2.7 Main Proof
Theorem statement here
Proof
The proof of this is detailed in a nice amount of detail in stack overflow, see code comment
Theorem statement here.
Let \(X\) be a smooth variety and \(D\) a (smooth?) divisor. Then
\[ (\omega _{X} \otimes \mathcal{O}_X( D))|_{D} \cong \omega _{D} \]
Proof
We have an exact sequence
\[ 0 \to \mathscr {I} / \mathscr {I}^{2}\to \Omega _X|_D \to \Omega _D \to 0 \]
from the conormal exact sequence, and it is exact on the left by Theorem 2.6.2. We apply fact that the determinant is “multiplicative” on short exact sequences,, concluding that
\[ \omega _D \otimes \mathscr {I} / \mathscr {I}^{2}\cong \omega _X|_D . \]
Note that \(\mathscr {I} / \mathscr {I}^{2}\) has rank one, so it’s determinant bundle is itself. Finally, by the alternate description of the conormal sheaf (reference), we tensor both sides by \(\mathcal{O}_X(D)\) which is the inverse of the conormal sheaf, and we conclude the thoerem.