There is no curve of degree 9 and genus 11 in \(\PP^3\).
Hint: Show that it would have to lie on a quadric surface, then use (6.4.1).
Solution
Suppose for contradiction that \[ X\subseteq\PP^3 \] is a curve of degree \(9\) and genus \(11\). Put \[ H=\OO_X(1), \] and let \(K\) be a canonical divisor on \(X\).
1. The curve \(X\) is contained in a quadric surface.
We have \[ \deg(2H)=18, \qquad \deg K=2g-2=20. \] If \(2H\) is nonspecial, Riemann–Roch gives \[ h^0(X,2H)=18+1-11=8. \]
Suppose instead that \(2H\) is special. By Clifford's theorem, \[ h^0(X,2H)\le\frac{18}{2}+1=10. \] Equality cannot occur. Indeed, since \(0<\deg(2H)<\deg K\), the equality case of Clifford’s theorem would imply that \(X\) is hyperelliptic and \[ 2H\sim 9A, \] where \(A\) is its unique \(g^1_2\). Let \[ f:X\longrightarrow\PP^1 \] be the degree-two morphism defined by \(\abs{A}\). Then \[ \OO_X(2H)\cong f^*\OO_{\PP^1}(9). \] Under the assumed equality, both \[ H^0(\PP^1,\OO_{\PP^1}(9)) \quad\text{and}\quad H^0(X,2H) \] have dimension \(10\), so pullback identifies them. Hence the complete map defined by \(\abs{2H}\) factors through the degree-two map \(f\) and cannot be an embedding.
But \(H\) is very ample because it is the hyperplane bundle of the given embedding \(X\subseteq\PP^3\). Hence \(2H\) is also very ample: the composition of this embedding with the quadratic Veronese embedding is defined by a linear subsystem of \(\abs{2H}\) and is already a closed immersion. This contradiction excludes equality. Thus in the special case \[ h^0(X,2H)\le9. \] In either case, \[ h^0(X,2H)\le9. \]
Now \[ h^0(\PP^3,\OO_{\PP^3}(2))=10, \] so the restriction map \[ H^0(\PP^3,\OO_{\PP^3}(2)) \longrightarrow H^0(X,2H) \] has nonzero kernel. A nonzero element of the kernel is a quadratic equation vanishing on \(X\). Hence \(X\) lies on a quadric surface \(Q\).
2. The quadric \(Q\) cannot be a union of planes or a double plane.
First, \(X\) is not contained in a plane. If it were, then it would be a nonsingular plane curve of degree \(9\), and the plane-curve genus formula would give \[ g(X)=\frac{(9-1)(9-2)}2=28, \] contrary to \(g(X)=11\).
If \(Q\) is a union of two planes, the irreducibility of \(X\) forces \(X\) to lie in one of them. If \(Q\) is a double plane, its underlying reduced surface is a plane containing \(X\). Both possibilities contradict the preceding paragraph.
3. The quadric \(Q\) cannot be nonsingular.
If \(Q\) is nonsingular, then \[ Q\cong\PP^1\times\PP^1, \] and FE-CRVQUAD writes the divisor class of \(X\) as \((a,b)\) with \[ \deg X=a+b, \qquad g(X)=(a-1)(b-1). \] Thus \[ a+b=9 \] and \[ (a-1)(b-1)=11. \] The second equation and \(a+b=9\) give \[ ab=19. \] Hence \(a\) and \(b\) would be roots of \[ t^2-9t+19=0, \] whose discriminant is \[ 81-76=5, \] not a square in \(\ZZ\). No integers \(a,b\) satisfy the required equations. Therefore \(Q\) is not nonsingular.
4. The quadric \(Q\) cannot be an irreducible quadric cone.
By step 2, a singular containing quadric that remains is an irreducible quadric cone. The cone calculation in FE-CRVQUAD says that an integral curve of odd degree \[ d=2a+1 \] on such a cone has arithmetic genus \[ p_a=a(a-1). \] Here \[ 9=2\cdot4+1, \] so any integral degree-\(9\) curve on the cone has \[ p_a=4\cdot3=12. \] Since \(X\) is nonsingular, its arithmetic and geometric genera agree, giving \[ g(X)=12, \] contrary to \(g(X)=11\).
5. Q.E.D.
Step 1 forces \(X\) onto a quadric. Steps <1>2–<1>4 exhaust the possible quadric surfaces and give a contradiction in every case. Therefore no curve of degree \(9\) and genus \(11\) exists in \(\PP^3\).