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It's well known for smooth curves that if $\pi:X\to Y$ is a finite map, $K_X=\pi^*K_Y+Ram(\pi)$, this is just the Riemann-Hurwitz formula at the level of line bundles. I've been told that this formula is true for a finite morphism of smooth varieties over $\mathbb{C}$, at least.

First, is the above true?

Second, how must the formula be corrected if $X$ and $Y$ are both smooth orbifolds/DM-stacks?

In particular, what if I want to use that $X$ and $Y$ are of the forms $\tilde{X}/G$ and $\tilde{Y}/H$ for groups $G,H$, and express the formula in terms of objects on $\tilde{X},\tilde{Y}$?

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@Charles: Sorry for the intrusion: $X$ and $Y$ were switched in the formula. I corrected it. – Sándor Kovács Mar 26 '11 at 18:38
@Charles: In your last question, do you assume that there is a map between $\overline X$ and $\overline Y$? – Sándor Kovács Mar 26 '11 at 20:03
@Sándor: Yes, the map I have does actually come from a map $\tilde{X}\to \tilde{Y}$. – Charles Siegel Mar 26 '11 at 20:26
For your last question, shouldn't this be automatic once you define the cotangent bundles and relative cotangent bundles and show that you have a short exact sequence 0 -> f^*Omega_Y -> Omega_X -> Omega_X/Y -> 0? – mdeland Mar 27 '11 at 14:53

In order for $K_X$ to exist you need some restriction on the singularities of $X$. See this MO answer for some thoughts on that. The most usual assumption to make is that $X$ is normal which implies that it is non-singular in codimension $1$.

In order to pull-back $K_X$ and still get a divisor, you need something more, the usual assumption is that it should be $\mathbb Q$-Cartier. See this MO answer for more on that.

If you have that $X$ and $Y$ are non-singular in codimension $1$, then you can forget about the second condition and just consider the formula on the non-singular part. (Strictly speaking you would want to restrict to the complement of the image of the singular part of $X$ on $Y$ so you'd still have a finite map).

So, let's say that both $X$ and $Y$ are smooth and $\pi$ is finite. Then localizing at height $1$ primes (or codimension $1$ points) reduces the statement to curves. Albeit not over the original field, but it's still OK.

I think this should cover two of your questions.

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Thanks a lot! Though seeing this answer, I think I'm realizing that the question I asked and the one I meant to ask are slightly different. Though the underlying varieties I have are normal, so all of this goes through, it's actually important for me to keep track of some stacky structure, because my group actions aren't faithful (at the least, I know that my group action on the domain has generic stabilizer nontrivial, and I think the action on the codomain might, but I haven't analyzed it completely yet.) – Charles Siegel Mar 26 '11 at 20:28

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