quadrics containing the tangential variety of a curve - MathOverflow

quadrics containing the tangential variety of a curve

2011-07-09T16:07:48Z

Let $C\subset\mathbb{P}^r$ be a smooth nondegenerate curve (not contained in any hyperplane) of degree $d$ genus $g>0$. Consider the tangential variety $X$ of $C$: $X=\cup_{p\in C}T_pC\subset \mathbb{P^r}$. This is a surface in $\mathbb{P}^r$ which is singular along $C$. My feeling is that $X$ can not be contained in any quadric hypersurface $Q$. Is this something reasonable to expect? A baby case is when $C$ is the rational normal curve in $\mathbb{P}^3, then X$ is a quartic hypersurface thus not contained in any $Q$. Any imput is welcome. Thanks a lot.

Answer by Yusuf Mustopa for quadrics containing the tangential variety of a curve

2011-07-09T17:04:43Z

There is an article of Eisenbud which discusses tangent developable surfaces to rational normal curves:

www.msri.org/~de/papers/pdfs/1992-007.pdf

It is shown in this paper that the tangent developable surface to a rational normal curve of degree 4 or greater is contained in a quadric hypersurface (see the bottom of p.13).

Answer by Dmitri for quadrics containing the tangential variety of a curve

2011-07-09T18:01:23Z

Edited.

Here is a construction of curves $C$ on a four-dimensional quadric $Q^4$ such that $TC\subset Q^4$. I am sure that this is a classical construction, (it might be I saw it previously and forgot).

Construction. Recall that $Q^4$ is isomorphic to $G(2,4)$ -- the Grassmanian of $2$-planes in a four-dimensional space, or equivalently to the space of in lines $\mathbb P^3$. The isomorphism is given by Plucker embedding of $G(2,4)$ to $\mathbb P^5$.

Now, take any curve $C'$ in $\mathbb P^3$ and associate to it a curve $C$ in $G(2,4)$ consisting of the collection of lines tangent to $C'$. I claim $TC\subset Q^4$ once we identify $G(2,4)$ with $Q^4$. The proof is left as an exercise.

PS. I think it will be more interesting to answer the following question: For each $n$, what is the maximal $k(n)$ such that $Q^n$ contains "non-degenerate" $k(n)$-dimensional subvariety $C^{k(n)}$ of arbitrary high degree, such that $TC^{k(n)}\subset Q^n$? I am pretty sure that the above construction can be generalised to show that $k(n)$ tends to infinity when $n$ tends to infinity. In fact from the very first glance it is not clear (for me) why the behaviour of such varieties $C$ should not have resemblance with algebraic Legendrian varieties about which you can read, for example, here : http://arxiv.org/abs/0805.3848 .