1
$\begingroup$

This is just a feeling that I had and I am curious if it is totally wrong or true to some extent.

Let $X\subseteq \mathbb{P}^r$ be an integral hypersurface of degree $r-1$, which is not a cone. In this choice of numerology, we know that through every point of $X$ there is a line that is contained in $X$.

Let $p\in X$ and assume that there is an $n-$dimensional family of lines in $X$ that pass through $p$. Is it true that if $n$ is large enough, it will force $p$ to be a singular point?

If so, do we have an estimate for that number $n$?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Just a comment: If $X$ is not a cone, the lines in $X$ through $p$ fill up at most a codimension-1 subset of $X$, hence $n$ is at most $r-3$. But for example there are smooth cubic threefolds in $\mathbf P^4$ with Eckardt points, i.e. points where the tangent hyperplane section is a cone (over a curve, ie $n=1$). So $n=r-3$ is not sufficient in general. $\endgroup$
    – Lazzaro Campeotti
    Commented Jul 8, 2016 at 12:18

1 Answer 1

Reset to default
2
$\begingroup$

To expand the comment of @potentially dense: the answer is $r-2$, and it does not depend on the degree of $X$. In general, suppose $X\subset \mathbb{P}^r$ is a hypersurface, and $p$ a smooth point of $X$. Lines passing through $p$ and contained in $T_p(X)$ are parametrized by a $\mathbb{P}^{r-2}$; since $T_p(X)\not\subset X$, we see that lines passing through $p$ and contained in $X$ form a family of dimension at most $r-3$. Now if (and only if) $p$ is a Eckart point, i.e. $X\cap T_p(X)$ is a cone, the dimension is exactly $r-3$. And it is easy to construct a smooth hypersurface with such a point, for instance $X$ given by $X_0^{d-1}X_1+X_1^{d}+\ldots+ X_r^d=0$, with $p=(1,0,\ldots ,0)$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .