Show that a nonsingular plane curve of degree 5 has no \(g_3^1\). Show that there are nonhyperelliptic curves of genus 6 which cannot be represented as a nonsingular plane quintic curve.
Solution
Let \[ C\subseteq\PP^2 \] be a nonsingular plane quintic.
1. The canonical embedding of \(C\) is the restriction of the secondVeronese embedding\[ v_2:\PP^2\hookrightarrow\PP^5. \]
Adjunction gives \[ K_C=(K_{\PP^2}+C)|_C=\OO_C(2). \] The restriction map \[ H^0(\PP^2,\OO(2))\longrightarrow H^0(C,K_C) \] is injective, since a nonzero conic cannot contain the integral quintic \(C\). Both vector spaces have dimension \(6\): the source is the space of quadratic forms in three variables, while \[ h^0(C,K_C)=g(C) = \frac{(5-1)(5-2)}2 =6. \] Thus restriction is an isomorphism, so the canonical map is \[ C\hookrightarrow\PP^2\xrightarrow{v_2}\PP^5. \] Write \[ S=v_2(\PP^2)\subseteq\PP^5 \] for the Veronese surface containing the canonical curve.
2. If \(C\) had a \(g^1_3\), every divisor in that pencil would span atrisecant line to the canonical curve.
Let \(D\) be a divisor in a hypothetical \(g^1_3\). Then \[ \deg D=3, \qquad h^0(D)=2. \] Riemann–Roch gives \[ h^0(K_C-D) = h^0(D)+g-1-\deg D =2+6-1-3 =4. \] Canonical hyperplanes containing the length-\(3\) subscheme \(D\) correspond to the four-dimensional subspace \(H^0(K_C-D)\subseteq H^0(K_C)\). Hence the canonical image of \(D\) spans a projective space of dimension \[ h^0(K_C)-h^0(K_C-D)-1 =6-4-1 =1. \] Thus \(D\) lies on a line \[ \ell\subseteq\PP^5, \] and \(\ell\cap C\) has length at least \(3\).
3. No line in \(\PP^5\) meets the Veronese surface \(S\) in a subscheme oflength at least \(3\). Therefore \(C\) has no \(g^1_3\).
In coordinates \[ [X_0:X_1:X_2:X_3:X_4:X_5] = [x^2:y^2:z^2:xy:xz:yz], \] the Veronese surface is cut out scheme-theoretically by the quadratic \(2\times2\) minors of the symmetric matrix \[ \begin{pmatrix} X_0&X_3&X_4\\ X_3&X_1&X_5\\ X_4&X_5&X_2 \end{pmatrix}. \] If a line \(\ell\) met \(S\) in length at least \(3\), every one of these quadrics would restrict to a degree-\(2\) form on \[ \ell\cong\PP^1 \] vanishing on a length-\(3\) subscheme. Every restriction would therefore be zero, so the scheme-theoretic quadratic description would force \[ \ell\subseteq S. \]
But \(S\) contains no lines. Indeed, if \(B=v_2^{-1}(\ell)\) were such a curve, then \[ \OO_S(1)|_\ell\cong\OO_{\PP^1}(1), \] whereas \[ v_2^*\OO_S(1)=\OO_{\PP^2}(2). \] Thus \[ 1=\deg\OO_\ell(1)=\deg\OO_B(2), \] which is impossible because the latter degree is twice the positive integer \(\deg\OO_B(1)\).
Step 2 would produce exactly such a trisecant line from any \(g^1_3\). Hence a nonsingular plane quintic has no \(g^1_3\).
4. There exist nonhyperelliptic genus-\(6\) curves with a \(g^1_3\).
Take a smooth member \[ X\in\left|\OO_{\PP^1\times\PP^1}(3,4)\right|. \] Such members exist because this complete linear system is base-point-free and a general member is smooth. If \(F_1,F_2\) are the two fibre classes, then \[ X\sim3F_1+4F_2, \qquad K_{\PP^1\times\PP^1}=-2F_1-2F_2. \] Adjunction gives \[ 2g(X)-2 = X\cdot(X+K_{\PP^1\times\PP^1}) =10, \] so \[ g(X)=6. \] One of the two projections \[ X\longrightarrow\PP^1 \] has degree \(3\), and hence gives a base-point-free \(g^1_3\).
The curve \(X\) is not hyperelliptic. If it also admitted a degree-\(2\) map to \(\PP^1\), the product of that map with the degree-\(3\) projection would have generic degree dividing both \(2\) and \(3\), hence would be birational onto its image in \(\PP^1\times\PP^1\). That image would have bidegree \((2,3)\) and arithmetic genus \[ (2-1)(3-1)=2, \] contradicting \(g(X)=6\).
5. The curves from step 4 cannot be represented as nonsingular planequintics.
The existence of a \(g^1_3\) is intrinsic to the curve. By step 3, a nonsingular plane quintic has no such pencil. Therefore the nonhyperelliptic genus-\(6\) curves constructed in step 4 admit no nonsingular plane quintic model.
6. Q.E.D.
Steps <1>1–<1>3 prove the first assertion, and steps <1>4–<1>5 prove the second.