3
$\begingroup$

Consider a smooth projective curve $C\subset\mathbb{P}^n$. Let $G(1,n)$ the Grassmannian of lines of $\mathbb{P}^n$. The variety $S_2(C)\subset G(1,n)$ parametrizing lines that are secant to $C$ (i.e., that intersect $C$ in at least two points) has dimension two.

Let $S_3(C)\subset S_2(C)\subset G(1,n)$ be the variety parametrizing lines that intersect $C$ in al least three points.

Question. Is there a formula (perhaps depending on $n$, the genus and the degree of $C$) for the dimension of $S_3(C)$? Under which assumption do we have that $\dim(S_3(C)) = 0$?

$\endgroup$
2
  • 1
    $\begingroup$ The answer depends very much on the embedding. Assuming that $C$ is embedded by a complete linear system $\lvert L\rvert$, $S_3(C)=\varnothing$ is equivalent to $h^0(L(-D_3))=h^0(L)-3$ for all effective divisors $D_3$ of degree 3. This is the case in particular if $\deg(C)\geq 2g+2$; it is also true for the canonical embedding ($L=K$) if $C$ is not hyperelliptic or trigonal. $\endgroup$
    – abx
    Commented Feb 17, 2021 at 20:46
  • $\begingroup$ The expected dimension is $4-n$. This should be what you see for a sufficiently general curve $C$ - dimension $1$ if $n=3$, $0$ if $n=4$, and empty for $n>4$. $\endgroup$
    – Will Sawin
    Commented Feb 18, 2021 at 0:06

1 Answer 1

4
$\begingroup$

(too long for a comment)

Let's consider the case $n=3$.

In $\mathbb P^3$, for a curve of genus $g$ and degree $d$, given a point on $x\in C$, lines through $x$ are parameterized by $\mathbb P^2$.

Lines that intersect another point in $C$ are the projection of $C$ from $x$. This is a curve of genus $g$ and degree $d-1$ in $\mathbb P^2$. (One point on this curve represents a tangent line and not a secant line).

Lines that intersect two other points in $C$ are nodes (or other multi-branch singularities) of this curves. The total number of singularities, counted by their contribution to the difference between the arithmetic and geometric genus, is $ (d-2)(d-3)/2 -g$.

In a sufficiently general situation, these will all be nodes, and so the set of lines (with a marked point) form a degree $(d-2)(d-3)/2-g$ cover of $C$. I think this is always positive unless $C$ is a plane curve (in which case this is trivial), $g=0$ and $d=3$, or $g=1$ and $d=4$.

How can we fail to be sufficiently general? Either if the projection map is generically not birational onto its image - that is if a generic line through two points on $C$ passes through a third (i.e. if the space of trisecants has dimension $2$), or if the general fiber has singularities other than nodes - which happens when a general point on $C$ lies on a tangent line (other than its own), or similar degeneracies. Probably one can eliminate these or classify when they happen, but I'm not sure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.