# 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?

• I think there is a typo, the sections $df_{i}/f_{i}$ glue to a global section, not the $df_{i}$. Sep 30, 2016 at 8:13

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 $[-V]$. )

• Dear @Georges Elencwajg,so now can we explain why "The book states then that the $df_i$ give a global section of $N_V^* \otimes [V]$" ?I still can't figure it out...Or is that just a typo? Feb 16, 2020 at 15:03
• Hello, I'm sincerely sorry because it's really an old answer, but since the normal vector w.r.t. $z^i_n$ is $\frac{\partial}{\partial z^i_n}$, thus by chain rule $\frac{\partial}{\partial z^i_n}=\frac{\partial}{\partial z^j_s}\frac{\partial z^j_s}{\partial z^i_n}$, so doesn't the cocycle $n_ij=\frac{\partial z^j_n}{\partial z^i_n}$? May 28, 2022 at 1:55

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

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.

Here is yet another way to look at this.

Since $$df_i = f_{ij} df_j$$ on $$V\cap U_i\cap U_j$$, we have that $$f_i^{-1} df_i = f_j^{-1} df_j$$ on $$V\cap U_i\cap U_j$$. This equality allows us to build a global meromorphic section of $$N_V^*$$, say $$s: V\to \mathcal{M}\otimes_{\mathcal{O}} N_V^*$$ with $$s|_{V\cap U_i}=f_i^{-1}df_i$$.

Now, the defining functions $$\{f_i,U_i\}$$ of $$V$$ are holomorphic and trivially respect the cocycle condition $$f_i=f_{ij}f_j$$ on $$U_i\cap U_j$$ so they induce a global section of $$[V]$$ in $$M$$. Call it $$f : M\to [V]$$. Note that $$f|_V$$ is a global section of $$[V]|_V$$. Also, recall that the way the section is induced is by composing each $$f_i$$ with the inverse of the trivialization on $$U_i$$, say $$\phi_i^{-1}$$, so that $$f|_{U_i} = \phi_i^{-1} f_i$$

We combine the sections $$s$$ and $$f|_V$$ into a meromorphic section of $$N_V^* \otimes [V]|_V$$ by tensoring : $$s\otimes_{\mathcal{O}}f|_V$$. The claim is that the section is holomorphic and non-vanishing. This should be clear as on each patch $$U_i$$, we have $$s\otimes_{\mathcal{O}}f|_V = f_i^{-1}df_i \otimes_{\mathcal{O}} \phi^{-1}_if_i = df_i\otimes_\mathcal{O} \phi_{i}^{-1}$$ (Since $$f_i$$ is holomorphic and $$\phi_i$$ is $$\mathbb{C}$$-linear on each fiber). Finally, $$\phi_i^{-1}$$ is non-vanishing (since it's invertible) and holomorphic.

So we have found a global section of $$N_V^* \otimes [V]|_V$$ that is both holomorphic and non-vanishing, implying that it must be the trivial bundle.