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By the Ritt's classification, for any pair of commuting polynomials (i.e. $f(g(z))=g(f(z))$) over $\mathbb C$ there is a common fixed point of them. My questions are:

  1. Is that true that this can be obtained rather simpler than with Ritt's classification?

  2. Is that true for any pair of commuting rational functions?

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2. Try $\frac 1 z$ and $-z$. – darij grinberg Dec 9 '10 at 20:53
Ok, You're right – zroslav Dec 9 '10 at 22:46
You might like Shields' 1964 Proc AMS paper "On fixed points of commuting analytic functions" where he shows that if f and g are commuting, analytic on the unit disc, and nice on the boundary, then f and g have a common fixed point. A 1973 Proc AMS paper of Behan "Commuting analytic functions without fixed points" and a 1984 Trans AMS paper of Cowen "Commuting analytic functions" have more results in this line. (Some of the conjectures in this last paper are disproved in Chalendar and Mortini's more recent "When do finite Blaschke products commute?"). Rational functions are subtle things. – anon Dec 9 '10 at 23:33
Dear Zroslav -- Thank you for posting about this. I have some work on existence of common fixed points for two commuting <i>automorphisms</i> of a projective variety. But I had been wondering about what happens with commuting self-maps, e.g., even a polynomial map from $\mathbb{P}^1$ to itself. I am glad to learn there is some literature on this subject. – Jason Starr Aug 5 '11 at 15:30
I have a question on Grinberg's (wonderful) example: For any pair of commuting automorphisms, such as in this example, there is an "elementary obstruction" (in the sense of Colliot-Th&eacute;l&egrave;ne and Sansuc) to the existence of a common fixed point: namely, for every invertible sheaf which is fixed by the automorphisms, there must be liftings of the two automorphisms to linearizations of the invertible sheaf which commute. Roughly this says that the rational maps are dehomogenizations of homogeneous polynomials on $A^2$ which commute on $A^2$. This "explains" Grinberg's example. contd – Jason Starr Aug 5 '11 at 15:36

I don't know the answers. Consider this as an extended comment. Let $P$ be the set of fixed points of $g$. Then $f$ maps $P$ into itself. Indeed let $x\in P$, so $g(x)=x$. Then $$f(x)=f(g(x))=g(f(x)),$$ which means that $f(x)\in P$. Now $P$ is finite, so $f$ must have a periodic point in $P$. So $f^m$ and $g$ have a common fixed point. Of course one can interchange $f$ and $g$ here.

With little more care, one can find a common fixed point of $f^m$ and $g^n$ with some $m,n$, which is REPELLING for both $F=f^m$ and $G=g^n$. Let this common repelling fixed point be $a$.

Evidently $F$ commutes with $G$. If two communing functions share a repelling fixed point, then they have the same Poincare function at this point. Poincare function $\phi$ is the ``linearizer'', that is the solution of the functional equation $$\phi(\lambda z)=F(\phi(z)), \quad \phi(0)=a,\quad\phi'(0)=1, $$ where $\lambda=F'(a)$.

Now if $F'(a)=G'(a)$ we easily conclude that $F=G$. Which means that $f$ and $g$ have a common iterate. If $F'(a)\neq G'(a)$ there is only a very restricted set of possibilities which were completely described and classified by Ritt.

Thus we come to a question:

Suppose that $f$ and $g$ have a common iterate. Must they have a common fixed point?

Or, perhaps those pairs which do not permit some explicit classification?

The question about rational functions that have a common iterate is interesting in itself and I am posting it separately, Rational functions with a common iterate

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