Suppose that $\gamma$ is a Jordan analytic curve on the Riemann sphere,
and there exist two rational functions $f$ and $g$ such that
$f$ maps $\gamma$ into a circle, and $g$ maps a circle into $\gamma$.
(All rational functions considered are of degree at least $2$).


Question: Does this imply that $\gamma$ is a circle?


By a "circle" I mean a circle on the Riemann sphere, that is a circle or a straight
line in the plane.


I have only one example of such situation, where $\gamma$ is not a circle,
but in this example, $\gamma$ is not Jordan,
it has the shape of figure 8. 


If one restrict to polynomials $f$ and $g$, the question can be answered relatively
easily, using Ritt's factorization theory for polynomials, or other tools.


For the origin of this problem, see arXiv:1110.6552.



Circles and rational functions
**Update:** I removed the links to the Sage code of the complicated explicit examples, because very much easier examples exist. See below for an example, and [this preprint](http://arxiv.org/abs/1502.07336) for more details concerning this answer and the computation of explicit examples.


**Answer:** The answer is no. In the following I'll describe how to find two rational functions $f,g\in\mathbb C(X)$, both of odd prime degree $\ell$, such that the following holds:


* $f(g(X))\in\mathbb R(X)$.
* $g(\mathbb R\cup\{\infty\})$ is a smooth closed Jordan curve in the complex plane.


Let $E$ be an elliptic curve, defined over the reals. For $p\in E(\mathbb C)$ we let $\bar p$ be the complex conjugate of $p$. Choose $E$ such that the following holds:


1. There is a point $w\in E(\mathbb R)$ with no $y\in E(\mathbb R)$ with $w=2y$.
- There is a point $z\in E(\mathbb C)$ of order $\ell$ with $\bar z\notin\langle z\rangle$. (Such a point always exists.)


Then $C=\langle z\rangle$ is a subgroup of order $\ell$ of $E$, and $E'=E/C$ is an elliptic curve over $\mathbb C$. Let $\phi:E\to E'$ be the associated isogeny, and $\phi':E'\to E$ be the dual isogeny. Then $\phi'\circ\phi:E\to E$ is the multiplication by $\ell$ map.


Let $\beta$ be the automorphism of order $2$ sending $p\in E(\mathbb C)$ to $w-p$. Similarly, define the involutory automorphisms $\beta'$ of $E'$ and $\beta''$ of $E$ by $\beta'(p')=\phi(w)-p'$ and $\beta''(p)=\ell w-p=\phi'(\phi(w))-p$.


Let $\psi$ be the degree $2$ covering map $E\to E/\langle\beta\rangle=P^1(\mathbb C)$, and define likewise $\psi':E'\to E'/\langle\beta'\rangle=P^1(\mathbb C)$ and $\psi'':E\to P^1(\mathbb C)$.


Let $f(X)$ and $g(X)$ be the rational functions defined implicitly by $\psi'\circ\phi=g\circ\psi$ and $\psi''\circ\phi'=f\circ\psi'$. Note that $\psi$, and $\psi''$ are defined over $\mathbb R$, while $\psi'$ is not.


As the multiplication by $\ell$ map $\phi'\circ\phi$ is defined over the reals, we obtain $f(g(X))\in\mathbb R(X)$.


We next claim that $g(X)$ is injective on $\mathbb R$. Suppose there are distinct real $u,v$ with $g(u)=g(v)$. Pick $p,q\in E(\mathbb C)$ with $\psi(p)=u$, $\psi(q)=v$. Then $\psi'(\phi(p))=g(u)=g(v)=\psi'(\phi(q))$, so $\phi(p)=\phi(q)$ or $\phi(p)=\phi(w)-\phi(q)$. Upon possibly replacing $q$ with $w-q$ we may assume $\phi(p)=\phi(q)$, so $p-q\in C$.


Next we study the effect of complex conjugation. As $\psi$ is defined over the reals, and $\psi(p)=u$ is real, we have $\psi(\bar p)=\psi(p)$, so $\bar p=p$ or $\bar p=w-p$. Likewise, $\bar q=q$ or $\bar q=w-q$. Recall that $p-q\in C$, and $\bar C\cap C=\{0\}$ by condition 2. So we can't have $(\bar p,\bar q)=(p,q)$, nor $(\bar p,\bar q)=(w-p,w-q)$.


Thus without loss of generality $\bar p=p$, $\bar q=w-q$. As $p-q$ and $\bar p-\bar q=p-w+q$ have order $\ell$, we see that $2p-w=r$ with $\ell r=0$ and $r\in E(\mathbb R)$. So $w=2(p+\frac{\ell-1}{2}r)$, contrary to condition 1. Furthermore, the function $g(X)$ behaves well at infinity by this geometric interpretation.


An explicit example, with $\omega$ a primitive third root of unity, is
\begin{align\*}
f(X) &= \frac{X^3 - 6(\omega + 1)X}{3X^2 + 1}\\
g(X) &= \frac{2X^3 + (\omega + 1)X}{X^2 - \omega}\\
f(g(X)) &= \frac{8X^9 - 24X^5 - 13X^3 - 6X}{12X^8 + 13X^6 + 12X^4
- 1}\in\mathbb Q(X).
\end{align\*}


First note that $g(\mathbb R\cup\{\infty\})$ is not contained in a circle, for instance because the points $g(0)=0$, $g(\infty)=\infty$, $g(1/2)=(1+2\omega)/3$, and $g(1)=(5+4\omega)/3$ do not lie on a circle.


Secondly, $g(\mathbb R\cup\{\infty\})$ is a Jordan curve, because $g$ is injective on $\mathbb R$: Suppose that $g(x)=g(x+\delta)$ for real $x,\delta$. A short calculation yields $\delta(8\delta^4 + 14\delta^2 + 49)=0$, so $\delta=0$.


