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The top exterior power of a rank-$r$ vector bundle is its determinant line bundle. A short exact sequence of vector bundles gives a canonical isomorphism of the middle determinant with the tensor product of the other two. Locally, a subbundle frame followed by lifts of a quotient frame defines it; changing lifts only adds off-diagonal blocks. This is the determinant step in adjunction for a smooth submanifold.

The derivative of a section defining a regular zero locus patches intrinsically along that locus because changes of trivialization introduce derivative terms multiplied by the vanishing section. Its tangent kernel and surjectivity give $N_V\cong E{|}_V$. Combining this with adjunction for a smooth submanifold gives $K_V=(K_M\otimes \det E){|}_V$.

The holomorphic normal bundle is the quotient $N_V=T^{1,0}M{|}_V/T^{1,0}V$, of rank $r$. The tangent sequence is a short exact sequence of holomorphic vector bundles:Taking determinant line bundles gives $\det (T^{1,0}M){|}_V\cong \det (T^{1,0}V)\otimes \det N_V$. Dualizing and rearranging proves adjunction for a smooth submanifold:The determinant identity follows locally by adjoining lifts of a quotient frame to a subbundle frame; changing the lifts adds only off-diagonal blocks, so does not change the determinant.

For a smooth divisor, choose reduced local defining functions $f_i$, with $f_i=u_{ij}{f}_j$ and $u_{ij}$ holomorphic and nowhere zero. The holomorphic line bundle associated to a divisor $[V]={\mathcal{O}}_M(V)$ has local frames $e_i=f_i^{-1}$, with $e_j=u_{ij}{e}_i$. The products $f_i{e}_i$ define its canonical section vanishing on $V$. Along $V$,Thus $v\mapsto d{f}_i(v)e_i$ patches to a map $T^{1,0}M{|}_V\to [V]{|}_V$, whose kernel is $T^{1,0}V$ and which is surjective because each reduced defining function has nonzero differential there. Hence $N_V\cong [V]{|}_V$. This is the normal bundle of a smooth analytic hypersurface. The divisor construction uses a closed hypersurface: an arbitrary nonclosed embedded hypersurface need not be a divisor. For example, a punctured line in ${\mathbb{C}}^2$ cannot be the support of a divisor, since its missing limit point would also lie in every local holomorphic zero set containing that line.

Let $s$ be the section of $E$ defining $V$. The intended hypothesis is that it is a regular zero locus: $ds$ has rank $r$ on $V$. In a local trivialization, differentiation of the component functions definesThis is intrinsic: differentiating a change-of-frame matrix introduces terms multiplied by $s$, which vanish on $V$. Its kernel is $T^{1,0}V$, so it induces an isomorphism of holomorphic normal bundles. ConsequentlyThis is the normal bundle of a regular zero locus. Here $\det E=L_1\otimes \cdots \otimes L_r$.

Regularity is essential if “vanishing” is read only as equality of underlying sets. On ${\mathbb{P}}^2$, the section $Z_0^2$ of $\mathcal{O}(2)$ vanishes set-theoretically on a smooth line $V$. Its derivative vanishes on $V$, andThus smoothness of the underlying zero set alone does not establish the claimed normal-bundle isomorphism. The preceding proof applies when the equations define the reduced smooth submanifold, equivalently are transverse to zero.

For Complex projective space, put $\mathcal{O}(a)=[aH]$. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Quotienting $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1})$ by its scalar maps into $\ell$ yields the Euler sequenceIts determinant gives $K_{{\mathbb{P}}^n}\cong \mathcal{O}(-n-1)=[-(n+1)H]$. For a smooth complete intersection defined regularly by the homogeneous equations, apply the normal-bundle result to $E={⨁}_i\mathcal{O}(d_i)$:This is the canonical bundle of a regular projective complete intersection. The same regularity qualification is needed here: the squared-line example would otherwise predict ${\mathcal{O}}_{{\mathbb{P}}^1}(-1)$ for its canonical bundle, whereas the line has $K_V={\mathcal{O}}_{{\mathbb{P}}^1}(-2)$.