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For a smooth complex submanifold $V\subset M$, the tangent exact sequence gives $\det TM{|}_V=\det TV\otimes \det N_V$. Dualizing yields $K_V=K_M{|}_V\otimes \det N_V$. This form of adjunction formula applies in any codimension.

For a smooth degree-$d$ hypersurface $X_d\subset {\mathbb{CP}}^n$, the adjunction formula and $[X_d]\cong \mathcal{O}(d)$ give

For a smooth degree-$d$ projective plane curve $X$, the adjunction formula gives $K_X\cong {\mathcal{O}}_X(d-3)$. A hyperplane section has degree $d$, so

The rank-two bundle of fibre spans is globally generated and splits as $\mathcal{O}(a_1)\oplus \mathcal{O}(a_2)$ with $a_1+a_2=g-2$. On its rational normal scroll, $L{|}_C$ is the degree-three pencil and $M{|}_C=K_C$. The adjunction formula then forces the displayed divisor class. Writing $a=a_2-a_1$ and $M=B+a_{2}L$ on the Hirzebruch surface gives $C\in |3B+(a+a_2+2)L|$.

For a line $L$ on a degree-$d$ smooth surface in ${\mathbb{P}}^3$, the adjunction formula gives $L^2=2-d$. If $d\ge 3$, all lines are negative curves, hence there are only countably many by countability of negative curves. Their projective Fano scheme of lines cannot have a positive-dimensional complex component, since such a component has uncountably many points. A zero-dimensional projective scheme of finite type has finitely many points, proving the assertion.

If a smooth connected complex curve $C\subset E(1)$ satisfies $C^2=k$ and has degree $d=C\cdot F$ over the base, the adjunction formula gives

For a smooth plane quartic $C\subset {\mathbb{P}}^2$, the adjunction formula gives $K_C\cong {\mathcal{O}}_C(1)$. Hence every line section $C\cap \ell$ is a canonical divisor.

On a smooth projective surface, fix an ample divisor $H$. For sufficiently large $m$, a general pencil in $|mH|$ is a Lefschetz pencil. The adjunction formula gives the displayed genus. If $m$ is divisible by four, both terms added to $1$ are even integers, so the genus is odd.

True for a smooth complex projective surface, the usual manifold convention in this question. Choose an ample divisor $H$. The auxiliary pencil-existence theorem states that, for all sufficiently large $m$, a general two-dimensional space of sections of ${\mathcal{O}}_X(mH)$ defines a Lefschetz pencil: its base points are transverse and finite, its general fibre is a smooth connected curve, and each singular fibre has exactly one ordinary node, with distinct critical values. High powers separate the required jets, so a generic pencil has precisely these singularities.

Choose such an $m$ divisible by four, and write $D=mH$. The adjunction formula for a smooth connected curve of class $D$ on a smooth projective surface is $2g-2=D\cdot (D+K_X)$. ConsequentlyWriting $m=4\ell$ gives $g=1+8{\ell }^2{H}^2+2\ell K_X\cdot H$, an odd integer. Thus a sufficiently positive four-divisible polarization gives an odd-genus Lefschetz pencil. The surface is taken to be smooth; a singular projective variety need not support a Lefschetz pencil in this smooth-manifold sense.

Work over $\mathbb{C}$, and let $C$ be a smooth projective curve of genus $g\ge 3$ and gonality three. A trigonal curve has a basepoint-free degree-three pencil, giving a finite morphism $f:C\to {\mathbb{P}}^1$ of degree three. It is not hyperelliptic, so its canonical map is an embedding. Let $D_t=f^{-1}(t)$, including its scheme multiplicities. The complete pencil has $h^0(D_t)=2$, and Riemann-Roch theorem givesThus the quotient of canonical sections evaluated on $D_t$ has dimension two, and $D_t$ spans a line in ${\mathbb{P}}^{g-1}$. This argument includes ramified fibres, rather than treating only three distinct points.

These two-dimensional quotients vary in a globally generated rank-two bundle $E$ on ${\mathbb{P}}^1$. More explicitly, the image ofhas constant rank two by the preceding dimension count, so it is a subbundle. The equality of the rank at every fibre justifies the bundle construction. On global sections it contains every canonical section, and therefore $H^0({\mathbb{P}}^1,E)=H^0(C,K_C)$. The Birkhoff–Grothendieck theorem givesGlobal generation forces nonnegative splitting degrees. Since $h^0(E)=a_1+a_2+2=g$,The natural map from $f^{\ast }E$ to $K_C$ is a surjection at every point, since the canonical system has no base points. By projectivization by quotients it gives a map $C\to F=\mathbb{P}(E)$ over ${\mathbb{P}}^1$, with $M{|}_C=K_C$. Its composition with the tautological map of $F$ is the canonical embedding of $C$, so $C\to F$ is itself a closed embedding. Each fibre of the ruling maps to the spanning line of $D_t$; their union is the canonical scroll of a trigonal curve, and the ruling restricts to the given $g_3^1$.

Put $a=a_2-a_1$. Tensoring $E$ by $\mathcal{O}(-a_2)$ identifies the abstract ruled surface with ${\mathbb{F}}_a=\mathbb{P}(\mathcal{O}(-a)\oplus \mathcal{O})$. The section coming from the quotient onto the smaller summand has self-intersection $-a$ and will be denoted by $B$. In this normalized presentation its tautological class is $B$; restoring the twist givesThis class gives the scroll map. It is an embedding if $a_1>0$. When $a_1=0$, it contracts $B$; thus the printed embedding description needs this positivity qualification. For example, in genus three the ruled model is ${\mathbb{F}}_1$ and its image is ${\mathbb{P}}^2$; in genus four the type $(0,2)$ has a quadric-cone image. In both cases the smooth curve and its canonical image are still described by the construction above.

As a divisor on $F$, the curve meets each fibre in degree three, so $C=3M+tL$ for an integer $t$. The scroll formula givesBy adjunction formula and $M{|}_C=K_C$,Taking degree on $C$ uses $\mathrm{deg}(L{|}_C)=3$ and gives $t=4-g$. ConsequentlyWriting $d=a+a_2+2$, this is

A divisor $rB+dL$ with $r\ge 0$ on ${\mathbb{F}}_a$ is basepoint-free when $d\ge ra$. Indeed, its pushforward along the ruling splits intoif each degree is nonnegative, sections generate these coefficient functions, and the fibre monomials generate at every point. In particular $3B+dL$ is basepoint-free when $d\ge 3a$. The Bertini smoothness theorem then makes its general member nonsingular, including the equality case, where the general member avoids $B$.

Conversely, if $d<3a$, then $(3B+dL)\cdot B=d-3a<0$, so every effective member contains $B$. Because the canonical scroll has $a_1\ge 0$,The residual system $|2B+dL|$ is basepoint-free. Its general member $R$ does not contain $B$ and intersects it inThus a general divisor in the original system is $B+R$ with intersecting components and is singular at an intersection point. The fixed component occurs exactly once for a general member, since $R$ is not forced to contain it. This proves both directions, with the canonical-scroll hypotheses retained:This is also the bound on the Maroni invariant. Merely observing a fixed component would not by itself prove singularity; the positive intersection with its residual divisor is essential.

For these linear systems of divisors, take a complex K3 surface $X$ and a nef big divisor $D$, so $D^2>0$. The exceptional terminology refers to the complete system $|D|$, rather than to an arbitrary subsystem.

A monogonal linear system on a K3 surface has the formwhere $|E|$ is an elliptic pencil on a K3 surface and $\Gamma$ is its section. Its intersection with an elliptic fibre is one, and its fixed part is $\Gamma$. For a concrete existence example, take the smooth Fermat quartic surface $X\subseteq {\mathbb{P}}^3$ and two skew lines on it. Such lines can be written aswith ${\zeta }^4=({\zeta }^{\prime }{)}^4={\eta }^4=({\eta }^{\prime }{)}^4=-1$, $\zeta \ne {\zeta }^{\prime }$ and $\eta \ne {\eta }^{\prime }$. The equations show that they are disjoint, and the four partial derivatives show that the quartic is smooth. The adjunction formula gives $K_X=0$, and the hypersurface exact sequence of sheaves $0\to {\mathcal{O}}_{{\mathbb{P}}^3}(-4)\to {\mathcal{O}}_{{\mathbb{P}}^3}\to {\mathcal{O}}_X\to 0$, together with the intermediate cohomology vanishing of these twists, gives $H^1(X,{\mathcal{O}}_X)=0$. Thus it is a K3 surface, and each line has square $-2$. If $H$ is the hyperplane class, the planes through ${\Gamma }_0$ cut out ${\Gamma }_0$ plus a residual cubic pencil $|E|$, $E=H-{\Gamma }_0$. This pencil has no base point off ${\Gamma }_0$. On ${\Gamma }_0$, its two local residual equations are the two normal derivatives of the quartic equation; they cannot vanish simultaneously because $X$ is smooth. Thus it is basepoint-free, withThe degree-one map of $\Gamma$ to the pencil base makes it a section and forces connected fibres. Bertini smoothness theorem and adjunction formula then make a general fibre a smooth genus-one curve.

Now choose $D=mE+\Gamma$. For every irreducible curve other than $\Gamma$, both $E$ and the effective curve $\Gamma$ have nonnegative intersection with it. Also $D\Gamma =m-2\ge 0$, so $D$ is nef, and $D^2=2m-2>0$. On a general elliptic fibre, the restriction of ${\mathcal{O}}_X(D)$ is the degree-one bundle of the section point. Its unique section vanishes at that point. Every global section of ${\mathcal{O}}_X(D)$ therefore vanishes along a dense subset of $\Gamma$, hence along $\Gamma$ itself. Removing this fixed divisor gives exactly $|mE|$. This verifies the monogonal example directly; its general divisor is a section plus $m$ fibres.

A hyperelliptic linear system on a K3 surface is a basepoint-free nef big system whose morphism is generically two-to-one. Equivalently its general smooth curve is hyperelliptic. Let $\pi :X\to {\mathbb{P}}^2$ be a double cover branched over a smooth sextic, and put $D={\pi }^{\ast }\mathcal{O}(1)$. The double-cover construction givesThus $X$ is a K3 surface. Further,since the second summand contributes $H^0(\mathcal{O}(-2))=0$. The map given by $|D|$ is precisely the double cover. A general member is the double cover of a line branched at six points, so Riemann-Hurwitz formula gives genus two. This is the hyperelliptic example.

A trigonal linear system on a K3 surface is a basepoint-free nonhyperelliptic nef big system whose general smooth curve has gonality three. For an example, take the same smooth quartic with a line ${\Gamma }_0$ and the hyperplane system $|H|$. Its general member is a smooth plane quartic of genus three; adjunction formula identifies its plane embedding with its canonical map, so it is not hyperelliptic. The elliptic pencil $E=H-{\Gamma }_0$ hasIts restriction to a general hyperplane curve is a basepoint-free degree-three pencil. The curve is neither rational nor hyperelliptic, so its gonality is exactly three. This verifies the trigonal example.

Thus the three examples exhibit respectively a fixed section with elliptic fibres, a double-plane map, and an embedded quartic with trigonal hyperplane sections. The last example also illustrates why low-genus trigonal sections should not be confused with a blanket numerical criterion for every higher-genus model.

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.

Let $X\subset {\mathbb{P}}^3$ have degree $d\ge 3$ and hyperplane class $H$. The adjunction formula gives $K_X=(d-4)H$. For a line $L\subset X$, the same formula on the smooth rational curve $L$ yields $(K_X+L)\cdot L=-2$, henceThus all lines are negative curves, and part (b) makes their set countable. The Fano scheme of lines is a closed projective scheme of $\mathrm{Gr}(2,4)$. A positive-dimensional complex algebraic component would have uncountably many points, so this scheme is zero-dimensional. A projective scheme of finite type has only finitely many zero-dimensional points. This proves the finiteness of lines on a smooth surface of degree at least three.

The adjunction formula on the degree-$d$ projective hypersurface gives $K_X=(d-4)H$. Since $C$ is a smooth rational curve, adjunction on $C$ gives $(K_X+C)\cdot C=2g(C)-2=-2$. Its degree is $H\cdot C=a$, so the smooth rational-curve self-intersection on a projective hypersurface formula is

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.

Along the smooth complex curve, there is an exact sequence of complex vector bundlesThe Whitney sum formula for Chern classes gives $c_1(TX){|}_{\Sigma }=c_1(T\Sigma )+c_1(N_{\Sigma /X})$. The complex tangent line has first Chern number $\chi (\Sigma )=2-2g(\Sigma )$. The normal complex line is an oriented real rank-two bundle, whose Euler number is the self-intersection of the curve: a generic normal section has signed zeros equal to the intersections with a small pushed-off copy. Thusand the adjunction formula isEquivalently, since $c_1(K_X)=-c_1(TX)$, it is $2g-2=\Sigma \cdot (\Sigma +K_X)$. The derivation applies to a compact smooth complex curve; its complex orientation fixes all signs.