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Let $f:X\to Y$ be a proper morphism of normal varieties (smooth as DM stacks, but I only care about the coarse spaces). The map $f$ is generically finite, but not flat (so no hope of smoothness and Grothendieck-Riemann-Roch applying) and I have a fairly detailed understanding of the fiber over any point, images of $f$ restricted to divisors and so forth.

Now, take a divisor $D$, and identify it with an invertible sheaf in the standard way. I'm looking for a way to compute the first Chern class of $f_*D$ on $Y$.

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I guess you mean $D$ is a Cartier divisor, right? (Otherwise you'll get a reflexive sheaf, but not an invertible one.) – Dave Anderson Dec 18 2010 at 22:11
Yes, $D$ can be taken as Cartier for this. – Charles Siegel Dec 18 2010 at 22:31
You can still use GRR. For example you can represent your map as a composition of a closed embedding of smooth DM stacks and a smooth map and apply GRR for each step separately. – Sasha Dec 19 2010 at 9:12

1 Answer

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Do you know the (generic) degree of your map $f$? As you probably know, standard intersection theory says $f_*[D] = n[f(D)]$ as classes in $A_{d-1}Y$, where $n$ is the degree of $f$ (restricted to $D$) and $d=\dim X = \dim Y$. No flatness or smoothness hypotheses on $f$ are needed for this; the sticky point is in identifying these divisors with line bundles. But since you're dealing with smooth DM stacks, that should be ok (over ${\Bbb Q}$ at least).

EDIT (incorporating the comments): For a proper map $f$, there is a map defined at the cycle level by $$f_*[D] = n\cdot [f(D)],$$ where $n$ is the degree of $D$ over $f(D)$ (i.e., degree of the induced field extension) when these have equal dimension, and $n=0$ when $\dim f(D)< \dim D$. This passes to rational equivalence, so defines a map $A_{d-1}X \to A_{d-1}Y$. In particular, if $f$ collapses a divisor $D$, then $f_*[D]=0$.

All this is in Fulton's Intersection Theory, Section 1.4.

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I do, but the divisors I'm most interested in are in the non-finite locus, and so have image of smaller dimension. – Charles Siegel Dec 18 2010 at 23:03
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Ah, that explains the question. Well, this just means the answer is zero. Unless I'm missing something obvious here... – Dave Anderson Dec 18 2010 at 23:14
PS: For reference, Fulton's Intersection Theory, section 1.4. (But I guess you must know this stuff.) – Dave Anderson Dec 18 2010 at 23:16
Is this always true, then? The homomorphism on cycles behaves nicely with respect to chern classes? My concern is that the higher codimension cycle may be the zero locus of a section of some vector bundle on $Y$, and I might have a nontrivial chern class for that. – Charles Siegel Dec 18 2010 at 23:45
Ah, I see now what your concern is. You really want to do the pushforward in K-theory, and then apply the Chern character... Fortunately, GRR doesn't require anything more than properness of $f$ -- and, for the basic versions, smoothness of $X$ and $Y$. I think the reduction to pushforward on Chow groups still goes through (for divisors), though, but should doublecheck this. – Dave Anderson Dec 19 2010 at 0:08
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