Relation between curvature of curve and dual curve?

Question

For a plane algebraic curve, does there exist a relationship between the curvature of the curve and the curvature of its dual curve?

Answer 1

There seems to be an inverse one.

Let's look at the conics in ${\Bbb A}^2\subset {\Bbb P}^2$.

Let ${\bf x}=\begin{pmatrix}x\\y\\z\end{pmatrix}$, ${\bf x}^TA{\bf x}=0$ has dual ${\bf x}^TA^{-1}{\bf x}=0$, where $A$ is non-degenerate symmetric.

So $(\frac{x}{a})^2\pm(\frac{y}{b})^2-z^2=0$ where $z=1$ has curvature between $a$ and $b$. The dual is $(\frac{x}{\frac{1}{a}})^2\pm(\frac{y}{\frac{1}{b}})^2-z^2$, which has curvature betweeen $\frac{1}{b}$ and $\frac{1}{a}$.

To track points we use a parametric representation: $x=x(t), y=y(t)$ dual to $p=p(t), q=q(t)$, where $p(t)x+q(t)y+1=0$ and $p(t)=\frac{-y'(t)}{x'(t)y(t)-x(t)y'(t)}, q(t)=\frac{x'(t)}{x'(t)y(t)-x(t)y'(t)}$.

Let's check $x(t)=a\frac{1-t^2}{1+t^2},y(t)=b\frac{2t}{1+t^2}$, and lo $p(t)=\frac{1-t^2}{a(1+t^2)}, q(t)=\frac{2t}{b(1+t^2)}$.

Looks promising.

The conics are the only smooth with smooth duals.

Let's try the cusp $x(t)=t^2,y(t)=t^3$. The formulas give $p(t)=\frac{3}{t^2},q(t)=-\frac{2}{t^3}$

Small values of $t$ are near the origin for the curve, but far away from the origin for the dual, indicating that we really should be working in ${\Bbb P}^2$, where at infinity there is a flex: $x^2z-y^3$ dehomogenised at $x$ is $z-y^3$, a flex.

We're led to thinking a flex has infinite curvature, a cusp $0$.

In addition for bitangent/node duality you'll have to ignore the other branch.

The singularities will only get worse.