# 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.

-
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

## 1 Answer

Answer edited, in response to the comment and a second wind for explaining mathematics:

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 $y_i$.

All that's left is to factor the polynomial, and kill all the linear factors, just throw them away, the reasons are explained in more computational detail on my blogpost, and there I also do explicit examples, but the method for calculating the equation of the dual curve is as above.

-
Charles: In general, try to avoid having all useful information in an answer contained off-site. – Larry Wang Jul 28 '10 at 0:25