# 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?

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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 $[-V]$. )

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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 $N_V^*\simeq [-V]|_V$.

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