Let $f$ and $g$ be two rational functions. To avoid trivialities, we suppose that their degrees are at least $2$. We say that they have a common iterate if $f^m=g^n$ for some positive integers $m,n$, where $f^m$ stands for the $m$-th iterate.

1. Can one describe/classify all such pairs?

This is probably very hard, and perhaps there exists no simple answer. But here is a simpler question:

2. Is there an algorithm which finds out whether two rational functions have a common iterate or not ?

I mean, I give you two rational functions, say with integer coefficients, and you tell me whether they have a common iterate or not. Perhaps using a super-computer...

Motivation. J. F. Ritt, (Permutable rational functions. Trans. Amer. Math. Soc. 25 (1923), no. 3, 399-448) gave a complete classification/description of all commuting pairs of rational functions (that is $f(g)=g(f)$)... except when they have a common iterate. I gave a completely different proof of Ritt's theorem, but again it does not apply to the case when $f$ and $g$ have a common iterate (MR1027462).

Polynomial pairs (commuting, or with a common iterate) are completely described in MR1501149 Ritt, J. F. On the iteration of rational functions. Trans. Amer. Math. Soc. 21 (1920), no. 3, 348-356, in the very end of this paper.

What is the exact relation between permutable pairs and pairs with a common iterate ?

3. If two functions have a common iterate, must they commute?

Or perhaps they must, but with explicitly listed exceptions? A positive answer to this will solve problem 2 above. See also my "answer" to on common fixed points of commuting polynomials (and rational functions) for an additional motivation.

EDIT. And one more question:

4. Can one describe commuting functions that have a common iterate?

This would complete Ritt's description of commuting functions.

  • 2
    $\begingroup$ Just to avoid having people make the same silly mistake I just made: recall that the degree of a rational function $P/Q$, where $P$ and $Q$ are relatively prime, is max(deg(P), deg(Q)). So standard counterexamples to question 3 involving fractional linear transformations don't count, these being of degree 1. $\endgroup$
    – Todd Trimble
    Dec 2, 2012 at 1:34
  • $\begingroup$ Yes, the degree of a rational function is the topological degree of the map of the Riemann sphere, and also the number of pre-images of a generic point, and also max{deg P,deg Q}, where P/Q is an irreducible representation. $\endgroup$ Dec 2, 2012 at 2:16
  • $\begingroup$ Tag open-problem is appropriate $\endgroup$ Dec 9, 2012 at 11:50
  • $\begingroup$ A criterion for rational functions of degree $d \geq 3$ to have a common iterate was given recently by Hexi Ye. See my edit below. $\endgroup$ Dec 11, 2012 at 22:16

3 Answers 3


Over ${\bf C}$, An easy counterexample to question 3 is $f(x) = x^2$, $g(x) = cx^2$ where $c$ is a nontrivial cube root of unity. Then $f(f(x)) = g(g(x)) = x^4$ but $f$ and $g$ do not commute. There are similar examples for higher iterates.

[Added later] A more exotic construction yields further examples, some defined over ${\bf Q}$, such as the degree-4 pair $$ f(y) = \frac{y^4+18y^2-47}{8y^3}, \phantom{\infty} g(y) = \frac{f-3}{f+1} = \frac{y^4-24y^3+18y^2-27}{y^4+8y^3+18y^2-27} $$ with $f \circ f = g \circ g$ but $f \circ g \neq g \circ f$. This is a "Lattès map" associated to the elliptic curve $E: y^2 = x^3 + 1$: the function $f$ comes from the doubling map $P \mapsto 2P$, and $g$ comes from $P \mapsto 2P+T$ where $T$ is the 3-torsion point $(0,1)$ (as the $(f,g)=(x^2,cx^2)$ example does on the multiplicative group). This elliptic curve yields examples of $f \circ f = g \circ g$ and $f \circ g \neq g \circ f$ with any degree $m^2+mn+n^2$ as long as that's not a multiple of 3, with $f,g \in {\bf Q}(y)$ if $n=0$. Other elliptic curves with complex multiplication yield further examples using the $x$-coordinate rather than the $y$-coordinate, e.g. $f(x) = -x(x^4+6x^2-3)^2 / (3x^4-6x^2-1)^2$ and $g = (f-1)/(f+1)$ from tripling on $y^2=x^3-x$.

  • $\begingroup$ Nevertheless, I have a hope that all counterexamples can be somehow explicitly listed. $\endgroup$ Dec 2, 2012 at 6:20
  • 1
    $\begingroup$ A nice treatment of iterative roots of a quartic polynomial can be found here: MR2772429 (2012i:30054) Strycharz-Szemberg, Beata; Szemberg, Tomasz Geometry of the locus of polynomials of degree 4 with iterative roots. (English summary) Cent. Eur. J. Math. 9 (2011), no. 2, 338–345. $\endgroup$ Dec 2, 2012 at 21:02
  • $\begingroup$ There is no question about POLYNOMIALS. Ritt answered all these questions for polynomials completely in the 1920-s. $\endgroup$ Dec 3, 2012 at 6:21
  • $\begingroup$ Interestingly, these examples are both associated to examples of commuting pairs of rational functions that do not have a common iterate. Might it be possible that all examples are associated in this way? $\endgroup$
    – Will Sawin
    Dec 4, 2012 at 6:42

I am replacing my previous incorrect answer by this one. I just learned about a recent preprint by Hexi Ye,


Among other things, he proves, for general $f$ with degree $d \geq 3$, that $\mu_f=\mu_g$ implies that $f$ and $g$ share an iterate (the converse is well known). The symbol $\mu_f$ denotes the unique $f$-invariant measure of maximal entropy for $f$ (and similarly for $g$). He also analyzes generic maps of degree $2$. The proof involves some holomorphic maps from $t \in \mathbb{C}$ to $f_t \in \rm{Rat}_d$, the set of rational functions of degree $d$ (not semigroups, which you point out to be impossible). As far as I can tell at the first glance, he does not seem to address the commutativity question.

  • $\begingroup$ Margaret, NO rational function of degree at least 2 can be embedded into an iterative semigroup. Even no meromorphic function. This is a very old result of I. N. Baker. $\endgroup$ Dec 3, 2012 at 6:19

Regarding the second question the algorithm is as follows:

  1. Find the flows (superfunctions) of the both functions in closed form

  2. See if they coincide at integer points.

For example, $f(x)=x^2$, $g(x)=x^4$.

The flows will be respectively, $C^{2^x}$, $C^{4^x}$.

Now we solve

$$C^{2^x} = C^{4^y}$$

and find $y=x/2$

This equation obviously has infinitely many integer solutions.

A more complicated case is when $f(x)=\frac{x+1}{x-1}$, $g(x)=\frac{x-1}{x+1}$

In this case the flows are:

$$f^*(x)=\frac{C \cos \left(\frac{3 \pi x}{4}\right)+\sin \left(\frac{3 \pi x}{4}\right)}{\cos \left(\frac{3 \pi x}{4}\right)-C \sin \left(\frac{3 \pi x}{4}\right)}$$

$$g^*(x)=\frac{\left(\left(\sqrt{2}-1\right) C-1\right) (-1)^x+\sqrt{2} C+C+1}{\left(-C+\sqrt{2}+1\right) (-1)^x+C+\sqrt{2}-1}$$

Solving equation f*(x) = g*(y) for integer x and y gives $x=4m, y=n$

  • $\begingroup$ why the downvote? $\endgroup$
    – Anixx
    Apr 26, 2013 at 4:30
  • 2
    $\begingroup$ @Anixx: Maybe because it is not clear at all what you are suggesting. (I didn't downvote though.) It seems to me that your approach more or less only works if you can compute a closed expression for the higher iterates - which is trivial in your examples - but impossible in general. $\endgroup$ Apr 26, 2013 at 14:11
  • $\begingroup$ @Peter Mueller I am sure that is the flows cannot be found, the question cannot be solved. $\endgroup$
    – Anixx
    Apr 26, 2013 at 16:26
  • 3
    $\begingroup$ This isn't an algorithm. $\endgroup$ Apr 26, 2013 at 23:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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