Algebraic composition operators (Q1194218)

Let \(F(X)\) be a linear space of complex valued functions on a set \(X\). Any self-map \(b:X\to X\) defines the automorphism \(C_ b: F(X)\to F(X)\) where \(C_ b u(x):= u(b(x))\). The paper deals with the following problem: For a given \(F(X)\) and a polynomial \(P(z)=z^ n+ p_{n-1}z^{n-1}+ \cdots+p_ 0\) is there a self-map \(b:X\to X\) such that: (i) \(C_ b: F(X)\to F(X)\) is an automorphism; (ii) \(P(C_ b)u=0\) for all \(u\in F(X)\); (iii) there exists no other polynomial with lower degree and the same property? If the answer to the question is affirmative, \(P(z)\) is called characteristic polynomial for \(F(X)\). It turns out that the supply of characteristic polynomials for a given space \(F(X)\) depends on the type of \(F(X)\). For the Hardy or Bergman spaces of functions analytic in the disk \(D\subset\mathbb{C}\) they are given by the infinite family \(z^ n-1\) \((n\geq 1)\) and the ``sporadic'' polynomial \(z^ 2-z\). The main result states: for \(C(X)\) all characteristic polynomials are of the form \(P(z)=z^ m \prod_{t\in G} (z-t)\), where \(m\geq 0\) is some integer and \(G\) is a finite union of finite subgroups of the unit circle \(\mathbb{T}\).