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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.

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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
    
@Noam: Sorry about the typo! Make that "Noam Elkies' deleted answer". I can't easily correct it now. –  Robert Bryant Aug 11 '12 at 14:35
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@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
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"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
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1 Answer

up vote 20 down vote accepted

Added: jc in his comment is right. $E(\mathbb R)$ also has subgroups of order $3$, by mistake I factored through such a group. Now I chose the correct subgroup, and the image $g(\mathbb R)$ isn't a circle anymore. In fact, $g(\mathbb R)$ is contained in a circle if and only if there is a linear fractional function $\lambda(X)$ over $\mathbb C$ such that $\lambda(g(X))\in\mathbb R(X)$. By the correct choice of the subgroup $C$ below this does not happen.

Data: In this file I provide Sage code which verifies the example given below. To use this program, one needs the data provided here.

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 $\bar z\notin\langle z\rangle$. (That is actually automatically satisfied at least for $\ell=3$.)

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 $\psi$, $\psi'$ and $\psi''$ are defined over $\mathbb R$.

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

I computed an explicit example in order to add evidence that these arguments are correct: I picked the elliptic curve $Y^2=X^3-91X+330$ and $w=(5,0)=2(5+4i,4-16i)=2y$ and $\ell=3$. The functions $g(X)$ and $f(X)$ are defined over a number field of degree $16$ (degree $8$ coming from $C$, and another degree $2$ extension from $i$), so I do not put them here. Instead they are given in the links at the beginning of the answer. The image $g([-100,100])$ is: alt text

Remark: This example came out of a group and Galois theoretic examination of the question. If there is some interest, I might add some more details about this aspect.

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This is very nice, though I still have to digest it. –  Alexandre Eremenko Dec 10 '12 at 14:02
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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
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