres are isomorphic. But this is impossible, since the pullback of U→P over a constant map is the trivial family.

and understand the general idea but I can't see why an isotrivial family would result in a constant map $B \rightarrow P$ which seems pretty crucial to that argument.

To add onto this I am trying to understand this in the specific case of the moduli functor of vector bundles (of rank $n$) over some nice scheme $X$ (and lets say everything is over $\mathbb C$). For simplicity if we take $X$ to just be a point and consider line bundles then for me the argument to show that this functor can't be representable would be something like:

"Representable functors are sheaves (in the etale topology say) but as by definition line bundles are Zariski (hence etale) locally trivial we fail one of the sheaf axioms."

But how does this argument make use of non-trivial automorphisms of line bundles over a point and how does it fit into the more general idea of the answer in the question I've linked? (I was thinking that perhaps non-trivial automorphisms of the line bundle over a point [ie. automorphisms of $\mathbb C$] allow us to construct non-trivial line bundles over arbitrary schemes $Y$ which then ensure the sheaf axioms are not satisfied- is this correct?)

Finally I would like to ask how do stacks help solve this issue? As in I would imagine it would go as follows: if we represent this moduli problem with a stack instead then the data of point of this stack is the single line bundle over a point along with the group of automorphisms $\mathbb C^*$. Then we can recover an arbitrary line bundle on a scheme $Y$ by taking an open cover and automorphisms of $\mathbb C$ for all the intersections in that cover. But specifically how could you define this idea as a functor/sheaf.

Sorry if this post is a bit jumbled, I'm having a bit of difficulty trying to identify and articulate exactly what the problems in my understanding are.


1 Answer 1


Whenever you have a family over $B$, you get a morphism $B\rightarrow P$ such that the pullback of the universal family to $B$ recovers your family. Now consider any point $b\in B$. Then because your family is isotrivial, the composition $b\hookrightarrow B \rightarrow P$ must be mapped to the same point $p$, for every $b\in B$. This implies that the morphism $B\rightarrow P$ factors through a point $B\rightarrow p \rightarrow P$. But we know that the family over $B$ can be recovered by pulling back the universal family, so since the map factors through a point we see that the family you started with must be trivial. So if you have an isotrivial but non-trivial family, then that cannot be recovered by pulling back the universal family.

I think what you say about line bundles is correct. After all transition functions are all about $GL_n(\mathbb{C})$!

I'm not quite sure what you mean when you say that the data of a point in a stack is the single line bundle over a point along with the group of automorphisms $\mathbb{C}^*$. In any case, here is some intuition why considering a sheaf in groupoids (which gives stacks) rather than a sheaf in sets (which gives schemes/algebraic spaces) might help with dealing with isotrivial / locally trivial families. In the usual definition of a sheaf of sets $\mathcal{F}$ - say you have a covering $\{U_i\}$ of $X$ and sections $s_i \in \mathcal{F}(U_i)$, then asking for $s_i$ to glue together is to require $s_i|_{U_i\cap U_j} = s_j|_{U_i\cap U_j}$ for all $i,j$

When you consider a sheaf in groupoids, the latter condition is relaxed to asking for isomorphisms $s_i|_{U_i \cap U_j} \stackrel{\sim}{\rightarrow} s_j|_{U_i\cap U_j}$ (and some more conditions). This should remind you of how one glues together a non-trivial vector bundle!

  • $\begingroup$ Hey thanks for your answer. It has definitely given me something to think about, although I'm afraid I still dont understand your 1st paragraph regarding why the map must factor through a point. Specifically if I think of our family of objects over $B$ (and respectively the universal family over $P$) and then fibre of family over a point $b \in B$ is the pullback of the fibre over the point $f(b) \in P$, then why can't isotriviality of the object over B for example follow from isotriviality of the universal family? Then it won't matter where b is sent as the pullback fibres will always be iso. $\endgroup$
    – Fromage
    Commented Mar 13, 2019 at 11:46
  • $\begingroup$ @Fromage I think it might be easier to think about if you consider e.g. the moduli space of curves of genus $g \ge 1$. Here you have a lot of different curves of the same genus (so your moduli space isn't just a point). Then if you have an isotrivial family over a base $B$, with fibers $C$, every point $b\in B$ must be mapped to the same point $[C]$ in the moduli space corresponding to the curve $C$. $\endgroup$
    – loch
    Commented Mar 13, 2019 at 15:35

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