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Given a morphism between normal varieties $f: X \to Y$, we can push forward a Cartier divisor $D$ to get a cycle $f_* D$. On the other hand, we can form the line bundle $\mathscr{L}(D)$ and push that forward to get $f_* \mathscr{L}(D)$.

Under what conditions can we get an identity of the form $f_* \mathscr{L}(D) = \mathscr{L}(f_* D)$?

Of course, we would need $f_* D$ to not be all of $Y$. Also, if $X$ is $n$ copies of $Y$ projecting down, then $f_* \mathscr{L}(D)$ is not a line bundle. So such an identity cannot hold in general.

But is there a useful set of circumstances where it does hold? For instance, if $f$ a birational contraction.

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  • $\begingroup$ (1) The Grothendieck-Riemann-Roch theorem gives an expression for the Chern classes of $f_* \mathcal L(D)$ in terms of $f_* D^n$ for various natural numbers $n$, in the case where the higher cohomology groups $R^i f_* \mathcal L(D)$ vanishes. $\endgroup$
    – Will Sawin
    Commented Mar 27, 2022 at 18:07
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    $\begingroup$ (2) For a birational contraction, I think this is true whenever $f_* \mathcal L(D)$ is a line bundle. Take a (meromorphic) global section of $\mathcal L(D)$ with divisor $D$, which gives a (meromorphic) global section of the puhsforward, and note that its order of vanishing / pole at each codimension one point of $Y$ is the same as over the inverse image in $X$, since birational contractions are an isomorphism over codimension one points of the base. $\endgroup$
    – Will Sawin
    Commented Mar 27, 2022 at 18:08
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    $\begingroup$ If $D$ is not effective, it seems to me that the result does not hold in general for birational contractions, since $f_*\mathcal{L}(D)$ is not necessarily a line bundle. For instance, if $f \colon Y \to X$ is the blow-up of a surface $X$ at a smooth point $p$, with exceptional divisor $E$, then $$f_* \mathcal{L}(-nE) = \mathfrak{m}^n_p$$ for all $n \geq 1$. $\endgroup$
    – Francesco Polizzi
    Commented Mar 28, 2022 at 8:06

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