OOPS, I used G to mean both the group and one of its generators! Call the group of homomgraphic functions H. I said, let
K := the open disk D minus the (proper) images of closure(D); O := closure(D) minus the (proper) images of D; and U := the union of all the circles. [...] Exercise: Exhibit an explicit point of K. (It's infinitely bigger than U, so how can you miss?-)
I believe a countable set of such points are the fixed points of the group elements other than W^n and G^n. These elements (and their positive powers) derange the circles of U. Since these don't overlap, a fixed point can't lie on one. The algebra is nicer if we switch to the strip gasket generated by 1 G(z) := z + 1, and W(z) := ----- z - i acting on the circle C := {|z + 1 - i/2| = 1/2} and its interior D := {|z + 1 - i/2| < 1/2}. We still have
Then the only relation is W^3 = 1. But obviously WWC = WC = C. Less obviously, G^n WGC = WGC, G^n WWGC = WWGC, W^n GGC = GGC, GWggC = WGWWGGGC.
although W^n C is less obvious, but the next two are more obvious since WGC is the line Imag(z)) = 1 and WWGC is the real axis. The simplest fixed points are now of 1 WG(z) := ---------, z + 1 - i namely [P,P'] := (sqrt(sqrt(5) - 2) + 1) %i sqrt(sqrt(5) + 2) + 1 [ -------------------------- - --------------------- ~ 0.74293 %i - 1.52909, 2 2 (1 - sqrt(sqrt(5) - 2)) %i sqrt(sqrt(5) + 2) - 1 -------------------------- + --------------------- ~ 0.25707 %i + 0.52909]. 2 2 But K is uncountable; where are the rest? The images of P under all the elements of H? Maybe we have to take a closure. Are infinite words (constant functions?) in H? --rwg