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Let $X$ be a complex variety with trivial canonical bundle $K_B=\mathcal{O}_X$, with a projection morphism $f$ onto a complex codimension one subspace $s$. Consider the sheaf $$ \mathcal{F} := f_*\left(\mathcal{O}_X(ns) \otimes \mathcal{O}_s\right) \, , $$ where $n$ is a positive integer, $\mathcal{O}_X(ns)$ is the sheaf of meromorphic functions with divisor greater than $-ns$, and $f_*$ is the pushforward map.

I have seen it stated without justification that $\mathcal{F}=K_s^{\otimes n}$, where $K_s$ is the canonical bundle of $s$. What I would like to see is a derivation of this result that explicitly uses the definitions of pushforwards and the sheaves $\mathcal{O}_X(ns)$ and $\mathcal{O}_s$.


Note: I have edited the post to say $\mathcal{O}_X(ns) \otimes \mathcal{O}_s$ instead of $\mathcal{O}_X(ns)|_s$. The latter notation is used in the reference where the result is stated, but seems to be an abuse of notation. Thanks to user347489 for pointing this out.

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  • $\begingroup$ Did you look here ? $\endgroup$ – user171326 May 31 '17 at 21:10
  • $\begingroup$ @N.H. Thanks for the link. Sorry if I'm being dense, but could you clarify how the adjunction formula implies this result about pushforwards? $\endgroup$ – diracula May 31 '17 at 21:47
  • $\begingroup$ Ok sorry I didn't think about the pushforward. My reasoning was to say that $\mathcal O_X(s)_{|s} = N_{s/X}$ and so $\mathcal O_X(s)_{s} = K_s$. But I don't see how to conclude that $f_*(\mathcal O_X(s)_{|s}) = K_s$, sorry my comment. $\endgroup$ – user171326 Jun 1 '17 at 7:30
  • $\begingroup$ On the other hand, $f_*(\mathcal O_X(s)_{|s})(U) = \mathcal O_X(s)_{|s}(f^{-1}(U)) = \mathcal O_s(s) (U) = K_s(U)$. So it seems that they indeed coincide. Does this make sense ? $\endgroup$ – user171326 Jun 1 '17 at 7:33
  • $\begingroup$ @N.H. Thanks for your reply. I'm not sure that $\mathcal{O}_s(s)$ makes sense since $s$ is not a divisor on $s$. However I think I can see using the adjunction formula that your second and fourth terms should be equal, which it seems effectively gives the result since the pushforward is quite trivial. However I would perhaps like to see the restriction done explicitly, to see how things work out as in the adjunction formula. $\endgroup$ – diracula Jun 1 '17 at 11:28

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