Continuous actions of the group of homeomorphisms of the circle on surfaces

0.32 MB·English

by  Emmanuel Militon

Continuous actions of the group of homeomorphisms of the
circle on surfaces
E. Militon ∗
28 janvier 2013
3
1
0
2 Abstract

n
a In this article we describe all the continuous group morphisms from the group of orientation
J preserving homeomorphisms of the circle to the group of homeomorphisms of the annulus or of the
5 torus.
2

]
S 1 Introduction
D
.
h
For a compact manifold M, we denote by Homeo(M) the group of homeomorphisms of M and by
t
a Homeo (M) the connected component of the identity of this group (for the compact-open topology). By
0
m
atheoremby Fischer(see [6]and[3]), the groupHomeo (M)is simple: the study ofthis groupcannotbe
0
[ reduced to the study of other groups. One natural way to have a better understanding of this group is to

look at its automorphism group. In this direction, Whittaker proved the following theorem.
2
v
Theorem. (Whittaker [14]) Given two compact manifolds M and N and for any group isomorphism
6
4 ϕ : Homeo0(M) → Homeo0(N), there exists a homeomorphism h : M → N such that the morphism ϕ is
8 the map f 7→h◦f ◦h−1.
0
.
1
This theorem was generalized by Filipkiewicz (see [5]) to groups of diffeomorphisms by using a pow-
1
erful theorem by Montgomery and Zippin which characterizes Lie groups among locally compact groups.
2
1 The idea of the proof of Whittaker’s theorem is the following: we see the manifold M algebraically by
v: considering the subgroup Gx of the group of homeomorphismof M consisting of homeomorphisms which
fix the point x in M. One proves that, for any point x in M, there exists a unique point y in N such
i x
X that ϕ(G ) is the group of homeomorphisms of N which fix the point y . Let us define h by h(x) = y .
x x x
r Then we check that h is a homeomorphism and that the homeomorphism h satisfies the conclusion of
a
the theorem. However, Whittaker’s proof uses crucially the fact that we have a group isomorphism and
cannot be generalized easily a priori to the cases of group morphisms. Here is a conjecture in this case.
Conjecture1.1. ForacompactmanifoldM,everygroupmorphismHomeo (M)→Homeo (M)iseither
0 0
trivial or induced by a conjugacy by a homeomorphism.
This conjectureis solvedin[11]inthe caseofthe circle. It is provedalsoin[8]inthe caseofgroupsof
diffeomorphismsof the circle. We canalso be interestedin morphisms fromthe groupHomeo (M) to the
0
group of homeomorphisms of another manifold Homeo (N). This kind of questions is adressed in [8] for
0
diffeomorphismsinthe casewhereN is acircleorthe realline. The followingconjecturelooksattainable.
Conjecture 1.2. DenotebyS−D aclosedsurfaceS withoneopen discremoved. DenotebyHomeo (S−
0
D) the identity component of the group of homeomorphisms S −D with support included in the interior
of this surface. Every group morphism from Homeo (S−D) to Homeo (S) is induced by an inclusion of
0 0
S−D in S.
∗supportedbytheFondationMathématiqueJacques Hadamard
1
In this article, we investigate the case of continuous group morphisms from the group Homeo (S1)
0
of orientation-preserving homeomorphisms of the circle to the group Homeo(S) of homeomorphisms of
a compact orientable surface S. By simplicity of the group Homeo (S1), such a morphism ϕ is either
0
one-to-oneor trivial. Moreover,as the quotient groupHomeo(S)/Homeo (S) is countable, anymorphism
0
Homeo (S1) → Homeo(S)/Homeo (S) is trivial. Hence, the image of the morphism ϕ is included in the
0 0
group Homeo (S). If the surface S is different from the sphere, the torus, the closed disc or the closed
0
annulus, then any compact subgroup of the group Homeo (S) is trivial by a theorem by Kerekjarto (see
0
[2]). Therefore,thegroupHomeo (S)doesnotcontainasubgroupisomorphictoSO(2)whereasthegroup
0
Homeo (S1)doesandthemorphismϕisnecessarilytrivial. Westudyinwhatfollowstheremainingcases.
0
Inthesecondsection,westateourclassificationtheoremofcontinuousactionsofthegroupHomeo (S1)
0
onthetorusandontheannulus. Anyactionwillbeobtainedbygluingactionswhichpreservealamination
bycirclesandactionswhicharetransitiveonopenannuli. Inthethirdsection,wedescribethecontinuous
actions of the group of compactly-supported homeomorphisms of the real line on the real line or on the
circle: this description is useful for the classification theorem and interesting in itself. The fourth and
fifth sections are devoted respectively to the proof of the classification theorem in the case of the closed
annulus and in the case of the torus. Finally, we discuss the case of the sphere and of the closed disc in
the last section.
2 Description of the actions
All the actions of the group Homeo (S1) will be obtained by gluing elementary actions of this group
0
on the closed annulus. In this section, we will first describe these elementary actions before describing
some model actions to which any continuous action will be conjugate.
TheeasiestactionofthegroupHomeo (S1)ontheclosedannulusA=[0,1]×S1preservesofafoliation
0
by circles of this annulus:
p: Homeo (S1) → Homeo(A)
0 .
f 7→ ((r,θ)7→(r,f(θ)))
The second elementary action we want to describe is a little more complex. Let π :R→R/Z=S1 be
the projection. For a point θ on the circle S1 = R/Z, we denote by θ˜a lift of θ, i.e. a point of the real
line whichprojects on θ. For a homeomorphismf of the circle,we denote by f˜a lift of f,i.e. an element
ofthe groupHomeoZ(R) ofhomeomorphismsof the realline whichcommute to integraltranslationssuch
that π◦f˜=f ◦π. A second elementary action is given by:
a : Homeo (S1) → Homeo(A)
− 0 .
f 7→ ((r,θ)7→(f˜(θ˜)−f˜(θ˜−r),f(θ))
Notice that the number f˜(θ˜)−f˜(θ˜−r) does not depend on the lifts θ˜and f˜chosen and belongs to the
interval [0,1]. An action analogous to this last one is given by:
a : Homeo (S1) → Homeo(A)
+ 0 .
f 7→ ((r,θ)7→(f˜(θ˜+r)−f˜(θ˜),f(θ)))
Noticethat the actionsa anda areconjugateviathe homeomorphismofA givenby(r,θ)7→(1−r,θ),
+ −
which is orientation-reversing,and via the orientation-preserving homeomorphism of A given by (r,θ)7→
(r,θ+r). Wenowdescribeanotherwaytoseetheactiona (anda ). WeseethetorusT2 astheproduct
− +
S1×S1. Let
aT2 : Homeo0(S1) → Homeo(T2) .
f 7→ ((x,y)7→(f(x),f(y)))
Itiseasilycheckedthatthisdefinesamorphism. Thisactionleavesthediagonal (x,x),x∈S1 invariant.
The action obtained by cutting along the diagonal is conjugate to a−. More pre(cid:8)cisely, let us d(cid:9)efine
h: A → T2
.
(r,θ) 7→ (θ,θ−r)
2
Then, for any element f in the group Homeo (S1):
0
h◦a−(f)=aT2(f)◦h.
Let G be the groupof homeomorphisms of the circle which fix a neighbourhoodof the point θ . For
θ0 0
a point θ of the circle, the image by the morphism a of the groupG leavesthe sets {(r,θ ),r ∈[0,1]}
0 − θ0 0
and{(r,θ +r),r ∈[0,1]}globallyinvariant. Oneachofthe connectedcomponents ofthe complementof
0
the union of these two sets with the boundary of the annulus, the action of G is transitive (this is more
θ0
easily seen by using the action aT2).
Let us describe now the model actions on the closed annulus which are obtained by gluing the above
actions. Take a non-empty compact subset K ⊂ [0,1] which contains the points 0 and 1 and a map
λ:[0,1]−K →{−1,+1}whichis constantoneachconnectedcomponentof [0,1]−K. Letus define now
an action ϕ of the group Homeo (S1) on the closed annulus. To a homeomorphism f in Homeo (S1),
K,λ 0 0
we associate a homeomorphism ϕ (f) which is defined as follows. If r ∈ K, we associate to a point
K,λ
(r,θ) ∈ A the point (r,f(θ)). If r belongs to a connected component (r ,r ) of the complement of the
1 2
compact set K and if λ((r ,r ))={−1}, we associate to the point (r,θ)∈A the point
1 2
r−r
((r −r )(f˜(θ)−f˜(θ− 1 ))+r ,f(θ))).
2 1 1
r −r
2 1
Thislastmapisobtainedbyconjugatingthehomeomorphisma (f)withthemap(r,θ)7→(ζ(r),θ)where
−
ζ is the unique linear increasing homeomorphism [0,1]→[r ,r ]. If r belongs to a connected component
1 2
(r ,r ) of the complement of the compact set K and if λ((r ,r )) = {+1}, we associate to the point
1 2 1 2
(r,θ)∈A the point
r−r
((r −r )(f˜(θ+ 1 )−f˜(θ))+r ,f(θ))).
2 1 1
r −r
2 1
Thislastmapisalsoobtainedafterrenormalizinga (f)ontheinterval(r ,r ). Thisdefinesacontinuous
+ 1 2
morphism ϕ : Homeo (S1) → Homeo(A). To construct an action on the torus, it suffices to identify
K,λ 0
the point (0,θ) of the closed annulus with the point (1,θ). We denote by ϕT2 the continuous action on
K,λ
the torus obtained this way. By shrinking one of the boundary components (respectively both boundary
components) of the annulus to a point (respectively to points), one obtains an action on the closed disc
(respectively on the sphere) that we denote by ϕD2 (respectively ϕS2 ).
K,λ K,λ
The main theorem of this article is the following:
Theorem 2.1. Any non-trivial continuous action of the group Homeo (S1) on the closed annulus is
0
conjugate to one of the actions ϕ . Any non-trivial continuous action of the group Homeo (S1) on the
K,λ 0
torus is conjugate to one of the actions ϕT2 .
K,λ
In particular, any continuous action of the group Homeo (S1) on the torus admits an invariant circle.
0
By analogy with this theorem, one can be tempted to formulate the following conjecture:
Conjecture 2.2. Any non-trivial continuous action of the group Homeo (S1) on the sphere (respectively
0
on the closed disc) is conjugate to one of the actions ϕS2 (respectively to one of the actions ϕD2 ).
K,λ K,λ
Notice that this theorem does not give directly a description of the conjugacy classes of such actions
as two actions ϕK,λ and ϕK′,λ′ may be conjugate even though K 6= K′ or λ 6= λ′. Now, let K ⊂ [0,1]
andK′ ⊂[0,1] be two compactsets which containthe finite set {0,1}. Let λ:[0,1]−K →{−1,+1}and
λ′ :[0,1]−K′ →{−1,+1}be twomapswhichareconstantoneachconnectedcomponentoftheirdomain
of definition. The following theorem characterizes when the actions ϕK,λ and ϕK′,λ′ are conjugate.
Proposition 2.3. The following statements are equivalent:
– the actions ϕK,λ and ϕK′,λ′ are conjugate;
– either there exists an increasing homeomorphism h : [0,1] → [0,1] such that h(K) = K′ and such
that λ′ ◦h = λ except on a finite number of connected component of [0,1]−K or there exists a
decreasing homeomorphism h:[0,1]→[0,1] such that h(K)=K′ and such that λ′◦h=−λ except
on a finite number of connected component of [0,1]−K.
3
Proof. Let us begin by proving that the second statement implies the first one.
If there exists an increasing homeomorphism h which sends the compact subset K ⊂ [0,1] to the
compact subset K′ ⊂ [0,1], then the actions ϕK,λ and ϕK′,λ◦h−1 are conjugate. Indeed, denote by ˆh the
homeomorphism [0,1] → [0,1] which coincides with h on K and which, on each connected component
(r ,r ) of the complement of K, is the unique increasing linear homeomorphism (r ,r ) → h((r ,r )).
1 2 1 2 1 2
Then the actions ϕK,λ and ϕK′,λ◦h−1 are conjugate via the homeomorphism
A → A
.
(r,θ) 7→ (hˆ(r),θ)
Similarly, suppose that h is an orientation-reversinghomeomorphism of the compact interval [0,1] which
sends the compact subset K ⊂ [0,1] to the compact subset K′ ⊂ [0,1]. As above, we denote by ˆh the
homeomorphism obtained from h by sending linearly each connected component of the complement of
K to each connected component of the complement of K′. Then the actions ϕK,λ and ϕK′,−λ◦h−1 are
conjugate via the homeomorphism
A → A
.
(r,θ) 7→ (hˆ(r),θ)
It suffices then to use the following lemma to complete the proof of the converse in the proposition.
Lemma 2.4. Let K ⊂ [0,1] be a compact subset which contains the points 0 and 1. If λ : [0,1] −
K → {−1,+1} and λ′ : [0,1] − K → {−1,+1} are continuous maps which are equal except on one
connected component (r1,r2) of the complement of K where they differ, then the actions ϕK,λ and ϕK,λ′
are conjugate.
Proof. Thislemmacomesfromthefactthattheactionsa anda areconjugateviathehomeomorphism
+ −
of the annulus (r,θ) 7→ (r,θ +r). More precisely, suppose that λ((r ,r )) = {+1}. Then the actions
1 2
ϕK,λ and ϕK,λ′ are conjugate via the homeomorphism h defined as follows. The homeomorphism h is
equal to the identity on the Cartesian product of the complement of (r ,r ) with the circle and is equal
1 2
to (r,θ)7→(r,θ+ r−r1 ) on the open set (r ,r )×S1.
r2−r1 1 2
Let us establish now the other implication. Suppose that there exists a homeomorphism g which
conjugatesthe actionsϕK,λ andϕK′,λ′. Now,foranyangleα,if wedenote byRα the rotationofangleα:
g◦ϕK,λ(Rα)=ϕK′,λ′(Rα)◦g.
The homeomorphism ϕ (R ) is the rotation of angle α of the annulus. This means that, for any point
K,λ α
(r,θ) of the closed annulus and any angle α∈S1:
g((r,θ+α))=g((r,θ))+(0,α).
In particular, the homeomorphism g permutes the leaves of the foliation of the annulus by circles whose
leaves are of the form {r}×S1. Fix now a point θ on the circle. The homeomorphism g sends the set
0
K×{θ }offixedpointsofϕ (G )(thepointswhicharefixedbyeveryelementofthisgroup)totheset
0 K,λ θ0
K′×{θ0} of fixed points of ϕK′,λ′(Gθ0). Fromthis and what is above,we deduce that for any r ∈K and
anyangleθ,g(r,θ)=(h(r),θ), whereh:K →K′ is ahomeomorphism. Moreover,ifthe homeomorphism
g is orientation-preserving, then the homeomorphism h is increasing and, if the homeomorphism g is
orientation-reversing, then the homeomorphism h is decreasing. We can extend the homeomorphism h
to a homeomorphism of [0,1] which sends K to K′. Notice that, for a connected component (r ,r ) of
1 2
the complement of K in [0,1], the homeomorphism g sends the open set (r ,r )×S1 onto the open set
1 2
(r′,r′)×S1, where (r′,r′)=h((r ,r )) is a connected component of the complement of K′ in [0,1].
1 2 1 2 1 2
Itsufficesnowtoestablishthattheconditiononthemapsλandλ′ issatisfiedforthehomeomorphism
h. Suppose, to simplify the proof, that the homeomorphism g is orientation-preserving and hence the
homeomorphism h is increasing: the case where the homeomorphism g is orientation reversing can be
treated similarly. Suppose by contradiction that there exists a sequence ((r1,n,r2,n))n∈N of connected
components of the open set [0,1]−K such that:
4
– λ((r ,r ))=+1;
1,n 2,n
– λ′((r′ ,r′ ))=−1, where (r′ ,r′ )=h((r ,r ));
1,n 2,n 1,n 2,n 1,n 2,n
– the sequence (r1,n)n∈N is monotonous and converges to a real number r∞.
We will prove then that either the curve {(r,θ0),r >r∞} (if the sequence (r1,n)n∈N is decreasing) or
the curve {(r,θ0),r <r∞} (if the sequence (r1,n)n∈N is increasing) is sent by the homeomorphism g to a
curve which accumulates on {r }×S1, which is not possible. The hypothesis λ((r ,r )) = −1 and
∞ 1,n 2,n
λ((r′ ,r′ ))=+1 for any n would lead to the same contradiction.
1,n 2,n
To achieve this, it suffices to prove that, for any positive integer n, the homeomorphism g sends the
curve {(r,θ ),r <r <r } onto the curve (r,θ + r−r1′,n ),r′ <r <r′ . Fix a positive inte-
0 1,n 2,n 0 r′ −r′ 1,n 2,n
n 2,n 1,n o
ger n. Observe that the restriction of the group of homeomorphisms ϕ (G ) to (r ,r )×S1 has
K,λ θ0 1,n 2,n
two invariant simple curves {(r,θ ),r <r<r } and (r,θ − r−r1,n ),r <r <r on which
0 1,n 2,n n 0 r2,n−r1,n 1,n 2,no
the action is transitive. The action of this group is transitive on the two connected components of
the complement of these two curves, which are open sets. Likewise, the restriction of the group of
homeomorphisms ϕK′,λ′(Gθ0) to (r1′,n,r2′,n) has two invariant simple curves (r,θ0),r1′,n <r <r2′,n
and (r,θ + r−r1′,n ),r′ <r <r′ on which the action is transitive and t(cid:8)he action of this grou(cid:9)p
0 r′ −r′ 1,n 2,n
n 2,n 1,n o
is transitive on the two connected components of the complement of these two curves, which are
open sets. Therefore, the homeomorphism g sends the curve {(r,θ ),r <r <r } to one of the
0 1,n 2,n
curves (r,θ ),r′ <r <r′ or (r,θ + r−r1′,n ),r′ <r <r′ . Let us find now which of them
0 1,n 2,n 0 r′ −r′ 1,n 2,n
is the i(cid:8)mage of the curve {((cid:9)r,θ ),nr <r <2,nr 1},n. We fix the orioentation of the circle induced by
0 1,n 2,n
the orientation of the real line R and the covering map R → S1 = R/Z. This orientation gives
rise to an order on S1 − {θ }. Take a homeomorphism f in G different from the identity such
0 θ0
that, for any x 6= θ , f(x) ≥ x. Then, for any r ∈ (r ,r ), p ◦ ϕ (f)(r,θ ) ≥ r, where
0 1,n 2,n 1 K,λ 0
p : A = [0,1] × S1 → [0,1] is the projection, and the restriction of the homeomorphism ϕ (f)
1 K,λ
to the curve {(r,θ ),r <r<r } is different from the identity. Likewise, for any r ∈ (r′ ,r′ ),
0 1,n 2,n 1,n 2,n
r−r′
p1◦ϕK′,λ′(f)(r,θ0) ≤ r and p1◦ϕK′,λ′(f)(r,θ0 + r′ −1r,′n ) ≥ r and the restrictions of the homeomor-
2,n 1,n
phismϕK′,λ′(f)to (r,θ0),r1′,n <r <r2′,n andto (r,θ0+ r′r−−r1′r,′n ),r1′,n <r <r2′,n aredifferentfrom
the identity. Moreo(cid:8)ver,the map (cid:9) n 2,n 1,n o
(r ,r ) → (r′ ,r′ )
1,n 2,n 1,n 2,n
r 7→ p ◦g(r,θ )
1 0
is strictly increasing as the homeomorphism g was supposed to be orientation-preserving. This implies
what we wanted to prove.
3 Continuous actions of Homeo (R) on the line
c
Let us denote by Homeo (R) the group of compactly supported homeomorphisms of the real line R.
c
In this section, a proof of the following theorem is provided.
Theorem 3.1. Let ψ : Homeo (R) → Homeo(R) be a continuous group morphism whose image has no
c
fixed point. Then there exists a homeomorphism h of R such that, for any compactly supported homeo-
morphism f of the real line:
ψ(f)=h◦f ◦h−1.
Remark: This theorem is true without continuity hypothesis. However, as the proof in this case is
more technical and as we just need the above theorem in this article, the theorem without continuity
assumption will be proved in another article.
Remark: This theorem also holds in the case of groups of diffeomorphisms, i.e. any continuous action
ofthe groupof compactly-supportedCr diffeomorphismsonthe realline is topologicallyconjugateto the
inclusion. The proof in this case is the same.
5
This theoremenables us to describe any continuous actionof the groupHomeo (R) onthe realline as
c
itsuffices to considerthe actiononeachconnectedcomponent ofthe complementof the fixed pointset of
the action.
Proof. Wefixamorphismψ asinthestatementofthetheorem. Duringthisproof,forasubsetA⊂R,we
willdenotebyG thegroupofcompactlysupportedhomeomorphismswhichpointwisefixaneighbourhood
A
of A and by F ⊂ R the closed set of fixed points of ψ(G ) (the set of points which are fixed by every
A A
element of this group). Let us begin by giving a sketch of the proof of this theorem. We will first prove
that, for a compact interval I with non-empty interior, the set F is compact and non-empty. As a
I
consequence, for any real number x, the closed set F is non-empty. Then we will prove that the sets F
x x
are single-point sets. Let {h(x)} = F . Then the map h is a homeomorphism and satisfies the relation
x
required by the theorem. Let us give details.
Notice that, as the group Homeo (R) is simple (see [6]), the image of ψ is included in the group of
c
increasinghomeomorphismsofR(otherwisetherewouldexistanontrivialgroupmorphismHomeo (R)→
c
Z/2Z) and the morphism ψ is a one-to-one map.
Lemma 3.2. For a compact interval I ⊂R with non-empty interior, the closed set F is non-empty.
I
Proof. Fix a compact interval I with non-empty interior. Take a non-zero vector field X : R → R
supported in I. The flow of this vector field defines a morphism
R → Homeo (R)
c .
t 7→ ϕt
Let us assume for the moment that the set F of fixed points of the subgroup {ψ(ϕt),t∈R} of Homeo(R)
is non-empty. Notice that this set is not R as the morphism ψ is one-to-one. As eachhomeomorphism ϕt
commutes with any element in G , we obtain that, for any element g in ψ(G ), g(F)=F. Moreover, as
I I
any element of the group G can be joined to the identity by a continuous path which is included in G
I I
and as the morphism ψ is continuous, any connected component C of F is invariant under ψ(G ). The
I
upper bounds and the lower bounds of these intervals which lie in R are then fixed points of the group
ψ(G ). This proves the lemma.
I
It remains to provethat the set F is non-empty. Suppose by contradictionthat it is empty. Then, for
any real number x, the map
R → R
t 7→ ψ(ϕt)(x)
isahomeomorphism. Indeed,ifitwasnotonto,thesupremumortheinfimumoftheimagewouldprovide
a fixed point for (ψ(ϕt))t∈R. If it was not one-to-one, there would exist t0 6= 0 such that ψ(ϕt0)(x) = x.
Then, for any positive integer n, ψ(ϕt0/2n)(x) = x and, by continuity of ψ, for any real number t,
ψ(ϕt)(x)=x and the point x would be fixed by (ψ(ϕt))t∈R.
Fix a real number x . Let T :G →R be the map defined by
0 x0 I
ψ(f)(x0)=ψ(ϕTx0(f))(x0).
The map T is a group morphism as, for any homeomorphisms f and g in G ,
x0 I
ψ(ϕTx0(fg))(x0) = ψ(fg)(x0)
= ψ(f)ψ(ϕTx0(g))(x0)
= ψ(ϕTx0(g))ψ(f)(x0)
= ψ(ϕTx0(g)+Tx0(f))(x0).
However, the group G , which is isomorphic to the group Homeo (R)×Homeo (R), is a perfect group:
I c c
any element of this group can be written as a product of commutators. Therefore, the morphism T is
x0
trivial. As the pointx is anypointin R,we deduce that the restrictionofψ to G is trivial,whichis not
0 I
possible as the morphism ψ is one-to-one.
6
Remark: if ψ(Homeo (R)) ⊂ Homeo (R), this lemma can be proved without continuity hypothesis.
c c
Indeed, consider anelement f in Homeo (R) supported in I. One ofthe connectedcomponents of the set
c
offixedpoints of ψ(f)is of the form(−∞,a] for some a in R. This intervalis necessarilyinvariantby the
group ψ(G ) which commutes with f. Hence, the point a is a fixed point for the group ψ(G ).
I I
During the proof, we will often use the following elementary result:
Lemma 3.3. Let I and J be disjoint compact non-empty intervals. For any homeomorphism g in
Homeo (R), there exist homeomorphisms g ∈G , g ∈G and g ∈G such that:
c 1 I 2 J 3 I
g =g g g .
1 2 3
Proof. Let g be a homeomorphism in Homeo (R). Let h be a homeomorphism in G which sends the
c 1 I
interval g(I) to an interval which is in the same connected component of R−J as the interval I. Let
h be a homeomorphism in G which is equal to g−1◦h−1 on a neighbourhood of the interval h ◦g(I).
2 J 1 1
Then, the homeomorphism h ◦h ◦g belongs to G . It suffices then to take g = h−1, g = h−1 and
2 1 I 1 1 2 2
g =h ◦h ◦g to conclude the proof of the lemma.
3 2 1
Before stating the next lemma, observe that, for compact intervals I with non-empty interiors, the
sets F are pairwise homeomorphic by an increasing homeomorphism. Indeed, let I and J be two such
I
intervals. Then there exists a homeomorphism λ in Homeo (R) such that λ(I)=J. Then λG λ−1 =G .
c I J
Taking the image by ψ, we obtain that ψ(λ)ψ(G )ψ(λ)−1 =ψ(G ) and therefore ψ(λ)(F )=F .
I J I J
Lemma 3.4. For a compact interval I ⊂R with non-empty interior, the closed set F is compact.
I
Proof. Fix a compact interval I with non-empty interior. Suppose by contradiction that there exists a
sequence(ak)k∈N ofrealnumbersinFI whichtendsto+∞(ifwesupposethatittends to−∞,weobtain
of course an analogous contradiction). Let us choose a compact interval J which is disjoint from I. By
the remark just before the lemma, there exists a sequence (bk)k∈N of elements in FJ which tends to +∞.
Take positive integers n , n and n such that a < b < a . Fix x < a . We notice then that for
1 2 3 n1 n2 n3 0 n1
any homeomorphisms g ∈G , g ∈G and g ∈G , the following inequality is satisfied:
1 I 2 J 3 I
ψ(g )ψ(g )ψ(g )(x )<a .
1 2 3 0 n3
Lemma 3.3 implies then that
{ψ(g)(x ), g ∈Homeo (R)}⊂(−∞,a ].
0 c n3
The greatest element of the left-hand set is a fixed point of the image of ψ: this is not possible as this
image was supposed to have no fixed point.
Lemma 3.5. The closed sets F , where x is a point of the real line, are non-empty, compact, pairwise
x
disjoint and have an empty interior.
Proof. Notice that, if an interval J is included in an interval I, then G ⊂ G and the same inclusion is
I J
true if we take the images of these subsets of Homeo (R) under the morphism ψ. That is why F ⊂ F .
c J I
Now,for any finite family of intervals (I ) whose intersectionhas non-empty interior,the intersectionof
n n
the closedsets F which contains the non-empty closed set F , is not empty. Fix now a point x in R.
In ∩nIn
Notice that
F = F ,
x I
\I
where the intersection is taken over the compact intervals I whose interior contains the point x. By
compactness, this set is not empty and it is compact.
Take two points x6=y in R et let us prove now that the sets F and F are disjoint. If they were not
x y
disjoint, the group generated by ψ(G ) and ψ(G ) would have a fixed point p. However, by Lemma 3.3,
x y
the groups G and G generate the group Homeo (R). Then, the point p would be a fixed point of the
x y c
group ψ(Homeo (R)), a contradiction.
c
7
Now,letusprovethatthesetsF haveanemptyinterior. Ofcourse,giventwopointsxandyofthereal
x
line, if h is a homeomorphismin Homeo (R) which sends the point x to the point y, then ψ(h)(F )=F .
c x y
Therefore, the sets F are pairwise homeomorphic. If the sets F had a non-empty interior, there would
x x
exist uncountably many pairwise disjoint open intervals of the real line, which is false.
Lemma 3.6. For any point x of the real line, any connected component C of the complement of F
0 x0
meets one of the sets F , with y 6=x.
y
Proof. Let us fix a point x in R. Let (a ,a ) be a connected component of the complement of F . It is
0 1 2 x0
possible that either a =−∞ or a =+∞.
1 2
Let us proveby contradictionthat there exists y 6=x such that F ∩(a ,a )6=∅. Suppose that, for
0 0 y0 1 2
any point y 6= x , F ∩(a ,a ) = ∅. For any couple of real numbers (z ,z ), choose a homeomorphism
0 y 1 2 1 2
h in Homeo (R) such that h (z )=z . We claim that the open sets ψ(h )((a ,a )), for y ∈R,
z1,z2 c z1,z2 1 2 x0,y 1 2
are pairwise disjoint, which is not possible as there would be uncountably many pairwise disjoint open
intervals in R. Indeed, if this was not the case, suppose that ψ(h )((a ,a ))∩ψ(h )((a ,a )) 6= ∅
x0,y1 1 2 x0,y2 1 2
for y 6= y . As the union of the closed sets F is invariant under the action ψ, then, for i = 1,2,
1 2 y
when a is finite, ψ(h )−1◦ψ(h )(a ) ∈/ (a ,a ), and ψ(h )−1◦ψ(h )(a ) ∈/ (a ,a ) so that
i x0,y1 x0,y2 i 1 2 x0,y2 x0,y1 i 1 2
ψ(h )(a )=ψ(h )(a ). Butthislastequalitycannotholdasthepointontheleft-handsidebelongs
x0,y1 i x0,y2 i
to F and the point on the right-hand side belongs to F and we observed that these two closed sets
y1 y2
were disjoint.
Lemma 3.7. Each set F contains only one point.
x
Proof. Suppose that there exists a real number x such that the set F contains two points p < p . By
x 1 2
Lemma 3.6, there exists a real number y different from x such that the set F has a common point with
y
the open interval (p ,p ). Take a point r < p . Then, for any homeomorphisms g in G , g in G and
1 2 1 1 x 2 y
g in G ,
3 x
ψ(g )◦ψ(g )◦ψ(g )(r)<p .
1 2 3 2
By Lemma 3.3, this implies that the following inclusion holds:
{ψ(g)(r),g ∈Homeo (R)}⊂(−∞,p ].
c 2
The supremum of the left-hand set provides a fixed point for the action ψ, a contradiction.
Take a homeomorphism f in Homeo (R) and a point x in R. Then the homeomorphism ψ(f) sends
c
the only point h(x) in F to the only point h(f(x)) in F . This implies that ψ(f)◦h=h◦f. Thus, it
x f(x)
suffices to use Lemma 3.8 below to complete the proof of Theorem 3.1.
Lemma 3.8. Let us denote by h(x) the only point in the set F . Then the map h is a homeomorphism.
x
Proof. By Lemma 3.5, the map h is one-to-one.
Fixx ∈Randletusprovethatthemaphiscontinuousatx . TakeacompactlysupportedC1 vector
o 0
fieldR→Rwhichdoesnotvanishonaneighbourhoodofx0 anddenoteby(ϕt)t∈R the flowofthisvector
field. Then, for any time t, h(ϕt(x ))=ψ(ϕt)(h(x )), which proves that the map h is continuous at x .
0 0 0
Finally,letusprovethatthemaphisonto. Noticethattheintervalh(R)isinvariantundertheaction
ψ. Hence, if sup(h(R)) < +∞ (respectively inf(h(R)) > −∞), then the point sup(h(R)) (respectively
inf(h(R))) would be a fixed point of the action ψ, a contradiction.
Proposition 3.9. Any action of the group Homeo (R) on the circle has a fixed point.
c
Hence, the descriptionofthe actionsofthe groupHomeo (R) onthe circleis givenby anactionofthe
c
group Homeo (R) on the real line which is homeomorphic to the circle minus one point.
c
8
Proof. AsH2(Homeo (R),Z)={0}(see[10]),theEulerclassofthisactionnecessarilyvanishessothatthis
b c
actionadmitsa fixedpoint(see [7]aboutthe Eulerclassofanactiononthe circleandits properties).
Finally, let us recall a result which is proved in [11].
Proposition 3.10. Any non-trivial action of the group Homeo (S1) on the circle is a conjugacy by a
0
homeomorphism of the circle.
4 Actions on the annulus
This section is devotedto the proof of Theorem2.1 in the case of morphisms with values in the group
of homeomorphisms of the annulus. Fix a continuous morphism ϕ:Homeo (S1)→Homeo(A).
0
First,asthegroupHomeo (S1)issimple(see[6]),suchagroupmorphismiseithertrivialorone-to-one.
0
We assume in the rest of this section that this group morphism is one-to-one. Recall that the induced
group morphism Homeo (S1) → Homeo(A)/Homeo (A) is not one-to-one as the source is uncountable
0 0
and the target is countable. Therefore, it is trivial and the image of the morphism ϕ is included in the
group Homeo (A).
0
Now,thesubgroupofrotationsofthecircleiscontinuouslyisomorphictothetopologicalgroupS1. The
image ofthis subgroupunder the morphism ϕ is a compactsubgroupofthe groupof homeomorphismsof
the closed annulus which is continuously isomorphic to S1. It is known that such a subgroupis conjugate
totherotationsubgroup (r,θ)7→(r,θ+α),α∈S1 ofthegroupofhomeomorphismsoftheannulus(see
[2]). This is the only pla(cid:8)ce in this proof where we (cid:9)really need the continuity hypothesis. After possibly
conjugating ϕ, we may suppose from now on that, for any angle α, the morphism ϕ sends the rotation of
angle α of the circle to the rotation of angle α of the annulus.
4.1 An invariant lamination
Fix a point θ on the circle. Recall that the group G is the group of homeomorphisms of the circle
0 θ0
which fix a neighbourhood of the point θ . Let us denote by F ⊂ A the closed subset of fixed points
0 θ0
of ϕ(G ) (i.e. points which are fixed under every homeomorphism in this group). The action ϕ induces
θ0
actionsonthetwoboundarycomponentsoftheannuluswhicharecircles. Theactionofϕ(G )oneachof
θ0
theseboundarycomponentsadmitsafixedpointbyProposition3.10. Therefore,thesetF isnon-empty.
θ0
For any angle θ, let α = θ−θ . Then G = R G R−1, where R denotes the rotation of the circle of
0 θ α θ0 α α
angle α. Therefore, ϕ(G ) = R ϕ(G )R−1, where R denotes here by abuse the rotation of angle α of
θ α θ0 α α
the annulus, and F =R (F ). Let
θ α θ0
B = F = R (F ).
θ α θ0
θ[∈S1 α[∈S1
ThissetisoftheformB =K×S1,whereK istheimageofF undertheprojectionA=[0,1]×S1 →[0,1].
θ0
The set K is compact and contains the points 0 and 1. Let F be the foliation of the set B whose leaves
are the circles of the form {r}×S1, where the real number r belongs to the compact set K.
Lemma 4.1. Each leaf of the lamination F is preserved by the action ϕ.
Proof. Fix an angle θ. Let us prove that, for any point x in F the orbit of the point x under the
θ
action ϕ is contained in R (x),α∈S1 ⊂A. First, for any orientation-preservinghomeomorphism f of
α
the circle which fixes the(cid:8)point θ, the h(cid:9)omeomorphism ϕ(f) pointwise fixes the set Fθ. Indeed, such a
homeomorphismofthecircleistheuniformlimitofhomeomorphismswhichpointwisefixaneighbourhood
ofθ andtheclaimresultsfromthecontinuityoftheactionϕ. Anyorientation-preservinghomeomorphism
of the circle g can be written g = R f, where f is a homeomorphism which fixes the point θ. Now, any
β
point x in F is fixed under ϕ(f) and and sent to a point in R (x),α∈S1 ⊂ A under the rotation
θ α
Rβ =ϕ(Rβ), which proves the lemma. (cid:8) (cid:9)
9
Lemma 4.2. Each closed set F intersects each leaf of the lamination F in exactly one point. Moreover,
θ
themap h which, to any point (r,θ) of K×S1 ⊂A, associates the only point of F on the leaf {r}×S1 is a
θ
homeomorphism of K×S1. This homeomorphism conjugates the restriction of the action ϕ to K×S1
K,λ
to the restriction of the action ϕ to K×S1 for any continuous map λ:[0,1]−K →{−1,+1}. Moreover,
this homeomorphism is of the form (r,θ)7→(r,η(r)+θ) where η :K →S1 is a continuous function.
Proof. By Lemma 4.1, the action ϕ preserves each set of the form {r}×S1 and induces on each of these
set an action on a circle. By Proposition 3.10, such actions are well-understood and the restriction of
this action to a subgroup of the form G admits exactly one fixed point (this action is non-trivial as the
θ
rotation subgroup acts non-trivially). This implies the first statement of the lemma.
Take a sequence (rn)n∈N of elements in K which converges to a real number r in K. Then, as the
point h(r ,θ ) belongs to {r }×S1 and to F , the accumulation points of the sequence (h(r ,θ ))
n 0 n θ0 n 0 n
belong to {r}×S1 and to F : there is only one accumulation point which is the point h(r,θ ). For
θ0 0
any angle θ, we have h(r,θ) = R ◦h(r,θ ). This implies that the map h is continuous. This map
θ−θ0 0
is one-to-one: if h(r,θ) = h(r′,θ′), this last point belongs to {r}×S1 = {r′}×S1 so that r = r′ and
Rθ−θ0 ◦h(r,θ0) = Rθ′−θ0 ◦h(r,θ0) so that θ = θ′. This map is onto by definition of the set B = K ×S1
which is the union of the F . As this map is defined on a compact set, it is a homeomorphism.
θ
Take any orientation-preserving homeomorphism f of the circle with f(θ) = θ′. Notice that Gθ′ =
fGθf−1. Therefore ϕ(Gθ′) = ϕ(f)ϕ(Gθ)ϕ(f)−1 and ϕ(f)(Fθ) = Fθ′. So, for any (r,θ) ∈ K ×S1, as the
actionϕ preserveseachleaf of the lamination F, the homeomorphismϕ(f) sends the point h(r,θ), which
isthe unique pointinthe intersectionFθ∩{r}×S1, tothe unique pointinthe intersectionFθ′∩{r}×S1,
which is h(r,θ′). This implies that
ϕ(f)◦h(r,θ)=h◦ϕ (f)(r,θ).
K,λ
Now, denote by η(r) the second projection of h(r,0). As the set F is the image by the rotation of angle
θ
θ of the set F , we have h(r,θ)=h(r,0)+(0,θ) and h(r,θ)=(r,η(r)+θ).
0
4.2 Action outside the lamination
Inthissection,westudytheactionϕoneachconnectedcomponentofA−K×S1. LetA=[r ,r ]×S1
1 2
be the closure of such a component. By the last subsection, such a set is invariant under the action ϕ.
This subsection is dedicated to the proof of the following proposition.
Proposition 4.3. The restriction ϕA of the action ϕ to A is conjugate to a (or equivalently to a ) via
+ −
an orientation preserving homeomorphism.
Proof. Notice that, for any point θ ofthe circle, the actionϕA admits no fixedpoint in the interiorofA
|Gθ
by definition of K×S1: we will often use this fact.
Let us begin by sketching the proof of this proposition. We prove that, for any point θ of the circle,
the morphism ϕA|Gθ can be lifted to a morphism ϕ˜Aθ from the groupGθ to the groupHomeoZ([r1,r2]×R)
of homeomorphisms of the closed band which commute to the translations (r,x)7→(r,x+n), where n is
an integer. Moreover, this group morphism can be chosen so that it has a bounded orbit. We will then
find a continuum G with empty interior which touches both boundary components of the band and is
θ
invariant under the action ϕ˜A. Then we prove that the sets G = π(G ), where π : [r ,r ]×R → A is
f θ θ θ 1 2
the projection,arepairwise disjointandaresimple paths whichjoin the two boundarycomponents ofthe
f
annulus A. We see that the group ϕ(Gθ ∩Gθ′), for θ 6= θ′, admits a unique fixed point a(θ,θ′) on Gθ.
This last map a turns out to be continuous and allows us to build a conjugacy between ϕA and a .
+
The following lemma is necessary to build the invariant sets G :
θ
Forahomeomorphismg inHomeo0(A),wedenotebyg˜theliftofg toHomeoZ([r1,r2]×R)(thismeans
that π ◦g˜ = g ◦π) with g˜((r ,0)) ∈ [r ,r ]×[−1/2,1/2). Let us denote by D ⊂ R2 the fundamental
1 1 2
domain [r ,r ]×[−1/2,1/2]for the action of Z on [r ,r ]×R.
1 2 1 2
10