# Projective duality

Given a curve how do you intuitively construct the picture of its projective dual? I know points --> lines, lines--> points but for something like the swallowtail this is not really obvious.

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I don't have a answer, but since no one else has said anything, have you tried breaking the curve into a series of lines to gain some intuition? –  Jonathan Fischoff Jul 23 '10 at 21:58

Let $F(x_0,x_1,x_2)=0$ be the equation for your curve, and take $(y_0,y_1,y_2)$ to be coordinates on $(\mathbb{P}^2)^*$. Also, assume that $F$ is irreducible and has no linear factors.
Then $y_0 x_0+y_1 x_1+y_2 x_2=0$ is the equation of a general line in $\mathbb{P}^2$ (recall, here $y_0,y_1,y_2$ are fixed, and the $x_i$ are the coordinates on the plane) and we look at the open set of $(\mathbb{P}^2)^*$ where $y_2\neq 0$. On this open set, we can solve the equation of the line for $x_2$, and look at $g(x_0,x_1)=y_2^n F(x_0,x_1,-\frac{1}{y_2}(y_0x_0+y_1x_1))$, a homogeneous polynomial of degree $n$ in $x_0,x_1$ with coefficients homogeneous polynomials in the $y_i$. This polynomial has zeros the intersections of our curve $C$ with the line $L$ we're looking at.
So we want to find points of multiplicity at least two. So how do we find multiple roots of a polynomial? We take the discriminant! Specifically, we do it for an affinization, and we get a homogeneous polynomial of degree $2n^2-n$ in the $y_i$.