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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?

zroslav
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  • Ok, You're right – zroslav Dec 09 '10 at 22:46
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    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 09 '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 automorphisms 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 05 '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-Thélène 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 05 '11 at 15:36
  • Contd. Are there examples of pairs of homogeneous polynomials (say of degrees $d$, resp. $e$), say $F=(F_0(X_0,X_1),F_1(X_0,X_1))$, resp. $(G_0(X_0,X_1),G_1(X_0,X_1))$ which commute in the sense that $( F_0(G_0,G_1), F_1(G_0,G_1) )$ equals $( G_0(F_0,F_1), G_1(F_0,F_1) )$, and yet there is no common fixed "point", i.e., line through the origin mapped to itself by both $F$ and $G$. The connection to rational maps is $f(z) = F_1(1,z)/F_0(1,z)$ and $g(z)=G_1(1,z)/G_0(1,z)$ so that the commutation condition above implies the usual commutation condition $f(g)=g(f)$. – Jason Starr Aug 05 '11 at 15:42
  • @Jason: do you have any reference to such a literature? It will be interesting to read. – zroslav Aug 09 '11 at 13:17