If \(C\) is an irreducible non-singular curve of degree \(d\) on the cubic surface, and if the genus \(g>0\), then \[ g \geqslant \begin{cases}\frac{1}{2}(d-6) & \text { if } d \text { is even, } d \geqslant 8 \\ \frac{1}{2}(d-5) & \text { if } d \text { is odd }, d \geqslant 13\end{cases} \] and this minimum value of \(g>0\) is achieved for each \(d\) in the range given.
Solution
Let \[ S=X_6=\operatorname{Bl}_{P_1,\ldots,P_6}\PP^2 \] be the cubic surface, and put \[ H=-K_S. \] If \(C\subseteq S\) is the given nonsingular irreducible curve, write \[ D=[C], \qquad d=H\cdot D. \] Adjunction gives throughout \[ \boxed{ g = 1+\frac{D^2-d}{2}.} \]
1. Since \(g>0\), the class \(D\) is in alternative (c) of V.4.8:\[ \boxed{ D\cdot L\ge0\text{ for every line }L, \qquad D^2>0.} \]
By P-AGH548IRREDCLASSES, an irreducible curve on the cubic surface is either a line, a square-zero conic, or has the displayed properties. Lines and conics are rational, hence have genus zero. Since \(g>0\), only case (c) can occur.
2. If a line \(E\) satisfies\[ D\cdot E=0, \]then contracting \(E\) preserves the degree, self-intersection, and genus ofthe curve.
Because \(C\) and \(E\) are distinct irreducible curves with intersection zero, they are disjoint. Contract \[ \sigma:X_r\longrightarrow X_{r-1} \] with exceptional curve \(E\). Then \[ D=\sigma^*D' \] for the image curve \(C'\), and \(C\to C'\) is an isomorphism because \(C\) avoids \(E\). Thus the genus is unchanged and \[ D^2=(D')^2. \]
The canonical formula for a point blowup is \[ -K_{X_r}=\sigma^*(-K_{X_{r-1}})-E. \] Since \(D\cdot E=0\), \[ (-K_{X_r})\cdot D = (-K_{X_{r-1}})\cdot D'. \] Thus the anticanonical degree \(d\) is unchanged as well.
3. Repeat step 2 while a zero-intersection line exists. The processeither reaches \(X_2\), or stops on some \(X_r\) with \(r\ge3\) at a class,still denoted \(D\), satisfying\[ D\cdot E>0 \]for every \((-1)\)-curve \(E\).
Each contraction lowers the Picard number by one, so after at most four contractions either the degree-seven del Pezzo \(X_2\) is reached or there is no line of intersection zero. The preceding step shows that the numerical quantities \(d,D^2,g\) are unchanged throughout.
The argument in V.4.8 also shows that after each contraction the descended class remains nonnegative on every \((-1)\)-curve of the new del Pezzo surface. Hence, if the process stops before \(X_2\), all these intersections are positive integers.
4. If the process stops on \(X_r\) with \(r\ge3\), then\[ \boxed{D^2\ge2d-6.} \]
Put \[ A=-K_{X_r}, \qquad A^2=9-r\le6. \] All intersections \(D\cdot E\) with \((-1)\)-curves are positive, so \[ N=D-A \] is nonnegative on every \((-1)\)-curve. By the global-generation lemma proved in P-AGH548IRREDCLASSES, \(N\) is globally generated. In particular it is nef, so \[ N^2\ge0. \] Indeed, if \(N\not\sim0\), choose two general effective members of \(|N|\) with no common irreducible component; their intersection number is \(N^2\ge0\). If \(N\sim0\), the same inequality is immediate.
Since \[ d=A\cdot D=A^2+A\cdot N, \] one has \[ \begin{aligned} D^2 &=(A+N)^2\\ &=A^2+2A\cdot N+N^2\\ &=2d-A^2+N^2\\ &\ge2d-A^2\\ &\ge2d-6. \end{aligned} \]
5. In the situation of step 4,\[ g\ge\frac{d-4}{2}, \]which is stronger than either lower bound required in the problem.
Adjunction and step 4 give \[ g = 1+\frac{D^2-d}{2} \ge 1+\frac{(2d-6)-d}{2} = \frac{d-4}{2}. \] For even \(d\) this is larger than \((d-6)/2\), and for odd \(d\) it is larger than \((d-5)/2\).
6. It remains only to analyze the case in which the contraction processreaches\[ X_2=\operatorname{Bl}_{P_1,P_2}\PP^2. \]Write\[ D = c h + b_1(h-e_1) + b_2(h-e_2), \]with\[ c,b_1,b_2\ge0, \]as in V.4.8. Put\[ s=b_1+b_2, \qquad p=b_1b_2. \]Then\[ \boxed{ d=3c+2s, \qquad D^2=c^2+2cs+2p.} \]
The anticanonical class of \(X_2\) is \[ A_2=3h-e_1-e_2. \] The intersection table gives \[ A_2\cdot h=3, \qquad A_2\cdot(h-e_i)=2. \] Hence \[ d=A_2\cdot D=3c+2b_1+2b_2=3c+2s. \]
Also \[ h^2=1, \qquad (h-e_i)^2=0, \qquad h\cdot(h-e_i)=1, \qquad (h-e_1)\cdot(h-e_2)=1. \] Expanding \(D^2\) gives \[ D^2=c^2+2c(b_1+b_2)+2b_1b_2 = c^2+2cs+2p. \]
7. In the notation of step 6, the genus is\[ \boxed{ g = 1+\frac{c(c-3)}2+(c-1)s+p.} \]
Substitute the formulas of step 6 into adjunction: \[ \begin{aligned} g &=1+\frac{D^2-d}{2}\\ &=1+ \frac{c^2+2cs+2p-(3c+2s)}2\\ &=1+\frac{c(c-3)}2+(c-1)s+p. \end{aligned} \]
8. Suppose \(d\) is even. Then \(c\) is even.
The formula \[ d=3c+2s \] shows \[ d\equiv c\pmod2. \] Thus even \(d\) forces even \(c\).
9. If \(d\) is even and \(c=0\), then\[ \boxed{ g=(b_1-1)(b_2-1).} \]Moreover \(g>0\) forces\[ b_1,b_2\ge2. \]
For \(c=0\), step 7 gives \[ g=1-s+p = 1-(b_1+b_2)+b_1b_2 = (b_1-1)(b_2-1). \]
Also \[ D^2=2b_1b_2>0, \] so \(b_1,b_2>0\). If either were \(1\), the displayed genus would be zero. Thus positive genus forces both to be at least \(2\).
10. If \(d\) is even, \(c=0\), and \(g>0\), then\[ \boxed{g\ge\frac{d-6}{2}.} \]
Here \[ d=2(b_1+b_2)=2s. \] By step 9, put \[ x=b_1-1\ge1, \qquad y=b_2-1\ge1. \] Then \[ x+y=s-2 \] and \[ g=xy. \] Since \[ xy-(x+y-1)=(x-1)(y-1)\ge0, \] we obtain \[ g\ge x+y-1=s-3 = \frac{d-6}{2}. \]
11. If \(d\) is even and \(c\ge2\), then again\[ \boxed{g\ge\frac{d-6}{2}.} \]
Subtract the desired lower bound from the genus formula. Using \[ d=3c+2s, \] one finds \[ \begin{aligned} g-\frac{d-6}{2} &= 4+ \frac{c(c-6)}2 + (c-2)s + p. \end{aligned} \]
The even integer \(c\) is at least \(2\). If \(c=2\), the right-hand side is \[ p\ge0. \] If \(c=4\), it is \[ 2s+p\ge0. \] If \(c\ge6\), every displayed term is nonnegative and the constant term is positive. Thus the difference is always nonnegative.
12. Therefore, for every even\[ d\ge8, \]one has\[ \boxed{g\ge\frac{d-6}{2}.} \]
If the contraction process stops before \(X_2\), step 5 is stronger. If it reaches \(X_2\), steps <1>8–<1>11 prove the displayed inequality. This covers every possibility.
13. Suppose \(d\) is odd. Then \(c\) is odd.
Again \[ d\equiv c\pmod2 \] by step 6, so odd \(d\) forces odd \(c\).
14. If \(d\) is odd and \(c=1\), then\[ g=p=b_1b_2. \]Since \(g>0\), one has \(b_1,b_2\ge1\), and therefore\[ \boxed{g\ge\frac{d-5}{2}.} \]
For \(c=1\), step 7 gives \[ g=p. \] Positive genus therefore forces both \(b_i\) to be positive. Since \[ (b_1-1)(b_2-1)\ge0, \] we have \[ p=b_1b_2\ge b_1+b_2-1=s-1. \] But \[ d=3+2s, \] so \[ s-1=\frac{d-5}{2}. \]
15. If \(d\) is odd and \(c\ge3\), then\[ g-\frac{d-5}{2} = \frac72+ \frac{c(c-6)}2 + (c-2)s + p. \]
This is a direct subtraction using the formulas of steps <1>6–<1>7: \[ \frac{d-5}{2} = \frac{3c+2s-5}{2}. \] Subtracting this from \[ 1+\frac{c(c-3)}2+(c-1)s+p \] gives the displayed expression.
16. If \(d\) is odd,\[ d\ge13, \]and \(c\ge3\), then\[ \boxed{g\ge\frac{d-5}{2}.} \]
If \(c=3\), then step 15 becomes \[ g-\frac{d-5}{2} = s+p-1. \] The degree condition gives \[ 13\le d=9+2s, \] so \(s\ge2\), and therefore the difference is nonnegative.
If \(c=5\), the difference is \[ 1+3s+p>0. \] For every odd \(c\ge7\), one has \[ \frac{c(c-6)}2>0, \qquad (c-2)s\ge0, \qquad p\ge0, \] so the difference is again positive.
17. Therefore, for every odd\[ d\ge13, \]one has\[ \boxed{g\ge\frac{d-5}{2}.} \]
5. On \(X_2\), step 14 treats \(c=1\) and step16 treats every odd \(c\ge3\).
We now construct curves attaining equality.
18. Contract the four pairwise disjoint exceptional lines\[ E_3,E_4,E_5,E_6\subseteq S. \]This gives a morphism\[ \pi:S\longrightarrow X_2 = \operatorname{Bl}_{P_1,P_2}\PP^2. \]If a divisor \(M\) on \(X_2\) has a nonsingular irreducible member avoiding thefour contraction points, its pullback is a nonsingular irreducible curve on\(S\) with the same degree and genus.
The four exceptional curves are disjoint \((-1)\)-curves, so they can be contracted successively. If a curve \(C_2\in|M|\) avoids the four image points, its inverse image is its strict transform and is isomorphic to \(C_2\).
If \(D=\pi^*M\), then \(D\cdot E_i=0\) for \(i=3,\ldots,6\). Applying the canonical blowup formula four times gives \[ H_S\cdot D = (-K_{X_2})\cdot M. \] Self-intersection is also preserved under pullback: \[ D^2=M^2. \] Hence adjunction gives the same genus on both surfaces.
19. Let \(d=2s\) be even with \(d\ge8\), so \(s\ge4\). On \(X_2\) set\[ \boxed{ M_{\mathrm{ev}} = 2(h-e_1) + (s-2)(h-e_2).} \]Then\[ \deg M_{\mathrm{ev}}=d, \qquad g(M_{\mathrm{ev}})=\frac{d-6}{2}. \]
This is the notation of step 6 with \[ c=0, \qquad b_1=2, \qquad b_2=s-2. \] Thus \[ d=2(2+s-2)=2s \] and, by step 9, \[ g =(2-1)((s-2)-1) = s-3 = \frac{d-6}{2}. \] Also \[ M_{\mathrm{ev}}^2=4(s-2)>0. \]
20. A general member of \(|M_{\mathrm{ev}}|\) is nonsingular andirreducible and can be chosen to avoid the four contraction points ofstep
18. Its pullback to \(S\) attains the even minimum.
The class \(M_{\mathrm{ev}}\) is nonnegative on the three \((-1)\)-curves of \(X_2\) and has positive square. By the lemma of V.4.8, a general member is nonsingular and irreducible. The same lemma makes the system globally generated, so requiring a member to pass through any one prescribed point is a proper hyperplane condition. Hence a general smooth member avoids the four contraction points.
Step 18 then pulls it back to a nonsingular irreducible curve on the cubic surface of degree \(d\) and genus \[ \frac{d-6}{2}. \]
21. Let \(d=2s+3\) be odd with \(d\ge13\), so \(s\ge5\). On \(X_2\) set\[ \boxed{ M_{\mathrm{odd}} = h + (h-e_1) + (s-1)(h-e_2).} \]Then\[ \deg M_{\mathrm{odd}}=d, \qquad g(M_{\mathrm{odd}})=\frac{d-5}{2}. \]
Here \[ c=1, \qquad b_1=1, \qquad b_2=s-1. \] Step 6 gives \[ d=3+2(1+s-1)=2s+3. \] Step 14 gives \[ g=b_1b_2=s-1 = \frac{(2s+3)-5}{2} = \frac{d-5}{2}. \] The square is \[ M_{\mathrm{odd}}^2 = 1+2s+2(s-1) = 4s-1>0. \]
22. A general member of \(|M_{\mathrm{odd}}|\) is nonsingular andirreducible and can be chosen away from the four contraction points. Itspullback to \(S\) attains the odd minimum.
21. V.4.8 therefore gives a generalnonsingular irreducible member and global generation. Choose such a memberoutside the four hyperplanes corresponding to the contraction points, andapply step
18.
23. Q.E.D.
Steps <1>1–<1>17 prove the lower bounds. Steps <1>18–<1>20 construct an equality curve for every even \(d\ge8\), and steps <1>21–<1>22 construct one for every odd \(d\ge13\).