Search the corpus

    Curves in projective space

    Contents

    A very ample divisor \(D\) on a smooth projective curve \(C\) gives a closed immersion \(\varphi_{\abs{D}}\colon C\to\PP^{\ell(D)-1}\) with \(\deg\varphi_{\abs{D}}(C)=\deg D\).

    The degree thresholds $2g$ and $2g+1$

    On a genus-$g$ curve, positive degree is ample, degree $2g$ frees base points, and degree $2g+1$ is very ample.Suggest a slogan

    Let \(D\) be a divisor on a smooth projective curve \(C\) of genus \(g\).

    • \(D\) is ample if and only if \(\deg D > 0\).

    • \(\abs{D}\) is base-point free if \(\deg D \geq 2g\).

    • \(D\) is very ample if \(\deg D \geq 2g+1\).

    The last two are sufficient, not necessary.

    All three come from running Riemann–Roch twice and watching the speciality term vanish. For base-point freeness one needs \(\ell(D-p) = \ell(D)-1\); with \(\deg D \geq 2g\) both \(D\) and \(D-p\) have degree \(> 2g-2\), both are nonspecial, and subtracting the two formulas gives the drop of exactly one. For very ampleness one needs the drop by two at every pair, and \(\deg D \geq 2g+1\) keeps \(D-p-q\) nonspecial as well. Ampleness follows because \(\deg nD > 0\) eventually exceeds \(2g+1\).

    The thresholds are not sharp, and the standard counterexample is the canonical divisor on a smooth plane quartic: \(g=3\), \(\deg K = 4 < 2g+1 = 7\), and \(K\) is very ample. Standard specializations are: on \(\PP^1\), ample, very ample, and \(\deg D\ge1\) coincide; on an elliptic curve, \(D\) is very ample exactly when \(\deg D\ge3\), giving the plane cubic model; on a genus-\(2\) curve, degree \(5\) gives a quintic model in \(\PP^3\). If \(D\) is very ample and \(\varphi\) is the resulting embedding, then \(\deg \varphi(C) = \deg D\).

    When is a divisor on a curve ample?
    When is a divisor on a curve base-point free?
    What degree guarantees a divisor on a curve is very ample?

    If \(\deg D\ge2g\), then \(D\) and \(D-p\) are nonspecial for every point \(p\), and Riemann–Roch gives \(\ell(D-p)=\ell(D)-1\). If \(\deg D\ge2g+1\), then \(D-p-q\) is nonspecial as well, and \(\ell(D-p-q)=\ell(D)-2\) for all points \(p,q\). Neither degree bound is necessary: on a smooth plane quartic, \(g=3\) and \(K\sim\OO_C(1)\) is very ample of degree \(4<2g+1\).

    Every curve embeds in $\PP^3$ and projects to a nodal plane curve

    Project until secants fill the room: every smooth projective curve embeds in $\PP^3$, and a general further projection gives only nodes.Suggest a slogan

    Every smooth projective curve embeds in \(\PP^3\), and is birational to a plane curve with at worst nodes.

    (The projection criterion)

    Let \(C \subseteq \PP^N\) and \(O \notin C\), and let \(\varphi\) be projection away from \(O\). Then \(\varphi\) is a closed immersion exactly when \(O\) lies on no secant and no tangent line of \(C\). The secant variety has dimension at most \(3\) and the tangent variety at most \(2\), so for \(N \geq 4\) such a point \(O\) exists. For \(N=3\), \(\varphi\) is birational onto a nodal image exactly when \(O\) lies on only finitely many secants, on no tangent, on no multisecant, and on no secant whose two tangent lines are coplanar.

    The dimension count is the whole argument and is the thing to say out loud: secants sweep at most a threefold, tangents at most a surface, and \(\PP^N\) has room to avoid both as soon as \(N \geq 4\). Projection then drops the ambient dimension one step at a time until it stops at \(\PP^3\), where the secant variety fills the space and further projection must lose injectivity.

    The consequence for genus is the nodal formula: if a plane curve of degree \(d\) has \(r\) nodes and nothing worse, its normalization has genus \[ g = \binom{d-1}{2} - r . \] So classifying all curves reduces to studying the family of plane curves of degree \(d\) with \(r\) nodes, which is nonempty exactly for \(0 \leq r \leq \binom{d-1}{2}\) and, by Severi and Harris, irreducible of dimension \(\tfrac{1}{2}d(d+3) - r\). The two extremes are Bertini at \(r=0\) and, at the top, the rational curve obtained by projecting the degree-\(d\) rational normal curve in \(\PP^d\) down to the plane.

    Can every curve be embedded in projective space, and in which one?
    Is every curve birational to a plane curve, and how singular can it be?
    What is the classification of curves in $\PP^n$?

    For \(C\subseteq\PP^N\) with \(N\ge4\), the secant variety has dimension at most \(3\) and the tangent variety at most \(2\), so some point \(O\) lies on neither, and projection from \(O\) embeds \(C\) in \(\PP^{N-1}\). Iterating gives an embedding in \(\PP^3\). In \(\PP^3\) the secant variety is all of \(\PP^3\), and projection from a general point is birational onto a plane curve whose singularities are nodes.

    Node, cusp, and tacnode; the delta invariant and the genus formula

    Let \(p\) be a singular point of a curve \(C\) with normalisation \(\nu : \tilde C \to C\). Then \[ \delta_p \da \dim_k \qty{ \qty{\nu_* \OO_{\tilde C} / \OO_C}_p } , \] a finite number, and \(r_p\) denotes the number of analytic branches of \(C\) at \(p\).

    (Node, cusp, and tacnode)

    Up to analytic isomorphism at the origin:

    name equation \(r_p\) \(\delta_p\)
    node, \(A_1\) \(y^2 = x^2\) \(2\) \(1\)
    cusp, \(A_2\) \(y^2 = x^3\) \(1\) \(1\)
    tacnode, \(A_3\) \(y^2 = x^4\) \(2\) \(2\)

    A node is a singular point of multiplicity \(2\) with two distinct tangent directions. More generally \(A_n : y^2 = x^{n+1}\) has \(\delta_p = \floor{(n+1)/2}\), and an ordinary point of multiplicity \(r\), meaning \(r\) smooth branches with distinct tangents, has \(\delta_p = \binom{r}{2}\).

    (Genus drop)

    \[ p_a(C) - g(\tilde C) = \sum_{p \in \Sing C} \delta_p . \] For a plane curve of degree \(d\) this reads \(g = \binom{d-1}{2} - \sum_p \delta_p\).

    (The genus formula with multiplicities)

    Let \(C \subseteq \PP^2\) be an integral curve of degree \(d\). Resolve its singularities by successively blowing up singular points of the strict transforms, and let \(m_i\) run over the multiplicities of all singular points met in this process, including the infinitely near points on the strict transforms. Then \[g(\tilde{C}) = \frac{1}{2}(d-1)(d-2) - \frac{1}{2} \sum_i m_i(m_i - 1).\] If every singular point of \(C\) is ordinary, no infinitely near singular points occur, and the sum runs over the singular points of \(C\).

    The tacnode \(y^2 = x^4\) has multiplicity \(2\) at the origin and \(\delta = 2\). One blowup, in the chart \(y = x y_1\), gives the strict transform \(y_1^2 = x^2\), which is a node at the origin: an infinitely near singular point of multiplicity \(2\). The two points contribute \(\frac{1}{2}(2 \cdot 1 + 2 \cdot 1) = 2 = \delta\), while the origin alone would give \(1\).

    (Invariants separating the normal forms)

    Node and cusp both have \(\delta_p = 1\) and are separated by the branch count: \(\nu^{-1}(p)\) has two points for the node and one for the cusp. A nodal and a cuspidal plane cubic both have geometric genus \(0\), and their groups \(\Pic^0\) are \(\GG_m\) and \(\GG_a\) respectively (FE-ADOKK).

    Node and tacnode both have two smooth branches. For a singular point with branches \(B_1,\ldots,B_r\), \(\delta_p=\sum_i\delta_p(B_i)+\sum_{i<j}(B_i\cdot B_j)_p\); for two smooth branches this is their intersection multiplicity, which is \(1\) for the transverse branches \(y=\pm x\) of the node and \(2\) for the tangent branches \(y=\pm x^2\) of the tacnode.

    The Milnor number is \(\mu_p = 2\delta_p - r_p + 1\): the node and the cusp have \(\mu = 1, 2\) and the tacnode \(\mu = 3\), which is the \(n\) in \(A_n\).

    (Singularities of general projections)

    By T-CRVEMBP3, projection of a smooth curve in \(\PP^3\) from a general point is birational onto a plane curve whose only singularities are nodes. The centres of projection that produce a cusp lie on a tangent line of the curve, and those that produce a tacnode lie on a secant line whose two tangent lines are coplanar; each set has dimension less than \(3\), so a general centre avoids them.

    Give the local normal forms distinguishing a node, a cusp and a tacnode.
    What is the delta invariant of a plane curve singularity?
    How does the arithmetic genus of a plane curve drop to the geometric genus?

    Projection from \(O\in\PP^3\setminus C\) produces a triple point when \(O\) lies on a trisecant line, a cusp when \(O\) lies on a tangent line, and a tacnode when \(O\) lies on a secant line whose two tangent lines at the points of \(C\) are coplanar.

    Curves on a quadric

    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?

    A curve of type \((g+1,2)\) on a smooth quadric has degree \(g+3\) and genus \(g\), so every genus occurs for curves on a quadric. The twisted cubic has type \((1,2)\), the elliptic quartic type \((2,2)\), and the canonical curve of genus \(4\) on a smooth quadric type \((3,3)\).

    Castelnuovo’s bound

    Castelnuovo's bound for space curves

    Castelnuovo is the balanced-quadric bound: maximal-genus nonplanar space curves lie on a quadric with bidegrees as equal as possible.Suggest a slogan

    Let \(C \subseteq \PP^3\) be a smooth curve of degree \(d\) and genus \(g\) not contained in a plane. Then \(d \geq 3\) and \[ g \leq \begin{cases} \tfrac{1}{4}d^2 - d + 1 , & d \text{ even} , \\[2pt] \tfrac{1}{4}(d^2-1) - d + 1 , & d \text{ odd} . \end{cases} \] The bound is attained for every \(d \geq 3\), and a curve attaining it lies on a quadric surface.

    The bound and its equality case are one statement: the extremal curves are the balanced curves on a quadric. Type \((a,a)\) on a smooth quadric has \(d = 2a\) and \(g = (a-1)^2 = \tfrac{1}{4}d^2 - d + 1\), and type \((a,a+1)\) has \(d = 2a+1\) and \(g = a^2 - a\), which is the odd formula. Thus the equality cases are exactly the balanced bidegrees on a quadric.

    The contrast with plane curves explains the nonplanarity hypothesis. A plane curve of degree \(d\) has \(g = \binom{d-1}{2} \approx \tfrac{1}{2}d^2\), roughly twice Castelnuovo’s bound: spreading a curve out into \(\PP^3\) costs genus, and the hypothesis that \(C\) is not planar is what makes the smaller bound apply.

    How large can the genus of a curve of degree $d$ in $\PP^3$ be?
    State Castelnuovo's bound.

    The bound is attained by curves of type \((a,a)\), with \(d=2a\) and \(g=(a-1)^2\), and of type \((a,a+1)\), with \(d=2a+1\) and \(g=a^2-a\), on a smooth quadric. A smooth plane curve of degree \(d\ge3\) has genus \(\binom{d-1}{2}\), which exceeds Castelnuovo’s bound.

    Curves of low degree in $\PP^3$

    By degree, for smooth curves in \(\PP^3\):

    • \(d=1\): a line, \(g=0\).

    • \(d=2\): a plane conic, \(g=0\).

    • \(d=3\): the plane cubic with \(g=1\), and the twisted cubic with \(g=0\).

    • \(d=4\): the plane quartic with \(g=3\); the rational quartic with \(g=0\); and the elliptic quartic with \(g=1\), which is the complete intersection of two quadrics.

    • \(d=5\): the plane quintic with \(g=6\), and curves with nonspecial hyperplane section of genus \(0, 1, 2\).

    • \(d=6\): the plane sextic with \(g=10\); nonspecial curves of genus \(0,1,2,3\); and the canonical curve of genus \(4\), the complete intersection of a quadric and a cubic.

    • \(d=7\): the plane septic with \(g=15\); nonspecial curves of every genus \(g \leq 4\); curves of genus \(5\); and curves of genus \(6\), the maximum allowed by Castelnuovo’s bound, which all lie on a quadric, for instance the curves of type \((3,4)\) on a smooth quadric.

    (Degree $7$ and genus $5$)

    Let \(X\) have genus \(5\) and \(P \in X\). Since \(\abs{K}\) is base-point free, \(D = K - P\) has degree \(7\) and \(h^0(D) = 4\). By the very-ampleness criterion and Serre duality, \(D\) is very ample exactly when \(h^0(P + Q + R) = 1\) for all \(Q, R \in X\). If \(X\) has no \(g^1_3\) this holds for every \(P\), and \(\abs{K - P}\) embeds \(X\) in \(\PP^3\) as a curve of degree \(7\); curves of genus \(5\) without a \(g^1_3\) exist, so \((d, g) = (7, 5)\) occurs. The hyperplane section \(K - P\) is special, in accordance with \(g \geq \tfrac{1}{2} d + 1\). [@Har10a]

    For \(g \geq 2\), a curve of genus \(g\) has a nonspecial very ample divisor of degree \(d\) exactly when \(d \geq g+3\), and then it embeds in \(\PP^3\). If the hyperplane section is special and \(C\) is not planar, then \(d \geq 6\) and \(g \geq \tfrac{1}{2}d + 1\); the only such curve with \(d=6\) is the canonical genus-\(4\) curve.

    Two curves in \(\PP^3\) with the same \((d,g)\) need not be deformations of each other: for degree \(9\) and genus \(10\) there are complete intersections of two cubics and curves of type \((3,6)\) on a smooth quadric. They are distinguished by \(h^0(\mci_C(2))\), which is \(0\) for the first and \(1\) for the second. By upper semicontinuity of \(h^0(\mci_C(2))\) in flat families, a complete intersection of two cubics is not a flat limit of curves of type \((3,6)\).

    Classify the curves of degree at most $7$ in $\PP^3$.
    Which pairs $(d,g)$ occur for curves in $\PP^3$?

    A nondegenerate curve \(C\subseteq\PP^3\) of degree \(d\) whose hyperplane section is nonspecial has \(4\le h^0(\OO_C(1))=d+1-g\), so \(g\le d-3\).

    Dual curves

    The dual curve, and the Plücker formula for its degree

    Let \(C \subseteq \PP^2\) be an irreducible plane curve over an algebraically closed field of characteristic \(0\) that is not a line. The dual curve \(C^\vee \subseteq (\PP^2)^\vee\) is the closure of the set of tangent lines \(T_p C\) at the smooth points \(p\) of \(C\). Its degree is the class of \(C\): the number of tangent lines to \(C\) through a general point of \(\PP^2\).

    1. (Biduality) \((C^\vee)^\vee = C\).

    2. (Plücker) If \(C\) has degree \(d\) and its only singularities are \(\delta\) nodes and \(\kappa\) ordinary cusps, then \[\deg C^\vee = d(d-1) - 2\delta - 3\kappa.\] In particular a smooth plane curve of degree \(d \geq 2\) has class \(d(d-1)\).

    The dual of a smooth conic is a smooth conic. The dual of a smooth plane cubic has degree \(6\); its \(9\) cusps correspond to the \(9\) flexes of the cubic, and it has no nodes because a smooth cubic has no bitangents. Its geometric genus is \(\binom{5}{2} - 9 = 1\), equal to the genus of the cubic, since the Gauss map \(C\to C^\vee\) is birational.

    What is the dual curve?

    What links to this