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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.

diracula
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  • Did you look here ? –  May 31 '17 at 21:10
  • @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? – diracula May 31 '17 at 21:47
  • 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. –  Jun 01 '17 at 07:30
  • 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 ? –  Jun 01 '17 at 07:33
  • @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. – diracula Jun 01 '17 at 11:28
  • You are right this does not make sense, it was an abuse of notation of mine (I usually write $O_Y(\dots) = O_X(\dots){|Y}$ but this is a mistake) but if instead I re-wrote the last line as $O_X(s){|s}(f^{-1}(U)) = O_X(s){|s}(U) = N{s/X}(U) = K_s(U)$ this time this is ok right ? –  Jun 01 '17 at 11:31
  • Adjunction is not really used here, I just used that $\mathcal O_X(s){|s} = N{s/X}$ (which is proved in the link in my first comment). Also, there are cumbersome notations and details, but basically pullback does nothing because you are pulling back something but really you just extend by zero and then restrict. So really nothing happens and you just end with $\mathcal O_X(s)_{|s}$ which was $K_s$. –  Jun 01 '17 at 11:34
  • @N.H. I believe adjunction is used to say $N_{s/X}=K_s$. (And actually this uses that $K_X=\mathcal{O}_X$, which it seems is required for the result I stated to be correct.) I agree the pushforward is trivial, but it would be nice to see the restriction working explicitly. – diracula Jun 01 '17 at 11:38
  • Ok yes of course you're right we are using adjunction. Sorry for my useless comments. –  Jun 01 '17 at 11:47
  • @N.H. Your comments weren't useless at all. It appears you are correct the adjunction formula essentially gives the specific result in my question. I will alter the question slightly to make it a bit more general and to emphasise that I would like to see the explicit restrictions. – diracula Jun 01 '17 at 11:53
  • Ok thanks ! I hope someone will able to give a complete answer, I would be interested by see the details. –  Jun 01 '17 at 11:57

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