quadrics containing the tangential variety of a curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:05:06Z http://mathoverflow.net/feeds/question/69885 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69885/quadrics-containing-the-tangential-variety-of-a-curve quadrics containing the tangential variety of a curve Jie Wang 2011-07-09T16:07:48Z 2011-07-10T22:56:32Z <p>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. </p> http://mathoverflow.net/questions/69885/quadrics-containing-the-tangential-variety-of-a-curve/69887#69887 Answer by Yusuf Mustopa for quadrics containing the tangential variety of a curve Yusuf Mustopa 2011-07-09T17:04:43Z 2011-07-09T17:04:43Z <p>There is an article of Eisenbud which discusses tangent developable surfaces to rational normal curves:</p> <p>www.msri.org/~de/papers/pdfs/1992-007.pdf</p> <p>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).</p> http://mathoverflow.net/questions/69885/quadrics-containing-the-tangential-variety-of-a-curve/69894#69894 Answer by Dmitri for quadrics containing the tangential variety of a curve Dmitri 2011-07-09T18:01:23Z 2011-07-10T21:10:09Z <p><strong>Edited.</strong> </p> <p>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).</p> <p><strong>Construction.</strong> 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 $\mathbb P^5$.</p> <p>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.</p> <p><strong>PS.</strong> I think it will be more interesting to answer the following <strong>question</strong>: <em>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 $TC^{k(n)}\subset Q^n$?</em> 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 : <a href="http://arxiv.org/abs/0805.3848" rel="nofollow">http://arxiv.org/abs/0805.3848</a> .</p>