Let \(X\) be a projectively normal curve in \(\PP^3\), not contained in any plane. If \(d=6\), then \(g=3\) or 4. If \(d=7\), then \(g=5\) or 6. Cf. (II, Ex. 8.4) and (III, Ex. 5.6).
Solution
Put \[ H=\OO_X(1). \]
1. One has\[ h^0(X,H)=4 \]and therefore\[ g\ge d-3. \]
Because \(X\) is not contained in a plane, no nonzero linear form on \(\PP^3\) vanishes identically on \(X\). Thus the restriction map \[ H^0(\PP^3,\OO_{\PP^3}(1)) \longrightarrow H^0(X,H) \] is injective. Projective normality makes the same restriction map surjective, so it is an isomorphism. Hence \[ h^0(X,H)=h^0(\PP^3,\OO_{\PP^3}(1))=4. \]
Riemann–Roch gives \[ h^0(X,H)-h^0(X,K-H)=d+1-g. \] Substituting \(h^0(X,H)=4\) yields \[ g=d-3+h^0(X,K-H)\ge d-3. \]
2. If \(d=6\), then\[ \boxed{g=3\text{ or }4}. \]
Step 1 gives \[ g\ge6-3=3. \] Since \(X\) is a nonplanar smooth space curve of degree \(6\), Castelnuovo's bound gives \[ g\le\frac{6^2}{4}-6+1=4. \] Thus \(3\le g\le4\).
3. If \(d=7\), then Castelnuovo and step 1 give\[ 4\le g\le6. \]
Step 1 gives \[ g\ge7-3=4. \] For odd degree \(7\), Castelnuovo's bound gives \[ g \le \frac{7^2-1}{4}-7+1 =6. \]
4. When \(d=7\), the case \(g=4\) is impossible.
Suppose \(d=7\) and \(g=4\). Then \[ \deg(2H)=14>2g-2=6, \] so \(2H\) is nonspecial. Riemann–Roch gives \[ h^0(X,2H)=14+1-4=11. \]
Projective normality requires the degree-two restriction map \[ H^0(\PP^3,\OO_{\PP^3}(2)) \longrightarrow H^0(X,2H) \] to be surjective. But \[ h^0(\PP^3,\OO_{\PP^3}(2))=10<11=h^0(X,2H), \] which is impossible. Hence \(g\ne4\).
5. If \(d=7\), then\[ \boxed{g=5\text{ or }6}. \]
Step 3 gives \(4\le g\le6\), and step 4 excludes \(g=4\).
6. Q.E.D.
Step 2 proves the degree-\(6\) assertion, and step 5 proves the degree-\(7\) assertion.