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Free lines for homeomorphisms of the open annulus
Lucien Guillou
To cite this version:
Lucien Guillou. Free lines for homeomorphisms of the open annulus. 2006. �hal-00020341�
ccsd-00020341, version 1 - 9 Mar 2006
FREE LINES FOR HOMEOMORPHISMS OF THE OPEN ANNULUS LUCIEN GUILLOU
Abstract. LetH be a homeomorphism of the open annulus S1×R isotopic to the identity and leth be a lift of H to the universal coverR×Rwithout fixed point. Then we show thathadmits a Brouwer line which is a lift of a properly imbedded line joining one end to the other in the annulus orH admits a free essential simple closed curve.
0. Introduction.
Our main result in this note is the following:
0.1 Theorem. Let H be a homeomorphism of the open annulus S1×Risotopic to the identity which admits a lift htoR2 without fixed point. Then
there exists an essential simple closed curve in the annulus free under H (and therefore it lifts to a Brouwer line for h)
or there exists a properly imbedded line in the annulus joining both ends which lifts to a Brouwer line for h.
A natural question, in caseH is fixed point free, is to ask, in the second case, for a properly imbedded line in the annulus joining both ends and free for H. It has been given a positive answer in many cases recently by P. B´eguin, S. Crovisier and F. Le Roux [BCL].
Note. A slightly weaker form of this theorem was communicated to a few people since 1994 but not published partly because I was unable to answer the preceding question. In his 2000 thesis, Sauzet [S]
reproved in his own way the exact statement above but this thesis stayed also unpublished. Recently, Addas- Zanatta [A] and B´eguin-Crovisier-Le Roux [BCL] used this result as a step towards interesting dynamical results, so we decided to publish these notes. Our original proof relied on hard to read original work of K´er´ekjart´o [K2], we tried here to give a proof starting from first principles.
Note further that, as remarked in [BCL], this theorem, for area preserving homeomorphisms, follows quickly from the much harder recent “version feuillet´e ´equivariante du th´eor`eme de Brouwer” of Le Calvez [L].
For comparison recall that in [G] we proved thatifH is a fixed point free orientation preserving homeo- morphism of the closed annulus, then there exists an essential simple closed curve in the annulus free under H or there exists an essential simple arc in the annulus free underH generalizing the celebrated Poincar´e- Birkhoff theorem. Part of the arguments given here can easily be adapted to prove this result too. Instead, these arguments will be used to prove:
0.2 Theorem. Let H be a homeomorphism of the torusT2 preserving orientation which admits a lift toR2without fixed point. Then there exists an essential simple closed curve on T2 which lifts to a Brouwer line of h.
Note that in the case of the torus, we cannot ask for an essential simple curve in T2 free under H: Bestvina and Handel have given an example of a fixed point free, homotopic to the identity, homeomorphism of the torus without any free essential simple closed curve [BH].
Finally, in section 5, we show how Theorem 0.1 can be used to solve a conjecture of Winkelnkemper about area preserving homeomorphisms of the 2-sphere with exactly two stable fixed points which generalises the classical Poincar´e-Birkhoff Theorem on the closed annulus.
2000 Mathematics Subject Classification. Primary 37E30
Key words and phrases. Brouwer homeomorphism, free line, fixed point, open annulus, torus, Poincar´e- Birkhoff Theorem.
Acknowledgements. I would like to thank F. Le Roux for useful comments on a preliminary version of this paper.
1. Basic facts about Brouwer homeomorphisms.
In this paragraphhwill be a fixed point free orientation preserving homeomorphism of the planeR2. Such an homeomorphism is called aBrouwer homeomorphism.
The most basic fact about these homeomorphisms is that they automatically no periodic points or even recurrent points. More generally we have the following lemma of J. Franks [F] for which we recall the following definition.
1.1 Definition. Letf be a homeomorphism ofR2. A subsetE ofR2 isfree(forf) ifET
f(E) =∅.
It iscriticalif intE is free butf(E)T E6=∅.
1.2 Lemma. Let(Bi)i=1,...,k,(k≥1)be a sequence of open discs inR2andf an orientation preserving homeomorphism of R2 such that:
1. BiT
Bj =∅ for alli6=j,1≤i, j≤k;
2. each Bi is free for f for each1≤i≤k;
3. there exist integers ni>0 such that fni(Bi)T
Bi+1 6=∅ for each i∈N wherei is taken modk.
Then there exists a simple closed curve of index 1 forf and, therefore, a fixed point forf.
We fix now a Brouwer homeomorphismh. As a consequence of Franks’ Lemma, every point in the plane is wandering underh.
1.3 Definition. A simple closed arcα⊂R2with endpointspandh(p) for somep∈R2is atranslation arcifαT
h(α) ={h(p)}. A translation arc will always be supposed to be oriented fromptoh(p).
1.4 Proposition. Every point of R2 is contained in a translation arc. If α is a translation arc, the trajectory S
n∈Zhn(α)is a submanifold of R2 homeomorphic toR.
Such a trajectoryl is not necessarily closed as a subset of R2 but, being a submanifold, ¯l\l is closed and we have:
1.5 Proposition. A trajectoryldivides the connected component ofR2\(¯l\l)containinglinto exactly two components.
We refer to [G] for a proof of the two preceding propositions.
2. Critical discs and Brouwer plane translation theorem.
The content of this section may be seen as a variation on [T].
Let stillhbe a Brouwer homeomorphism.
A disc will always be acloseddisc in that paper, otherwise we say “open disc”.
2.1 Lemma. If the diskD is free or critical, then DT
hn(D) =∅ for everyn∈Z\ {0,±1}.
Proof. Let us suppose by contradiction that there exists a point x∈D such thathn(x)∈D for some n >1. Any arc fromxtohn(x) contained in intDexcept, eventually, for its endpoints, would be free sinceh has no periodic point. Then, one could find a open disc, neighborhhod of this arc which contradicts Franks’
Lemma (withi= 1).
2.2 Lemma. Let D be a critical disc. Then the boundary of D decomposes into two disjoint arcs (eventually degenerated, that is reduced to a point) containing respectively h−1(D)T
D and DT
h(D) and two translations arcs, these four arcs having disjoint interiors.
Proof. Because of Jordan’s theorem, h−1(D)T
D and DT
h(D) are not linked on ∂D (that is, are contained in disjoint intervals of ∂D). Letλbe the smallest closed sub-arc (eventually reduced to a point) of ∂D containing h−1(D)T
D and disjoint from h(D). According to the preceding Lemma and Jordan’s Theorem, h(D)T
D is contained either in intλor ∂D\λ. In the first case, the frontier of the unbounded component ofR2\(h−1(D)S
D) is a Jordan curve send into the disc it bounds byh, which gives a fixed point forh: a contradiction. Therefore, the second case holds and the arc µ, smallest closed sub-arc (eventually reduced to a point) of∂D containingDT
h(D), is disjoint fromλ. The end points ofµ andh(λ) coincide and, sincehpreserves orientation, the two sub-arcs of∂D which are the components of∂D\(intλS
intµ) are translation arcs.
2.3 Definition. A translation arc as in the preceding lemma, oriented as said before, will be saidhigh ifDis on the right of the translation arc andlow ifDis on the left of the translation arc.
CONVENTION. From now on, unless said otherwise, all discs will beeuclideandiscs.
2.4 Definition. Let us consider two critical discsD′ andD which satisfy:
1.∂DT
D′ is a non degenerated sub-arc of a translation arc αof∂D.
2. h−1(D)T
intD′ =∅= intD′T h(D).
D′ will be called anextensionofD (high or low according to the type of the arcα).
2.5 Remarks.
1. D′ is a high extension ofD if and only if D is a low extension ofD′. Indeed, it is enough to check that the low translation arc ofD′ contains∂D′T
D. But, if this were not true, intD would contain a point ofD′T
h(D′) or ofD′T
h−1(D′) in contradiction to point 2 of the definition.
2. DS
D′ is still a critical topological disc when D′ is a high extension ofD.
2.6 Lemma. LetDandD′be two critical discs andα−,α+the two arcs of translations of∂D. Suppose that intD′T
∂D is not empty, meetsα+ and that D′T
(h−1(α+)S
h(α+)) =∅.
Then,DS
D′ is critical. Therefore, D′ is a high extension of D.
2.7 Remark. Of course, we have an analogous Lemma obtained by exchangingα+andα−and “high”
and “low”.
Proof. We will first prove:
2.8 Assertion. Let D be any critical or free topological disc and α any translation arc such that intDT
α6=∅ andDT
(h−1(α)S
h(α)) =∅, thenDT¯l=DT
αwherel is the trajectory generated byα.
Given this assertion and the fact that l separates the connected component ofR2\(¯l\l) containingl into two components invariant underh, one concludes easily the proof of the Lemma.
Proof of the Assertion. Suppose first thatDT
hn(α)6=∅for somen∈Z,|n|>1. We can find an arcγ in intD, except perhaps for one end point, joining a pointx∈intαT
intDto a pointhn(y) for somey∈α.
Since intD and the subarc [x, y] of α are free, γ is free also. Then, small enough open topological discs, neighborhoods of [h(x), h(y)] andγ respectively, would satisfy the hypothesis of Franks’ Lemma giving rise to a fixed point forhand a contradiction.
If DT¯l 6= ∅, we can suppose that DT
l = ∅ from what precedes and we can find a free arc γ inside intD, except perhaps for one end point, joining a pointx∈intαT
intDto a pointz∈¯l\l. We choose then
a small free discD′ aroundz such thatD′T
(h−1(γ)S
h(γ)) =∅. ThenD′S
γis free and a free topological disc, small enough neighborhood ofD′S
γ, verifies the same hypothesis asDin the first paragraph.
2.9 Lemma. If(Di)i∈I (whereIis a finite set{0,1, . . . , n}orNorZ) is a sequence of critical discs such thatDi+1be a high extension ofDi, thenS
i∈IDiis a critical disc. Furthermore,(Di\Di−1)T
(Dj\Dj−1) =∅ if i6=j.
Proof. Let li be the translation line generated by the high translation arc of Di. By Assertion 2.8, the high translation arc ofDi+1 does not meetli and we define Σi as the component ofR2\(¯li\li) which contains the high translation arc of Di+1. It follows from the preceding Lemma that Σi ⊃ Σi+1. Using the invariance of these Σi under hone gets the Lemma by an easy induction (and, in caseI=Z, a second induction going fromitoi−1 and using the low translation arc ofDi).
To each pointxin the plane, we now associate the radius rx of the euclidean critical disc with center atx.
2.10 Lemma. The mapx7→rx is continuous.
Proof. This follows from the fact that, ifd(x, y)< ǫ, thenB(y, rx−ǫ)⊂B(x, rx)⊂B(y, rx+ǫ).
2.11 Lemma. Let D be a critical disc andαbe one of the two translation arcs of ∂D. There exists a critical discD′ with center onα such that D′ is an extension (high or low according to the choice of α) of D.
Proof. Let C+ (resp. C−) be the set of pointsxin αsuch that the critical disc with center atxmeets h(intD) (resp. h−1(intD)). To prove the Lemma it is enough to prove that C+S
C− is not all ofα. This will follow from the connectivity ofαand the following three points.
•C+T
C− =∅: LetD′be an euclidean critical disc inC+T
C−, that is, meetingh(intD) andh−1(intD).
Then there exists a topological disc ∆⊂intD′\intDmeeting bothh(intD) andh−1(intD) which gives with intD two discs satisfying the hypothesis of Franks’ Lemma and therefore a contradiction.
•C+ andC− are not empty: it is enough to consider discs centered on the end points ofα.
•C+ andC− are open: this follows from the continuity of the mapx7→rx.
Under the hypothesis and notations of the preceding Lemma, since D′ is critical, one can repeat the construction using this time the translation arc on∂D′ of the same type thanα. One gets in this way an infinite sequence D0 = D, D1 = D′, D2, . . . , Dn, . . . of critical discs such that by Lemma 2.9, for every n, Si=n
i=0Di is a critical disc and (Di\Di−1)T
(Dj\Dj−1) =∅ ifi6=j.
2.12 Lemma. In this situation, the sequence{Dn}converges to infinity, that is, given any compactum K⊂R2, there existsN such that the connected set S
i≥N Di does not meet K.
Proof. Since the Dn are critical, the radius of those meetingK is bounded below as well as their area.
By the Lemma allDn\Dn−1are disjoint and, by euclidean geometry, have area greater than half the area ofDn. Therefore only a finite number of discs Dn can meetK.
It is now easy to prove the Brouwer plane translation theorem.
2.13 Definition. A Brouwer line L for the Brouwer homeomorphism h is the image of a proper embedding ofRintoR2such that
1. Lis free underh.
2. Lseparatesh(L) andh−1(L).
2.14 Plane Translation Theorem (Brouwer). Let hbe a Brouwer homeomorphism. Then every point inR2 belongs to a Brouwer line.
Proof. Let x be the point in consideration and D0 the critical disc centered at x. One constructs as above a sequence D0, D1, . . . , Dn, . . . of critical discs converging to infinity, and also, starting from the other translation arc ofD0, another sequenceD0, D−1, . . . , D−n, . . . of critical discs converging to infinity.
Note that Sk=j
k=iDk is critical for i < j where i, j ∈ Z since one can see that finite sequence as obtained by the preceding construction fromDi and thereforeS
i∈ZintDi is free. It is then easy to construct inside S
i∈ZintDi a proper embedding LofRjoining linearly the centers of theDi’s. It has to be free and also to satisfy property 2 of a Brouwer line since one can find inD0an simple arc joiningh−1(D0) toh(D0) crossing Lonly atxand ifLwhere not separatingh−1(L) andh(L), one could complete this arc into a simple closed curve crossingL only atxin contradiction to the properness ofL.
3. Further results on euclidean critical discs.
We still fix a Brouwer homeomorphismh.
3.1 Definition. Let D be a critical disc. We know that its boundary is the union of two translation arcs α+ and α− (high and low respectively) and of two disjoints arcs, λ and µ, smallest sub-arcs of∂D containingh−1(D)T
D and DT
h(D) respectively. Chooseǫ so that 0< ǫ≤ 1
8inf(l(α+), l(α−), l(λ), l(µ)) wherel(.) is the length of the corresponding arc and where we omitl(λ) andl(µ) ifλ(or equivalentlyµ) is degenerated. We haveǫ < π
8rifr is the radius ofD.
ThenD without the discs of radiusǫ centered at the end points ofα− is a topological disc D+ andD without the discs of radiusǫ centered at the end points ofα+ is also a topological discD−. If λ(or µ) is not degeneratedD+ andD− are topological critical discs and otherviseD+=D− is a free topological disc.
In any case, area(D±)> 2π 3 r2.
So, to each pointxof the plane we may associate three critical topological discs: the euclidean critical discD centered at xand, ifλand µare non degenerated, twoderived topological discsD+ andD−. We will also say thatD is derived fromD+orD−. These three discs will be calledgeneralized critical discs.
We will also say thatD is derived from D+ or D−. The centers, radii and translation arcs of the derived discs are those ofDby definition. We will extend this vocabulary to the case whereλandµare degenerated:
in that caseD+=D− is free and not critical but, having its area bounded below as said above, it will not cause any problem below.
The euclidean critical discD will be saidthe underlying discofD+ andD−and alsoDon occasion.
We will begin to improve the extension Lemma as follows.
3.2 Lemma. Let D be a critical disc and αone of the two translation arcs of∂D. Then there exists a generalised critical disc D′ such that
1. The underlying disc ofD′ is an extension ofD.
2. D′T
∂D=D′T intα 3. h−1(D)T
D′ =∅=D′T h(D)
3.3 Definition. Given two critical discs D1 and D2 admitting derived discs D and D′ which satisfy conditions 1, 2 and 3 above we will say thatD2 is astrict extension ofD1 (high or low according to the type ofα).
3.4 Remarks. 1. DS
D′ is a topological critical disc (or free ifD andD′ are free).
2. IfD2 is a strict high (resp. low) extension of D1, thenD1 is a strict low (resp. high) extension of D2.
Proof. LetA+ (resp. A−) be the subset ofαof pointsx∈αsuch that the critical disc with center atx meetsh(D) (resp. h−1(D)). Then
•A+ and A− are not empty: it is enough to look at critical discs centered at the end points ofα.
•A+ and A− are closed: this follows from the continuity of the mapx7→rx.
•IfA+T
A−=∅, the connectivity ofαimplies that an honest critical disc satisfies the Lemma.
•If A+T
A− 6=∅, then this intersection is reduced to only one disc: the one centered at the midpoint ofαand with the end points ofαon its boundary.
Indeed, if there were a discD′ meeting h−1(D),h(D) and at most one end point ofα, one could find a free arc δcontained inD′\D except for its endpointsa∈h−1(D) andb ∈h(D). One could also find a free arcγin D, disjoint fromδand joiningh(a) toh−1(b) and therefore, by fattening, a contradiction with Franks’Lemma.
Note that the same argument shows that a critical discDwhich contains the two end points ofαcannot meeth−1(D) orh(D) except in these end points.
Then that one discD′ admits a derived disc (D′+or D′− according to the type ofα) which satisfies the first case of the Lemma.
3.5 Lemma. Analogs of Lemmas 2.9 and 2.12 are now equally true with extension replaced by strict extension and with the same proofs (for the second Lemma we use the fact that the free discs eventually met have an area bounded below as said above).
It is only to get the following Lemma that we had to consider strict extensions and derived discs. But of course this Lemma will be crucial in section 4.
3.6 Lemma. LetKbe a compactum in the plane. There exists a finite number nof pairs of generalised discs (D1i, Di2),1≤i≤n, such that every critical disc centered on K admits one of these pairs as a couple of strict extensions: high and low.
Proof. IfDis an euclidean critical disc centered atxand (D1, D2) a pair of euclidean generalised critical discs such thatD1 is a strict high extension of D and D2 a strict low extension of D, then there exists a neighborhoodVxofxsuch that the euclidean critical discs centered onVxall admitD1andD2as strict high and low extensions (sincex7→rxis continuous and the arcs ∂DT
D1 and∂DT
D2 being free, neighboring arcs are still free). A finite sub-covering of the covering ofK by theVx’s leads to the sought after pairs.
4. Homeomorphisms of the torus and the open annulus.
We continue to callha Brouwer homeomorphism of the plane.
We will denote bysandtthe applications (x, y)7→(x, y+ 1) and (x, y)7→(x+ 1, y) ofR2. Also,πwill be the projectionR2→R/Z×R.
4.1 Definition. IfLis a Brouwer line forh, we denote byD(L) the component ofR2\Lwhich contains h(L) and byG(L) the one which containsh−1(L).
4.2 Lemma. Let (Ln)n∈Z be a locally finite family of Brouwer lines for h such that S
nD(Ln) is connected and R2\S
nD(Ln)is a connected, unbounded open setR. Then, the frontier of R is a Brouwer line for h.
Proof. Given the local finiteness of the family,S
nD(Ln) =S
nD(Ln) and a classical result of K´er´ekjart´o [K1, page 87] implies thatR is a submanifold of the plane. R2being unicoherent andS
nD(Ln) connected, FrRis connected. But the frontier ofRis free, because ifx∈FrR, there isnsuch thatx∈Lnand therefore, h(x) ∈D(Ln) ⊂R2\R. Since h−1(R) ⊂R, it cannot be a circle and so it is a proper free line. It is a Brouwer line sinceh−1(R)⊂R andh(R)⊂R2\R.
4.3 Lemma. Supposehcommutes with t and let be given a Brouwer line Lfor hwhich projects onto and properly on they-axis. Then there exists a Brouwer line L′ for h which projects onto and properly on the y-axis and free fort.
Furthermore, if h commutes also with s and if L is invariant under s, one can find L′ also invariant under s.
Proof. Since L projects down properly onto the y-axis, there exists N > 0 such that G(L) or D(L) contains [N,+∞)×{0}. Exchanginghandh−1we can suppose thatD(L) contains a half line [N,+∞)×{0}
and consider the unbounded connected componentR ofR2\S
k≥0tk(D(L)) and its frontierL′. For everyA >0 there isn(A) such that forn≥n(A),tn(LT
R×[−A, A])T
R×[−A, A] =∅, therefore the family (tk(L))k≥0 is locally finite and L′ projects onto and properly on the y-axis. Furthermore, by Lemma 4.2 , L′ is a Brouwer line and it satisfies t(L′) ⊂ D(L′) since if x∈ L′, there exists n such that x∈tn(L) and sot(x)∈tn+1(D(L)⊂R2\R=D(L′).
Notice that, sincesandtcommute, if s(L) =Lthen alsos(L′) =L′.
Therefore to conclude that, in this second case, exists a Brouwer line going down in the annulus on a free line joining the two ends, it is enough to apply the following Sublemma toL′.
4.4 Sublemma. Let L be a Brouwer line for hwith a proper onto projection on the y-axis such that t(L) ⊂ D(L). Then there exists a Brouwer line L′ with a proper onto projection on the y-axis such that t(L′)⊂D(L′).
Furthermore, ifhcommutes withs ands(L) =L, then one can chooseL′ such thats(L′) =L′. Proof. Let δ : L →]0,+∞[ be a function such that the δ-neighborhood W of L does not meet h−1(W)S
h(W) and define L0 = ∅ and, for all k > 0, Lk = LT
(R×[−k, k]). Since Lk is compact, there existsn=n(k) such thatLk
Ttn(Lk) =∅. Let us suppose thatLk−1T
t(Lk−1) =∅ and let us show that there exist a Brouwer line L′ with a proper onto projection on the y-axis such thatL′k−1 =Lk and L′kT
tm(L′k) =∅for some m,0< m < n, ifn >1. (This is enough to finish the proof of the Sublemma).
DefineXi=LkT . . .T
ti(Lk) ifi≥0. By hypothesis,t(Xn−1)T
Lk =Xn =∅and Xn−1 is a compact ofLk disjoint fromLk−1 as well as theδ-neighborhood of Xn−1(by diminishing δnearXn−1 if necessary).
Theδ-neighborhood ofXn−1inLis a finite union of open intervals inLk\Lk−1. Let us replace these intervals by arcs with the same end points but included in the intersection ofG(L) and of the δ-neighborhood of L.
We thus get a Brouwer lineL′ which projects onto and properly on they-axis and such thatL′k−1=Lk−1. We are left to check thatXn−1′ =∅.
To do that, notice thatt(Lk)T
L′k⊂X1\Xn−1and thatt(L′k) andt(Lk) differ only in theδ-neigborhood of t(Xn−1). For δ small enough, this neighborhood does not meet Lk or L′k and therefore L′kT
t(L′k) = L′kT
t(Lk) ⊂ X1 = LkT
t(Lk). We deduce that ti(L′k)T
ti+1(L′k) ⊂ ti(Lk)T
ti+1(Lk) for i ≥ 0 and Xn−1′ ⊂Xn−1 and conclude thatXn−1′ =L′kT
t(L′k)T
Xn−1′ ⊂L′kT
t(L′k)T
Xn−1=L′kT
t(Lk)T
Xn−1⊂ (X1\Xn−1)T
Xn−1=∅.
In case s(L) = L, one can choose the function δ such that δs =δ so that s(W) =W. We forget all Lk, k ≥0, consider instead onlyLand note that sinces(L) =Lthere existsn >0 such thatLT
tn(L) =∅.
We define analogously Xi = LT . . .T
ti(L) for i ≥ 0. Since s and t commute, each Xi, i ≥ 0, satisfies s(Xi) =Xi, therefore we can choose the family of arcs changingLtoL′ so thats(L′) =L′.
4.5 Theorem. LetH be a homeomorphism of the torusT2preserving orientation which admits a lifth toR2without fixed point. Then there exists an essential simple closed curve on T2 which lifts to a Brouwer line of h.
Proof. Applying Lemma 3.6, we get a finite number of generalised critical discs{D1, . . . , Dr}such that every critical disc centered on [0,1]×[0,1] admits one of theDi’s, 1≤i≤ras high strict extension. Sinceh commutes withsandt, we have in fact a finite set of euclidean generalised critical discs{D1, . . . , Dr}such that for every critical discDofR2there existsn, m∈Zandibetween 1 andrsuch thatsntm(Di) is a high strict extension ofD.
We choose then a critical discD. To this disc we associate a discD′from{sntm(Di)|n, m∈Z,1≤i≤r}
as just said. Since the underlying disc toD′ is critical, there exists alsoD′′ in the same set which is a high extension ofD′. In that way, we construct an infinite sequence D′, D′′, . . . , D(n), . . .of generalised critical discs such that Si=n
i=1D(i) is a critical topological disc for every n (according to Lemma 3.5). Now, the
