Using the method of (4.13.1), verify that there are nonsingular curves in \(\PP^3\) with \(d=8, g=6,7\); \(d=9, g=7,8,9\); \(d=10, g=8,9,10,11\). Combining with (IV, §6), this completes the determination of all possible \(g\) for curves of degree \(d \leqslant 10\) in \(\PP^3\).
Solution
Let \[ S=\operatorname{Bl}_{P_1,\ldots,P_6}\PP^2 \] be a nonsingular cubic surface, with the six points in general position. Write \[ h=\pi^*\OO_{\PP^2}(1), \qquad e_1,\ldots,e_6 \] for the usual basis of \(\Pic S\). Its hyperplane class is \[ H=-K_S=3h-e_1-\cdots-e_6. \]
1. If\[ D=ah-\sum_{i=1}^6b_i e_i, \]then a nonsingular member \(C\in|D|\) has\[ \boxed{ \deg C=3a-\sum_i b_i, \qquad g(C)=1+\frac{a^2-\sum_i b_i^2-\deg C}{2}.} \]
The intersection form is \[ h^2=1, \qquad e_i^2=-1, \qquad h\cdot e_i=e_i\cdot e_j=0 \quad(i\ne j). \] Hence \[ H\cdot D=3a-\sum_i b_i \] and \[ D^2=a^2-\sum_i b_i^2. \] Since \(H\) is the hyperplane class, \(H\cdot D\) is the degree of \(C\) in \(\PP^3\). Adjunction, using \(K_S=-H\), gives \[ 2g(C)-2 = D\cdot(D+K_S) = D^2-H\cdot D, \] which is the stated genus formula.
2. The following nine divisor classes have exactly the requesteddegree-genus pairs:\[ \begin{aligned} D_{8,6} &=8h-4e_1-3e_2-3e_3-2e_4-2e_5-2e_6,\\ D_{8,7} &=9h-4e_1-3e_2-3e_3-3e_4-3e_5-3e_6,\\ D_{9,7} &=8h-4e_1-3e_2-3e_3-2e_4-2e_5-e_6,\\ D_{9,8} &=8h-4e_1-3e_2-2e_3-2e_4-2e_5-2e_6,\\ D_{9,9} &=9h-4e_1-3e_2-3e_3-3e_4-3e_5-2e_6,\\ D_{10,8} &=8h-4e_1-3e_2-3e_3-2e_4-e_5-e_6,\\ D_{10,9} &=8h-4e_1-3e_2-2e_3-2e_4-2e_5-e_6,\\ D_{10,10} &=8h-4e_1-2e_2-2e_3-2e_4-2e_5-2e_6,\\ D_{10,11} &=9h-4e_1-3e_2-3e_3-3e_4-2e_5-2e_6. \end{aligned} \]
For these classes, respectively, \[ \begin{aligned} \left(\sum b_i,\sum b_i^2\right) &=(16,46),(19,61),(15,43),(15,41),(18,56),\\ &\qquad(14,40),(14,38),(14,36),(17,51). \end{aligned} \] Step 1 therefore gives \[ \begin{aligned} (H\cdot D,D^2) &=(8,18),(8,20),(9,21),(9,23),(9,25),\\ &\qquad(10,24),(10,26),(10,28),(10,30). \end{aligned} \] Applying the genus formula in step 1 gives, in the same order, \[ (d,g) = (8,6),(8,7),(9,7),(9,8),(9,9), (10,8),(10,9),(10,10),(10,11). \]
3. Every class in step 2 intersects each of the 27 lines on \(S\)positively.
The 27 line classes on the cubic surface are \[ E_i=e_i, \qquad L_{ij}=h-e_i-e_j, \qquad Q_i=2h-\sum_{j\ne i}e_j \] for \(1\le i<j\le6\) and \(1\le i\le6\).
For a class \[ D=ah-\sum_i b_i e_i \] from step 2, every \(b_i\) is positive, so \[ D\cdot E_i=b_i>0. \] In every displayed class the sum of the two largest multiplicities is at most \(7\), while \(a\ge8\). Hence \[ D\cdot L_{ij} = a-b_i-b_j >0. \]
Finally, if \(d=H\cdot D\), then \[ \sum_i b_i=3a-d, \] so \[ \begin{aligned} D\cdot Q_i &=2a-\sum_{j\ne i}b_j\\ &=2a-(3a-d-b_i)\\ &=d-a+b_i. \end{aligned} \] For eight of the nine classes one has \(d-a\ge0\). The only exception is \(D_{8,7}\), where \(d-a=-1\) but every \(b_i\ge3\). Thus \[ D\cdot Q_i>0 \] for every \(i\) in every case.
4. Each class in step 2 contains a nonsingular irreducible curve.
Step 2 shows that every one of the nine classes has positive self-intersection. Step 3 shows that it has nonnegative intersection with every line on the cubic surface. Therefore alternative (c) of Exercise V.4.8 applies: each class contains a nonsingular irreducible curve.
For such a curve, step 2 computes its degree and genus. Since the cubic surface is already embedded in \(\PP^3\) by \(|H|\), these are nonsingular curves in \(\PP^3\) with precisely the degree-genus pairs requested in the problem.
5. Q.E.D.
Step 2 constructs a divisor class for each requested pair, step 3 checks the line-intersection hypotheses needed for the cubic-surface existence criterion, and step 4 supplies the required nonsingular curves.