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

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 $$N_V^* \otimes [V]$$?

$$df_i$$ 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 $$N^*_V\otimes [V]$$.
• Thanks professor @Ted Shifrin ,is the last statement in your answer due to this fact:every global meromorphic 1-form $u$ with the property $div(u)+V\geqq 0$ one-to-one corresponds to $u\in H^0(M,N_V^*\otimes [V]|_V)$? Feb 25, 2020 at 8:26
• No, it corresponds to an element of $H^0(V,\Omega^1_M\otimes [V]|_V)$. But our particular $1$-form annihilates $TV$ and hence is a section of $N^*_V$. Feb 25, 2020 at 17:17