Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
Example with figure 8 I mentioned is this. Let $k=2\pi i/n$, Let $g(z)=(kz+1/(kz))/2$. It maps the real line into a hyperbola $\gamma$; then $f$ is the $n$-th Chebyshev polynomial; it maps this hyperbola back on the real line. Hyperbola has the shape of figure 8 on the sphere. This example was shown to me by Fedor Pakovich. – Alexandre Eremenko Aug 5 '12 at 8:38
I think you should be more explicit about what you mean by 'maps into' and 'Jordan analytic curve'. After all, if you take $g(z) = z^2+ iz^{-2}$, this maps the real line into one branch of the hyperbola $xy=1$ (the one with $x$ and $y$ nonnegative), and this closes to be a (nonsmooth) Jordan analytic curve $\gamma$ on the Riemann sphere. Then the map $f(w) = w^2-2i$ maps $\gamma$ into the real line again. Of course, the composition $f\circ g$ isn't a covering map, but you didn't ask for that. (Credit: This is a slight modification of Noam Elkie's deleted answer.) – Robert Bryant Aug 11 '12 at 14:32
@Alexandre: Do you also require the curve $\gamma$ to be closed as well as smooth everywhere? By modifying the above construction, I could arrange to have $g$ map the real axis into a proper subset of the hyperbola $xy=1$, one that avoids the corner. Then taking $\gamma $ to be the hyperbola without the corner at infinity, it would be smooth and all your conditions would be satisfied. – Robert Bryant Aug 12 '12 at 13:10
"Into" isn't really "into" here, is it? I mean, you are, most likely, not looking for an idiotic counterexample with two constant maps. So, what exactly does "into" stand for? – fedja Dec 1 '12 at 19:25
Fedja, "into" is the same as "to". $X\to Y$. Nothing else. – Alexandre Eremenko Dec 3 '12 at 6:12

1 Answer 1

up vote 26 down vote accepted

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 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 $w=2y$.
  2. 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 $\delta=0$.

share|cite|improve this answer
This is very nice, though I still have to digest it. – Alexandre Eremenko Dec 10 '12 at 14:02
By eye it seems this curve is quite hard to distinguish from an ellipse and I guess it probably looks quite like a circle (the radius in the x-direction seems to be about 45 and the radius in the y-direction maybe 47), though it apparently is not. Could you show a plot with aspect ratio of 1? – j.c. Dec 10 '12 at 17:39
@jc: Thanks for the observation, I fixed the mistake. (Indeed, the original picture was an exact circle ...) – Peter Mueller Dec 10 '12 at 20:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.