Let $C \subset \mathbb{P}^n$ be a projective curve. What is the dual curve in general?

According to Wikipedia, for a smooth hypersurface $X$ defined by a homogeneous polynomial $f,$ the dual curve $\check{X}$ is the image of $X$ under the mprphism

$$\mathbb{P}^n \to \check{\mathbb{P}}^n$$

$$[x_0: \cdots :x_n] \mapsto [\partial{f}/\partial{x_0}: \cdots :\partial{f}/\partial{x_n}]$$

So if $C$ is (smooth) complete intersection, one can somehow understand $\check{C},$ but what if $C$ is not a complete intersection curve? is there then a generalized notion of dual curves (including the singular curves)?


I was thinking about the following question:

Let $C$ be a projective curve of degree $d \leq n.$ Show that there is a linear subspace of $\mathbb{P}^n$ of dimension $d$ containing $C.$

I wanted to invoke induction on $n$ to prove it. The cases $n=1,2$ are trivial, now for $n$ if I can show that there is a hyperplane $H \cong \mathbb{P}^{n-1}$ containing $C$ then I will be done, so I thought if I look at the dual curve $\check{C}$ (if exists!) the existence of $H$ is equivalent to the fact that $\check{C}$ is not empty (contains a point, since points are dual to hyperplanes). Also my second idea, was to project $C$ from a point $p\not \in C$ into a hyperplane which I don't think if I could proceed further.

BTW I'd like to see different approaches for this question.


For any smooth curve $C$ in $\mathbb P^n$, the dual is the set of the hyperplanes $H\in (\mathbb P^n)^{\vee}$ which are tangent to some point of $C$. If $C$ is not smooth (but geometrically reduce so the smooth locus is open and dense in $C$), we take closure of the hyperplanes tangent to some smooth point of $C$.

For the problem which motivates this question, I don't think your argument is correct (a $H$ is tangent to some point of $C$). The direct proof can be as follows:

Consider the invertible sheaf $L=O_{\mathbb P^n}(1)|_C$ on $C$. It has degree $d$. So $\dim H^0(C, L)\le d+1$. Therefore the canonical map $$ H^0(\mathbb P^n, O_{\mathbb P^n}(1))\to H^0(C, L)$$ is not injective by dimension comparison. Let $s\ne 0$ be an element in the kernel. It defines a hyperplane $H$ such that $C\subseteq H$: for all $x\in C$, $s(x)=0$. Now we can do induction on $n$.

Edit A different approach using linear systems. As above, we only need to show $C$ is contained in some hyperplane. Suppose the contrary. Then for any hyperplane $H$, $H|_C$ is a well defined effective divisor on $C$, so fixing a hyperplane $H_0$ and noting $D_0=H_0|_C$, we get a linear map $$ |H_0|\to |D_0|.$$ Now the lefthand side term is a projective space of dimension $n$, and the rhs is a projective space of dimension $\le d$. This map is not constant because by every point of $C$ passes a hyperplane. Now it is known that this can happen only when $n\le d$. This second prof looks too complicate, hope someone can give a better answer.

  • $\begingroup$ very nice dear @QiL. Is there a way to avoid sheaves for this question? $\endgroup$ – Ehsan M. Kermani Jan 14 '13 at 21:09
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    $\begingroup$ Dear @ehsanmo: use linear systems if you prefer. Replace the $H^0$ by $|\ \ |$. $\endgroup$ – user18119 Jan 14 '13 at 21:36
  • $\begingroup$ Dear @QiL, I would very much appreciate it if you could write it as the second approach. I have learned a lot from many of your comments so far, thank you. $\endgroup$ – Ehsan M. Kermani Jan 14 '13 at 22:03
  • $\begingroup$ Dear @QiL, thank you very much again. $\endgroup$ – Ehsan M. Kermani Jan 14 '13 at 23:57

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