TITLE: Self-dual plane curves
QUESTION [10 upvotes]: Suppose that $C\subset \mathbb P^2$ is a plane projective curve (base field is $\mathbb C$) and $C^*\subset (\mathbb P^2)^*$ is its dual. What are the known examples in which $C$ is projectively (i.e., linearly) isomorphic to $C^*$? Besides the non-degenerate conic, I am aware of the cuspidal cubic. Anything else?
Thank you in advance,
Serge

REPLY [12 votes]: There are also the limaçons, which are rational curves of degree 4 with one node and two cusps:  There is a $1$-parameter family of these up to projective equivalence (the parameter is $a\not= \pm1,\pm2$):
$$
[x_0,x_1,x_2] 
= \bigl[2t^2{+}at^2(1{-}t^2),\ 2t^2+a(1{-}t^2),\ (1{+}a)t+(1{-}a)t^3\bigr].
$$
See L. E. Wear, Self-dual plane curves of the fourth order, American Journal of Mathematics 42 (1920), 97–118.  Up to projective equivalence, these are the only irreducible self-dual curves of degree $4$ having only traditional singularities (i.e., nodes and cusps).
Remark 1: Also, regarding the question of whether there are self-dual curves of positive genus: The quintic
$$
8\,(x^5-10x^3y^2+5xy^4)-15(x^2{+}y^2)^2+10(x^2{+}y^2)-3 = 0
$$
has genus $1$ and is self-dual.  It has 5 cusps, spaced evenly around the unit circle $x^2+y^2=1$ and including $(1,0)$, and no other singularities.  (The Plücker formulae imply that an irreducible self-dual curve of degree $5$ and genus $1$ and whose only singularities are nodes and cusps must have $0$ nodes and $5$ cusps. By Bezout, no three of the cusps can lie on a line, and hence the $5$ cusps must lie on a nonsingular conic.  If one assumes that the conic is $x^2+y^2=1$ and that the cusps are equally spaced around the unit circle, one easily finds that the above example is the only possibility.  Fortunately, it works out to be self-dual.)
Remark 2:  Here is another interesting family of self-dual rational curves: $C_{n,m}$ given by the parametrization $f_{n,m}:\mathbb{P}^1\to \mathbb{P}^2$, where
$$
f_{n,m}\bigl([1,t]\bigr) 
= \bigl[\bigl(t^n+(n{-}2m)^2\bigr),\ \bigl(t^n+(n{+}2m)(n{-}2m)\bigr)t^m,\ \bigl(t^n+(n{+}2m)^2\bigr)t^{2m}\bigr].
$$
For relatively prime positive integers $m$ and $n$, the curve $C_{n,m}$ is of degree $d=n{+}2m$ (except for the case $(n,m)=(2,1)$, when the parametrization degenerates to a conic). The curves $C_{n,1}$ (for $n\not=2$) have only traditional singularities, with $\kappa = n$ cusps (where $t^n = n^2-4 $) and $\delta = \tfrac12 n(n{-}1)$ nodes.  An interesting thing about these curves is that, for fixed $n$, the automorphism $\psi_n:\mathbb{P}^1\to\mathbb{P}^1$ of the curve $C_{n,m}$ is given by $\psi_n\bigl([1,t]\bigr) = \bigl([1,\epsilon t]\bigr)$, where $\epsilon^n=-1$ and the isomorphism $\phi_{n,m}:\mathbb{P}^2\to(\mathbb{P}^2)^*$ that defines the self-duality via $f_{n,m}^{(1)}\circ\psi_n = \phi_{n,m}\circ f_{n,m}$ depends only on the congruence class of $m$ modulo $2n$, because the tangential mapping is given by
$$
f^{(1)}_{n,m}\bigl([1,t]\bigr) 
= \left[\begin{matrix}
\bigl(-t^n+(n{+}2m)^2\bigr)t^{2m}\\ 
-2\bigl(-t^n+(n{+}2m)(n{-}2m)\bigr)t^m\\ 
\bigl(-t^n+(n{-}2m)^2\bigr)
\end{matrix}\right].
$$
Thus, this provides examples of countable families of self-dual rational curves of arbitrarily high degree that share the same automorphism $\psi_n$ and correlation mapping $\phi_{n,m}$. In particular, the curves in such a family cannot all be solutions of a nondegenerate first order differential equation.