Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

To explain, I will use the following concrete example: Let $\mathcal{M}_g$ be the functor for the moduli problem of classifying genus $g$ smooth projective curves (taking a scheme $S$ to the set of ways that $S$ parametrizes genus $g$ curves). This, as is well known, has a coarse moduli space: $M_g$.

Let $\sigma \in Aut(\mathbb{C})$, and say that we begin with a specific $\mathbb{C}$-curve of genus $g$. This $\sigma$ may act in two ways on this curve (by act I mean turn it into a different curve, not act as meaning an automorphism): 1. By taking a fiber product over the automorphism $Spec(\mathbb{C}) \rightarrow Spec(\mathbb{C})$ (this is the usual action people use, and has nothing to do with the moduli space) 2. By letting $\sigma$ act on $M_g(\mathbb{C})$, the point corresponding to our curve, will be taken to another $\mathbb{C}$-point, which in turn corresponds to another curve.

The question is: do these two actions agree in general? Are there conditions that need to be fulfilled for this to be true?

As an aside, this is true if the moduli space is fine. For example, if the moduli of genus $g$ curves were fine (which it isn't), then the curve corresponding to $Spec(\mathbb{C})\rightarrow M_g$ will be given simply by taking the fiber product with the universal family $E_g\rightarrow M_g$. Since fiber products commute, we get that the two actions above agree. This reasoning breaks down, however, if the moduli space is not fine.

share|improve this question
I've already edited once a few minutes ago, so I don't want to edit again, but I wanted to add that if the actions in general don't agree, then I would be very interested in results like "this is true if the moduli problem has a fine moduli algebraic stack" (for example, is THAT true?). –  Makhalan Duff Jun 28 '11 at 23:28
add comment

2 Answers

up vote 4 down vote accepted

If $X$ is the coarse moduli scheme associated to a functor $F$ on schemes, then in particular, there is a natural transformation $F\to h_X$, where $h_X$ is the functor of points of $X$.

Unless I am missing something, if you apply the naturality of this transformation to the map $\mathrm{Spec}(\mathbb{C}) \to \mathrm{Spec}(\mathbb{C})$ associated to your automorphism $\sigma$, this implies that the two actions coincide in your case, since the associated map $F(\mathrm{Spec}(\mathbb{C})) \to F(\mathrm{Spec}(\mathbb{C}))$ is given by the fiber product of the curve, and the associated map $X(\mathrm{Spec}(\mathbb{C})) \to X(\mathrm{Spec}(\mathbb{C}))$ is just the action on the points. I wish I could make the obvious 2-by-2 commutative diagram...

share|improve this answer
What you write is true, but it deals only with the $2^{nd}$ action I was talking about and not the $1^{st}$. This is because the way $\sigma$ acts on $F(Spec(\mathbb{C}))$ is not (a priori) related to the action of taking the object over $\mathbb{C}$ and base changing to $\mathbb{C}$ via $\sigma$. I'm saying this awkwardly, but hopefully you'll understand what I mean. –  Makhalan Duff Jun 29 '11 at 0:31
Isn't this the way the functor $\mathcal{M}_g$ is set up? That is, if $S\to T$ is a map of schemes, isn't the associated map $\mathcal{M}_g(T)\to \mathcal{M}_g(S)$ given by fiber product? –  Ramsey Jun 29 '11 at 0:51
I agree with Ramsey. –  Martin Brandenburg Jun 29 '11 at 1:07
Uhm... wait. I now think that you're right. –  Makhalan Duff Jun 29 '11 at 1:08
add comment

There are curves (with automorphisms) for which the field of definition is not the field of moduli. Concretely, there are curves $C$ with corresponding point $P \in M_g$, such that $C$ cannot be defined over $K =\mathbb{Q}(P)$, only over larger fields. That means that there are automorphisms of $\mathbb{C}$ (fixing $K$) whose action don't change $P$ but change $C$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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