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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.