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Let $\mathcal{X}$ be a canonical stack (edit: I forgot to say I also want $\mathcal{X}$ smooth), and $\pi : \mathcal{X}\to X$ its coarse moduli space morphism. Let also $D$ be a prime divisor (i.e. just one reduced component) on $X$, and $\mathcal{D}=\pi^{-1}(D)$ the pulled back divisor on $\mathcal{X}$, which will also be prime.

Is the isomorphism of $\mathcal{O}_X$-modules

$$\pi_* \mathcal{O}(\mathcal{D})\cong \mathcal{O}(D)$$

always true?

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  • $\begingroup$ What do you mean when you say that $\mathcal{D}$ will be prime? For example if $\mathcal{X}$ is the $n$-th root stack of $D$ (say everything is smooth for simplicity), then $\pi^{-1}(D)$ is $n\cdot \mathcal{D}$, where $\mathcal{D}$ is the universal root. (I don't know exactly what you mean by ``canonical stack'', but I guess the above is an example...) $\endgroup$
    – Mattia Talpo
    Commented Nov 13, 2015 at 2:12
  • $\begingroup$ Another comment is that if $\mathcal{X}$ is tame, then there is a projection formula for coherent sheaves for $\pi$, that in your case implies exactly what you wrote down, since $\mathcal{O}(\mathcal{D})=\pi^*\mathcal{O}(D)$ (where $\mathcal{D}$ is as in your notation, not as I used it in my comment above). $\endgroup$
    – Mattia Talpo
    Commented Nov 13, 2015 at 2:16
  • $\begingroup$ @Mattia: a root stack is not a canonical stack as it has isotropy in codimension 1. $\endgroup$
    – Qfwfq
    Commented Nov 13, 2015 at 10:50
  • $\begingroup$ A canonical smooth DM stack is -as per Fantechi-Mann-Nironi- one that has isotropy only in codimension at least 2. $\endgroup$
    – Qfwfq
    Commented Nov 13, 2015 at 11:00

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I'm writing an answer that expands on my last comment:

I assume that your $D$ on $X$ is an effective Cartier divisor, that I see as an invertible sheaf $\mathcal{O}(D)$ on $X$ with a global section $1_D$.

Then its pullback $\mathcal{D}$ as a divisor will be given by $(\pi^*\mathcal{O}(D),\pi^*1_D)$. If the stack $\mathcal{X}$ is tame, for example by Lemma 2.8 of http://arxiv.org/pdf/0811.1949v2.pdf there is an isomorphism $\pi_*\mathcal{O}(\mathcal{D})=\pi_*\pi^*\mathcal{O}(D)\cong \mathcal{O}(D)$.

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  • $\begingroup$ Thanks for the answer! My divisor is in fact only $\mathbb{Q}$-Cartier, but I think this is not a problem (for other reasons) if I take a multiple. $\endgroup$
    – Qfwfq
    Commented Nov 13, 2015 at 12:31
  • $\begingroup$ No problem! After writing the first comment I suspected that a root stack was not an example. $\endgroup$
    – Mattia Talpo
    Commented Nov 13, 2015 at 14:24

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