Let $X$ be a Noetherian, integral, regular scheme and let $D$ be a divisor on $X$. Moreover let $\mathcal L$ be an invertible sheaf on $X$ (a line bundle). One can be define the invertible sheaf $$\mathcal L(D):=\mathcal L\otimes_{\mathcal O_X}\mathcal O_X(D)$$ Now restrict our attention to the quotient sheaf $\mathcal L(D)/\mathcal L$. Often authors give a particular notation for this quotient and they put: $$\mathcal L(D)|_D:=\mathcal L(D)/\mathcal L\,,$$ so it seems that $\mathcal L(D)/\mathcal L$ should be some kind of "restriction of $\mathcal L(D)$ to the divisor $D$". Could you explain what is the intuition behind this notation? I don't see the naive geometry behind it.


Part of the point is that the inclusion of which the quotient is the cokernel is an isomorphism away from $D$. I would guess that you know this, but in case not: this setup comes from taking the sequence $$ 0 \to \mathcal O_X(-D) \to \mathcal O_X \to \mathcal O_D \to 0 $$

and twisting by $\mathcal L(D)$ to obtain $$ 0 \to \mathcal L \to \mathcal L(D) \to \mathcal L|_D(D) \to 0, $$

where I have been somewhat loose with the notation for the last sheaf because we can restrict before or after twisting by a line bundle.

  • $\begingroup$ Can you explain in detail the following point? " the quotient is the cokernel is an isomorphism away from D" $\endgroup$ – Dubious Oct 14 '17 at 12:29
  • $\begingroup$ @Dubious, $\mathcal O_D$ is the cokernel of the inclusion $0 \to \mathcal O_X(-D) \to \mathcal O_X$. For points not lying on $D$, the induced map on stalks is an isomorphism, so the cokernel vanishes away from $D$. $\endgroup$ – Tabes Bridges Oct 14 '17 at 21:09

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