The line, the conic, the twisted cubic curve and the elliptic quartic curve in \(\PP^3\) have no multisecants. Every other curve in \(\PP^3\) has infinitely many multisecants.
Hint: Consider a projection from a point of the curve to \(\PP^2\).
Solution
Here a multisecant line means a line not contained in \(X\) whose scheme-theoretic intersection with \(X\) has length at least \(3\).
1. A line and a nonsingular plane conic have no multisecants.
For a line \(X\), every other line meets \(X\) in length at most \(1\).
Let \(X\) be a nonsingular plane conic. Any line not contained in \(X\) meets the degree-\(2\) plane curve \(X\) in scheme-theoretic length at most \(2\) by Bezout. Hence no such line is a multisecant.
2. The twisted cubic has no multisecants.
By Exercise IV.3.4, the homogeneous ideal of the twisted cubic is generated by quadrics. Suppose a line \[ \ell\subseteq\PP^3 \] met the twisted cubic in length at least \(3\). Every quadric vanishing on the curve restricts to a section of \[ \OO_\ell(2) \] vanishing on a subscheme of length at least \(3\). Such a section must be zero, so every defining quadric contains \(\ell\). Since those quadrics generate the ideal of the twisted cubic, their scheme-theoretic intersection is the curve itself. Thus \(\ell\) would be contained in the twisted cubic, impossible for a line having finite intersection with it.
3. The elliptic quartic has no multisecants.
By Exercise IV.3.6, an elliptic quartic is a complete intersection \[ X=Q_1\cap Q_2 \] of two quadrics in \(\PP^3\).
If a line \(\ell\) met \(X\) in length at least \(3\), then each quadratic equation \(Q_i|_\ell\) would be a section of \(\OO_\ell(2)\) vanishing on a subscheme of length at least \(3\). Hence both restrictions would vanish identically. Thus \[ \ell\subseteq Q_1\cap Q_2=X, \] again contradicting the requirement that a multisecant have finite intersection with \(X\).
4. Suppose that a curve \(X\subseteq\PP^3\) has only finitely manymultisecants. If \(X\) is planar, then \(X\) is a line or a conic.
Let \(X\) lie in a plane and have degree \(d\). If \(d\ge3\), then a general line in that plane meets the smooth curve \(X\) transversely in \(d\) distinct points. The set of transverse lines is a nonempty open subset of the dual plane, so there are infinitely many such lines. Each is a multisecant, contrary to the finiteness hypothesis.
Therefore \(d\le2\). A smooth integral plane curve of degree \(1\) is a line, and one of degree \(2\) is a conic.
5. Now suppose \(X\) is nonplanar and has only finitely many multisecants.There is a point\[ P\in X \]through which no multisecant passes.
Each multisecant has finite intersection with the integral curve \(X\). The union of the intersections with the finitely many multisecants is therefore a finite set of points of \(X\). Choose \(P\) outside that set.
6. Projection from the point \(P\) of step 5 extends to a morphism\[ \pi_P:X\longrightarrow\PP^2 \]defined by the base-point-free linear system cut out by planes through \(P\)after subtracting the fixed point \(P\). Its line bundle is\[ H-P, \qquad H=\OO_X(1). \]
Planes through \(P\) cut out a three-dimensional space of sections of \(H\), all vanishing at \(P\). Dividing by the fixed zero at \(P\) gives sections of \(H-P\).
At a point \(Q\ne P\), some plane through \(P\) avoids \(Q\), so these sections have no base point there. At \(P\), choose a plane through \(P\) not containing the tangent line \(T_PX\); its section vanishes to order exactly one at \(P\), and hence gives a nonzero value after subtracting \(P\). Thus the induced linear system on \(H-P\) is base-point free and defines the stated morphism.
7. The morphism \(\pi_P\) separates points of \(X\).
For distinct points \(Q,R\ne P\), equality \[ \pi_P(Q)=\pi_P(R) \] means that \(P,Q,R\) are collinear. Their common line would meet \(X\) in at least three distinct points and hence be a multisecant through \(P\), contrary to step 5.
The image of \(P\) under the extended projection is the tangent direction \(T_PX\). If some \(Q\ne P\) had \[ \pi_P(Q)=\pi_P(P), \] then the line \(PQ\) would equal \(T_PX\). Its intersection with \(X\) would have length at least \(2\) at \(P\) and would also contain \(Q\), again making it a multisecant through \(P\). Thus \(\pi_P\) separates all points.
8. The morphism \(\pi_P\) separates tangent vectors of \(X\).
Let \(Q\ne P\). The differential of projection at \(Q\) fails to be injective exactly when the center \(P\) lies on the embedded tangent line \(T_QX\). In that case the line \(PQ=T_QX\) meets \(X\) with length at least \(2\) at \(Q\) and also contains \(P\), so it is a multisecant through \(P\), impossible by step 5.
At \(P\), choose affine coordinates centered at \(P\) in which the tangent line is the \(x\)-axis and a local parameter \(t\) on \(X\) has expansions \[ x=t+O(t^2), \qquad y=a_2t^2+O(t^3), \qquad z=b_2t^2+O(t^3). \] After cancelling the common factor \(t\) in the projection from \(P\), the local map is represented near \(P\) by \[ [1+O(t):a_2t+O(t^2):b_2t+O(t^2)]. \] Its differential at \(P\) vanishes exactly when \[ a_2=b_2=0. \] That condition says that both transverse coordinates vanish to order at least \(3\), so the tangent line meets \(X\) at \(P\) with length at least \(3\). It would therefore be a multisecant through \(P\), again impossible. Hence \(\pi_P\) separates tangent vectors everywhere.
9. The projection \(\pi_P\) is a closed immersion, and its image is anonsingular plane curve of degree \(d-1\), where \(d=\deg X\).
Steps <1>7–<1>8 show that the base-point-free linear system in step 6 separates points and tangent vectors. By the closed-immersion criterion for a linear system, \(\pi_P\) is a closed immersion.
Moreover, \[ \pi_P^*\OO_{\PP^2}(1)\cong\OO_X(H-P), \] whose degree is \[ d-1. \] Because \(\pi_P\) is an embedding, its plane image therefore has degree \(d-1\) and is isomorphic to the smooth curve \(X\).
10. Under the assumptions of step 5,\[ g(X)=\frac{(d-2)(d-3)}2. \]
By step 9, \(X\) is isomorphic to a nonsingular plane curve of degree \(d-1\). The plane-curve genus formula gives \[ g(X) = \frac{((d-1)-1)((d-1)-2)}2 = \frac{(d-2)(d-3)}2. \]
11. Castelnuovo’s bound and step 10 force\[ d=3\quad\text{or}\quad d=4. \]
The curve is nonplanar, so Castelnuovo's bound applies.
If \(d\) is even, steps 10 and Castelnuovo give \[ \frac{(d-2)(d-3)}2 \le \frac{d^2}{4}-d+1. \] After multiplying by \(4\) and simplifying, \[ (d-2)(d-4)\le0. \] Since a nonplanar curve has degree at least \(3\), the even case gives \[ d=4. \]
If \(d\) is odd, the corresponding inequality is \[ \frac{(d-2)(d-3)}2 \le \frac{d^2-1}{4}-d+1, \] which simplifies to \[ (d-3)^2\le0. \] Hence \[ d=3. \]
12. A nonplanar curve with only finitely many multisecants is either thetwisted cubic or the elliptic quartic.
If \(d=3\), Exercise IV.3.4(d) says that a nonplanar degree-\(3\) curve in \(\PP^3\) is the twisted cubic.
If \(d=4\), step 10 gives \[ g(X)=1. \] By Exercise IV.3.6, a nonplanar degree-\(4\) curve of genus \(1\) is the elliptic quartic, equivalently a complete intersection of two quadrics.
13. Every curve in \(\PP^3\) other than the line, conic, twisted cubic,and elliptic quartic has infinitely many multisecants.
Suppose instead that a curve had only finitely many multisecants. If it were planar, step 4 would make it a line or a conic. If it were nonplanar, step 12 would make it a twisted cubic or an elliptic quartic.
Thus any curve outside the four listed cases cannot have finitely many multisecants. Since the set of multisecants is either finite or infinite, it must be infinite.
14. Q.E.D.
Steps <1>1–<1>3 prove that the four stated curves have no multisecants, and step 13 proves that every other curve has infinitely many.