1
$\begingroup$

Let $C$ a projective curve in $\mathbb{P}^2(\mathbb{C})$. We say a line $L \subset \mathbb{P}^2$ is mulltiple tangent of $C$ if there are $P_1, \dots P_k$ points on $C$ such that $L$ is the tangent of $C$ at $P_i$ for every $i$ and $k \geq 2$ and none of them is an inflection point.

We have the Gauss map: $\nu :C \to C^*$ from the curve to its dual.Now, we have $\nu(P_i)=Q$ for every $i$. I would like to get a proof of the fact that $Q$ is an ordinary $k$ fold point of $C^*$.

I think that the fact that we are on $\mathbb{C}$ and we have the analytic topology could be used to make things simpler, but I do not really understand how the dual curve is made.

EDIT: I tried to write an explicit parametrization of the dual curve, but I still can't conclude.(I'm gonna work over $\mathbb{C}$ because I do not know anything about completion. Let's say for the sake of simplicity, $k=2$ and $P_1=[1 ;0 ;0],P_2=[0; 1 ; 0]$ and $L=\{z=0\}$. Locally, near $P_1$ the curve has a parametrization of the kind $(x,z(x))$ in the standard affine coordinates, such that $z'(0)=0,z''(0) \neq 0$. Now , the points in the tangent curve near the image of $P_1$ should be written as $[z'(x) ; -xz'(x)+z(x) ; 1]$. The problem is that I do not know how to find the tangents to the dual curve having written this.

$\endgroup$
  • $\begingroup$ This is not true unless you require that the curve has no inflection points (consider $V(y-x^3)$, for instance). And the analytic topology is unnecessary - one can do everything in completions and it will work just fine for any field of characteristic not $2$. The first thing to do is to get a handle on how the dual curve works: you may wish to peruse the wikipedia page about it, for instance. From there, look at what happens to the equation of the tangent line near $P_i$ that map to the same $Q$ - since it's not an inflection point, there's a specific relation that must be satisfied... $\endgroup$ – KReiser Jan 20 at 8:06
  • $\begingroup$ I edited for what about inflection points thank you $\endgroup$ – Tommaso Scognamiglio Jan 20 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.