Understand why $dT$ give a global section of $N_V^* \otimes [V]$

Question

This question had been discussed in this post:Adjunction formula (Griffiths & Harris proof) ,however,I still can't get why $dT$ give a global section of $N_V^* \otimes [V]$.

The adjunction formula on Griffiths & Harris book (p. 146).states that if $V \subset M$ is a smooth analytic hypersurface then we have an isomorphism $N^*_V \simeq [-V]|_V$, where $N_V$ is the normal bundle of $V$ and $[-V]$ the line bundle associated to the divisor $-V$.

The strategy is to show that $N_V^* \otimes [V] \simeq \mathcal{O}_Y$ (the trivial line bundle over $Y$) by constructing a nonvanishing global section.

If $V$ is defined by $T=\left\{f_i \right\}$ on $U_i$ then the cocycles of $[V]$ are $f_{ij}=f_i/f_j$ and $df_i$ is a section of $N^*_V$. On the other hand, using the product rule for the derivative one gets that $df_i = f_{ij} df_j$ and hence glue to a section of $[V]$. The book states then that the $dT$ give a global section of $N_V^* \otimes [V]$. Why is that?I know a line bundle is trivial iff it admits a nowhere zero global section.The smoothness of hypersurface gurantees $df_i\not=0$ on $U_i\cap V$,so how can we explain $dT$ give a global section of $N_V^* \otimes [V]$?

Thanks in advance!

Answer 1

$df_i$ gives a non-vanishing holomorphic $1$-form on $U_i$. Since its restriction to $V\cap U_i$ vanishes, it is a section of $N^*_V\big|_{U_i}$. Because $df_i = f_{ij} df_j$, these transform by the transition functions of $[-V]$, which is to say that $N^*_V \cong [-V]\big|_V$.

Here's another way to look at it. Note that $\dfrac{df_i}{f_i} = \dfrac{df_j}{f_j}$ gives a global meromorphic $1$-form with a pole along $V$, and so we have a non-vanishing global (holomorphic) section of $N^*_V\otimes [V]$.