Given f E Homeo+(S1), we say that f is Mobius if there exists some homeomorphism g of S1 such that gfg-1 EM. However, Mobius topology, like maps, is more than just fixed points. On the other hand, the group generated by any Mobius-like map h is a convergence group.
Suppose G is a non-cyclic Mobius-like group with one or two global fixed points. In the case where G is a Mobius-like group without global fixed points, we don't have such nice results. In chapter 3 we prove the characterization of the Mobius-like groups with two global fixed points, as mentioned in the theorem above.
In Chapter 4 we do the same for Mobius-like arrays with a global fixed point. Finally, we also give a conjecture for the characterization of Mobius-like arrays without global fixed points.
Preliminary Definitions and Observations
1} serves as a model for the hyperbolic plane H 2• The group M is precisely the set of isometries that preserve the orientation of H 2• Every element of M preserves the S1 limit of D, so we have the following faithful inclusion. As for fn elevators; 's, condition a) means that the graphs of fn; 's approaches a type of step function consisting of jumps and planes of length 1.
Observation 1.4 In the above definition one can replace conditions a) and b) by
Groups With Two Global Fixed Points
In fact, we can assume that G is a real Mobius group, since the condition L(G)=S1 is preserved under conjugation. Since we have assumed that G is not cyclic, it follows that G is non-discrete, and therefore L( G)=S1•. Therefore, after moving to a subsequence, we can assume that {/n} is strictly monotonic on both (a, b) and (b, a), only in the opposite sense.
Note that the action of G on 81 has the property that for any pair of distinct elements ft,j2 of G, if z ::j; a, b then j1(z) # fz(z), so we will never encounter any problems defining j1 , j2 at the inserted intervals. It is easy to check that G is a subgroup of Homeo+ ( 81) and that every element of G in Homeo( 81) is conjugate with a Mobius transform on 81. In the above example we have perturbed the operation of G by intervals to be inserted only at points of (a, b).
It's clear, s; is homeomphic with S1. Since U~1 [xi, Yi] is invariant under the action of G, we conclude that there is an induced action of G on S.!. In both cases we get a contradiction in the fact that c is approximated by points of S1-L(G).
Groups With One Global Fixed Point
The graphs of lifts toR of different elements of G are disjoint everywhere except where they all intersect, so this case can be done in much the same way as in the proof of Theorem 2.1. We can also transition to a subsequence so that all fn's move the points of S1 - {a} in the same direction. This completes the proof under the assumption that the fn's move the points of S1 - {a} counterclockwise.
First we show that, after moving to a subsequence, we can get all difference functionsf,::::;1fn's with m > n of the same type, that is, all parabolic or all hyperbolic. Then remove the finitely many elements of {fn} that produce 2's in the first row of the matrix of {fn}. Let fn2 be the next element among the remaining elements. As before, discard infinite elements of the sequence that produced 2's in the second row (that is, continue to a subsequence).
Now both the first and second rows consist of only 1, but the rest of the rows have new assignments. Now we return to our starting sequence {fn}· As shown above, after we move to a subsequence, we can assume that all /;;.1 fn are of the same type. In other words, the graphs of fn are pairwise everywhere except at point a where they all intersect.
We will now complete the proof by contradiction; i.e., we will prove that h as claimed does indeed exist, for every interval J = ( u, v) that does not contain a. From the definition of the intervals (xi, Yi) it is clear that each of them represents a maximal interval of the flatness of the boundary function f in the sense that if f is flat in an open interval I C (c, a) , then I C (x , Yi) for some i. This concludes the list of technical claims we need for the rest of the argument.
This comes as a consequence of the maximality of (xj, Yi) as a flatness interval of f.9n(I)~9n+t(I) for n large, which gives a contradiction in the same way as in the proof of the Simplicity condition . Once this is established, the rest of the proof proceeds in exactly the same way as the corresponding part of the proof of Theorem 3.4.
Groups with no global fixed points
Let g be a genuine hyperbolic Mobius transformation on S1 such that g(X) = Y and the axis of g crosses X and Y. Now we need to see how G acts on S1, and to do so we denote by C the region bounded by H -orbits X and Y (shaded area in Figure 5.1). Our main guide is the idea that we should first fully understand how G acts on the C region; once this is accomplished, you should be able to understand how G acts on S1.
So the set G defines a kind of disk tiling, the tilings are translations of the region C from the elements of G. But then there is a translation of C that lives in the "gap" between Z and f(Z), which decreases d. So we have proved Claim 4 and thus achieved our goal: G is a Mobius-like group, but it is not a convergence group because H <-.t G is not a convergence group.
One must be careful when "blowing up" the H trajectories: we should not insert intervals at points that are fixed points of some element of H. It is not difficult to see that one can find a subsequence {fnJ such that the group H generated by all the fn;s is a discrete Mobius group.12. In other words, the axes of the elements of the sequence fn;hf~\ of conjugates of h, limit themselves to the axis {oo, 1}.
Now choose a point x E R which is not fixed by any element in H and whose H path does not contain oo. This could certainly be done if we knew that oo is not a fixed point of any element of H. When you "explode" the circuit of oo, define the induced effect of p on a new circle S1 to be hyperbolic so that its axis is determined by the endpoints of the interval I inserted at the point oo.14 See Figure 5.9.
Now, the definition of the induced action on I for the rest of the elements of H is done in the most natural way: given a point y from the H orbit of oo, let f. We therefore define the group G acting on S1 just as before, and show that it defines a tiling of the disc. Just look at the limit function of the sequence {fn; }: it consists of two jumps and two flats.
