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Let $Y$ be a smooth complex analytic hypersurface through the blowup centre with $[-Y]\cong K_X$. Its multiplicity at that point is one, so ${\sigma }^{\ast }Y=\tilde{Y}+E$. Combining this with the canonical bundle formula for a point blowup gives $[-\tilde{Y}]\cong K_{\tilde{X}}\otimes [E{]}^{\otimes (2-n)}$. The nontrivial powers of the exceptional line bundle of a point blowup prove the equivalence for $n\ge 2$. On a compact complex curve the premise cannot hold for an irreducible hypersurface: it is one point, so its negative divisor has degree $-1$, whereas the canonical bundle has even degree $2g-2$.

Here a Calabi-Yau threefold is a compact complex three-dimensional Kähler manifold with trivial canonical bundle and $H^1(M,\mathbb{C})=0$. The Hodge decomposition theorem for compact Kähler manifolds gives $H^1(M,{\mathcal{O}}_M)=0$, and Serre duality for compact complex manifolds then gives $H^2(M,{\mathcal{O}}_M)=0$. This supplies a positive integral Kähler class and hence a projective embedding.

For the blowup of a complex manifold at a point in complex dimension $n$, the differential gives a section of $K_{\tilde{M}}\otimes ({\varphi }^{\ast }{K}_M{)}^{-1}$ whose divisor on a complex manifold is $(n-1)E$. Indeed, a local chart has $x_i=u$ and $x_j=u{t}_j$ for $j\ne i$, so its Jacobian determinant is $\pm u^{n-1}$. The holomorphic line bundle associated to a divisor then gives the formula for the canonical bundle.

Over any field, the canonical bundle of projective space is $\mathcal{O}(-n-1)$. Indeed, dualizing the Euler sequence gives $0\to {\Omega }_{{\mathbf{P}}^n}^1\to \mathcal{O}(-1{)}^{\oplus (n+1)}\to \mathcal{O}\to 0$; taking determinants gives the result. Hence Serre duality identifies $H^i({\mathbf{P}}^n,\mathcal{O}(m))$ with the dual of $H^{n-i}({\mathbf{P}}^n,\mathcal{O}(-m-n-1))$.

Over $\mathbb{C}$, the four partial derivatives show that this projective surface is smooth. Adjunction makes its canonical bundle trivial, and the hypersurface exact sequence gives $H^1({\mathcal{O}}_X)=0$, so it is a K3 surface. Lines given by $x_0=\zeta x_1$, $x_2=\eta x_3$, with ${\zeta }^4={\eta }^4=-1$, lie on it. Choosing both roots differently gives skew lines, one of which is a section of the elliptic pencil defined by planes through the other.

A section of the cube of the canonical bundle, locally $c=\varphi (z)d{z}^3$ with holomorphic coefficient. The flat coordinates obtained by integrating cube roots have transitions $w\mapsto \zeta w+b$, ${\zeta }^3=1$; general real-linear deformations do not preserve these as holomorphic transitions.

On a Riemann surface, a holomorphic differential form is a holomorphic section of the canonical bundle. In a local complex coordinate $z$ it has the form $f(z)dz$ with $f$ holomorphic.

A section $\omega$ of the canonical bundle of a Riemann surface, locally $\omega =\varphi (z)dz$ with holomorphic $\varphi$. A nonzero form on a compact genus-$g$ surface has total zero order $2g-2$. Its integral supplies flat coordinates away from zeros.

A complex K3 surface is a compact, simply connected complex surface with trivial canonical bundle. For projective algebraic surfaces, one can equivalently require a trivial canonical bundle and $H^1(X,{\mathcal{O}}_X)=0$. Nonprojective complex K3 surfaces are also included, which is essential for the full moduli space of Ricci-flat K3 metrics.

On a compact Kähler manifold $X$, if $P$ is a positive holomorphic line bundle, then $H^i(X,K_X\otimes P)=0$ for $i>0$. The canonical-bundle twist is part of the hypothesis and conclusion; positivity alone does not remove it on a general manifold. For a complex torus, the canonical bundle is trivial, so higher cohomology of a positive line bundle vanishes without an additional twist.

On a Riemann surface, a meromorphic n-differential is a meromorphic section of the $n$th power of the canonical bundle. In a coordinate it is $F(z)(dz{)}^n$, with coefficients transforming by the $n$th power of the derivative. At a ramification coordinate $w=t^e$, its pullback has order $e{\mathrm{ord}}_w\omega +n(e-1)$. At a modular cusp, the exponential coordinate gives $dz=(h/(2\pi i))dq/q$, so a weight-$2n$ form of cusp order $v$ produces a differential of order $v-n$.

Let $C$ be the smooth algebraic curve. A nonzero regular section of a line bundle has an effective zero divisor, so its degree of a line bundle is nonnegative. Degrees add under tensor products and change sign on taking the dual. We use genuine line subbundles, so their quotients in $E$ are locally free. Both bounds can be proved constructively.

For the upper bound, take a rational frame of $E^{\ast }$ over the function field of $C$. Its two members are rational sections with only finitely many poles. Choose an effective divisor $D$ large enough to bound the pole orders of both rational sections. They become global sections of $E^{\ast }\otimes {\mathcal{O}}_C(D)$ and define a sheaf mapwhich is an isomorphism at the generic point. It is therefore injective as a sheaf map: a section in its kernel would vanish generically in a locally free sheaf, which is a torsion-free sheaf. For any line subbundle $L\subset E$, at least one of its two component maps into ${\mathcal{O}}_C(D)$ is nonzero. Such a map is a nonzero global section of ${\mathcal{O}}_C(D)\otimes L^{-1}$. Its zero divisor on an algebraic curve is effective, soThis upper degree bound for line subbundles on a smooth curve depends only on a choice made for $E$, and applies to every one of its line subbundles.

For the lower-bound question we will actually construct subbundles ${\mathcal{O}}_C(-m{p}_0)$ for every sufficiently large integer $m$, where $p_0$ is a fixed point. First prove the globally generated vector bundle property rather than assuming that negative-degree inclusions automatically are subbundles.

Apply the upper-bound argument to the rank-two vector bundle $F=E^{\ast }\otimes K_C$, where $K_C$ is the canonical bundle. Let $B$ be its upper degree bound. Choose $m$ with $m-1>B$. For any point $p\in C$, a nonzero global section of $F(-m{p}_0+p)$ would give a nonzero mapIts image can be saturated to a saturated line subbundle on a smooth curve $M\subset F$. Locally the local ring of a smooth algebraic curve is a discrete valuation ring: dividing out the common power of a uniformizer makes the image a primitive rank-one summand, and its quotient is free. Thus the saturation is indeed a subbundle. The original map gives a nonzero global section of $M\otimes {\mathcal{O}}_C(-m{p}_0+p)$ and consequently $\mathrm{deg}M\ge m-1$, contradicting the upper bound. Hence $H^0(F(-m{p}_0+p))=0$ for every $p$. By Serre duality,The exact sequenceand its long exact sequence in sheaf cohomology show that evaluation $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ is onto for every $p$. This proves the global generation of positive twists on a smooth curve with one bound on $m$ that works at every point.

Write $W=H^0(E(m{p}_0))$ and $N=\dim W$. The evaluation map $W\otimes {\mathcal{O}}_C\to E(m{p}_0)$ is surjective with kernel bundle of rank $N-2$. Consider the incidence subsetIt is the total space of this kernel bundle, so $\dim Z=1+(N-2)=N-1$. Its projection into $W$ is closed because $C$ is proper, and its image has dimension at most $N-1<N$. Choose $s$ outside this proper closed subset. It vanishes nowhere. In a local trivialization at any point, one component of $s$ is a unit, so it can be completed to a basis. Therefore ${\mathcal{O}}_C\stackrel{s}{\to }E(m{p}_0)$ is a subbundle with a locally free rank-one quotient. Twisting back givesThis proves the unbounded negative degrees of line subbundles. The nowhere-vanishing construction guarantees a subbundle at every point, including the zeros that might otherwise have occurred in a generic meromorphic inclusion. If the curve has several connected components, apply these constructions on each component and add the degrees; the same upper-bound and no-lower-bound conclusions follow.

For a complex K3 surface, $L=H^2(X,\mathbb{Z})$ is torsion-free and has its integral cup product pairingThe K3 intersection lattice isthe even unimodular integral quadratic lattice of signature $(3,19)$. Here $U$ is the hyperbolic plane with Gram matrixand $E_8(-1)$ is the negative of the positive-definite E8 lattice. Thus the rank is $3\cdot 2+2\cdot 8=22$ and the pairing has determinant of absolute value one.

These invariants can be checked from the topology and Hodge theory. The trivial canonical bundle and $H^1(X,{\mathcal{O}}_X)=0$ give $\chi ({\mathcal{O}}_X)=2$. The Noether formula gives $c_2(X)=24$, which is the topological Euler characteristic. Since $X$ is simply connected, $b_1=b_3=0$, so $24=2+b_2$ and $b_2=22$. Integral Poincare duality gives unimodularity; the universal coefficient sequence and $H_1(X,\mathbb{Z})=0$ give torsion-freeness of $H^2(X,\mathbb{Z})$.

Its second Stiefel–Whitney class is the reduction modulo two of $c_1(X)$, hence zero. The characteristic-class identitymakes the lattice even. Finally the real and imaginary parts of a holomorphic two-form give two positive directions, a Kähler class gives a third, and the primitive real $(1,1)$ classes are negative definite by the Hodge–Riemann bilinear relations. Therefore the signature is $(3,19)$. The classification of indefinite even unimodular lattices then identifies the displayed lattice uniquely. In particular the Picard group is only the integral $(1,1)$ part; it should not be confused with this rank-twenty-two topological lattice.

The Torelli theorem concerns a connected smooth projective curve over $\mathbb{C}$ and its canonical principal polarization of a complex torus. Its conclusion isThe polarization of a complex torus is part of the hypothesis. We will reconstruct the curve intrinsically from the Jacobian variety and the theta divisor, using the Gauss map of a theta divisor, and prove the reconstruction in both the nonhyperelliptic and hyperelliptic curve cases. The needed biduality and ordinary-tangency facts about projective dual varieties are the general projective-duality results permitted here.

First explain why the polarized data determine the theta divisor up to translation. On $J=\mathrm{Jac}(X)$, take integral one-form classes ${\alpha }_i,{\beta }_i$ dual to the symplectic cycle basis. Pullback by the degree-one Abelian sum map identifies them with the corresponding classes on $X$; the Riemann bilinear relations giveOn $X^{g-1}$, the sum map pulls each one-form back to the sum of its pullbacks from the factors. It has generic degree $(g-1)!$ onto $W_{g-1}$: the Abel theorem for divisors and the existence of classes with $h^0=1$ give a unique effective unordered representative, and ordering gives the factorial. Integrating a wedge of $2g-2$ basis one-forms over the product, only terms assigning a matching pair ${\alpha }_i,{\beta }_i$ to each factor survive. Dividing by $(g-1)!$ gives integral one when precisely one matching pair is omitted, and zero for all other basis wedges. Hence the Poincaré dual class of $W_{g-1}$ isThis is exactly the integral form of the canonical principal polarization of a complex torus constructed by periods.

Its associated line bundle has one-dimensional section space. One can check this without any theta multiplicity assumption: in normalized period coordinates a representative has sections satisfyingExpand the first periodicity in a Fourier series $F(z)={\sum }_{k\in {\mathbb{Z}}^g}{c}_k{e}^{2\pi i{k}^{t}z}$. The second periodicity forces $c_k=c_0{e}^{\pi i{k}^t\tau k}$. Positive definiteness of $\mathrm{Im}\tau$ makes this series normally convergent, so the section space has dimension exactly one. Other line bundles giving the same polarization of a complex torus differ by a degree-zero line bundle. The map $x\mapsto T_x^{\ast }L\otimes L^{-1}$ is an isomorphism from $J$ to its dual abelian variety for a principal polarization of a complex torus: on lattices it is the unimodular alternating form. Thus all these line bundles are translates, and so are their unique zero divisors. In particular the given polarized data recover the reduced divisor on an algebraic curve $W_{g-1}$ up to translation. Translation does not change tangent hyperplanes after identifying tangent spaces with $T_{0}J$.

Put $V=T_{0}J=H^0(K_X{)}^{\ast }$ and $P=\mathbb{P}(V^{\ast })$. At a smooth class $L={\mathcal{O}}_X(D)\in \Theta$, the Riemann singularity theorem says $h^0(D)=1$. The derivative of the Abelian sum map identifies the translated tangent hyperplane with $H^0(K_X(-D){)}^{\perp }$. Thus the intrinsic Gauss map of a theta divisor isIn particularWe will see that this map is dominant and generically finite. Its function-field extension therefore determines an intrinsic finite cover: normalize $P$ in $\mathbb{C}(\Theta )$, obtaining $Z\to P$. Normalization in a finite extension is finite here. Let $B\subset P$ be its reduced codimension-one branch locus. Using this finite cover avoids counting omitted points of the rational Gauss map of a theta divisor as ramification. Everything in this construction is determined by the polarized Jacobian variety.

Suppose first that $g\ge 3$ and $X$ is nonhyperelliptic. Its canonical map embeds it as a nondegenerate curve $C\subset \mathbb{P}(V)$. For clarity, this follows directly from the Riemann-Roch theorem: $h^0(p)=1$ on a positive-genus curve, so $K_X$ has no base point. If a length-two effective divisor $E$ had $h^0(E)\ge 2$, a nonconstant meromorphic function with poles bounded by $E$ would give a map of degree at most two to ${\mathbb{P}}^1$. Degree one forces genus zero, and degree two is the hyperelliptic curve case. Hence $h^0(E)=1$, and the Riemann-Roch theorem gives $h^0(K_X(-E))=g-2$. The canonical sections therefore separate every pair of points and every tangent direction, proving the embedding.

Let $U\subset P$ be the open set of hyperplanes transverse to $C$. A hyperplane in $U$ cuts out $n=2g-2$ distinct points. These points span the hyperplane: if a second independent canonical section vanished at all of them, its zero divisor would have to equal the same degree-$2g-2$ divisor, and the ratio of the two sections would be a constant. Consequently some $g-1$ of the intersection points are independent. For the divisor $D$ formed by such a subset, $h^0(K_X(-D))=1$ and $h^0(D)=1$.

We need to know that every $(g-1)$-subset of a general hyperplane section has this property. Here is the monodromy argument, with its essential steps. The incidence space choosing one intersection point is a projective bundle over $C$ before restricting to $U$, so it is irreducible. The incidence space choosing two distinct ordered intersection points is likewise a projective bundle over $C\times C$ minus its diagonal, with fibre ${\mathbb{P}}^{g-3}$; it too is irreducible. Restricting to transverse sections preserves irreducibility and hence connectedness of these covers. Thus the monodromy action of a covering space on the $n$ intersection points is two-transitive. A general tangent hyperplane has just one contact of order two, by characteristic-zero projective biduality theorem; locally its intersection equation is $z^2=t$ together with the other simple points. A small loop in $t$ exchanges exactly those two points. Two-transitivity conjugates this transposition to every transposition, so the monodromy is the full symmetric group ${\mathfrak{S}}_n$.

It follows that the finite étale cover of $U$ choosing a $(g-1)$-subset is connected and irreducible. The condition of dependent evaluations is closed on that cover. It is proper because we have exhibited an independent subset; its image under the finite map is therefore a proper closed subset of $U$. Outside that image every subset is independent. By the Abel theorem for divisors, distinct such subsets have distinct Abelian sums: $h^0(D)=1$ means there is only one effective representative of its class. Thus the subset cover is birational to $\Theta$ with its Gauss map of a theta divisor, andMore precisely it is exactly the restriction of $Z\to P$ over $U$, since this étale cover is normal and has the same function field.

There can consequently be no branch component over $U$. Its complement is the irreducible hypersurface $C^{\ast }$ of tangent hyperplanes, the projective dual variety of $C$. Around a general point of $C^{\ast }$, the transposition of two roots acts nontrivially on $(g-1)$-subsets: choose a subset containing exactly one of those two roots. Hence the finite cover is ramified there. This proves the exact identityThe quoted biduality theorem now recovers $C=(C^{\ast }{)}^{\ast }=B^{\ast }$ as an embedded curve, and the canonical map identifies $X$ with $C$. This proves reconstruction in the nonhyperelliptic case, including special curves rather than only a general curve in moduli.

Now suppose that $X$ is a hyperelliptic curve of genus $g\ge 3$. Let $\pi :X\to {\mathbb{P}}^1$ be its degree-two map. The Riemann-Hurwitz formula gives $2g+2$ branch points $b_1,\dots ,b_{2g+2}$. Choose an affine coordinate with no branch at infinity and writeThe forms $dx/y,xdx/y,\dots ,x^{g-1}dx/y$ are holomorphic and independent; their orders at the two points at infinity show there are exactly $g$ of them. Thus the canonical map factors as $\pi$ followed by the degree-$g-1$ rational normal curve embeddingWrite $R=\nu ({\mathbb{P}}^1)$.

A general hyperplane section of $R$ consists of $g-1$ distinct points $x_1,\dots ,x_{g-1}$ avoiding the branch points. The corresponding canonical divisor on $X$ consists of the $g-1$ pairs ${\pi }^{-1}(x_j)$. Choosing one lift from each pair gives an effective divisor $D$ with $h^0(D)=1$: the evaluation matrix of the canonical forms has Vandermonde rank $g-1$. If instead a degree-$g-1$ sub-divisor contains an entire pair, it has at least two sections, because that pair is a fibre of $\pi$ and moves in a pencil. Every choice not taking exactly one point from each pair contains an entire pair. Therefore the generic smooth theta points over the hyperplane are exactly the independent choices of one lift in each pair, andThe cover of these choices is again the restriction of $Z$ to the open set of transverse sections of $R$ avoiding the $b_i$: the identification on a dense open follows from the Abel theorem for divisors, and the choice cover is finite étale and normal. Its components would correspond to components of the function field of the irreducible theta divisor, so it is connected.

There are precisely two types of codimension-one branching. If a simple root $x_j$ crosses a branch point $b_i$, its two lifts exchange, giving nontrivial monodromy on the choices. This yields the hyperplaneIf two roots collide away from the $b_i$, a local transverse parameter gives roots $x_0\pm \sqrt{t}$. Trivialize the two sheets of $\pi$ near $x_0$. Choices selecting different sheets for the two roots are exchanged when the roots exchange; hence this also causes branching. Its locus is $R^{\ast }$, the hypersurface of tangent hyperplanes to $R$. Away from these two loci all roots and all their lifts vary locally without branching, so there are no other branch components. ConsequentlyFor $g\ge 3$, $R^{\ast }$ is irreducible and nonlinear. Indeed, hyperplane sections of the degree-$g-1$ rational normal curve are binary forms of degree $g-1$, and the repeated-root hypersurface is their irreducible discriminant of degree $2g-4\ge 2$. Thus $B$ intrinsically separates into its single nonlinear component $R^{\ast }$ and its $2g+2$ linear components. Biduality recovers $R$, while dualizing $H_i$ recovers the point $\nu (b_i)$ on $R$.

These data recover the double cover uniquely. Identify $R$ with ${\mathbb{P}}^1$ and move infinity away from the recovered branch set. Its function field isAny other quadratic extension with the same simple branch points has a defining rational function whose ratio to this product has even valuation everywhere. On ${\mathbb{P}}^1$ such a rational function is a constant times a square, and every nonzero complex constant is a square. The quadratic extensions are therefore isomorphic; their smooth projective models are isomorphic. This proves the hyperelliptic reconstruction. The branch locus also distinguishes the two cases: the nonhyperelliptic locus has one irreducible branch hypersurface, whereas the hyperelliptic locus has the additional $2g+2$ linear components.

Finally handle the small genera. In genus two, the theta divisor is $W_1$, the image of $X$ itself. The Abelian sum map on $X$ is injective because a principal divisor $p-q$ with $p\ne q$ would give a degree-one map to ${\mathbb{P}}^1$. Its derivative is nonzero since the canonical bundle has no base point, so this compact injective immersion identifies $X$ with $W_1$. In genus one the same argument identifies $X$ with its Jacobian variety after choosing an origin. In genus zero the Riemann-Roch theorem supplies a function with one simple pole, giving $X\cong {\mathbb{P}}^1$. These cases complete the proof: the canonically polarized Jacobian determines the curve in every genus.

Choose holomorphic coordinates centered at the point $p$ and a polydisc $U\subset {\mathbb{C}}^n$ about the origin. The local blowup of a complex manifold at a point is the incidence manifoldAway from $z=0$ the direction is forced to be $[\ell ]=[z]$, giving a biholomorphism with $U\setminus \{0\}$. Glue this punctured part to $X\setminus \{p\}$. The charts below make the result a complex manifold. The projection is proper near $p$, since it is the restriction of projection from a closed subset of $U\times {\mathbb{CP}}^{n-1}$; globally the modification is proper and is an isomorphism off $p$. Its exceptional divisor is

On ${\ell }_i\ne 0$, let $t=z_i$ and $u_j={\ell }_j/{\ell }_i$ for $j\ne i$. ThenThese are coordinates on the open domain where $t$ and all $t{u}_j$ lie in the original polydisc; at $t=0$ all finite $u_j$ are allowed. Their inverse reads the direction ratios and $z_i$ from the incidence point. In this chart $E$ is the nonsingular hypersurface $t=0$. On overlap with chart $k\ne i$, the holomorphic coordinate changes areThey are holomorphic with holomorphic inverses where $u_k\ne 0$, and the $n$ charts cover every point of $E$. Intrinsically, changing the original coordinates by $F$ lifts via $[\ell ]\mapsto [A(z)\ell ]$, where $F(z)=A(z)z$, $A(0)=d{F}_0$, and $A$ is holomorphic and invertible near zero. One can take $A(z)={\int }_0^{1}d{F}_{sz}ds$ on a sufficiently small polydisc. Away from the origin the lift is uniquely forced; this proves coordinate-independence of the gluing near $E$.

The canonical bundle is $K_X={\Lambda }^n{\Omega }_X^1$. In the $i$th blowup chart the top-form Jacobian iswhere the $u_j$ occur in increasing order. Every term with a second $dt$ vanishes in the wedge product of differential forms. Thus the differential gives a map ${\sigma }^{\ast }{K}_X\to K_{\tilde{X}}$ vanishing to order $n-1$ along $E$ and nowhere else. To express this through the supplied meromorphic-section hypothesis, take a nonzero meromorphic section $s$ of $K_X$ and put $D=\mathrm{div}(s)$. Locally $s=f(z)d{z}_1\wedge \cdots \wedge d{z}_n$, soHere ${\sigma }^{\ast }D$ is the total divisor pullback, including any additional exceptional multiplicity from $f$; it is not merely the strict transform. A line bundle with a nonzero meromorphic section is isomorphic to the line bundle of that section's divisor, as its local coefficient ratios are its frame transitions. Pullback of divisor line bundles and additivity of analytic divisor line bundles therefore give the canonical bundle formula for a point blowup:For $n=1$ the blowup is already an isomorphism, consistently with the zero exponent.

For the projective-plane construction, an invertible linear change of homogeneous coordinates puts $x=[1:0:0]$. Lines through $x$ are parametrized by $[a:b]\in {\mathbb{CP}}^1$, with equation $b{Z}_1-a{Z}_2=0$. Thus the required map on the punctured plane isIt is well-defined under rescaling and holomorphic on the covering sets $Z_1\ne 0$ and $Z_2\ne 0$, where its target coordinates are $Z_2/Z_1$ and $Z_1/Z_2$. These sets cover the complement of $x$.

The graph closure isOver the complement of $x$ it is the graph of $f$. In the affine chart $Z_0\ne 0$, its equation is exactly the two-dimensional local incidence model for the blowup of a complex manifold at a point, so its first projection identifies it with the required blowup globally. The second projection is a holomorphic map $\tilde{f}:S\to {\mathbb{CP}}^1$ extending $f\circ \sigma$. Explicitly, the two blowup charts giveThese formulas agree when $v=1/u$ and remain holomorphic at $t=0$. Hence the pencil extends over the exceptional divisor, where it records the projective tangent direction. This is the projective pencil resolved by a point blowup.

On a complex manifold, a divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{m}_{Y}Y$, where $Y$ ranges over irreducible complex analytic hypersurfaces and $m_Y\in \mathbb{Z}$. The local algebra needed here is that the local ring ${\mathcal{O}}_{X,x}$, identified in coordinates with convergent power-series germs, is a regular local ring and a unique factorization domain. In particular, each height-one prime is generated by one irreducible germ, and a nonzero meromorphic germ is a quotient of holomorphic germs. Reduced local equations of the same hypersurface differ by a holomorphic unit. These facts include local branches when a globally irreducible hypersurface has several branches at a singular point.

The order of a meromorphic function along a hypersurface is the exponent of its reduced local equation in the quotient's factorization. Equivalently, at a generic smooth point of $Y$, write $f=t^{m}u$, with $t$ a transverse local equation of $Y$ and $u$ a holomorphic unit; then ${\mathrm{ord}}_Y(f)=m$. Changing a reduced equation multiplies it by a unit and leaves $m$ unchanged. These local orders agree on overlaps and are constant along the generic smooth locus of an irreducible hypersurface. A nonzero global meromorphic function therefore defines its principal divisor on a complex manifoldPositive coefficients record zeros and negative coefficients record poles.

A local defining function $f_i$ for $D$ on $U_i$ is a nonzero meromorphic function whose orders there equal the coefficients of $D$. It can be constructed as the product of local reduced hypersurface equations to the prescribed integer powers. On an overlap, $f_i/f_j$ has no zero or pole along any hypersurface. The unique factorization domain property then makes it a nowhere-zero holomorphic function, after choosing representatives of the germs. The holomorphic line bundle associated to a divisor is obtained by gluing holomorphic frames usingThese ratios satisfy the cocycle condition, and the expressions $f_i{e}_i$ agree on overlaps. They form the canonical meromorphic section of a holomorphic line bundle with divisor $D$.

If $D=\mathrm{div}(f)$ globally, $f_i/f$ is a holomorphic unit on $U_i$, and $(f_i/f)e_i$ is a global nowhere-zero holomorphic section. Hence $[D]$ is holomorphically trivial. Conversely, if $[D]$ has a nowhere-zero global holomorphic frame $e$, write $e_i=a_{i}e$ with $a_i$ holomorphic units. The coefficients $f_i{a}_i$ of the canonical meromorphic section of a holomorphic line bundle agree on overlaps and give a nonzero global meromorphic function $f$. Since multiplication by a unit does not change orders, $\mathrm{div}(f)=D$. Therefore

For a smooth hypersurface $Y\subset X$, the adjunction formula isIndeed, the holomorphic conormal sequence has first term $N_{Y/X}^{\ast }\cong [-Y]{|}_Y$; taking its determinant yields $K_X{|}_Y\cong N_{Y/X}^{\ast }\otimes K_Y$, which gives the formula. Here the canonical bundle is the top exterior power of the holomorphic cotangent bundle.

Apply this first to $W\subset {\mathbb{P}}^4$. Its degree-$d_1$ equation is a section of $\mathcal{O}(d_1)$, so $[W]\cong \mathcal{O}(d_1)$. Since $K_{{\mathbb{P}}^4}\cong \mathcal{O}(-5)$, we obtain $K_W\cong {\mathcal{O}}_W(d_1-5)$. The restricted second equation cuts out $S$ in $W$ and has associated bundle ${\mathcal{O}}_W(d_2)$. A second application of adjunction formula gives the canonical bundle of a regular projective complete intersection

It remains to prove that a nonzero exponent cannot yield a trivial holomorphic line bundle on $S$. Choose a connected component $S_0$ and a hyperplane meeting but not containing it. Such a hyperplane exists: a positive-dimensional compact projective submanifold cannot lie entirely in an affine chart, since all its affine coordinate functions would be global holomorphic functions and hence constant. A hyperplane through one point and not another consequently meets $S_0$ properly. Its equation gives a holomorphic section of ${\mathcal{O}}_{S_0}(1)$ that has both a zero and a nonzero value. For $m>0$, the $m$th power has the same two properties. Were ${\mathcal{O}}_{S_0}(m)$ trivial, this section would be represented by a global holomorphic function, contradicting its constancy on compact connected $S_0$. For $m<0$, triviality would imply triviality of its positive dual power. This proves nontrivial hyperplane powers on a compact projective submanifold and henceThe degrees are positive because the two equations define a hypersurface and a surface of the specified dimensions.

The calculation uses “non-singular” in the usual smooth complete-intersection sense: the equations define reduced smooth divisors, with independent differentials along $S$. If only their reduced zero sets are required to be smooth, polynomial multiplicities cannot determine the canonical bundle. For example, $Q_1=Z_0^m$ and $Q_2=Z_1^4+Z_2^4+Z_3^4+Z_4^4$ have the same smooth quartic surface in the hyperplane $Z_0=0$ for every $m\ge 1$. That surface has trivial canonical bundle, while the displayed polynomial degree sum is $m+4$. Thus unreduced equations must not be substituted into the adjunction formula for the reduced surface.

The canonical bundle of a K3 surface is trivial, so $c_1(TX)=0$. Apply the adjunction formula with $g=3$:This is the Self-intersection of a smooth curve on a K3 surface.