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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.

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up vote 3 down vote accepted


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 : .

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Thanks. But I think my curve is nondegenerate, which means it is not contained in any hyperplane. – Jie Wang Jul 9 '11 at 18:07
The answer is completely rewritten. – Dmitri Jul 10 '11 at 18:35
Thanks. That example will work.I apprciate it. What about if I modify my question by supposing $C$ is a general curve embedded in $mathbb{P}^r$ by a general $g^r_d$? – Jie Wang Jul 11 '11 at 13:39
What is $g^r_d$? Do I understand correctly that you want to know if each non-degenerate curve $C$ in $\mathbb P^n$ can be deformed so that $TC$ is not contained in any quadric? If this is indeed the question, I have to think more about it. You might consider again to make this assumption explicit in your question (or ask a follow up question). – Dmitri Jul 11 '11 at 14:49
Yes. By $g^r_d$ I mean a linear system of (projective) dimesion $r$ and degree $d$. This linear system maps $C$ into $\mathbb{P}^r$ (assuming it is very ample). So by general $g^r_d$ I mean exactly what you said: you could deform it in $\mathbb{P}^r$ so that $TC$ is not contained in a quadric. Thanks – Jie Wang Jul 11 '11 at 19:28

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

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).

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Thanks for letting me know. What about curves of $g>0$? – Jie Wang Jul 9 '11 at 17:32
I'm not aware offhand of any results on tangent developable surfaces associated to higher-genus curves. However, the tangent developable surface to a given curve C is contained in the secant variety of C, so it might be a good idea to look into some papers on that. – Yusuf Mustopa Jul 9 '11 at 17:45
The first natural case to consider after rational normal curves is elliptic normal curves; the secant varieties of these have been studied by v.Bothmer-Hulek in and also by T. Fisher in – Yusuf Mustopa Jul 10 '11 at 2:10

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