By Ian Chiswell

The idea of R-trees is a well-established and critical zone of geometric team idea and during this e-book the authors introduce a development that offers a brand new standpoint on workforce activities on R-trees. They build a gaggle RF(G), outfitted with an motion on an R-tree, whose components are definite capabilities from a compact actual period to the crowd G. additionally they research the constitution of RF(G), together with a close description of centralizers of components and an research of its subgroups and quotients. Any staff performing freely on an R-tree embeds in RF(G) for a few collection of G. a lot is still performed to appreciate RF(G), and the huge record of open difficulties incorporated in an appendix may perhaps most likely bring about new tools for investigating workforce activities on R-trees, relatively unfastened activities. This publication will curiosity all geometric workforce theorists and version theorists whose examine comprises R-trees.

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2 Let (X, d) be an R-tree. Then (X, d) is metrically complete if and only if every monotone increasing Cauchy sequence with respect to some base-point converges. 43. 2 we can now prove the following. 3 The R-tree XG associated with RF (G) via the length function L is metrically complete. Proof If {yμ }μ∈N with yμ = f μ , L( f μ ) is a Cauchy sequence in XG that is monotone increasing with respect to x0 then, by the construction of XG described in the previous section, we have αμ := L( f μ ) α and (after changing the values of the functions f μ at the right-hand endpoints if necessary) μ < ν implies that f μ = fν |[0,L( fμ )] .

Then there exist f1 , g1 , u ∈ RF (G) such that f = f1 ◦u, g = u−1 ◦g1 , and f g = f1 ◦ g1 . Proof If ε0 := ε0 ( f , g) = 0 then our conclusion is satisfied for f1 := f , g1 := g, and u := 1G ; hence, we may assume that ε0 > 0. 14 we can find reduced functions f1 : [0, L( f ) − ε0 ] → G, g1 : [0, L(g) − ε0 ] → G, u : [0, ε0 ] → G, and v : [0, ε0 ] → G such that f = f1 ◦ u and g = v ◦ g1 . By the definition of ε0 plus the fact that ε0 > 0, we have f (L( f ) − δ ) = (g(δ ))−1 , 0 ≤ δ < ε0 . 10) gives u−1 (δ ) = u(ε0 − δ ) −1 = v(δ ), 0 ≤ δ < ε0 .

Making use of the equations g2 = w ◦ g3 , g = g2 ◦ v, and g = u−1 ◦ g1 , we find that w(ξ ) = g2 (ξ ) = g(ξ ) = u−1 (ξ ), 0 ≤ ξ < L(u), and, defining w(L(u)) := u−1 (L(u)), we obtain w = u−1 , as required. Note that L(g3 ) = L(g2 ) − L(u) > 0 by the case assumption. 18 (associativity of the circle product) in the last step. 1. 18 now yields f g = f1 ◦ g1 = f1 ◦ (g3 ◦ v) = ( f1 ◦ g3 ) ◦ v; in particular, ε0 ( f1 , g3 ) = 0. 16 (visible cancellation) gives ( f g)h = (( f1 ◦ g3 ) ◦ v)(v−1 ◦ h1 ) = ( f1 ◦ g3 )h1 .

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