We build examples of conjugators h by building the function g == hv^h~1. We build g by building a Markov partition for g. Let OQ, a^ . . . , a^-i be positive integers with sum S. The integers represent the lengths in units of 1/S of intervals in counter-clockwise order of a Markov partition of S1 where the first interval (with length flo/S) has left endpoint 0. The action of g is to fix 0 and take each interval affinely to the union of two consecutive intervals. This corresponds to the structures in Section 6 if n is an integral power of 2. However for any n, a Markov structure for Vg based on n initial intervals can be built by starting with intervals of length Ifri for the first partition and then pulling back endpoints under iterates of Vg"1 for the finer partitions. Once a string of positive integers is written out for the first partition for g, then the remaining partitions
are similarly obtained by pulling back endpoints under iterates ofg~1. The only question that remains is whether the endpoints in the Markov structure for g are dense in S1. That this is the case is seen by noting that each interval is subdivided into two intervals that are shorter than the original by some fraction. This fraction is of the form
^/(^i + ^214-1) or ^i+i/^i + ^21+1) where subscripts are treated cyclically mod n.
Among these ratios is one that is largest. Now we observe that all intervals shrink under repeated subdivision at least as fast as powers of this largest ratio. Once the density of the endpoints is established, then there is a unique conjugating homeomorphism h: S1 -> S1 that is an isomorphism of the Markov structure for Vg to that of g.
Once the values of the ^ are known, then it is a straightforward task to check if g is ^-compatible, to calculate values of gb and S( ) at the endpoints, and so forth.
Whether h is PL is detected by the equal pairs condition according to Lemma 8.1.
This condition is easier to interpret if n is even, so our examples will all use an even number of initial intervals.
Ifg turns out to be ^-compatible and n is an integral power of 2, then conjugation by h is at least an endomorphism ofT into T. This is because the endpoints of the Markov partition for Vg will be all of Z[^] and the endpoints of the Markov partition for g will be a (perhaps proper) subset of Z[^]. This gives A(Z[^J) c Z[^J and the rest follows from Lemma 3.2, from (3.1) and from Lemma 3.3.
Example 1. — The sequence 2, 2, 3, 1, 4, 2, 1, 1, 2, 2, 3, 1, 2, 2, 2, 2 has 16 lengths with sum 32. It is easily checked that the function g defined by this partition is ^-compa-tible. The equal pairs condition is violated, so h is not PL. The results in Part III were discovered by studying this example. This gives an example of a non-PL homeomorphism h that conjugates T into T and has lifts that conjugate F into F. In addition, the function g has zero break at its fixed point 0. This makes the function S( ) well defined. Since
(10.1) g\x)==^x)-Wx))
it also makes the non-linearity of g a <c coboundary ". This shows that (10.1) is not sufficient to reach the conclusion of Theorem 2. The right hypotheses to replace (10.1) seems to be that the following statement fails for only finitely many pairs x andj in S1:
(») If x and y have ^(.v) = gQ(Jy) for some positive integers p and q, then
W W = W j O .
Example 2. — The sequence 2, 2, 1, 1, 1, 1 has 6 lengths with sum 8. Since it satisfies the equal pairs condition, the conjugating homeomorphism h is PL. Since the number of intervals is not a power of 2, there are endpoints of the Markov partition for Vg that are not in Z[j]. Thus A(Z[^]) q Z[^], h conjugates T into but not onto T, and h~1
conjugates T onto a group of PL homeomorphisms of Si that properly contains T.
Remark. — We do not have an example of a non-PL homeomorphism h so that g is ^-compatible and h~1 conjugates v^ to a PL function. Thus we do not know whether
the hypothesis that h~l^h be r-compatible in Theorem 2 can or cannot be replaced by the hypothesis that h~l^h be PL. The point is that under the weaker hypothesis, Lemma 5.3 and its consequences Lemma 6.2, Lemma 7.2 and (7.7) do not seem to be available.
Example 3. — The sequence 1, 1, 3, 1, 2, 4, 1, 1, 1, 1, 6, 2, 2, 2, 2, 2 has 16 lengths with sum 32. Here the conjugator h fails to have A(Z[^J) = Z[^] in spite of the fact that there are 24 intervals. It is easy to check that the midpoint of the unique interval of length 4 is a point of period 2 under the action of g and cannot be the image of any element of Z[y.
Example 4. — The sequence 1, 3, 4, 2, 1, 3, 1, 1 has 8 lengths with sum 16. There is no well defined function 2( ) since the break at the fixed point is not zero.
Example 5. — This sequence 1, 1, 3, 1, 2, 1, 3, 1, 2, 2, 1, 3, 2, 1, 1, 3, 2, 2 has 18 lengths with sum 32. Two of the endpoints (sixth and twelfth) form a cycle of period 2 and have breaks — 2 and + 1. Even though g\0) === 0, we cannot define 2( ) because the sum of the breaks on a future orbit that contains these endpoints does not converge.
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Department of Mathematical Sciences
State University of New York at Binghamton Binghamton, NY 13902-6000
Manuscrit regu Ie 31 janvier 1995.