Adjunction formula (Griffiths & Harris proof) I'm having trouble understanding the proof of the adjunction formula on Griffiths & Harris book (p. 146).
The formula 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 $f_i$ on $U_i$ then the cocicles 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 $df_i$ give a global section of $N_V^* \otimes [V]$. Why is that?
Is this statement true? I see that if one has sections $s$ of $L$ and $s'$ of $L'$ then $s \otimes s'$ is a section of $L \otimes L'$. In our case we know that $df_i$ is a section of both $N_V^*$ and $[V]$ and so we get that $df_i \cdot df_i$ (and not $df_i$) is a section of $N_V^* \otimes [V]$. What am I missing here?
 A: The normal bundle
Choose on each $U_i$ a coordinate system $(z_1^{i},...,z_n^{i})$ such that $U_i\cap V$ is given by the equation $z_n^{i}=0$, so that $f_i=z_n^{i}$ .
The normal bundle $N_V$ on $V \;$ is then given by the cocycle  
$$n_{ij}=\frac {\partial z_n^{i}}{\partial z_n^{j}} \in \mathcal O^*(V \cap U_{ij})$$ Note carefully that this bundle is defined only on $V$ , and not on $M$. 
The bundle associated to $V$
In the same notation the bundle $\mathcal O(V)=[V]$ is defined by a cocycle $g_{ij}\in \mathcal O^*(U_{ij})$ satisfying
$z^i_n=z^j_n.g_{ij}$ on $U_{ij}$.
 Taking partial derivatives with respect to $z^j_n$ yields $\frac {\partial z_n^{i}}{\partial z_n^{j}} = g_{ij}+z^j_n.\frac {\partial g_{ij}}{\partial z^j_n} $.
Restricting this to $V \cap U_{ij}$, we get a cocycle for $[V]|V$ ( remember that $z^j_n=0$ on $V\cap U_j$) $$\frac {\partial z_n^{i}}{\partial z_n^{j}} = g_{ij} \in \mathcal O^*(V \cap U_{ij})$$ 
The two displayed equations prove that $n_{ij}=g_{ij}$ and thus that
 $$N_V \simeq[V]|V$$ 
(As you see, I prefer to  avoid differential forms and dualization : no $N_V^*$, no $[-V]$. )
A: The $df_i$ define local frames for $N_V^*$. These induce holomorphic local trivializations of $N_V^*$ which, because $df_i = f_{ij}df_j$ on V, have transition functions $f_{ij}^{-1}$, suitably restricted. This indicates that $N_V^*\simeq [-V]|_V$. 
A: Let me explain this problem more explicitly.
We first should observe the following fact:Let $\varphi=(\varphi^1,...,\varphi^n)$ be a holomorphic map $U\rightarrow\mathbb C^n$ of some connected open neighbourhood $U\subset\mathbb C^n$of the origin,such that $\varphi ^n(z_1,...,z_{n-1},0)=0$ for   all $(z_1,...,z_{n-1},0)\in U$.Then using identify theorem,we have $\varphi ^n(z)=z_n\cdot h(z_1,...,z_n)$,where $h(z_1,...,z_n)$ is a power series in $z_1,...,z_n$.Hence,
$$\frac {\partial\varphi^n} {\partial z_k} (z_1,...,z_{n-1},0)=\begin{cases} 0&\mbox{for } k\mbox  { $=1,...,n-1$}\\h(z_1,...,z_{n-1},0) &\mbox{for }k=n  \end{cases} $$
Now,let us apply this to our situation. We may fix local coordinates $\varphi _i:U_i\cong\varphi_{i}(U_i)\subset\mathbb C^n$,such that $\varphi_{i}(U_i\cap V)= \left\{\ (z_1,...,z_n)\in\varphi_i(U_i)\, |\,z_n=0  \right\}$.Since $V$ is defined by $f_i$ on $U_i$,we get $f_i=\varphi_i^{n}$.
Also,it is easy to verify that the transition maps $\varphi_{ij}:\varphi_j(U_i\cap U_j)\cong\varphi_i(U_i\cap\ U_j)$ have the above property!
So,on $V\cap\ U_i\cap\ U_j$,$$df_i(x)=d(\varphi_{ij}\circ\varphi_j)^n(x)=\frac {\partial\varphi_{ij}^{n}} {\partial z_n}dz_n\circ \varphi_j(x)=h(z_1,...,z_{n-1},0)dz_n\circ \varphi_j(x)$$ is a section of $\mathcal N_V^*$, we have $f_i/f_j=f _{ij}$,so $df_i=f_{ij}df_j+f_{j}df_{ij}=f_{ij}df_j$.Using cocycle condition,we can solve this problem.
