# Do only finitely many bisecants of a canonical curve intersect two distinct codimension 2 spaces simultaneously?

## Setup & question

Let $C \hookrightarrow \mathbb{P}^{g-1}$ be a general canonical curve of genus $g \ge 4$ and let $Y_1,Y_2 \subset \mathbb{P}^{g-1}$ be codimension 2 linear subspaces such that $Y_i \cap C = \emptyset$ for $i=1,2$. Given any pair of points $(p,q) \in C^2$ we will denote by $L_{p,q} \subset \mathbb{P}^{g-1}$ the bisecant line through $p$ and $q$ (or the tangent line if $p=q$).

Are there only finitely many pairs $(p,q) \in C^2$ such that $L_{p,q}$ intersects both $Y_1$ and $Y_2$?

## A remark on the generality assumption

Let me emphasize that $Y_i$'s are not general except for the fact that they do not intersect $C$. In particular, $Y_i$'s can be skew (i.e. they lie in a common hyperplane). However, $C$ is assumed to be general.

In fact the generality of $C$ is mandatory: Take $g=4$ and $C$ "special'' in the sense that the quadric surface containing $C$ is a cone. Denote the vertex of this cone by $x$. There is a pencil of lines through $x$ which are trisecants (these are the fibers of the projection from $x$). If we take $Y_1$ and $Y_2$ to be (non-skew) lines intersecting at $x$ then we have our counter-example.

For every curve $C$ of genus $g\geq 5$ that is neither hyperlliptic nor trigonal and that admits no morphism of degree greater than 1 to a curve of positive genus that has general moduli, there exists no such pair $(Y_1,Y_2)$ of distinct linear spaces.

For every point $p$, for every codimension $2$ linear space $Y$ that does not contain $p$, the span of $p$ and $Y$ is a hyperplane. This hyperplane intersects $C$ in only finitely many points. Thus, there are only finitely many secant lines to $C$ that contain $p$ and intersect $Y$. Thus, for $Y_1$ and $Y_2$, there is no point $p$ such that infinitely many secant lines to $C$ contain $p$ and intersect both $Y_1$ and $Y_2$. So if there are infinitely many secant lines that intersect both $Y_1$ and $Y_2$, then for a general $p$ in $C$, there exists at least one such secant line that contains $p$.

For each of the two linear subspaces, consider the corresponding linear projection $$\pi_j: \mathbb{P}^{g-1}\setminus Y_j \to \Pi_j, \ \ j=1,2,$$ where $\Pi_j$ is a copy of $\mathbb{P}^1$. Consider the morphism $$\pi:C\to \Pi_1\times \Pi_2, \ \ \pi(p)=(\pi_1(p),\pi_2(p)).$$ By the previous paragraph, $\pi$ maps $C$ to its image as a morphism of degree $d\geq 2$. The image cycle in $\Pi_1\times\Pi_2$ has bidegree $(2g-2,2g-2)$. By hypothesis, $C$ admits no finite, surjective morphism to a curve of genus $h$ with $1\leq h\leq g-1$. Thus, the image curve must be a genus $0$ curve. Either the image curve is a smooth curve of bidegree $(1,1)$, or it is a singular curve.

If the image curve is a smooth curve, then it is the graph of an isomorphism from $\Pi_1$ to $\Pi_2$. Thus, the associated $2$-dimensional linear systems of canonical divisors on $C$ are equal, i.e., $Y_1$ equals $Y_2$ and the morphism $\pi$ factors through the diagonal.

Edit. Since the curve has general moduli, the gonality of $C$ is $[g/2]+1$. Thus, the image curve either has bidegree $(2,2)$ or $(3,3)$. If the image curve has bidegree $(2,2)$, then projection away from a singular point defines an isomorphism to a plane conic. In other words, $C$ has a theta characteristic $L$ with $h^0(C,L) \geq 2$. This cannot happen if $C$ has general moduli by a Theorem of Montserrat Teixidor, cf. the following MathOverflow answer and reference: Theta characteristics of genus$\geq3$ curve

MR0887499 (88e:14037)
Teixidor i Bigas, Montserrat(E-BARU)
Half-canonical series on algebraic curves.
Trans. Amer. Math. Soc. 302 (1987), no. 1, 99–115.
14H10
http://www.jstor.org/stable/2000899?origin=crossref&seq=1#page_scan_tab_contents

I will have to think further about whether a general curve could have an invertible sheaf $L$ with $L^{\otimes 3}\cong \omega_C$ and $h^0(C,L)\geq 2$, but it seems even less likely than having a theta characteristic with $h^0(C,L)\geq 2$.

Second edit. Emre and Gregor Bruns finished the proof in the comments. Let me just summarize. The normalization $B$ of the image of $\pi$ is a smooth rational curve. For the associated morphism $$\widetilde{\pi}:C\to B,$$ $\widetilde{\pi}^*\mathcal{O}_B(1)$ is an invertible sheaf $\mathcal{L}$ on $C$ that has a basepoint free pencil of sections (defining the morphism to $B\cong \mathbb{P}^1$). The image of $B$ in $\Pi_1\times \pi_2$ is a curve of bidegree $(d_1,d_2)$ for integers $d_1,d_2\geq 0.$ Thus, $L^{\otimes d_j}$ equals $\widetilde{\pi}^*(\text{pr}_j^*\mathcal{O}_{\Pi_j}(1))$. This equals $\pi_j^*\mathcal{O}_{\Pi_j}(1)$, which in turn equals $\omega_C$ since $\pi_j$ is a linear projection of a canonical curve (and the center $Y_j$ of the projection is disjoint from $C$). Since $\mathcal{L}^{\otimes d_1}$ is isomorphic to $\mathcal{L}^{\otimes d_2}$, also $d_1$ equals $d_2$; call this common integer $d$. Since the gonality of a general curve is $[g/2]+1$, also the degree $(2g-2)/d$ of $\mathcal{L}$ is at least $[g/2]+1$. Thus $d$ equals $1,$ $2,$ or $3.$

We want to prove that $d$ equals $1$, for then the pencil of sections of $\omega_C$ coming from $\pi_1$ and $\pi_2$ are both the same: just the pencil of sections coming from $\widetilde{\pi}.$ This implies that the zero loci of those pencils are equal, i.e., $Y_1$ equals $Y_2$. So the strategy is to rule out $d=2$ and $d=3$ by contradiction.

By the theorem of Montserrat Teixidor, for a general curve there is no invertible sheaf $\mathcal{L}$ on $C$ with both $h^0(C,\mathcal{L})\geq 2$ and with $\mathcal{L}^{\otimes 2}\cong \omega_C$, i.e., $\mathcal{L}$ is a theta characteristic. This rules out $d=2$.

Finally, as Gregor Bruns proves, also we cannot have an invertible sheaf $\mathcal{L}$ with both a basepoint free pencil of sections and with $\mathcal{L}^{\otimes 3}\cong \omega_C$ for a curve of general moduli (hyperelliptic curves of genus $g = 3h+1$ do have such linear systems).

Here is my interpretation of Gregor's argument, but I hope that Gregor will correct me if I am wrong. For any invertible sheaf $\mathcal{L}$ on $C$ with a basepoint free pencil of sections, $$\text{span}(s_0,s_1)\subset H^0(C,\mathcal{L}),$$ setting $\mathcal{F}=\omega_C\otimes\mathcal{L}^\vee$, we can apply the Basepoint Free Pencil Trick (p. 126 of Arbarello-Cornalba-Griffiths-Harris) to see that the kernel of the following cup-product pairing equals $H^0(C,\omega_C\otimes (\mathcal{L}^\vee)^{\otimes 2}) \cong H^1(C,\mathcal{L}^{\otimes 2})^\vee$: $$\text{span}(s_0,s_1)\otimes H^0(C,\omega_C\otimes\mathcal{L}^\vee) \to H^0(C,\omega_C).$$ By the Gieseker-Petri Theorem (which implies smoothness of the parameter spaces of $\mathfrak{g}^r_d$s), for a curve $C$ of general moduli, this map is injective. Thus $H^1(C,\mathcal{L}^{\otimes 2})$ is zero. On the other hand, if $\mathcal{L}^{\otimes 3}\cong \omega_C$, then this group equals $H^1(C,\omega_C\otimes \mathcal{L}^\vee) \cong H^0(C,\mathcal{L})^\vee$, by Serre duality. So the hypothesis that $\mathcal{L}$ has a basepoint free pencil of divisors leads to the contradictory conclusion that $\mathcal{L}$ has only the zero global section. This rules out $d=3$.

Altogether, this proves that for a general canonical curve $C$, there cannot exist infinitely many secant lines to $C$ that intersect both $Y_1$ and $Y_2$ unless $Y_1$ equals $Y_2$.

• Even if the fiber over a point $\pi(p) \in \Pi_1\times\Pi_2$ is of degree $\ge 3$ this doesn't imply that C has trisecant lines. Because the preimage $\pi^{-1}\pi(p)$ consists of points in the triple intersection $\langle Y_1, p \rangle \cap \langle Y_2, p \rangle \cap C$. Here $\langle Y_i, p \rangle$ denotes the span of $p$ and $Y_i$. Therefore the points of $\pi^{-1}\pi(p)$ lie in a codimension two locus, not necessarily a line. Or am I missing something? – Emre Apr 16 '17 at 14:25
• As you mentioned, the image curve is singular. The preimage of a singularity will consist of $2d$ points where $d$ is the degree of $\pi|_C$. These $2d$ points lie in a codimension 2 locus. But Brill-Noether rules this out when $d=g-1$. When $d=\frac{2g-2}{3}$, this can unfortunately happen when $g\ge 10$ and $g \equiv 1 (3)$ but not otherwise. I'm not sure how to get rid of this second condition but we seem to be close. – Emre Apr 19 '17 at 16:23
• Note that on a general curve, for every $L$ with $h^0(L)\geq 2$ we have that $2L$ is nonspecial (this follows from Petri's theorem and the base point free pencil trick). Now apply this to the case where $3L = K_C$: Since $h^0(L) = h^1(2L) \geq 2$, the curve is not general (in fact the curve will be rather special). – Gregor Botero Apr 19 '17 at 18:27
• Just a subcomment to Gregor's remark. As Gregor wrote, one does not need that the pencil is base point free to apply the argument: if $s_0,s_1\in H^0(L)$ are linearly independent and $t\in H^0(K-2L)$ is a nonzero, then $s_0 \otimes s_1t - s_1 \otimes s_0 t$ is a nonzero element in the kernel of the multiplication map $H^0(L)\otimes H^0(K-L) \to H^0(K)$. Moreover, Gregor's remark gives also that on a general curve there is no theta characteristic $L$ with $h^0(L)\geq 2$, without using Teixidor's theorem. – Daniele A Apr 20 '17 at 13:45
• What an adventure, thanks a lot for solving this problem! Let me just add that the method of argument works just as well even if $Y_i \cap C$ are non-empty: we just have that $\mathcal{L}$ squares or cubes to $\omega_C$ minus base points, which is impossible in general because of Petri. However, if $Y_1 \cap Y_2 \cap C \neq \emptyset$ then of course our pencils $\Pi_i$'s don't see that these base points intersect so we can not initiate the argument. In other words, as long as $Y_1 \cap Y_2 \cap C = \emptyset$ then there are only finitely many bisecants (or tangents) intersecting both $Y_i$'s. – Emre Apr 20 '17 at 14:30