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



Self-dual plane curves
I believe that this is true for all plane curves parameterized by a monomial map $\mathbb{P}^1\to \mathbb{P}^2$. Denote by $[s,t]$ homogeneous coordinates on $\mathbb{P}^1$, and denote by $[u,v,w]$ homogeneous coordinates on $\mathbb{P}^2$. Up to permuting $u$, $v$, and $w$, every birational monomial map is given by $$f:\mathbb{P}^1\to \mathbb{P}^2, \ \ f^\*[u,v,w] = [\alpha\cdot s^b,\beta \cdot s^{b-a}t^a, \gamma\cdot t^b],$$ for positive, relatively prime integers $b\geq a$ and for invertible elements of the base field $\alpha$, $\beta$ and $\gamma$ (there is one exception if $b$ equals $1$ and $\beta$ is zero). 


If you think of the associated morphism $\mathbb{A}f: \mathbb{A}^2\to \mathbb{A}^3$, it is easier to compute that, with respect to the standard trivializations of the tangent (or, better, cotangent) sheaves on affine space associated to the structure of additive group, the derivative map is just, $$df = \left[ \begin{array}{ccc} b\alpha\cdot s^{b-1} & 0 \\ (b-a)\beta\cdot s^{b-a-1}t^a & a\beta\cdot s^{b-a}t^{a-1} \\ 0 & b\gamma\cdot t^{b-1} \end{array} \right].$$ Thus, with respect to the standard dual ordered basis $u^\vee,v^\vee,w^\vee$ of the dual projective plane $\left(\mathbb{P}^2\right)^\vee$, the tangent map is $$f^\vee:\mathbb{P}^1\to \left(\mathbb{P}^2\right)^\vee, \ \ (f^\vee)^\*[u^\vee,v^\vee,w^\vee] = [(b-a)\beta\gamma\cdot t^b, -b\alpha\gamma\cdot s^at^{b-a}, a\alpha\beta\cdot s^b].$$ Up to permuting $s$ and $t$, this is projectively equivalent to the image of $f$.


The Plücker relations might rule out examples that have positive geometric genus.


**Edit.** The defining equation of the image of $f$ is the binomial $\beta^b\cdot u^{b-a}w^a - \alpha^{b-a}\gamma^a\cdot v^b=0$. Note that the dehomogenizations with respect to $u$ and $w$ are still homogeneous with respect to an appropriate weighting of the two variables. Perhaps you can relate your question to Saito's characterization of unicritical polynomials on affine space . . .


**~~Second Edit.~~** **What I wrote below about the number of moduli is wrong!** Robert Bryant points out that there exist other examples of self-dual curves, including self-dual curves of positive genus. There is one more observation. Except for plane conics, every self-dual plane curve admits a nontrivial automorphism. For the normalization of the plane curve, $f:C\to \mathbb{P}^2$, for the associated tangency morphism $f^{(1)}:C\to (\mathbb{P}^2)^\vee$, and for an isomorphism $\phi:\mathbb{P}^2 \to (\mathbb{P}^2)^\vee$ such that $\phi(\text{Image}(f))$ equals $\text{Image}(f^{(1)})$, for the unique automorphism $\psi:C\to C$ such that $f^{(1)}\circ \psi$ equals $\phi \circ f$, $\psi$ is **not** the identity automorphism. If $\psi$ has a fixed point, then up to specifying the vanishing sequence of $f$ at the fixed point, self-duality will give a first-order differential equation for $f$ (as an analytic arc) that should have only one constant of integration. So families of self-dual curves with a fixed point have bounded moduli (either $1$ modulus, or perhaps $1+2=3$ moduli if we include the coordinates of the image point, or perhaps $3+8=11$ moduli if we also allow $\phi$ to vary).


