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1040-7294/02/1000-0829/0 © 2002 Plenum Publishing Corporation

Journal of Dynamics and Differential Equations, Vol. 14, No. 4, October 2002 (© 2002)

Extending Circle Mappings to the Annulus

Fernando Oliveira1

1Departamento de Matematica, Universidade Federal de Minas Gerais, Av. Antonio Carlos6627, 30123-970 Belo Horizonte MG, Brasil. E-mail: fernando@mat.ufmg.br

Received March 7, 2001

We show that any monotone degree one C. mapping of the circle can berealized as the induced mapping on the boundary by a C. area preservingdiffeomorphism of the open annulus. The circle mapping may have criticalpoints, allowing the possibility of Denjoy behavior on the boundary.

KEY WORDS: annulus; embedded invariant circle; Aubry–Mather set; primeend; circle map; Denjoy.

1. INTRODUCTION

It is not known whether there exists a C. area preserving diffeomorphismof a surface with an embedded Denjoy curve, more precisely, a C. areapreserving diffeomorphism of a surface S and a continuous injective func-tion c: S1Q S, such that if C=c(S1), then f(C)=C and the restriction off to C has irrational rotation number and a wandering interval. By theDenjoy theorem, c can not be of class C2.This question is related to some other interesting ones.It is not known whether there are Aubry–Mather sets contained in

invariant curves. Herman (1983) constructed examples of area preservingdiffeomorphisms of the annulus exhibiting such phenomena, but hisexamples are only of class C3− e.Another, is a problem of interest in the study of generic properties of

area preserving diffeomorphisms, as to whether the induced map on primeends of invariant simply connected open sets is minimal. If this is generi-cally true, then, by an argument shown to me by John Franks, it would

follow that generically the closure of invariant manifolds of hyperbolicperiodic points are topologically transitive sets.The work on counter examples to the Seifert conjecture, that every

flow on the 3-sphere without singular points has a periodic orbit, led somepeople to produce examples of Denjoy behavior on invariant curves. Knill(1981) made a very beautiful geometric construction, embedding a Denjoycurve in a C. diffeomorphism of the annulus, but his example is not areapreserving. See also Harrison (1989), where a Denjoy curve is embedded asa fractal in a C2+d diffeomorphism of the annulus.Our result is the following:

Theorem 1. Let a be a monotone degree one C. mapping of thecircle S1. Then, there exists a continuous function h: S1×RQ S1×R withthe following properties:

(1) h(x, 0)=(a(x), 0).

(2) The restrictions of h to S1×(0,.) and S1×(−., 0) are C. areapreserving diffeomorphisms.

(3) h satisfies the twist condition “h1“y > 0.

(4) If a is a homeomorphism, then h is a homeomorphism of S1×R.

(5) If a is a diffeomorphism, then h is a diffeomorphism of S1×R.

Note that this result can not be improved to obtain a C1 function hsatisfying properties (1) and (2). In fact, by (1) we have that det hŒ(x, 0)=0at points where aŒ(x)=0, and by (2) we have det hŒ(x, y)=1 if y ] 0.When a maps intervals to points, we have some kind of degenerated

Denjoy behavior, but this is not so interesting, since in this case the restric-tion of h to S1×(0,.) does not extend to a homeomorphism of its lowerprime end compactification. Hall (1981) constructs an example of a C.

homeomorphism a: S1Q S1 with exactly one critical point, which hasirrational rotation number and a wandering interval. It follows from theabove result that this mapping can be extended to a homeomorphism ofS1×[0,.), which is an area preserving diffeomorphism of S1×(0,.), andwhich fails to be differentiable at exactly one point. In this case, a is theinduced mapping on prime ends.

2. PROOF OF THE THEOREM

Firstly, we would like to remark that if a has no critical points, then itis easy to find a C. area preserving diffeomorphism h: S1×RQ S1×Rsuch that h(x, 0)=(a(x), 0), namely h(x, y)=(a(x), yaŒ(x)).

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Let us proceed to the general case. We are going to use lower caseletters to denote functions of S1 or S1×R, and upper case to denote theirliftings to the universal cover. Unless stated otherwise, all functions are C..The construction makes use of generating functions in a way slightly

different from the usual generating function for a twist map of the annulus.If U is an open subset of R2 and S(x, z) is a function from U to R withSxz(x, z) ] 0 for every (x, z) ¥ U, then, one way S can be used to define areapreserving diffeomorphisms is the following. Let F(x, z)=(x, Sx(x, z)) andG(x, z)=(Sz(x, z), z). It follows that

det FŒ(x, z)=det GŒ(x, z)=Sxz(x, z) ] 0

for every (x, z) ¥ U, and therefore F and G are local diffeomorphisms.Hence, wherever (x, z)=F−1(x, y) is defined, the function H=G p F−1

satisfies detHŒ(x, y)=1.a has a lifting A: RQ R, which satisfies A(x1) [ A(x2) if x1 < x2 and

A(x+1)=A(x)+1.Consider S: R2Q R given by S(x, z)=xu(z)+A(x)(z−u(z))+b2 z

2,where b will be specified when we verify the twist condition, and the func-tion u: RQ R satisfies the following conditions:

(6) u(0)=0.

(7) uŒ(0)=0.

(8) 0 < uŒ(z) < 1 if z ] 0.

(9) limzQ. u(z)=..

(10) limzQ −. u(z)=−..

(11) |uœ(z)| < 1 for all z ¥ R.

For example, u(z)=z−arctan z satisfies the above conditions.Let us now consider functions F and G: R2Q R2, defined by

F(x, z)=(x, Sx(x, z))=(x, u(z)+AŒ(x)(z−u(z))),

G(x, z)=(Sz(x, z), z)=(xuŒ(z)+A(x)(1−uŒ(z))+bz, z).

The functions F and G satisfy the following conditions:

(12) F(x, 0)=(x, 0) and G(x, 0)=(A(x), 0).

(13) F(x+1, z)=F(x, z)+(1, 0) and G(x+1, z)=G(x, z)+(1, 0).

(14) det FŒ(x, z)=det GŒ(x, z)=uŒ(z)+AŒ(x)(1−uŒ(z)) > 0, if z ] 0.

(12) follows from (6) and (7). (13) follows from A(x+1)=A(x)+1.(14) follows from (8) and the fact that AŒ(x) \ 0.

Extending Circle Mappings to the Annulus 831

From (13), we see that F and G are liftings of C. functions of S1×R,which we denote by f and g, respectively. By (12), we have f(x, 0)=(x, 0)and g(x, 0)=(a(x), 0), and by (14), the restrictions of f and g toS1×(0,.) and S1×(−., 0) are local diffeomorphisms.Now, let us show that F is a bijection of R2, and that the restrictions

of G to R×(0,.) and R×(−., 0) are also bijections. Note that F fixes allvertical lines {x}×R, and when restricted to each of them, F has the formf(z)=u(z)+AŒ(x)(z−u(z)). The derivative of this function is given by(14), and therefore f is strictly increasing. By (9) and (10), we have that f isa bijection of {x}×R. The argument for G is similar. G fixes all horizontallines R×{z}, and when restricted to each of them, G has the form k(x)=xuŒ(z)+A(x)(1−uŒ(z))+bz. The derivative of this function is given by(14), implying that k is strictly increasing when z ] 0. Since k(x+1)=k(x)+1, we have that k is a bijection of R×{z}, when z ] 0.It follows that the restrictions of the functions f and g to S1×(0,.)

and S1×(−., 0) are bijections, and since they are local diffeomorphisms,they are diffeomorphisms. Furthermore, since f(x, 0)=(x, 0), f is ahomeomorphism of S1×R.Consider now h=g p f−1 and H=G p F−1. h is a continuous function,

and its restrictions to S1×(0,.) and S1×(−., 0) are diffeomorphisms.By (14), detHŒ(x, y)=1 if y ] 0, implying that h is area preserving. Sincef(x, 0)=(x, 0) and g(x, 0)=(a(x), 0), we have that h(x, 0)=(a(x), 0).Since g(x, 0)=(a(x), 0), when a is a homeomorphism we have that g,

and therefore h, are homeomorphisms of S1×R. If besides that, a has nocritical points, we have that

det FŒ(x, z)=det GŒ(x, z)=uŒ(z)+AŒ(x)(1−uŒ(z)) > 0

for any (x, z) ¥ R2, and h is a diffeomorphism of S1×R.It remains to prove that h satisfies the twist condition. Firstly, we are

going to show that “H1“y (x, y) > 0 when y ] 0, where H1 denotes the first

coordinate of H. Recall that we are using the notation (x, z)=F−1(x, y).Differentiating H1 p F=G1 with respect to z gives (

“H1“y p F) Sxz=Szz, and

since Sxz(x, z) > 0 if z ] 0, it is enough to show that Szz(x, z) > 0 for every(x, z). We have that Szz(x, z)=(x−A(x)) uœ(z)+b, and by (11), forSzz(x, z) to be positive at every (x, z), it suffices to choose the lifting A andthe constant b so that 0 < x−A(x) < b for every x.Finally, we would like to say what happens to the image of vertical

lines at points H(x, 0)=(A(x), 0). It is easy to see that

“H1“y(x, 0)=˛ (x−A(x)) uœ(0)+bAŒ(x)

> 0 if AŒ(x) > 0

+. if AŒ(x)=0

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and

“H2“y(w, 0)=˛ 1AŒ(x) > 0 if AŒ(x) > 0

+. if AŒ(x)=0

showing that, for each x fixed, the derivative of yMH(x, y) at y=0 hasnorm+. if AŒ(x)=0. But that is not a problem, since a simple calculationgives

limyQ 0

H2(x, y)−H2(x, 0)H1(x, y)−H1(x, 0)

=1

(x−A(x)) uœ(0)+b

showing that the images of vertical lines have a well defined tangent line ofpositive slope at points (A(x), 0), independently of AŒ(x).

ACKNOWLEDGMENTS

This work was supported by Grant 200992/87-2 from CNPq, a branchof the Brazilian government devoted to scientific and technological devel-opment. I am also grateful to John Franks, Clark Robinson and KeithBurns, for the hospitality during my visit to the Department of Mathema-tics at Northwestern University during the winter of 2001.

REFERENCES

Hall, Glen R. (1981). A C. Denjoy counterexample. Ergod. Th. Dynam. Sys. 1, 261–272.Harrison, J. (1989). Denjoy fractals. Topology 28(1), 59–80.Herman, Michael R. (1983). Sur les courbes invariants par les difféomorphismes de l’anneau.Astérisque, Vol. 103–104. Société Mathématique de France, Paris.

Knill, R. J. (1981). A C. flow on S3 with a Denjoy minimal set. J. Diff. Geom. 16, 271–280.

Extending Circle Mappings to the Annulus 833