Arboreal Group Theory: Proceedings of a Workshop Held by Juan M. Alonso (auth.), Roger C. Alperin (eds.)

By Juan M. Alonso (auth.), Roger C. Alperin (eds.)

During the week of September thirteen, 1988 the Mathematical Sciences learn Institute hosted a 4 day workshop on Arboreal crew concept. This quantity is the manufactured from that assembly. this system based with regards to the idea of teams performing on timber and a number of the functions to hyperbolic geometry. issues comprise the speculation of size services, constitution of teams appearing freely on timber, areas of hyperbolic constructions and their compactifications, and moduli for tree actions.

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Ii). Let F = (g,AA(p),c(A» and D = (j,sq,c(F»), with q E p(F). viii). ii). iv), it follows 8(A) ~ 0, as required. vii)'. Let g E E(y, z). v), it remains to show that d:= Ic(j,st-l,Ii)1 = Ilfl-lhll iff 8(j,s,g) = 8(h,t,g) = O. viii), ([g], d) becomes a A-metric space and O:g is an isometry. Let A = (f, s, g), B = (h,t,g), C = (c(A),p-lq,c(B)) with p E peA), q E pCB), F = (g,q,c(B», D = (j, st-I, c(F» and H = (ly, rlB];\ Ii). A. Basarab and hence D = (f,srl,h). i). Consequently, o(C) = le(A)1 +d-le(B)1 = d-Ifl + Ihl since o(A) = o(B) = 0 by assumption, and oeD) = If I + d - Ihl.

Some questions. Growth Functions of Amalgams 33 Fonnula (27) was first computed by Floyd and Plotnick [F-Pl] using very different methods. Using instead a minimal set of generators, Cannon found [C]: G[n+lJ = 1- (4n + 2z + 2z2 + ... + 2z2n+l + z2n+2 + 2)z - (4n + 2)z 2 _ ... - (4n + 2)z2n+1 + z2n+2 1 Thus, enlarging the set of generators has produced a dramatic simplification. Can one obtain an even 'simpler growth function for G[n+lJ? From our point of view, (27) can be "explained" as follows: (28) where the Fi are free groups, and the growth functions of free groups (with respect to a basis) are a quotient of linear polynomials.

3. 7r1 (X, x) is a group with respect to the composition law above. PROOF: Let a = (s,j,t), (3 = (p,g,q)" = (r,l,u) E S. We have to show that (a 0 (3) 0 , == 000 ((3 0 , ) . Let A = (f,tp,g), B = (g,qr,l), C = (c:(A), A~ qr, I), D = (f, tPPB, c:(B»). By definition we get (a 0 (3) 0 , = (SPAPC,C:(C), A~U) and ao({3o,) = (sPD,c:(D), A~A~U). v). iii). D The next lemma is immediate. A. 4. The maps Gx -- S : s 1-+ (s, lx, 1) and E(x, x) -- S : f 1-+ (1, f, 1) identify G x with a subgroup Of7rl(X, x) and E(x, x) with a subset of 7rl(X, x).

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