A curve of degree 5 and genus 2 in \(\PP^3\) is contained in a unique quadric surface \(Q\). Show that for any abstract curve \(X\) of genus 2, there exist embeddings of degree 5 in \(\PP^3\) for which \(Q\) is nonsingular, and there exist other embeddings of degree 5 for which \(Q\) is singular.
Solution
Write line-bundle classes additively, and let \(K\) denote a canonical divisor. For an embedding \(X\subseteq\PP^3\), put \(H=\OO_X(1)\).
1. Every curve of degree \(5\) and genus \(2\) in \(\PP^3\) is contained in aquadric surface.
Since \(\deg H=5\), \[ \deg(2H)=10. \] Riemann–Roch gives \[ h^0(X,2H)-h^0(X,K-2H)=10+1-2=9. \] Now \[ \deg(K-2H)=2-10=-8, \] so \(h^0(X,K-2H)=0\) and therefore \[ h^0(X,2H)=9. \] On the other hand, \[ h^0(\PP^3,\OO_{\PP^3}(2))=10. \] Thus the restriction map \[ H^0(\PP^3,\OO_{\PP^3}(2)) \longrightarrow H^0(X,2H) \] has nonzero kernel. A nonzero element of that kernel is the equation of a quadric containing \(X\).
2. The containing quadric is unique.
First, \(X\) is not contained in a plane. Otherwise its image would be a nonsingular plane quintic, whose genus is \[ \frac{(5-1)(5-2)}2=6, \] contrary to \(g(X)=2\).
Suppose that two linearly independent quadrics \(Q_1,Q_2\) contain \(X\). If they had a common surface component, that component would have to be a plane. Writing their equations as \[ q_1=\ell m_1, \qquad q_2=\ell m_2, \] the integrality of \(X\) and the fact that \(X\not\subseteq V(\ell)\) would force both \(m_1\) and \(m_2\) to vanish on \(X\). Since \(q_1,q_2\) are independent, \(m_1,m_2\) are independent, so \(X\) would lie on the line \(V(m_1,m_2)\), impossible.
Hence \(Q_1\) and \(Q_2\) have no common surface component. Their intersection is then a complete-intersection curve of degree \[ 2\cdot2=4. \] Every irreducible curve component of that intersection has degree at most \(4\), so it cannot contain the degree-\(5\) curve \(X\). This contradiction proves uniqueness.
3. Every line bundle \(L\) of degree \(5\) on an abstract genus-\(2\) curvedefines a degree-\(5\) embedding\[ \varphi_L:X\hookrightarrow\PP^3. \]
By Exercise IV.3.1, every divisor of degree \(5\) on a genus-\(2\) curve is very ample. Moreover, \[ \deg(K-L)=2-5=-3, \] so Riemann–Roch gives \[ h^0(X,L)=5+1-2=4. \] Thus the complete linear system \(\abs{L}\) embeds \(X\) in \(\PP^3\). Its hyperplane bundle is \(L\), so the image has degree \(5\).
4. Let \(L\) have degree \(5\), and put\[ M=L-2K, \qquad D=L-K. \]If \(M\) is non-effective, then the unique quadric containing\(\varphi_L(X)\) is nonsingular.
Here \[ \deg M=1, \qquad \deg D=3. \] Since \(\deg(K-D)=-1\), Riemann–Roch gives \[ h^0(X,D)=2. \] We claim that \(\abs{D}\) is base-point free. If \(P\) were a base point, then \[ h^0(X,D-P)=h^0(X,D)=2. \] The divisor \(E=D-P\) has degree \(2\), and Riemann–Roch gives \[ h^0(X,E)-h^0(X,K-E)=1. \] Hence \(h^0(X,K-E)\ge1\). But \(K-E\) has degree \(0\), so it must be linearly equivalent to \(0\). Thus \[ D-P\sim K, \] and therefore \[ M=D-K\sim P, \] contradicting the assumption that \(M\) is non-effective. Hence \(D\) is a base-point-free pencil.
By Exercise IV.1.7, \(\abs{K}\) is also a base-point-free pencil. Its evaluation sequence, tensored by \(\OO_X(D)\), is \[ 0 \longrightarrow \OO_X(D-K) \longrightarrow H^0(X,K)\otimes\OO_X(D) \longrightarrow \OO_X(K+D) \longrightarrow0. \] On global sections, the kernel of multiplication \[ H^0(X,K)\otimes H^0(X,D) \longrightarrow H^0(X,L) \] is therefore \(H^0(X,D-K)=H^0(X,M)=0\). Both source and target have dimension \(4\), so multiplication is an isomorphism.
Consequently the complete map \(\varphi_L\) is the composite \[ X \xrightarrow{(\varphi_K,\varphi_D)} \PP^1\times\PP^1 \xrightarrow{\text{Segre}} \PP^3. \] Its image is therefore contained in the smooth Segre quadric. By step 2, this is the unique quadric containing \(\varphi_L(X)\).
5. If \(M=L-2K\) is effective, then the unique quadric containing\(\varphi_L(X)\) is singular.
Since \(\deg M=1\), effectivity gives \[ M\sim P \] for some point \(P\in X\). Let \(s\) be a nonzero section of \(\OO_X(P)\), and let \(u,v\) be a basis of \(H^0(X,K)\).
The canonical system defines the degree-two surjection \[ \varphi_K:X\longrightarrow\PP^1 \] from Exercise IV.1.7. Pullback therefore injects \[ H^0(\PP^1,\OO_{\PP^1}(2)) \hookrightarrow H^0(X,2K). \] Both spaces have dimension \(3\), since Riemann–Roch gives \[ h^0(X,2K)=4+1-2=3. \] Hence \[ u^2,\quad uv,\quad v^2 \] form a basis of \(H^0(X,2K)\).
Multiplication by \(s\) embeds this space in \[ H^0(X,2K+P)=H^0(X,L), \] so \[ su^2,\quad suv,\quad sv^2 \] are linearly independent. Extend them by a section \(w\) to a basis of \(H^0(X,L)\). In the corresponding coordinates \([x_0:x_1:x_2:x_3]\) on \(\PP^3\), every point of \(\varphi_L(X)\) satisfies \[ x_0x_2-x_1^2=0. \] Thus \(\varphi_L(X)\) lies on the quadric cone \[ Q_0=V_+(x_0x_2-x_1^2), \] which is singular at \([0:0:0:1]\). Step 2 says that there is only one containing quadric, so that quadric is \(Q_0\) and is singular.
6. For every abstract genus-\(2\) curve, both kinds of degree-\(5\) embeddingexist.
For a singular containing quadric, choose any point \(P\in X\) and set \[ L_{\mathrm{sing}}=2K+P. \] This has degree \(5\), so step 3 gives an embedding, while \[ L_{\mathrm{sing}}-2K\sim P \] is effective. Step 5 makes its unique containing quadric singular.
For a nonsingular containing quadric, fix \(P_0\in X\). By T-CRVJACFUN, the Jacobian \(\Jac(X)\) has dimension \(2\), while the Abel–Jacobi image \[ X\longrightarrow\Jac(X), \qquad P\longmapsto\OO_X(P-P_0) \] has dimension at most \(1\). Choose \[ A\in\Pic^0(X) \] outside that image, and put \[ M=A\otimes\OO_X(P_0). \] Then \(\deg M=1\). It is not effective: otherwise \(M\cong\OO_X(P)\) for some \(P\), which would give \[ A\cong\OO_X(P-P_0), \] contrary to the choice of \(A\). Now set \[ L_{\mathrm{sm}}=\OO_X(2K)\otimes M. \] Again \(\deg L_{\mathrm{sm}}=5\), so step 3 gives an embedding, and \(L_{\mathrm{sm}}-2K=M\) is non-effective. Step 4 makes its unique containing quadric nonsingular.
7. Q.E.D.
Steps <1>1–<1>2 prove existence and uniqueness of the containing quadric. Step 3 realizes every degree-\(5\) line bundle as an embedding, steps <1>4–<1>5 determine the quadric from \(L-2K\), and step 6 constructs both possibilities on every abstract genus-\(2\) curve.