# Ordinary r-fold point on the dual curve

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.

• 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... – KReiser Jan 20 at 8:06
• I edited for what about inflection points thank you – Tommaso Scognamiglio Jan 20 at 8:19