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In the preprint "Smooth toric DM stacks", Fantechi, Mann and Nironi define the stacks of their title, and show that each of these can be obtained through the following sequence of steps:

1) start with a scheme (the coarse moduli scheme) with at worst finite quotient singularities, and take the associated canonical stack;

2) use a root stack construction to possibly add some extra stack structure to divisors (given by an integer for each divisor);

3) finally add a gerbe.

Not all smooth DM stacks can be obtained this way, e.g. for $n>3$ take the global quotient $\mathbb{C}^n/S_n$, where the symmetric group $S_n$ acts by permuting the factors of $\mathbb{C}^n$. The coarse moduli scheme is smooth here, and there is no gerbe, but the stack doesn't seem to arise as a root stack.

Are there conditions known for reasonable (finite type over a field,...) smooth DM stacks under which the stack can be obtained by the "bootstrapping" procedure described above (or a similar one)?

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1 Answer 1

up vote 6 down vote accepted

Notation: Let $x$ be a point of a smooth separated finite type DM stack $\mathcal X$ over a field. Suppose
• $G$ is the stabilizer of $x$,
• $V$ is the tangent space of $x$ (which comes equipped with an action of $G$),
• $G^\textrm{triv}\subseteq G$ is the subgroup which acts trivially on $V$,
• $H = G/G^\textrm{triv}$ (note $H$ acts on $V$),
• $K$ is the subgroup of $H$ generated by pseudoreflections on $V$, and
• $K'$ the commutator subgroup of $K$.

$\mathcal X$ can be expressed as you described (in an étale neighborhood of $x$) if and only if $K'$ is trivial.

I'll now unpack that answer. Any smooth separated finite type DM stack over a field can be (canonically!) obtained from its coarse space with the following steps (this is basically Theorem 6.1 of my paper with Matt Satriano, Torus Quotients as Global Quotients by Finite Groups):

  1. take the canonical stack of the coarse space,
  2. do a root stack construction along the ramification divisor of the coarse space map, rooting each component of the ramification divisor by the degree of ramification,
  3. take the canonical stack again (the root stack may not be smooth any more, but it will have quotient singularities!), and
  4. add a gerbe.

Your question is "when can we skip step 3?" That is, when is the root stack from step 2 already smooth?

Using the notation above, and looking formally locally around $x$ (so we can assume $\mathcal X=[V/G]$; you can describe it étale locally too, but it's clearer this way), the above steps are:

  1. $\bigl[(V/K)/(H/K)\bigr]$ is the canonical stack of the coarse space $V/H = V/G$,
  2. $\bigl[(V/K')/(H/K')\bigr]$ is a root stack of $\bigl[(V/K)/(H/K)\bigr]$,
  3. $[V/H]$ is the canonical stack of $\bigl[(V/K')/(H/K')\bigr]$, and
  4. $\mathcal X = [V/G]$ is a $G^\textrm{triv}$-gerbe over $[V/H]$.

Note that step 1 is the familiar way of building the canonical stack of a space with quotient singularities (in this case $V/H$) by expressing it as a quotient by a finite group somehow, and then quotienting out the subgroup generated by pseudoreflections, with the Chevalley-Shephard-Todd theorem ensuring that you don't lose smoothness. This description tells us that the canonical stack is any description of the space as a quotient where the group acts without pseudoreflections.

Note that in step 3, we're quotienting out by $K'$, which has no pseudoreflections since it's a commutator subgroup (all its elements must therefore act with determinant 1, and there are no pseudoreflections of determinant 1). So it makes sense that this step is a canonical stack.

It's pretty clear that step 4 is a (trivial) gerbe.

Seeing that step 2 is a root stack is a bit more complicated. Matt and I will soon post an updated paper which explains this. I'll just link it here once it's up.

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Thanks! Could you illustrate this for [C^n/S_n]? –  Johan Jan 26 at 16:27
@Johan: I've updated my answer with the local description of the bootstrapping procedure, which I think makes things much clearer. In the case of $[\mathbb C^n/S_n]$, you have that $G=H=K=S_n$ and $K'=A_n$. Please let me know if something is still unclear. –  Anton Geraschenko Jan 27 at 16:06
Thanks Anton! Frankly, I had forgotten about this question, but it is great to have such a complete answer now. –  Johan Jan 30 at 8:52

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