By Ciro Ciliberto
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N (z)). Then Ψ is an analytic function from Ck into Cn . We claim that ψj (z) = 0 for j > l. Indeed, since j > l, the map γ −1 τj γ is not in F. Hence there exists a net (xλ )λ∈Λ in V converging to x0 so that γs−1 τj γs (xλ ) = σ1 (xλ ) for all λ ∈ Λ. 15, Φ(xλ , z)(tj h) is diagonal for all λ in Λ. First consider the case when σ1 (x0 ) = x0 . Then A(x, z) is continuous, and so Φ(x, z) is point–norm continuous. Taking limits, we conclude that Φ(x0 , z)(tj h) is diagonal; whence ψj (z) = 0.
G. KATSOULIS Proof. If E is empty, then σ1 = σ2 and there is nothing to prove. 17, the sets Eα are open and Eid ∩ E(12) ∩ E = ∅. Arguing as before using the connectedness of E, it follows that there is some α so c that Eα contains E. But clearly both Eid and E(12) contain E = Z(X, σ)◦ . Thus Eα = X and so X and Y are conjugate. At the other end of the spectrum, total disconnectedness makes piecewise conjugacy very tractable. 20. Let X be a totally disconnected compact Hausdorﬀ space. Fix a homeomorphism γ of X onto another space Y .
2) C0 (X) ×σ F+ n is semisimple. 2. SEMISIMPLICITY 45 (3) There are no non-empty generalized wandering sets. When X is metrizable, these are also equivalent to (4) Each σi is surjective and the v-recurrent points are dense in X for every v ∈ F+ n. Proof. Assume ﬁrst that there exists a nonempty open set U ⊂ X and u, v ∈ F+ n so that −1 σuwv (U ) ∩ U = ∅ for all w ∈ F+ n. Let h = 0 be continuous function with support contained in U . Then (h ◦ σuwv )h = 0. We will show that A(X, σ) is not semisimple.