# Self-dual plane curves

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?

Serge

• "Besides the non-degenerate conic, I am aware of the nodal cubic." The dual curve of the nodal plane cubic is a degree $4$ curve with three cusps. Commented Dec 30, 2015 at 10:40
• @Jason: of course! I mean cuspidal, thank you for the correction. Commented Dec 30, 2015 at 10:54

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).

• Concerning the differential equation that you mention. I think you will find, when you write it out carefully, that this differential equation is quite degenerate and winds up being an underdetermined system, so that the space of local solutions $f$, once you fix $\phi$ and the involution $\psi$ that has a(n isolated) fixed point, does not turn out to be finite dimensional after all. I can give details if requested. In particular, the space of local self-dual curves with fixed $\phi$ and $\psi$ depends on two 'arbitrary' functions of one variable, not on any finite number of constants. Commented Jan 2, 2016 at 21:28
• I guess I was misled by the solution in case the automorphism is the identity map. I will strike out that edit! Thank you. Commented Jan 2, 2016 at 21:30
• Yes, the automorphism $\psi$ can be the identity only in the case of a conic. (This is even true locally.) Note that 'the' automorphism need not be uniquely defined if the curve itself has ambient automorphisms, as is the case with your 'monomial curves' $f(z) = [1, z^a, z^b]$, where $0<a<b$ are relatively prime, which have a 1-parameter ambient symmetry group. Even in the case of the quintic I describe below, the group of ambient automorphisms has order 10, so there is not a unique $\psi$. Commented Jan 4, 2016 at 8:54

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.