Curves of type $(a,b)$ on a quadric surface

Let $k$ be algebraically closed and let $Q\subseteq\PP_k^3$ be a smooth quadric. The Segre identification $Q\cong\PP_k^1\times\PP_k^1$ gives $\Pic Q\cong\ZZ^2$, generated by the two ruling classes $A,B$. They satisfy $A^2=B^2=0$ and $A\cdot B=1$: different fibres of the same projection are disjoint and linearly equivalent, while one fibre of each projection meets transversely in one point. For a nonempty effective divisor $C$ of type $(a,b)$, \[ \deg C=a+b,\qquad p_a(C)=(a-1)(b-1), \] as computed in P-AGH356QUADRICCURVES. Thus a hyperplane section of the curve has length $a+b$, with multiplicities retained. If $C$ is nonsingular and integral, its geometric genus equals this arithmetic genus by P-AGH353ARITHGENUS and Serre duality [@Har10a, Exercise III.5.3 and Chapter IV, §1]. The product differential formula gives $K_Q=(-2,-2)$, and adjunction yields the same genus identity $2g-2=C\cdot(C+K_Q)$. Nonsingular irreducible curves of every type $a,b\ge1$ exist by P-AGH356QUADRICCURVES, part (b).

(The split hypothesis over other fields)

The product description and the two ruling classes also apply over any field to the split quadric $xy=zw$. They do not apply to every smooth quadric over an arbitrary field. For example, the real quadric $x_0^2+x_1^2+x_2^2+x_3^2=0$ is smooth but has no real point, while $\PP_\RR^1\times\PP_\RR^1$ has real points. Hence those real schemes cannot be isomorphic.

(The quadric cone)

Let $k$ be algebraically closed, let $Q_0=V_+(x_0x_2-x_1^2)\subseteq\PP_k^3$, and let $L=V_+(x_0,x_1)$ be a ruling line. Then $\Cl(Q_0)\cong\ZZ[L]$, and the hyperplane class is $2[L]$. For an integral curve $C\subset Q_0$ of degree $d$, the arithmetic genera are \[ p_a(C)= \begin{cases} (a-1)^2,&d=2a,\\ a(a-1),&d=2a+1. \end{cases} \] When $d=2a$, the curve is the scheme-theoretic complete intersection of $Q_0$ with a surface of degree $a$. These are arithmetic-genus statements; they are geometric-genus statements when the curve is nonsingular.

The cone is integral: its equation is an irreducible primitive polynomial linear in $x_0$ over $k[x_1,x_2,x_3]$. Outside its vertex, the charts $x_0\ne0$ and $x_2\ne0$ are affine planes. It is a hypersurface and hence Cohen–Macaulay; since the only possible nonregular point has codimension two, Serre’s criterion makes it normal [@Har10a, Theorems II.8.21A and II.8.22A]. Its complement of $L$ is the affine plane $D_+(x_0)$. The divisor restriction sequence therefore generates its class group by $[L]$. There is no relation: a rational function with divisor supported on $L$ is a unit on that affine plane, hence is a nonzero constant and has zero divisor [@Har10a, Proposition II.6.5]. On the chart $x_2\ne0$, the hyperplane equation is $x_0/x_2=(x_1/x_2)^2$, so the hyperplane divisor is $2L$.

Write $[C]=m[L]$. On a normal scheme, the ideal of an integral prime divisor is the sheaf of rational functions whose divisor is at least that divisor; regularity elsewhere is checked at the other codimension-one points. This follows from the codimension-one intersection description for normal domains, as in the reflexive-module criterion. Thus linear equivalence identifies the corresponding ideal sheaves by multiplication by a rational function. For $m=2a$ this gives $\mathcal I_C\cong\OO_{Q_0}(-a)$, and for $m=2a+1$ it gives $\mathcal I_C\cong\mathcal I_L\otimes\OO_{Q_0}(-a)$. The hypersurface sequence gives $\chi(\OO_{Q_0}(t))=\binom{t+3}{3}-\binom{t+1}{3}=(t+1)^2$, while $\chi(\OO_L(t))=t+1$ [@Har10a, Theorem III.5.1]. These are polynomial identities at every integer twist by P-AGH352HILBPOLY. Hence $\chi(\mathcal I_L(t))=t(t+1)$ by Euler additivity. The sequence $0\to\mathcal I_C(t)\to\OO_{Q_0}(t)\to\OO_C(t)\to0$ gives respectively \[ P_C(t)=(t+1)^2-(t-a+1)^2, \qquad P_C(t)=(t+1)^2-(t-a)(t-a+1). \] Their leading coefficients are $2a$ and $2a+1$, so $m=d$ in either case. Their constant terms give the displayed genera by $p_a(C)=1-P_C(0)$. In the even case, the inclusion $\mathcal I_C\hookrightarrow\OO_{Q_0}$ is a section of $\OO_{Q_0}(a)$; it lifts to a degree-$a$ form on $\PP^3$ by P-AGH355COMPINT, part (a). Its zero scheme on $Q_0$ is exactly $C$, proving the complete-intersection assertion.

(Specified genera and low-degree curves)

Over an algebraically closed field, for every $g\ge0$ a nonsingular curve of type $(g+1,2)$ exists on a smooth quadric and has degree $g+3$ and genus $g$. The types $(1,2)$ and $(2,2)$ give the twisted cubic and an elliptic quartic. A nonsingular curve of type $(3,3)$ has degree six and genus four, and adjunction gives $\omega_C\cong\OO_C(1,1)$, so its given embedding is canonical. The two rulings restrict to degree-three pencils on it [@Har10a, Chapter II, §§7 and 8]. They are distinct: the divisor sequence twisted by $\OO_Q(1,-1)$ and the vanishing $H^1(Q,\OO_Q(-2,-4))=0$ from P-AGH356QUADRICCURVES give $H^0(C,\OO_C(1,-1))=0$, so the two degree-three line bundles cannot be isomorphic.

(Degree nine and genus ten)

A nonsingular curve of type $(3,6)$ on a smooth quadric has degree nine and genus ten by P-AGH356QUADRICCURVES. The exact sequence \[ 0\to\mathcal I_Q(2)\to\mathcal I_C(2)\to\OO_Q(-1,-4)\to0 \] has $\mathcal I_Q(2)\cong\OO_{\PP^3}$ and $H^0(Q,\OO_Q(-1,-4))=0$, so $h^0(\PP^3,\mathcal I_C(2))=1$. In contrast, a nonsingular intersection of two cubic surfaces has degree nine and genus ten by P-AGH284COMPINT, and its saturated homogeneous ideal is generated by the two cubics. It has no degree-two equation, so $h^0(\PP^3,\mathcal I_C(2))=0$. Thus the dimension of the space of containing quadrics distinguishes these two constructions with the same degree and genus.

What are the curves on a smooth quadric surface?

Compute the genus of a curve of type $(a,b)$.

What happens on a quadric cone?

Dependencies

T-SRFADJThe adjunction formula on a surfaceuses

Backlinks

T-CRVCASTCastelnuovo's bound for space curvesuses
FE-CRVDEGSCurves of low degree in $\PP^3$uses

What links to this

Curves in projective space