In defining $\overline{M_{g,n}}$, we say that an $S$point of it is a nodal curve $X \to S$, and $n$ sections $s_i : S \to X$, $i = 1,\cdots, n$ and plus stability conditions. Implicitly, we are ordering these sections. What if we don't order them and take the following as definition. An $S$point is a nodal curve $X \to S$ together with an effective relative Cartier divisor $D$ on $X$ of degree $n$, plus the same stability condition on each geometric fiber. What kind of moduli space shall we get with this definition? Is it just the usual $\overline{M_{g, n}}$ quotient by the order $n$ symmetric groups?

$\begingroup$ With the usual approach to the moduli space, you do not order punctures/marked points. Incidentally, you should also decide if you want to consider moduli stack or coarse moduli space. $\endgroup$– MishaJul 15, 2013 at 4:01

$\begingroup$ You get the quotient if you assume that the Cartier divisor is fibrewise multiplicity free. $\endgroup$– nafJul 15, 2013 at 5:24

$\begingroup$ Are you working with $\overline{M}_{g,n}$ as a coarse moduli (algebraic) space (in fact a projective scheme, as it turns out) or as a "fine moduli" stack (DeligneMumford, as it turns out)? This will make a difference. The coarse moduli space for your "moduli problem" will be the $S_n$quotient of the usual coarse moduli space $\overline{M}_{g,n}$. However, with the most natural definition of stack, your stack will have larger "inertia" than the DeligneMumford stack of stable marked curves. $\endgroup$– Jason StarrJul 15, 2013 at 9:25

$\begingroup$ @Misha The definition given in Knudsen's original paper ("The projectivity of the moduli space of stable curves II: the stacks $M_{g,n}$" Math. Scand 52 (1983) 161199) places an order on the sections. $\endgroup$– S. Carnahan ♦Jul 17, 2013 at 2:59

$\begingroup$ Scott: And before Knutson were Ahlfors, Bers et al. $\endgroup$– MishaJul 17, 2013 at 3:02
1 Answer
I believe the answer is yes, both if you think of stacks and coarse spaces. I don't understand Jason Starr's comment, but maybe I am missing something obvious.
To prove it on the level of stacks, the biggest difficulty is maybe sorting out the formalism. Let me try to spell out some details here.
Consider the functor $F$ which maps a scheme $S$ to the groupoid whose objects are $n$pointed stable curves $(X,\sigma_1,...,\sigma_n)$ over $S$ and whose morphisms are pairs $(f,\pi)$ with $f : X \to X'$ an isomorphism over $S$ and $\pi \in \mathbb S_n$ a permutation, such that $f \circ \sigma_i = \sigma_{\pi(i)}'$ for all $i$. In other words $F(S)$ is the action groupoid of $\mathbb S_n$ acting on $\overline M_{g,n}(S)$. Then the stackification of $F$ is $[\overline M_{g,n}/\mathbb S_n]$.
Let $G$ be the functor you propose, $G(S)$ is the groupoid whose objects are pairs $(X,D)$ of a curve over $S$ and a relative effective multiplicity free Cartier divisor of degree $n$, satisfying the obvious stability condition. Morphisms are isomorphisms $f \colon X \to X'$ over $S$ with $f(D)=D'$.
There is a natural transformation $F \to G$ given by letting $D$ be the union of the schemetheoretic images of the sections $\sigma_i$. Since $G$ is a stack we get an induced map from the stackification of $F$ to $G$. The crucial point in showing that this is an isomorphism is the following: pick $(X,D)$ in $G(S)$. Since $D \to S$ is finite étale, there is an open cover $S' \to S$ such that $D \times_S S' \to S'$ is trivial. Then the pullback of $(X,D)$ to $S'$ is in the essential image (in fact in the image) of $F(S') \to G(S')$.

$\begingroup$ Regarding my comment: there are many $n$pointed, genus $g$ stable curves that have many more selfmaps in your stack than they have in the usual stack. So they have bigger inertia, i.e., 2fibered product of the diagonal with itself over the 2fibered product of the stack. $\endgroup$ Jul 19, 2013 at 18:08

$\begingroup$ Can you give an example of an $(X,D)$ with larger automorphism group than the corresponding point in $\overline{M_{g,n}}/\mathbb S_n$? $\endgroup$ Jul 19, 2013 at 19:59

1$\begingroup$ Let $X$ be a general hyperelliptic curve with hyperelliptic involution $i$. Let $p$ be any point of $X$ with $i(p)\neq p$. Then the $2$pointed curve $(X,p,i(p))$ has trivial automorphism group in the stack $\mathcal{M}_{g,2}$, yet it has nontrivial automorphism group in the stack $[\mathcal{M}_{g,2}/\mathfrak{S}_2]$ that you defined above. $\endgroup$ Jul 23, 2013 at 13:32