Varieties with few trisecant lines

Question

Let $X\subset\mathbb{P}^N$ be an irreducible projective variety. Let's denote by $\mathcal{T}$ the following property: through a general point $x\in X$ there is no line intersecting $X$ in at least three points counted with multiplicity (but not contained in $X$). I am assuming that $X$ is not a hypersurface.

Is there some natural condition on $X$ implying that $\mathcal{T}$ holds? For instance, if $X$ is linearly normal does then $\mathcal{T}$ necessarily hold? Is there a classification of the varieties $X$ for which $\mathcal{T}$ does not hold?

Answer 1

You can have a look at Ingrid Bauer's paper

Bauer, I., The classification of surfaces in $\mathbb{P}^5$ having few trisecants, Rend. Semin. Mat., Torino 56, No. 1, 1-20 (1998). ZBL0965.14029.

It turns out that, if a smooth surface $X \subset \mathbb{P}^5$ satisfies $\mathcal{T}$, then $\deg X \leq 10$. Moreover, a fine classification of these surfaces is provided (they belong to eight families).

As a consequence, if $\deg X \geq 11$ then $\mathcal{T}$ does not hold (and so the answer to your first question is negative).

Answer 2

Assume that $X$ is set theoretically defined by quadrics: $ X = Q_1\cap\dots\cap Q_r$.

If $L$ is a line trisecant to $X$ then $L$ is trisecant to $Q_i$ for all $i$ and hence $L\subset Q_i$ for all $i$. So $L\subset X$.

Hence $X$ does not have any trisecant line besides the lines it containes.

Answer 3

A consequence of Gruson-Peskine $k$-secant lemma is the following : if $2N -3n-1>0$ then the trisecants of $X$ do not fill the ambiant space. In particular, if you know that the secant variety of $X$ fills the ambiant space, then you have a numerical condition that guarantees that a general secant is not a trisecant.