If \(D\) is any divisor of degree \(d\) on the cubic surface (4.7.3), show that \[ p_a(D) \leqslant \begin{cases}\frac{1}{6}(d-1)(d-2) & \text { if } d \equiv 1,2\, (\bmod 3) \\ \frac{1}{6}(d-1)(d-2)+\frac{2}{3} & \text { if } d \equiv 0\, (\bmod 3)\end{cases} \] Show furthermore that for every \(d>0\), this maximum is achieved by some irreducible nonsingular curve.
Solution
Let \[ S\subseteq\PP^3 \] be the nonsingular cubic surface, and let \(H\) denote its hyperplane class. The anticanonical description of the cubic surface gives \[ \boxed{H=-K_S,\qquad H^2=3.} \] For a divisor \(D\) of degree \(d\) we therefore have \[ d=D\cdot H. \]
1. Adjunction gives\[ \boxed{ p_a(D)=1+\frac{D^2-d}{2}.} \]
On a nonsingular surface, \[ 2p_a(D)-2=D\cdot(D+K_S). \] Since \(K_S=-H\) and \(D\cdot H=d\), \[ D\cdot(D+K_S)=D^2-d. \] Solving for \(p_a(D)\) gives the formula.
2. The self-intersection of \(D\) satisfies\[ \boxed{3D^2\le d^2.} \]
The hyperplane class \(H\) is ample. The Hodge index inequality P-AGH519HODGEINDEX gives \[ (D^2)(H^2)\le(D\cdot H)^2. \] Substituting \[ H^2=3, \qquad D\cdot H=d \] gives \[ 3D^2\le d^2. \]
3. One also has the parity condition\[ \boxed{D^2\equiv d\pmod2.} \]
Step 1 shows that \[ D^2-d=2p_a(D)-2 \] is even. Hence \(D^2\) and \(d\) have the same parity.
4. If\[ d=3m, \]then\[ \boxed{D^2\le3m^2.} \]
Step 2 gives \[ D^2\le\frac{d^2}{3}=3m^2. \] No parity correction is needed because \[ 3m^2\equiv3m=d\pmod2. \]
5. If\[ d=3m+1, \]then\[ \boxed{D^2\le3m^2+2m-1.} \]
3. The next smaller integer has the required parity, giving\[ D^2\le3m^2+2m-1. \]
6. If\[ d=3m+2, \]then\[ \boxed{D^2\le3m^2+4m.} \]
Step 2 gives \[ D^2\le\frac{(3m+2)^2}{3} = 3m^2+4m+1+\frac13, \] so \[ D^2\le3m^2+4m+1. \] The difference \[ (3m^2+4m+1)-(3m+2) = 3m^2+m-1 \] is odd. Hence the top integer allowed by Hodge has the wrong parity, and step 3 forces \[ D^2\le3m^2+4m. \]
7. If \(d\equiv0\pmod3\), then\[ \boxed{ p_a(D) \le \frac16(d-1)(d-2)+\frac23.} \]
Write \(d=3m\). By steps 1 and 4, \[ \begin{aligned} p_a(D) &\le 1+\frac{3m^2-3m}{2}\\ &= \frac{3m^2-3m+2}{2}. \end{aligned} \] On the other hand, \[ \frac16(d-1)(d-2)+\frac23 = \frac{(3m-1)(3m-2)+4}{6} = \frac{3m^2-3m+2}{2}. \]
8. If \(d\equiv1\pmod3\), then\[ \boxed{ p_a(D) \le \frac16(d-1)(d-2).} \]
Write \(d=3m+1\). Steps 1 and 5 give \[ \begin{aligned} p_a(D) &\le 1+ \frac{(3m^2+2m-1)-(3m+1)}2\\ &= \frac{m(3m-1)}2. \end{aligned} \] Since \[ \frac16(d-1)(d-2) = \frac{(3m)(3m-1)}6 = \frac{m(3m-1)}2, \] this is the required bound.
9. If \(d\equiv2\pmod3\), then\[ \boxed{ p_a(D) \le \frac16(d-1)(d-2).} \]
Write \(d=3m+2\). By steps 1 and 6, \[ \begin{aligned} p_a(D) &\le 1+ \frac{(3m^2+4m)-(3m+2)}2\\ &= \frac{m(3m+1)}2. \end{aligned} \] Also \[ \frac16(d-1)(d-2) = \frac{(3m+1)(3m)}6 = \frac{m(3m+1)}2. \] This proves the numerical inequality in every residue class.
10. Choose one of the \(27\) lines\[ L\subseteq S. \]There is a nonsingular irreducible conic \(Q\subseteq S\) with\[ \boxed{H=L+Q,\qquad Q\sim H-L.} \]
Planes in \(\PP^3\) containing \(L\) form a pencil. For such a plane \(\Pi\), the plane section \(S\cap\Pi\) has degree \(3\) and contains \(L\), so scheme-theoretically \[ S\cap\Pi=L+Q_\Pi \] for a residual conic \(Q_\Pi\).
A singular plane conic is reducible or nonreduced. In either case its support contains a line. Hence a singular residual conic would make \(\Pi\) contain a line on \(S\) other than \(L\) (or give a doubled line). Since a nonsingular cubic surface has only finitely many lines, only finitely many planes in the pencil can give such a residual conic. A general plane through \(L\) therefore gives a nonsingular irreducible conic \(Q\). Its divisor class satisfies \[ L+Q\sim H. \]
11. The line and conic classes satisfy\[ \boxed{ L^2=-1, \qquad H\cdot L=1, \qquad Q^2=0, \qquad H\cdot Q=2.} \]
The first two equalities are the defining numerical properties of a line on the cubic surface FE-SRFCUBIC. Since \(Q=H-L\), \[ H\cdot Q = H^2-H\cdot L = 3-1 =2, \] and \[ Q^2 =(H-L)^2 = 3-2-1 =0. \]
12. The residual conics cut out by planes through \(L\) form a base-pointfree pencil in \(|H-L|\). Consequently\[ \boxed{\OO_S(Q)\text{ is globally generated}.} \]
Every residual conic has class \(H-L=Q\), so these residual conics form a one-dimensional linear subsystem of \(|Q|\).
If \(x\notin L\), choose a plane through \(L\) not containing \(x\); its residual conic does not contain \(x\).
Now let \(x\in L\). Choose local homogeneous coordinates in which \[ I_L=(u,v). \] Because the cubic equation vanishes on \(L\), it can be written \[ F=uA+vB \] with quadratic forms \(A,B\). Smoothness of \(S\) at \(x\) implies \[ (A(x),B(x))\ne(0,0). \] A plane through \(L\) has equation \[ \alpha u+\beta v=0. \] After removing the component \(L\), its residual conic contains \(x\) exactly when \[ \alpha B(x)-\beta A(x)=0. \] This is one linear condition on \([\alpha:\beta]\in\PP^1\), not every member of the pencil. Hence some residual conic avoids \(x\). Thus \(|Q|\) has no base point anywhere on \(S\): already this residual-conic subpencil has no base point. Therefore \(\OO_S(Q)\) is globally generated.
13. The line bundle\[ \OO_S(H+L) \]is globally generated.
The section cutting out \(L\) gives the exact sequence \[ 0 \longrightarrow \OO_S(H) \longrightarrow \OO_S(H+L) \longrightarrow \OO_L(H+L) \longrightarrow0. \] Since \[ (H+L)\cdot L=1-1=0, \] and \(L\cong\PP^1\), one has \[ \OO_L(H+L)\cong\OO_{\PP^1}. \]
Also \[ H^1(S,\OO_S(H))=0. \] Indeed, twisting the cubic hypersurface sequence by \(\OO_{\PP^3}(1)\) gives \[ 0 \longrightarrow \OO_{\PP^3}(-2) \longrightarrow \OO_{\PP^3}(1) \longrightarrow \OO_S(H) \longrightarrow0, \] and the intermediate cohomology of projective space gives the desired vanishing.
Therefore restriction \[ H^0(S,\OO_S(H+L)) \longrightarrow H^0(L,\OO_L) \] is surjective. Choose a section restricting to the nonzero constant \(1\); it has no zero on \(L\).
Off \(L\), the subsystem obtained by multiplying hyperplane sections by the section of \(L\) has no base point because \(H\) is very ample. Hence \(\OO_S(H+L)\) is globally generated everywhere.
14. The morphism defined by \(|H+L|\) is birational onto its image.
On the open set \(S\setminus L\), the subsystem \[ s_L\,H^0(S,\OO_S(H)) \subseteq H^0(S,\OO_S(H+L)) \] has the same ratios as the hyperplane system \(|H|\). Since \(|H|\) is the given embedding \[ S\hookrightarrow\PP^3, \] these sections separate general points of \(S\setminus L\). The complete system \(|H+L|\) therefore defines a generically one-to-one morphism, hence a birational morphism onto its image.
15. For every \(m\ge1\), a general member of\[ |mH| \]is nonsingular and irreducible.
The hyperplane bundle \(\OO_S(H)\) is very ample, and every positive power of a very ample bundle is very ample. Thus \(|mH|\) is the hyperplane system of an embedding. Bertini’s hyperplane theorem gives a nonsingular general member; since \(S\) is irreducible of dimension two, a general hyperplane section is connected, hence a nonsingular connected curve is irreducible.
16. For every \(n\ge1\), the class\[ nH-L \]contains a nonsingular irreducible curve.
For \(n=1\) this is the nonsingular irreducible conic \[ Q\sim H-L \] from step 10.
12. Exercise II.7.5(d), proved inP-AGH275AMPLEPROPS, says that the tensor product of a very ample bundleand a globally generated bundle is very ample. Hence \(\OO_S(nH-L)\) is veryample. A general member is therefore nonsingular and irreducible by the sameBertini argument as in step
15.
17. For every \(n\ge1\), the class\[ nH-Q \]contains a nonsingular irreducible curve.
Since \(Q=H-L\), \[ nH-Q=(n-1)H+L. \] For \(n=1\) this is the line \(L\) itself.
For \(n=2\), the class is \[ H+L. \] By step 13 its complete linear system is base-point free, and by step 14 the associated morphism is birational, hence separable. Bertini’s theorem for a base-point-free separable linear system therefore gives a nonsingular general member. Since the associated morphism has two-dimensional image, Bertini’s irreducibility theorem for a base-point-free linear system also gives an irreducible general member.
13. Again P-AGH275AMPLEPROPS makes their tensor product very ample,and a general member is nonsingular and irreducible.
18. If\[ d=3m>0, \]the class\[ \boxed{D_m=mH} \]has degree \(d\), reaches the upper bound, and contains a nonsingularirreducible curve.
7. Step 15 supplies a nonsingular irreduciblemember.
19. If\[ d=3m+1>0, \]the class\[ \boxed{D_m=mH+L=(m+1)H-Q} \]has degree \(d\), reaches the upper bound, and contains a nonsingularirreducible curve.
Put \(n=m+1\). Since \(d>0\), one has \(n\ge1\). By step 11, \[ H\cdot D_m = 3m+1 =d, \] and \[ \begin{aligned} D_m^2 &=(mH+L)^2\\ &=3m^2+2m-1. \end{aligned} \] This is equality in step 5, so step 8 gives the maximal arithmetic genus. Since \[ D_m=nH-Q, \] step 17 gives a nonsingular irreducible member.
20. If\[ d=3m+2>0, \]the class\[ \boxed{D_m=mH+Q=(m+1)H-L} \]has degree \(d\), reaches the upper bound, and contains a nonsingularirreducible curve.
Put \(n=m+1\ge1\). Using step 11, \[ H\cdot D_m = 3m+2 =d, \] and \[ \begin{aligned} D_m^2 &=(mH+Q)^2\\ &=3m^2+4m. \end{aligned} \] Thus equality holds in step 6, and step 9 gives the maximal genus. Because \[ D_m=nH-L, \] step 16 supplies a nonsingular irreducible member.
21. Q.E.D.
Steps <1>1–<1>9 prove the stated sharp genus bound for every divisor of degree \(d\). Steps <1>10–<1>17 construct the smooth line/conic auxiliary classes and prove the needed linear systems contain nonsingular irreducible curves. Steps <1>18–<1>20 realize equality for every positive degree in the three residue classes modulo \(3\).