Assume \(X\) is not hyperelliptic.
The curves of genus 5 whose canonical model in \(\PP^4\) is a complete intersection \(F_2 . F_2 . F_2\) form a family of dimension 12.
\(X\) has a \(g_3^1\) if and only if it can be represented as a plane quintic with one node. These form an irreducible family of dimension 11. Hint: If \(D \in g_3^1\), use \(K-D\) to map \(X \to \PP^2\).
* In that case, the conics through the node cut out the canonical system (not counting the fixed points at the node). Mapping \(\PP^2 \to \PP^4\) by this linear system of conics, show that the canonical curve \(X\) is contained in a cubic surface \(V \subseteq \PP^4\), with \(V\) isomorphic to \(\PP^2\) with one point blown up (II, Ex. 7.7).
Furthermore, \(V\) is the union of all the trisecants of \(X\) corresponding to the \(g_3^1\) (5.5.3), so \(V\) is contained in the intersection of all the quadric hypersurfaces containing \(X\). Thus \(V\) and the \(g_3^1\) are unique.
Note. Conversely, if \(X\) does not have a \(g_3^1\), then its canonical embedding is a complete intersection, as in (a). More generally, a classical theorem of Enriques and Petri shows that for any nonhyperelliptic curve of genus \(g \geq 3\), the canonical model is projectively normal, and it is an intersection of quadric hypersurfaces unless \(X\) has a \(g_3^1\) or \(g=6\) and \(X\) has a \(g_5^2\). See Saint-Donat.
(Erratum to part (b))
The word “node” in part (b) is too restrictive. The correct statement is that \(X\) is trigonal if and only if it is the normalization of a plane quintic having exactly one singularity of delta invariant \(1\). That singularity can be an ordinary node or an ordinary cusp. In part (c), “the node” should accordingly be replaced by “the double point”.
The cuspidal case really occurs. Over a field of characteristic zero, the plane quintic \[ C=V\left(y^2z^3-x^3z^2+x^5+y^5\right) \subseteq\PP^2 \] has an ordinary cusp at \([0:0:1]\) and no other singularity. Step <1>9 checks that its normalization is a nonhyperelliptic trigonal curve of genus \(5\) and, using the uniqueness proved below, cannot have a nodal plane quintic model. Thus replacing “node” by “node or ordinary cusp” is necessary, not merely a weakening of the proof.
Solution
Let \(K\) denote a canonical divisor on \(X\). We work over the algebraically closed ground field of Chapter IV.
1. Smooth complete intersections of three quadrics in \(\PP^4\) form anirreducible family of canonically embedded genus-\(5\) curves of dimension\[ \boxed{12}. \]
Put \[ W_2=H^0\bigl(\PP^4,\OO_{\PP^4}(2)\bigr), \qquad \dim W_2=15. \] A complete intersection of three quadrics is determined by the three-dimensional subspace of \(W_2\) spanned by its equations. Hence such complete intersections are parametrized by the open subset \[ U\subseteq\operatorname{Gr}(3,W_2) \] on which the common zero locus is a smooth curve of codimension \(3\). This open set is nonempty and irreducible, and \[ \dim U=3(15-3)=36. \]
For \(C\in U\), adjunction gives \[ K_C = \bigl(K_{\PP^4}+2H+2H+2H\bigr)|_C =H|_C. \] Moreover \(\deg C=2^3=8\), so \[ 2g(C)-2=\deg K_C=8, \] and therefore \(g(C)=5\). Thus the displayed embedding is the canonical embedding.
Conversely, for a canonical curve which is such a complete intersection, its three-dimensional space of quadratic equations recovers the point of \(\operatorname{Gr}(3,W_2)\). Isomorphic canonical curves differ by an element of \(\PGL_5\). Since \[ \dim\PGL_5=25-1=24 \] and a smooth curve of genus \(5\) has a zero-dimensional automorphism group, the family of isomorphism classes has dimension \[ 36-24=12. \] Irreducibility follows from that of \(U\).
2. If \(A\) is a \(g^1_3\) on \(X\), then \(A\) is base-point-free and\[ L:=K-A \]is a base-point-free \(g^2_5\).
If \(A\) had a base point \(P\), then \[ h^0(A-P)=h^0(A)=2, \] so \(A-P\) would be a \(g^1_2\), contradicting the hypothesis that \(X\) is not hyperelliptic. Thus \(A\) is base-point-free.
Since \(\deg K=8\), one has \(\deg L=5\). Riemann–Roch gives \[ h^0(L)-h^0(A)=\deg L+1-g=1, \] and hence \[ h^0(L)=3. \]
Suppose \(P\) were a base point of \(L\). Then \(h^0(L-P)=3\), while Riemann–Roch applied to \(L-P=K-A-P\) gives \[ h^0(L-P)=h^0(A+P). \] Thus the degree-\(4\) divisor \(A+P\) would satisfy \(h^0(A+P)=3\), attaining equality in Clifford’s theorem. Since \(0<\deg(A+P)<2g-2\), the equality case of Clifford’s theorem would make \(X\) hyperelliptic, a contradiction. Therefore \(L\) is base-point-free, and its degree and number of sections make it a \(g^2_5\).
3. The morphism\[ \phi_L:X\longrightarrow\PP^2 \]is birational onto an integral plane quintic whose only singularity hasdelta invariant \(1\).
Because \(|L|\) is complete and has dimension \(2\), its image is not contained in a line. If \(e\) is the generic degree of \(\phi_L\) and \(d\) is the degree of its image, then \[ 5=\deg L=ed. \] Since \(5\) is prime and \(d>1\), necessarily \[ e=1, \qquad d=5. \] Thus \(\phi_L\) is birational onto an integral plane quintic \(C\).
The arithmetic genus of a plane quintic is \[ p_a(C)=\frac{(5-1)(5-2)}2=6, \] whereas its normalization \(X\) has genus \(5\). Hence \[ \sum_{q\in\operatorname{Sing}C}\delta_q =p_a(C)-g(X)=1. \] There is therefore exactly one singular point, and its delta invariant is \(1\). Such a plane double point is either an ordinary node or an ordinary cusp. This proves the corrected forward implication in part (b).
4. Conversely, the normalization of an integral plane quintic havingexactly one singularity of delta invariant \(1\) carries a \(g^1_3\).
Let \[ \nu:X\longrightarrow C\subseteq\PP^2 \] be the normalization and let \(p\) be the unique singular point. Its multiplicity is \(2\). Blow up \(p\): \[ \pi:S=\operatorname{Bl}_p\PP^2\longrightarrow\PP^2. \] For a node or an ordinary cusp, the strict transform \(\widetilde C\) is smooth and is naturally isomorphic to \(X\).
Write \(H\) for the pullback of a line and \(E\) for the exceptional curve. Then \[ \widetilde C\sim5H-2E. \] The pencil of lines through \(p\) has strict-transform class \(H-E\), and on \(\widetilde C\) it has degree \[ (5H-2E)\cdot(H-E)=3. \] It has no base point on the blowup. Hence it cuts out a base-point-free \(g^1_3\) on \(X\). This proves the corrected converse in part (b).
5. The conics through the unique double point cut out the completecanonical system on \(X\), and they embed the blowup \(S\) as a cubic scroll\[ V\subseteq\PP^4. \]
On \(S\) one has \[ K_S=-3H+E, \qquad \widetilde C\sim5H-2E. \] Adjunction therefore gives \[ K_X = \bigl(K_S+\widetilde C\bigr)|_{\widetilde C} = (2H-E)|_{\widetilde C}. \] The divisor \(2H-E\) is exactly the strict-transform system of conics through \(p\).
Its space of global sections has dimension \(5\). The restriction map to \(X\) is injective, since \[ (2H-E)-\widetilde C=-3H+E \] has no nonzero section. Since \(h^0(X,K_X)=5\), restriction is an isomorphism. Thus the conics through \(p\) cut out the complete canonical system.
By Exercise II.7.7(c), the complete system \(|2H-E|\) embeds \(S\) in \(\PP^4\) as a smooth cubic scroll \(V\). Its restriction to \(X\) is the canonical embedding, so the canonical curve lies on \(V\). This argument applies equally to the nodal and cuspidal cases.
6. The surface \(V\) is the union of the trisecant lines corresponding tothe \(g^1_3\).
4. The ruling fibres cover\(S\), so their image lines cover \(V\). Hence \(V\) is exactly the union ofthese trisecants.
7. The common zero locus of all quadrics containing the canonical curveis exactly \(V\). Consequently both \(V\) and the \(g^1_3\) are unique.
6. The restriction of theequation of \(Q\) to\[ \ell\cong\PP^1 \]has degree \(2\), but it vanishes on the length-\(3\) subscheme\(\ell\cap X\). It must therefore vanish identically. Hence every quadriccontaining \(X\) contains every ruling line, and therefore contains \(V\):\[ H^0(\PP^4,\mci_X(2)) \subseteq H^0(\PP^4,\mci_V(2)). \]
Exercise II.7.7(c) gives \(V\) scheme-theoretically as the common zero locus of three independent quadrics; equivalently, \[ h^0(\PP^4,\mci_V(2))=3. \] On the other hand \[ h^0(\PP^4,\OO(2))=15, \qquad h^0(X,2K)=16+1-5=12, \] so the restriction map to \(H^0(X,2K)\) has kernel of dimension at least \(15-12=3\). Therefore \[ h^0(\PP^4,\mci_X(2))\ge3. \] The preceding inclusion forces equality: \[ H^0(\PP^4,\mci_X(2)) = H^0(\PP^4,\mci_V(2)). \] Thus the intersection of all quadrics through \(X\) is precisely \(V\).
Now start with any \(g^1_3\) on \(X\). Steps <1>2–<1>6 construct from it a cubic scroll, and the argument just given identifies that scroll with the same intrinsic base locus of the quadrics through \(X\). Hence the scroll is unique. A smooth cubic scroll \(\operatorname{Bl}_p\PP^2\) has a unique ruling by lines, and intersecting that ruling with \(X\) recovers the \(g^1_3\). The \(g^1_3\) is therefore unique as well.
8. The trigonal genus-\(5\) curves form an irreducible family of dimension\[ \boxed{11}. \]The nodal plane-quintic models form a dense open subfamily; cuspidal modelsform a proper subfamily.
Consider pairs \[ (p,C), \qquad p\in\PP^2, \] where \(C\) is a plane quintic singular at \(p\). For fixed \(p\), the condition that a quintic be singular at \(p\) imposes three independent linear conditions on \[ \PP H^0(\PP^2,\OO(5))\cong\PP^{20}. \] Thus these pairs form an irreducible projective bundle of dimension \[ 2+17=19. \]
The locus on which \(p\) is an ordinary node and \(C\) has no other singularity is nonempty and open, hence irreducible of dimension \(19\). An ordinary cusp is obtained when the quadratic tangent cone acquires a double factor while the cubic transverse term remains nonzero. This is one additional condition, and a small deformation of that quadratic term splits the double tangent into two distinct tangents. Since smoothness away from \(p\) is open, the cuspidal locus lies in the closure of the nodal locus. Therefore the locus of quintics having exactly one delta-one singularity is irreducible of dimension \(19\).
Steps <1>3–<1>4 identify its normalizations with the trigonal genus-\(5\) curves. By step 7 the \(g^1_3\) is unique, so the residual \(g^2_5=|K-A|\) is unique; choosing its three sections only changes coordinates in \(\PP^2\). Hence two such plane models represent the same curve exactly up to the action of \(\PGL_3\). Stabilizers are zero-dimensional because a genus-\(5\) curve has a zero-dimensional automorphism group. Since \[ \dim\PGL_3=8, \] the family of isomorphism classes is irreducible of dimension \[ 19-8=11. \] The nodal locus is the nonempty dense open described above; the cuspidal locus is its codimension-one specialization.
9. The cuspidal quintic displayed in the erratum gives a genuinecounterexample to the literal nodal assertion in part (b).
Assume the ground field has characteristic zero and set \[ F=y^2z^3-x^3z^2+x^5+y^5. \] At \[ p=[0:0:1], \] the affine equation is \[ y^2-x^3+x^5+y^5=0. \] On the blowup chart \(y=xu\), its strict transform has equation \[ u^2-x+x^3(1+u^5)=0. \] This is smooth at \((x,u)=(0,0)\) because its \(x\)-derivative there is \(-1\), and it meets the exceptional curve \(x=0\) there with multiplicity \(2\). Thus \(p\) is an ordinary cusp and \(\delta_p=1\).
There is no singular point on \(z=0\), because there \[ F_x=5x^4, \qquad F_y=5y^4. \] On the chart \(z=1\), a singular point must satisfy \[ x^2(-3+5x^2)=0, \qquad y(2+5y^3)=0, \qquad 3y^2-2x^3=0. \] If \(x=0\), the last equation forces \(y=0\). If \(y=0\), it similarly forces \(x=0\). In the remaining case \[ x^2=\frac35, \qquad y^3=-\frac25, \qquad 3y^2=2x^3. \] Cubing the last equality and substituting the first two gives \[ \frac{108}{25}=\frac{648}{625}x, \] so \(x=25/6\), contradicting \(x^2=3/5\). Thus \(p\) is the only singular point. Since its germ at \(p\) is unibranch, the quintic is integral: distinct projective plane components would meet by Bezout, producing either another singular point or more than one branch at \(p\).
Let \(X\) be its normalization. The genus formula gives \[ g(X)=6-1=5, \] and projection from \(p\) gives a degree-\(3\) map to \(\PP^1\). The curve is not hyperelliptic: if it also had a degree-\(2\) map to \(\PP^1\), the product of the degree-\(2\) and degree-\(3\) maps would be birational onto a curve of bidegree \((2,3)\) in \(\PP^1\times\PP^1\), because its generic degree divides both \(2\) and \(3\). Such a curve has arithmetic genus \[ (2-1)(3-1)=2, \] contradicting \(g(X)=5\).
Finally, let \(M\) be the line bundle of any birational plane quintic model of \(X\). Then \(\deg M=5\) and \(h^0(M)\ge3\). Riemann–Roch gives \[ h^0(K-M)=h^0(M)-1\ge2, \] so \(K-M\) is a degree-\(3\) pencil. Clifford’s theorem forces \(h^0(M)=3\), and step 7 says that this \(g^1_3\) is the unique one. Hence \[ M=K-A \] for the unique trigonal pencil \(A\), so the plane quintic model is unique up to projective coordinates. The model above is cuspidal, and therefore this \(X\) has no nodal plane quintic model. This proves the erratum asserted above.
10. Q.E.D. for the corrected statement.
Step 1 proves part (a). Steps <1>2–<1>4 prove the corrected equivalence in part (b), and step 8 proves its irreducibility and dimension claim. Steps <1>5–<1>7 prove all the cubic-scroll, trisecant, and uniqueness assertions in part (c). Step 9 proves that the printed word “node” cannot be retained in general.