On each member of a cover, choose a reduced <local defining function of a complex analytic hypersurface> $f_i$ for $Y$. On overlaps $f_i=g_{ij}f_j$ for $g_{ij}\in\mathcal O_X^*$. Since $f_j|_Y=0$,
$$
df_i|_Y=g_{ij}df_j|_Y.
$$
Each $df_i$ annihilates $TY$ and is nonzero in the normal direction because $Y$ is smooth. If $e_i=g_{ij}^{-1}e_j$ are the frames of the <holomorphic line bundle associated to a divisor> $\mathcal O(Y)$, then
$$
[v]\longmapsto df_i(v)e_i
$$
is a well-defined nowhere-zero holomorphic map of line bundles. Thus the <normal bundle of a smooth analytic hypersurface> is
$$
\boxed{N_{Y/X}\cong\mathcal O(Y)|_Y.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 118 / 1 / c / Solution

{title2=$N_{Y/X}\cong\mathcal O(Y)|_Y$}

For a smooth <complex analytic hypersurface> $Y\subset X$, its <holomorphic normal bundle> is naturally the restriction of its <holomorphic line bundle associated to a divisor>:
$$
N_{Y/X}\cong\mathcal O(Y)|_Y.
$$
If reduced local defining functions satisfy $f_i=g_{ij}f_j$, then $df_i|_Y=g_{ij}df_j|_Y$. Pairing a normal vector with $df_i$ and using the reciprocal transition functions of $\mathcal O(Y)$ gives the isomorphism.

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