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Geometriae Dedicata (2022) 216:29
https://doi.org/10.1007/s10711-022-00684-9
ORIGINAL PAPER
Classification of partially hyperbolic surface endomorphisms
Layne Hall1·Andy Hammerlindl1
Received: 15 November 2020 / Accepted: 7 February 2022
© The Author(s) 2022, corrected publication 2022
Abstract
We show that in the absence of periodic centre annuli, a partially hyperbolic surface endo-
morphism is dynamically coherent and leaf conjugate to its linearisation. We proceed to
characterise the dynamics in the presence of periodic centre annuli. This completes a classi-
fication of partially hyperbolic surface endomorphisms.
Keyword Partial hyperbolicity ·Non-invertible dynamics ·Dynamical coherence
Mathematics Subject Classification 37C05 ·37D30 ·57R30
1 Introduction
The dynamics of non-invertible surface maps are less understood than their invertible coun-
terparts. In [6], it is shown that certain classes of partially hyperbolic surface endomorphisms
are leaf conjugate to linear maps. Such a comparison to linear maps cannot be achieved in
general, with [6,8,12] all constructing examples which do not admit centre foliations. Further,
[8] introduces the notion of a periodic centre annulus as a geometric mechanism for failure
of integrability of the centre direction. In this paper, we show that periodic centre annuli are
the unique obstruction to dynamical coherence, and give a classification up to leaf conjugacy
in their absence. We further give a characterisation of endomorphisms with periodic centre
annuli, completing a classification of partially hyperbolic surface endomorphisms.
Before stating our results, we recall preliminary definitions. A cone family C⊂TM
consists of a closed convex cone C(p)⊂TpMat each point p∈M. A cone family is
D f -invariant if Dpf(C(p))is contained in the interior of C(f(p)) for all p∈M.Amap
f:M→Mis a (weakly) partially hyperbolic endomorphism if it is a local diffeomorphism
and it admits a cone family Cuwhich is Df -invariant and such that 1 <Dfvufor all unit
vectors vu∈Cu.WecallCuan unstable cone family. Let Mbe a closed oriented surface. A
This work was partially funded by the Australian Research Council.
BLayne Hall
layne.hall@monash.edu
Andy Hammerlindl
andy.hammerlindl@monash.edu
1School of Mathematical Sciences, Monash University, Clayton, VIC 3800, Australia
0123456789().: V,-vol 123
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29 Page 2 of 19 Geometriae Dedicata (2022) 216:29
partially hyperbolic endomorphism on Mnecessarily admits a centre direction, that is, a Df-
invariant line field Ec⊂TM[4, Section 2]. The existence of this line field and orientability
in turn imply that M=T2.
The homotopy classes of endomorphisms play a key role in their classification. To each
partially hyperbolic endomorphism f:T2→T2, there exists a unique linear endomorphism
A:T2→T2which is homotopic to f. We call Athe linearisation of f.Ifλ1and λ2are
the (not necessarily distinct) eigenvalues of A,thenAis one of three types:
•if |λ1|<1<|λ2|,wesay Ais hyperbolic if,
•if 1 <|λ1|≤|λ2|,wesay Ais expanding,and
•if 1 =|λ1|<|λ2|,wesay Ais non-hyperbolic.
Regardless of homotopy class, there exists a unique semiconjugacy H:R2→R2from
a lift of fto Awhich is both uniformly continuous and a finite distance from the identity [1,
8.4.1]. This map does not necessarily descend to a semiconjugacy T2→T2, and this is a
nuance that we discuss in more detail in Sect. 4. We will be concerned about the particular
case of when Ais expanding, and in this case, the semiconjugacy descends to the compact
manifold by Theorem 2 of [15].
The existing classification of surface endomorphisms relies on integrability of the centre
direction. We say a partially hyperbolic endomorphism of T2is dynamically coherent if
there exists an f-invariant foliation tangent to Ec. Otherwise, we say that the endomorphism
is dynamically incoherent. Though an endomorphism does not necessarily admit a centre
foliation, it always admits a centre branching foliation:an f-invariant collection of C1
curves tangent to Ecwhich cover T2and do not topologically cross. Branching foliations
tangent to a general continuous distribution are constructed in §5 of [2].
Suppose that f,g:T2→T2are partially hyperbolic endomorphisms which are dynam-
ically coherent. We say that fand gare leaf conjugate if there exists a homeomorphism
h:T2→T2which takes centre leaves of fto centre leaves of g, and which satisfies
h(f(L)) =g(h(L))
for every centre leaf Lof f.In[6], it is shown that if f:T2→T2is a partially hyperbolic
surface endomorphism whose linearisation Ais hyperbolic, then fis dynamically coherent
and leaf conjugate to A.
There also exist dynamically incoherent partially hyperbolic surface endomorphisms. The
first known examples were constructed in [6,12] and are homotopic to non-hyperbolic linear
maps. Examples homotopic to expanding linear maps were later constructed in [8], leading
to the result that every linear map on T2with integer eigenvalues λ1,λ
2satisfying |λ2|>
1 is homotopic to an incoherent partially hyperbolic surface endomorphism. A geometric
mechanism for incoherence called a periodic centre annulus was introduced in [8]. A periodic
centre annulus is an immersed open annulus X⊂T2such that fk(X)=Xfor some k>0
and whose boundary, which must consist of either one or two disjoint circles, is C1and
tangent to the centre direction. We further require that a periodic centre annulus is minimal,
in the sense that there is no smaller annulus YXwith the same properties.
Of the incoherent examples constructed in [8], an interesting class are those homotopic to
linear maps which are homotheties or non-trivial Jordan blocks, which themselves as maps
on T2are not partially hyperbolic. Our first result is that such linearisations are, in a sense,
defective.
Theorem A Let f :T2→T2be a partially hyperbolic endomorphism with linearisation
A:T2→T2. If f does not admit a periodic centre annulus, then the eigenvalues of A have
distinct magnitudes.
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In light of the preceding theorem, the absence of a periodic centre annulus implies that
the linearisation is itself a partially hyperbolic endomorphism. In this absence, it then makes
sense to talk of leaf conjugacies of a map to its linearisation. A main result of the current
paper is that periodic centre annuli are the unique obstruction to classification up to leaf
conjugacy.
Theorem B Let f :T2→T2be partially hyperbolic. If f does not admit a periodic centre
annulus, then f is dynamically coherent and leaf conjugate to its linearisation.
To give a comprehensive classification of partially hyperbolic surface endomorphisms,
we look to understand dynamics of endomorphisms with periodic centre annuli. Our setting
for this classification bears some similarity to diffeomorphisms on T3, where centre-stable
tori are known to be the unique obstruction to both dynamical coherence and leaf conjugacy
[9,14]. The analogue of a periodic centre annulus in this other setting is a region between
centre-stable tori. A classification of such diffeomorphisms is given in [11], where it is shown
that there are finitely many centre-stable tori and the dynamics on the regions between such
tori take the form of a skew product. The finiteness of periodic centre annuli also holds in
our setting, and while not difficult to prove, is essential for our classification.
Theorem C Let f :T2→T2be a partially hyperbolic endomorphism. Then f admits at
most finitely many periodic centre annuli X1,...,Xn.
We can follow the same procedure as [11] to show that the dynamics on a periodic centre
annulus also takes the form of a skew product. This result is stated as Proposition 5.1 and
discussed in Sect. 5, though as it is a straightforward adaption, its complete proof is deferred
to an auxiliary document [7].
Unlike diffeomorphisms on T3, the fact that the dynamics on a periodic centre annulus
takes the form of a skew product is not a complete classification in our setting, as it is not
necessarily true that the annuli cover T2. This is seen in the examples of [8]. However, this
is at least true for the examples in [6,12], and in fact holds for any endomorphism with
non-hyperbolic linearisation, as stated in the following result.
Theorem D Let f :T2→T2be a partially hyperbolic endomorphism with non-hyperbolic
linearization A and with at least one periodic centre annulus. Then T2is the union of the
closures of all of the periodic centre annuli of f .
It then remains to characterise the dynamics when the linearisation is expanding, where
the collection of closed periodic centre annuli will not cover T2. This is our final result,
completing a classification of partially hyperbolic surface endomorphisms.
Theorem E Let f :T2→T2be a partially hyperbolic endomorphism which admits a
periodic centre annulus, and let X 1,..., Xnbe the collection of all disjoint periodic centre
annuli of f . Let A :T2→T2be the linearisation of f and suppose A is an expanding linear
map, so that there is a unique semiconjugacy H :T2→T2from f to A. Then:
•The set
=
i
k≥0
f−k(Xi)
is dense in T2.If X is a connected component in this set, then X is an annulus, and
H(X)is either a periodic or preperiodic circle under A.
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•The complement
T2\
is a union of disjoint circles tangent to the centre direction of f . If S is a connected
component of T2\, then S and H(S)are circles.
•The endomorphism f does not admit an invariant partially hyperbolic splitting T T2=
Ec⊕Eu.
This result can be phrased as saying that either a point eventually lies in a periodic centre
annulus, where we understand the dynamics as a skew product due to our earlier discussion,
or it lies on an exceptional set of centre circles. The orbits of the annuli and exceptional circles
themselves are understood through the semiconjugacy. The final property demonstrates that
the behaviour of these endomorphisms is distinctly non-invertible.
The current paper is structured as follows. We begin in Sect. 2by studying how periodic
centre annuli manifest within centre branching foliations. In particular, a Reeb component or
what we call a tannulus give rise to periodic centre annuli. This characterisation is fundamental
to the rest of the paper, and in this process we prove Theorem C.
We next establish Theorem Ain Sect. 3. This is done by introducing rays associated to
branching foliations as a tool for relating the dynamics of fto its linearisation A,anidea
that is used only within that section.
Section 4contains the proof of Theorem B. We use topological properties of the branching
foliation that were established in the preceding sections, paired with the Poincaré–Bendixson
theorem, to establish dynamical coherence. A leaf conjugacy is then constructed with an
averaging technique, similar to [6].
We conclude the paper by proving both Theorems Dand Ein Sect. 5.
2 Tannuli and Reeb components
In this section, we characterise how periodic centre annuli can arise in branching foliations.
These are fundamental to proving the main theorems of the paper, and we shall prove Theorem
Cin the process.
Let f:T2→T2be a partially hyperbolic surface endomorphism. Lift fto a diffeomor-
phism ˜
f:R2→R2and recall that ˜
fadmits an invariant splitting Eu⊕Ec[13]. For an
unstable segment Ju⊂R2,defineU1(J)={p∈R2:dist(p,J)<1}. The following result,
which is the basis of ‘length versus volume’ arguments, is proved for the endomorphism
setting in Section 2 of [6].
Proposition 2.1 There is K >0such that if J ⊂R2is either an unstable segment or a centre
segment, then
volume(U1(J)) > Klength(J).
Recall that an essential circle C⊂T2has a slope, which can be defined in terms of the
first homology group of T2.
Lemma 2.2 Let X1,X2⊂T2be periodic centre annuli of f . Then the boundary circles of
X1and X2have the same slope. Moreover, these circles do not topologically cross.
Proof Let X1and X2be two periodic centre annuli. By replacing fwith an iterate, we may
assume that X1and X2are f-invariant.
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First suppose that X1and X2have different slopes. Let ˜
X1,˜
X2⊂R2be lifts of X1and
X2such that at least one is by a lift ˜
fof f. Then the image of ˜
X1∩˜
X2under iterates of ˜
f
is always some translation of ˜
X1∩˜
X2by an element of Z2. Since the annuli have distinct
slopes, then ˜
X1and ˜
X2each lie a bounded distance from lines L1,L2⊂R2with different
slopes. Then ˜
X1∩˜
X2lies inside a finite parallelogram bounded by edges parallel to L1and
L2. Thus, all forward iterates of ˜
X1∩˜
X2have uniformly bounded volumes. However, a small
unstable curve inside this region grows exponentially in length. By Proposition 2.1,thisisa
contradiction.
Now knowing that the boundary circles of the annuli have the same slope, assume that a
boundary circle of X1topologically crosses a boundary circle of X2.Let p:R2→T2be the
covering projection. By lifting a point x∈X1∩X2, we can find a point ˜x∈˜
X1whose orbit
under ˜
flies in ˜
X1∩p−1(X2)for all time. Since only finitely many connected components
of p−1(X2)intersect ˜
X1, this means that there is a connected component ˜
X2of p−1(X2)
which is periodic under ˜
f. Replacing fby an iterate, we may assume that ˜
f(˜
X2)=˜
X2.
Since the annuli X1and X2are of the same slope, the ˜
f-invariant set ˜
X1∩˜
X2lies within
a bounded neighbourhood of some line L. By making a linear coordinate change on R2,we
may assume that Lis vertical, and that the lifts ˜
X1and ˜
X2are both invariant under vertical
translation. Let L1and L2be lifts of boundary circles of ˜
X1and ˜
X2, respectively, that cross.
Denote the connected components of R2\L1as U+and U−. The existence of a topological
crossing means there is a compact interval or point Swith S⊂L1∩L2, such that L2
passes from U−about a neighbourhood of one endpoint of S, and then enters U+about a
neighbourhood of the other endpoints of S. Since both L1and L2are invariant under vertical
translation, then L2also crosses from U−to U+locally at S+(1,0). By a connectedness
argument, L2must then have crossed back from U+to U−at some point qon L1between
Sand S+(1,0). Since both L1and L2are C1, then there must be a disc bounded by a bigon
consisting of a segment along one of each of L1and L2. We can also see that such a bigon
whose vertices are at crossings of L1and L2must have uniformly bounded volume: it must
be bounded by 1 in the vertical direction, and since it lies a uniform distance from a vertical
line L, it is also uniformly bounded in the horizontal direction. Since ˜
fis a diffeomorphism,
then the disc Vbounded by this bigon is mapped to a disc bounded by another one of these
bigons. Thus ˜
fn(V)must have uniformly bounded volume for all n. By considering a small
unstable curve Ju⊂V, then Proposition 2.1 gives a contradiction.
Let X⊂T2be an open annulus. If the boundary of Xconsists of two disjoint circles, then
the width is the Hausdorff distance between the two boundary circles. However, we must
handle the case where Xis a dense subset of T2and its boundary is a single circle. In this
case, lift Xto a strip ˜
Xon the universal cover and define the width of Xto be the Hausdorff
distance between the two boundary components of ˜
X. This is independent of the choice of
lift since the lifts are all translates of each other. The preceding lemma allows us to prove
Theorem C.
Proof of Theorem CWe use what is basically the idea of local product structure, but instead
in the context of an unstable cone. Since Cuand Ecare a bounded angle apart, then there
is δ>0 such that if a periodic centre annulus Xcontains an unstable curve of length 1,
then Xmust have width at least δ.Since Xis periodic, then by iterating a small unstable
curve within X,weseethat Xmust contain an unstable curve of length 1. Thus any periodic
centre annulus has width at least δ. But by the preceding lemma, all periodic centre annuli
are disjoint, so by compactness of T2there can only be finitely many.
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29 Page 6 of 19 Geometriae Dedicata (2022) 216:29
Periodic centre annuli will be easier to understand when we can consider them as a part
of a branching foliation.
Lemma 2.3 There exists an invariant centre branching foliation Fc
bran of T2which contains
the boundary circles of all periodic centre annuli.
Proof Start with any invariant centre branching foliation Gof T2, which, as mentioned in the
introduction, exists by the arguments in Section 5 of [2]. We modify the leaves of Gto create
Fc
bran.
Let X⊂T2be a periodic centre annulus, and consider the set k≥0f−k(@X)of orbits
of the boundary circles of X, which necessarily consists of centre circles. Let Kbe the
closure of this set under the compact-open topology. Note that if a sequence of centre circles
converges in this topology, then its limit too will be a centre circle by a Haefliger compact
leaf argument. So, Kconsists of centre circles. Suppose we have a leaf L∈Gwhich crosses
acirclein K.CutLinto segments at each point it intersects K, and complete these segments
into a new centre leaf by continuing them along the circles of Kthat they each intersect. The
new leaves cover L, and they now only lie either ‘inside’ or ‘outside’ the orbit of the annulus
X. This implies that the new leaves do not cross on K, nor do they cross the circles in K.
Thus, if we add both the new leaves and the circles of Kto the foliation while removing the
leaf L, we have a branching foliation tangent to the centre direction. Doing this for all such
leaves Lyields a new invariant centre branching foliation, G1, which contains the boundary
circles of X.
Since all the periodic annuli have disjoint interiors, we can continue this process for each
periodic centre annulus. There are finitely periodic centre annuli by Lemma 2.2, so the process
terminates to give us the desired branching foliation Gn=: Fc
bran.
For the remainder of this section, we will take Fc
bran to be an invariant centre branching
foliation containing all periodic centre annuli. Recall that a branching foliation also has an
associated approximating foliation Fc
"as in [2]. Since Fc
bran contains a periodic centre annulus
then Fc
"necessarily has rational slope, and so either the foliation is the suspension of some
circle homeomorphism or contains a Reeb component. In the dynamically incoherent example
of [6,12], the periodic centre annuli correspond to Reeb components in the approximating
foliation. In the incoherent example of [8] and the coherent but not uniquely integrable
example of [12], the centre curves form what we call an tannulus. This name is because of
the similarities of the annulus to a collection of translated graphs of the tangent function.
Rigorously, a tannulus is a foliated closed annulus such that every leaf in the interior is
homeomorphic to Rand which when lifted to a foliation on a strip I×R⊂R2,issuch
that the leaves tend to one boundary component of I×Rwith y-coordinate approaching ∞
when followed forwards, and then tending toward the other component with y-coordinate
approaching −∞ followed backwards. We say that an annulus in Fc
bran is a Reeb component
or tannulus if its approximation in Fc
"is such an annulus.
Recall that if Ais the linearisation of f,thenFc
bran has leaves that lie a bounded distance
from an A-invariant linear foliation A.Letπ:R2→Rbe a linear projection which maps
each leaf of Aonto R.
Lemma 2.4 The image and preimage of a Reeb component in Fc
bran under f is a Reeb
component. Similarly, the image and preimage of a tannulus in Fc
bran is a tannulus.
Proof Reeb components and tannuli in Fc
bran cannot map to annuli foliated by circles, so it
remains to show that a Reeb component cannot map to tannulus and vice versa. Lift Fc
bran
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Geometriae Dedicata (2022) 216:29 Page 7 of 19 29
to R2. Observe that if Lis a leaf of Fc
bran contained in a Reeb component, then π(L)is an
interval of the form [a,∞)for some finite constant a.However,ifLis contained within a
tannulus, then π(L)is all of R.ThenπA(L)is a half-bounded interval, and πA(L)is all
of R.Since fis a finite distance from A, we cannot have f(L)=Lor f(L)=L.
We now show that the presence of a Reeb component or tannulus necessitates the existence
of a periodic centre annulus.
Lemma 2.5 If X is a tannulus or Reeb component of a branching foliation Fc
bran, then there
is k ≥0such that f k(X)is periodic.
Proof If Fc
bran contains a Reeb component, then as Fc
bran can contain only finitely many Reeb
components, there is a periodic Reeb component. If Fc
bran contains a tannulus, denote this
tannulus as X.IfJu⊂Xis an unstable curve, then fk(Ju)grows unbounded in length,
implying that fk(X)must have width bounded below by some δ>0 for all sufficiently
large k. However, it is clear from their definition that all tannuli can intersect only on their
boundary. Thus, there can only be finitely many tannuli of width δ,sothat fk(X)is periodic
for all sufficiently large k.
Corollary 2.6 If there exists a Reeb component or tannulus in Fc
bran, there exists a periodic
centre annulus.
Since Reeb components and tannuli are the only components of a rational-slope branching
foliation which could have non-circle leaves, we also have the following.
Corollary 2.7 If f does not admit any periodic centre annuli and A has rational eigenvalues,
the leaves of any invariant centre branching foliation are circles.
The converse of this characterisation is also true—that a periodic centre annulus is neces-
sarily a centre Reeb component or tannulus—though the ideas used to see this are precisely
those we use to prove dynamical coherence in Sect. 4. Since the converse is not needed for
our results, we do not include a proof in this paper.
3 Degenerate linearisations
In this section, we prove Theorem A. To relate the dynamics of the endomorphism to its
linearisation, we introduce the notion of a ray associated to a branching foliation, which will
only be used in this section.
Let f0:T2→T2be a partially hyperbolic endomorphism, and for the remainder of this
section, suppose that f0does not admit a periodic centre annulus. Let Abe the linearisation
of f0.IfAhas real eigenvalues, then by replacing f0by an iterate, we may assume that
eigenvalues of Aare positive. Proving Theorem Athen amounts to ruling out the possibilities
that Ahas complex eigenvalues, is a homothety, or admits a non-trivial Jordan block, where
in the latter two cases we may assume the eigenvalues of Aare positive.
Lift f0to a diffeomorphism f:R2→R2,andletAalso denote the lift of the linearisation.
Further, lift the centre direction and unstable cone of f0to R2, and denote them Ecand Cu.
Up to taking finite covers, we can assume that Ecand Cuare orientable, and by replacing f
by f2,thatDf preserves these orientations. Similarly, we assume that Ais an orientation
preserving linear map.
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Recall that there exists an invariant centre branching foliation of R2which descends to
T2, as is constructed for surfaces in Section 5 of [2]. Moreover, given small " >0, the leaves
of this branching foliation lie less than " in C1-distance from an approximating foliation Fc
"
which also descends to T2.ThisimpliesthatFc
"lies a finite distance from a linear foliation
of R2, and thus, so does Fc
bran. Moreover, from Lemma 2.7, the absence of periodic centre
annuli implies that Fc
"descends to the suspension of a circle homeomorphism. This is already
enough to rule out the case of complex eigenvalues.
Lemma 3.1 The linearisation A of f has real eigenvalues.
Proof An invariant centre branching foliation Fc
bran which descends to T2is a finite distance
from a linear foliation of R2.Since fand Aare a finite distance apart, this linear foliation must
be A-invariant. A linear map of R2with complex eigenvalues preserves no such foliation.
Now we begin to consider to the homothety and Jordan block cases.
Lemma 3.2 The linearisation A must have an eigenvalue greater than 1.
Proof Let U⊂R2be an open ball. If the Ahas no eigenvalue greater than 1, then the volume
of An(U)is bounded. Since fis a finite distance from A, the volume of fn(U)can grow at
most polynomially. But Ucontains some small unstable curve Ju, and so by Proposition 2.1
implies that the volume of fn(U)must grow exponentially, a contradiction.
Using the orientation on Ec, we can talk about forward centre curves emanating from a
point. Given a branching foliation Fc
bran and a leaf L∈Fc
bran through the origin, we call the
forward half of Lto be all points on Lforward of the origin. Recall that a ray is a subset of
R2given by {t·v∈R2:t≥0}for some non-zero vector v∈R2.
Lemma 3.3 Let Fc
bran be an invariant branching foliation. Then there exists a unique ray
R(Fc
bran)emanating from the origin such that if L∈Fc
bran passes through the origin, the
forward half of Llies a finite distance from R(Fc
bran).
Proof Let Fc
"be an approximating foliation of Fc
bran.BytakingTFc
"to be transverse to Cu,
TFc
"has a natural orientation that is consistent with that of Ec.SinceFc
"descends to T2,it
lies a finite distance from a linear foliation of R2. Thus the forward half of the leaf Fc
"(p)lies
close to some ray R(Fc
")emanating from the origin. Let L∈Fc
"and J⊂Lbe a subcurve
obtained by taking all points forward of some point q∈L. Since we know Fc
"descends to a
suspension, then Jlies a finite distance from R(Fc
").
Since the leaves of Fc
bran and Fc
"are a finite distance apart on the universal cover and they
have consistent orientations, then any forward half a leaf of Fc
bran also lies a finite distance
from R(Fc
bran):= R(Fc
").
Recall that there can be many invariant centre branching foliations, and by using the explicit
constructions of [2], we have two concrete examples as follows. Using the orientations on
Cuand Ec, we may roughly view Culocally as vertical and Echorizontal. One can obtain
a branching foliation by taking the maximal highest forward centre curve through a point
q∈R2and gluing it to the lowest backward centre curve through q. We denote this branching
foliation as Fmax. A second (though not necessarily distinct) branching foliation can be
obtained by flipping this construction, and gluing the lowest forward to the highest backward
curves. We denote this branching foliation Fmin.
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Consider now the space Sof rays through the origin in R2. Since each ray may be associated
to a unit vector in R2, the space Sis homeomorphic to a circle. We specifically work with
Srather than the projective space RP1in order to handle the case when Ahas a non-trivial
Jordan block. Observe that a line on R2quotients down to a pair of antipodal points on S,so
that the complement of this line in Shas two connected components.
Lemma 3.4 There exists a line γ⊂R2such that one connected component S+of S \γ
contains R(Fc
bran)for any centre branching foliation. In particular, S+contains both R (Fmax)
and R(Fmin ).
Proof Since an invariant centre branching foliation in the absence of periodic centre annuli
is approximated by a suspension, there exists a circle C0⊂T2passing through (0,0)∈T2
which is transverse to the centre direction. The lift C⊂R2of C0passing through the origin
in R2then lies close to a line γ. The line γdivides the ray space Sinto two connected
components S−and S+,andthecurveCdivides R2into two half spaces −and +.
If a leaf of the lifted centre branching foliation intersected the lifted circle Ctwice, then
by transversality, a leaf of a close approximating foliation would also intersect Ctwice. Since
Cis transverse to the centre direction, a leaf of this approximating foliation cannot intersect
Ctwice by a Poincaré–Bendixson argument, see for example our argument given in later in
Proposition 4.2.
Thus, as a forward centre curve through the origin intersects Cat the origin, any such
curve must lie entirely in either −or +. The orientations on the transverse Ecand Cu
imply that all forward centre curves from the origin all must lie in only one of these half
spaces, which we take to be +. Now if a ray lies a finite distance from a forward centre
curve, then it necessarily lies in S+, or is one of the two points corresponding to γitself.
However, every forward centre curve lies close to a curve obtained by lifting a circle in a
suspension on C0, so such a ray cannot be one of the points corresponding to γ.
By giving an orientation on Swe obtain the usual notion of ‘clockwise’ and ‘counter-
clockwise’ on the circle. We assume without loss generality that the path from R(Fmax)
to R(Fmin)that is contained in S+⊂Spoints in the clockwise direction. From a natural
perspective, this means that the ray R(Fmin )lies clockwise from R(Fmax ).
Lemma 3.5 There are smooth foliations Fcw and Fccw of R2which descend to T2such that:
1. both T Fcw and T Fccw are a small distance from the centre direction, and are transverse
to Cu
2. with respect to the orientation given by Cu,TFcw lies below Ecand T Fccw above Ec
3. the rays R(Fcw )and R(Fccw)are contained in S +⊂S
4. the path from R(Fmin)to R(Fcw)that is contained in S+goes in the clockwise direction,
while the path from R(Fmax)to R(Fccw )that is contained in S+goes in the counter-
clockwise direction.
5. if A has distinct eigenvalues, then R (Fcw)and R(Fccw)are not tangent to the A -invariant
directions.
Proof We shall start by finding Fcw.LetFc
"be an approximating foliation associated to
Fc
bran.ThenFc
"descends to a suspension of homeomorphism on some circle C0⊂T2which
is transverse to Ec. Now take a smooth distribution Ecw to lie sufficiently close to Ecso
that it remains transverse to the unstable cone, but that with respect to the orientation on
the unstable cone, lies strictly below Ec. For concreteness, this construction may be done as
follows. Let Xcbe the (continuous) vector field consisting of unit vectors that point in the
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29 Page 10 of 19 Geometriae Dedicata (2022) 216:29
Fig. 1 A depiction of the ray
space Sand the points we are
considering in S
R(Fcw)
R(Fccw)
R(Fmax)
R(Fmin)
γ
γ
direction of Ecwith the same orientation as Ec,andlet Xube any vector field consisting
of unit vectors lying inside of the cone family Cuand with the same orientation as the cone
family. If " >0 is small, then the vector field defined by X=Xc−"Xuconsists of vectors
close to Ecbut rotated slightly clockwise away from Cu. This vector field is continuous and
not smooth in general. By a small perturbation (much smaller than "), we can approximate
Xby a smooth vector field Xcw such that its vectors are close to Ecbut rotated slightly
clockwise away from Cu. This smooth vector field then defines Ecw.
As a smooth distribution, Ecw integrates to a smooth foliation Fcw . The return map of
the flow along TFcw to the circle C0must then have a lower rotation number than that
of the return map of TFc
", implying that R(Fcw)lies a small clockwise distance in Sfrom
R(Fmin). With this, we obtain the third and fourth properties. The last property can be satisfied
by refining our perturbation: if R(Fcw )happened to point in the expanding Adirection, then
another small clockwise perturbation of TFcw will integrate to the desired foliation.
By instead perturbing Ecto Eccw which lies strictly above Ec, the resulting distribution
integrates to the desired Fccw by a similar argument.
The foliations Fcw and Fccw are thought of as clockwise and counterclockwise pertur-
bations to the centre direction, respectively. To illustrate to the reader the objects we are
considering in S,belowinFig.1is the ray space Sindicating the relationships we have
established from the rays of Fmax,Fmin ,Fcw and Fccw.
The next step is to iterate these perturbed foliations backward. Given any foliationFof R2
which descends to T2, define a foliation f−n(F)of R2by f−n(F)( p)=f−n(F(fn(p))).
Observe that since the linearisation Ais an orientation preserving linear map, it induces an
orientation preserving map S→S, which we also denote as A.
Lemma 3.6 Let Fbe either Fcwor Fccw, and let Ln∈f−n(F)be a leaf of f −n(F).Then
the forward half of Lnlies a bounded distance from a translate of the ray R(f−n(F)) :=
A−n(R(F)). This distance is uniformly bounded for all n and choices of Ln.
Proof If L⊂R2is the line containing R(F), then using the fact that Ais a finite distance
from f, it is straightforward to see that any leaf of f−n(F)is a bounded distance from
A−n(R(F)). This distance is naturally bounded across all choices of Ln. To see uniform
boundedness for all n, we refer the reader to §2 of [6], where it is shown there under the
assumption that Ais hyperbolic. The argument relies on the finite distance between fand A.
In our setting, when Ahas eigenvalues of magnitude equal to or greater than 1, the proof is
nearly identical. In repeating the argument, one can replace the fact that R(F)is not tangent
to the contracting eigendirection of Awith the final property of Lemma 3.5.
Suppose further now that TFis transverse to Ecand Cu, as is the case for Fcw and Fccw .
Then the tangent lines Tf−n(F)converge to the centre direction as n→∞.Since fis
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Geometriae Dedicata (2022) 216:29 Page 11 of 19 29
A=λ10
0λ2
A=λ0
0λA=λ0
1λ
Fig. 2 The dynamics of the map A:S→S. From left to right are the cases when Ahas distinct eigenvalues,
Ais a homothety, and when Ahas a non-trivial Jordan block
partially hyperbolic, there exists a centre cone family, that is, a cone family transverse to
Cuwhich contains both TFand Ec, and that is invariant under Df−1. This cone family
inherits a Df −1-invariant orientation from the orientation on Ec. Using this orientation, we
consider the sequence of curves Lnemanating from the origin given by the forward half of
the leaf of f−n(F)through the origin. By an Arzela-Ascoli argument, this sequence has a
convergent subsequence in the compact open topology. The limit of such a subsequence is
then a complete forward centre curve emanating from the origin. We use this to prove the
following.
Lemma 3.7 We have limn→∞(A−n(R(Fcw))) =R(Fmin )and limn→∞(A−n(R(Fccw ))) =
R(Fmax).
Proof We prove the first limit, the second follows by a similar argument. Let Lnbe the forward
half of the leaf in f−n(Fcw )that passes through the origin. By the discussion preceding the
lemma, the sequence Lnhas a convergent subsequence, and the limit Lof this subsequence
is a complete forward centre curve through the origin. By Lemma 3.6, the limit Llies a
uniformly bounded distance from limn→∞ A−n(R(Fcw)), which we note must exist since A
cannot have complex eigenvalues. Thus, if we can show that Lis actually the forward half
of a leaf in Fmin, we will have proved the result.
Recall that TFcw lies strictly below Ecwith respect to the orientation on Cu. Since this
orientation is preserved by Df ,thenTf−n(F)lies strictly below Ecfor all n. Thus the limit
Lmust in fact be the lowest forward centre curve through the origin. This is, by definition,
the forward half of a leaf of Fmin, giving the result.
The final observation we need to make is that the dynamics of the map A:S→Sis
determined by the eigenvalues of the linearisation. An invariant line of Aquotients down to an
antipodal pair of fixed points in S.IfAhas distinct eigenvalues, the eigenline corresponding to
the greater eigenvalue corresponds a pair of attracting fixed points, while the other eigenline
becomes two repelling fixed points. If Ais a homothety, the induced map on Sis the identity.
When Ahas a non-trivial Jordan block, the map has just one pair of fixed points which both
have the property of being attracting on one side and repelling on the other. Each of these
behaviours is demonstrated in Fig. 2, and they allow us rule out the remaining cases.
Proof of Theorem AThe possibility of Aadmitting complex eigenvalues was ruled out in
Lemma 3.1.
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29 Page 12 of 19 Geometriae Dedicata (2022) 216:29
Next suppose that Ais a homothety, so that its induced map S→Sis the identity. Thus
A−n(R(Fcw)) =R(Fcw )= R(Fmin), which contradicts the fact that limn→∞(R(Fcw)) =
R(Fmin)by Lemma 3.7.
Finally, suppose that Ahas a non-trivial Jordan block. Then the map A:S→Shas
only two fixed points, each of which is repelling on one side and attracting on the other, as
illustrated in Fig. 2. Only one of these points can lie in the component S+of S\γ,and
so R(Fmax)and R(Fmin )must both be this fixed point. Then one of R(Fcw )and R(Fccw)
lies on the side of this fixed point that is attracting under A, and the other, the side that is
repelling. Without loss of generality we may assume then that A−n(R(Fcw )) converges to
R(Fmax)=R(Fmin ),butthat A−n((R(Fccw ))) converges to the point antipodal to R(Fmin ).
This again contradicts Lemma 3.7.
With Theorem Aproved, we conclude the section by using the tools we have developed
to establish a property that will be used in the construction of a leaf conjugacy. We relax
our assumption that Ais expanding, so that we are assuming that Ais any linearisation f
of a partially hyperbolic endomorphism without periodic centre annuli. We have shown that
Amust have eigenvalues of distinct magnitude, so let Acdenote the linear foliation of T2
associated to the smaller eigenvalue of A.
Lemma 3.8 Given any invariant centre branching foliation Fc
bran, its leaves lie a bounded
distance to the lines of Acon the universal cover.
Proof We have shown that Amust have eigenvalues of distinct magnitude. The map A:S→
Sthen has two pairs of fixed points, each pair corresponding to the real eigenvalues λcand
λuof A, with |λc|<|λu|. By the last item of Lemma 3.5, the rays R(Fcw)and R(Fccw )are
not any of the fixed points of A:S→S.ThenR(Fmax )=limn→∞(A−n(R(Fcw ))) and
R(Fmin)=limn→∞(A−n(R(Fccw))) are both fixed points associated to λc. Only one such
fixed point lies in S+,soR(Fmax )=R(Fmin). Since any forward centre curve must lie in
between a cone region bounded by the lowest and highest forward centre curves, and this
region must lie a finite distance from R(Fmax ), it follows that R(Fc
bran)=R(Fmax )for any
invariant centre branching foliation Fc
bran.Since R(Fmax)is a fixed point associated to λc,
then Fc
bran is a finite distance from Ac.
4 Coherence and leaf conjugacy
In this section, we prove Theorem B.Let f0:T2→T2be partially hyperbolic. We assume
for the remainder of this section f0does not admit a periodic centre annulus. Lift f0to
f:R2→R2, and similarly lift the linearisation of f0to A:R2→R2.Asanabuseof
notation, we denote by Fc
bran and Fc
"the lifts of the branching and approximating foliations
respectively. Since f0does not admit any periodic centre annuli, then by Theorem A, the linear
map Ahas real eigenvalues with distinct magnitudes. Let Acand Aube the linear foliations
of R2associated to the eigenvalues of Awith smaller and larger magnitude, respectively.
Define a projection πu:R2→Rto be a linear map which maps each line of Auonto Rand
whose kernel is the leaf of Acthrough the origin. We will later use a projection πc:R2→R
which is defined analogously.
Lemma 4.1 There is C >0such that if q ∈Fc
bran(p),then|πu(p)−πu(q)|<C.
Proof The branching foliation Fc
bran has leaves that lie a finite distance from lines in Acby
Lemma 3.8, which implies that leaves of Fc
bran are a uniformly bounded distance in the πu
direction.
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Geometriae Dedicata (2022) 216:29 Page 13 of 19 29
Recall from the introduction that there exists an unstable foliation Fuon R2.Akey
step in proving that Fc
bran is in fact a true foliation will be showing that leaf segments of
Fugrow large in the πu-direction under forward iteration. When Ais hyperbolic or non-
hyperbolic (cases defined in the introduction), this property of Fucan be proved by using the
Poincaré–Bendixson Theorem and ‘length versus volume’ arguments, as is done in Section
2of[6]. When Ais expanding, it becomes difficult to compare growth rates of centre and
unstable curves, so length versus volume arguments are less practical. Instead, we will use
the Poincaré–Bendixson Theorem paired topological properties of Fc
", the latter of which are
now well understood due to the results in Sect. 2.
Proposition 4.2 If L∈Fc
",thenLcan intersect each leaf of Fuat most once.
Proof Suppose that a curve in Fc
bran intersects a leaf of Fumore than once. Since this curve
of Fc
bran is transverse to Fu, then by the Poincaré–Bendixson Theorem, Fumusthavealeaf
which is a circle. As Fuis a foliation consisting of lines, this is a contradiction.
Since Fc
bran contains no periodic centre annuli, then Fc
"contains no Reeb components or
tannuli, so Fc
"is the lift of a suspension. This implies the following property.
Lemma 4.3 The leaf space R2/Fc
"of Fc
"is homeomorphic to R. Moreover, the correspond-
ing homeomorphism can be chosen to satisfy the following property: There exists a deck
transformation τ∈Z2such that if Lccorresponds to l ∈Rin the leaf space, then τn(Lc)
corresponds to l +n.
We will frequently identify a leaf in Fc
"with its corresponding representative in R.
For a set X⊂R2and η>0, define Uη(X)={x∈R2:dist(x,X)<η}.
Lemma 4.4 If L∈Fc
", then each of the connected components of Uη(L)\Lcontains a leaf
of Fc
".
Proof The only foliations lifted from T2which do not satisfy the desired property are those
which admit Reeb components and tannuli. If Fc
bran contained either of these, fwould admit
a periodic centre annulus.
ACu-curve is a curve tangent to the cone family Cu; we will use Cu-curves instead of
curves tangent to an unstable direction to utilise the compactness of T2. Since the unstable
cone and centre direction are a bounded angle apart, then a Cu-curve must ‘progress’ some
amount in the leaf space of Fc
". This is captured in the following lemma.
Lemma 4.5 There is δ>0such that if J uis a Cu-curve of length 1with an endpoint on L,
then the endpoints of J ulie on leaves which are at least δapart in the leaf space of Fc
".
Proof We begin by establishing the property for a single leaf. Let η>0 be small. Since Ec
and Cuare a bounded angle away from each other, if p∈Land Juis a unit length Cu-curve
with pas an endpoint, then Juis not contained within Uη(L). By Lemma 4.4,Juintersects
a leaf ˆ
L.Ifˆ
lcorresponds to L, then define δl=|
ˆ
l−l|.
Now consider [0,1]⊂Rin the leaf space. Since [0,1]is compact, the existence of δlfor
each l∈[0,1]implies that there is δ>0 such that all leaves in [0,1]satisfy the property in
the proposition. Now suppose that L∈Fcand that Juis a Cu-curve of unit length with an
endpoint on L.Ifτis as in Lemma 4.3,thenτnL∈[0,1]for some n, so the claim holds for
τnL. Since the unstable cone commutes with deck transformations, τn(Ju)is a Cu-curve of
unit length with an endpoint on τnL, so the property holds for L.
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29 Page 14 of 19 Geometriae Dedicata (2022) 216:29
This progression of unstable curves in the leaf space of Fc
"then implies that large unstable
curves will be large in the πu-direction.
Lemma 4.6 If q ∈Fu(p)and q = p, then supn|πufn(p)−πufn(q)|=∞.
Proof Recall that the leaves of Fc
"are uniformly bounded in the πu-direction. This implies
that for any constant D>0, there is d>0 such that if pand qare on leaves that are a
distance dapart in the leaf space, then |πu(p)−πu(q)|>D.Let Jube an unstable curve
with endpoints pand q. For large enough n, the iterate fn(Ju)has length greater than d/δ.
We can then decompose fn(Ju)into at least d/δ subcurves of unit length. Each subcurve,
by Lemma 4.5, progresses δin the leaf space of Fc
".Since fn(Ju)cannot intersect a leaf of
Fc
"twice, then fn(p)and fn(q)must lie on leaves which are at least a distance dapart in the
leaf space. Hence |πufn(p)−πufn(q)|>D.SinceDwas arbitrary, the claims follows.
Now given two leaves of Fc
bran which are connected by an unstable curve, the preceding
lemma implies that forward iterates of these two leaves under fmust grow apart in the
πu-direction. We use this argument to show that leaves of Fc
bran do not intersect, establishing
coherence.
Proposition 4.7 Let f0:T2→T2be a partially hyperbolic endomorphism. If f0does not
admit a periodic centre annulus, then it is dynamically coherent.
Proof Suppose there are two leaves L1,L2∈Fc
bran which coincide at a point p∈R2. Con-
sider the intervals πufn(L1)and πufn(L2). Since both of these intervals contain πufn(p),
their union is an interval, Lemma 4.1 implies that the length of this interval is at most 2C.
In particular, if q1∈L1and q2∈L2are points connected by an unstable segment, then
|πufn(q1)−πufn(q2)|is bounded. This contradicts Lemma 4.6.
The branching foliation Fc
bran gives a partition of T2tangent to Ec, and therefore is
actually a genuine foliation Fcby Remark 1.10 in [3]. This foliation is f-invariant. Since
Fc
bran descends to a branching foliation of T2,thenFcdescends to the desired foliation on
T2. Hence f0is dynamically coherent.
Let Fcbe the invariant centre foliation from the preceding proof.
Proposition 4.8 The foliations Fcand Fuof R2have global product structure.
Proof Proposition 4.2 gives uniqueness of an intersection between leaves of the foliations.
Observe that since Fchas no Reeb components, then πc(Lc)=Rfor all Lc∈Fc.One
can use Lemma 4.5 to show that if Lu∈Fu,thenπu(Lu)=R.ByanIntermediateValue
Theorem argument, Lcmust intersect Lu.
With dynamical coherence established, we complete the proof of Theorem Bby construct-
ing a leaf conjugacy from f0to A. We note that this construction was done for the case of
a hyperbolic linearisation in Section 3 of [6], which made use of the Franks semiconjugacy.
However, there is a subtle issue in the proof of Theorem B of [6] pointed out to us by Marisa
Cantarino and which we clarify here. If f0:T2→T2is an endomorphism whose lineariza-
tion A:T2→T2is hyperbolic, then there is a semiconjugacy H:R2→R2between lifted
maps on R2, but the semiconjugacy does not necessarily quotient down to T2.ThatHexists
and is a finite distance from the identity on R2can be seen by adapting the proof of (2.2) in
[5]. A full statement and proof is also given as Theorem 8.2.1 in [1]. The semiconjugacy is
known to descend in the case that Ais expanding due to the work in [15], a fact we will use
later in Sect. 5.
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Theorem B of [6] still holds, as to establish it, we only need the following property of Hon
the universal cover: if L⊂R2is a centre leaf of the lifted map f,thenH(L+z)=H(L)+z
for any z∈Z2. To see this, one can use an argument similar to that which we shall use in the
proposition that follows this discussion, Proposition 4.9.
In the context of this paper, we do not use the semiconjugacy itself to construct the leaf
conjugacy between fand A. Instead, we use a map that arises using the semiconjugacy
construction of [5], which has only the key properties we need for the averaging procedure.
Proposition 4.9 There exists a unique map H c:R2→Acwhich sends a point p ∈R2to
the unique line in L ∈Acwhich satisfies
sup
n≥0
dist(fn(p), An(L)) < ∞.
This map H csatisfies Hcf=AHc, where A is the action of the map on its invariant linear
foliation Ac, and the induced map R2→R2/Acis continuous. Moreover, H ccommutes
with deck transformations.
Proof This map can be constructed by using the arguments used to build the Franks semicon-
jugacy in the proof of (2.2) in [5]. The proof relies only on the fact that fand Ainduce the
same homomorphism on π1(T2), and that the greatest eigenvalue of Ahas magnitude greater
than 1. This is the case for all partially hyperbolic surface endomorphisms. The argument is
also sketched in Theorem 3.1 of [10].
The fact that Hccommutes with deck transformations is the only property listed that is
not given immediately by the construction. To see that it holds, note that since
An(Hc(L)+z)=AnHc(L)+Anz
and
fn(L+z)=fn(L)+Anz,
it follows that An(Hc(L)+z)stays a uniformly bounded distance away from fn(L+z)
as n→+∞.By uniqueness, Hc(L+z)=Hc(L)+z. Here, we used the property that
f(x+z)=f(x)+A(z)which holds because Ais the linear part of f.
Now we establish properties of Hcto show it is suitable for using an averaging technique.
Lemma 4.10 We have Hc(q)=Hc(p)if and only if q ∈Fc(p).
Proof First, suppose that q∈Fc(p).Then|πufn(p)−πufn(q)|<C.ThenifL∈Ac
is such that dist(fn(p), An(L)) is bounded, then so is dist(fn(q), An(L)). Hence Hc(p)=
Hc(q).
Now suppose that Hc(q)=Hc(p)but that q/∈Fc(p).Then fn(p)and fn(q)must both
lie close to An(L)for some L∈Ac. This implies that |πufn(p)−πufn(q)|is bounded. By
Lemma 4.8 we have global product structure, so there exists a point ˆq∈Fu(p)∩Fc(q).Since
centre leaves are bounded in the πc-direction, then |πufn(ˆq)−πufn(q)|must bounded, so
that |πufn(p)−πufn(ˆq)|is also bounded. This contradicts that ˆq∈Fu(p),sowehave
q∈Fc(p).
For q∈Fc(p),letdc(p,q)be the distance of the leaf segment from pto q.
Lemma 4.11 There exists T >0such that if q ∈Fc(p)and dc(p,q)>T, then |πc(p)−
πc(q)|>1.
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Proof This follows immediately from the fact that Fcis equivalent to the suspension of a
circle homeomorphism with the same slope as Ac.
Finally, we construct the leaf conjugacy.
Proof of Theorem BFix L∈Fcand let α:R→R2be an arc length parametrisation of L.
For p∈L,lets=α−1(p).LetTbe as in Lemma 4.11 and define h(p)as the unique point
in Hc(L)which satisfies
πch(p)=1
TT
0
πcα(s+t)dt.
Define hon each leaf of Fcto obtain a map h:R2→R2.Thismaphis a homeomorphism,
and descends to the desired leaf conjugacy on T2. The details of why is the case follow an
argument are identical to that of Section 3 and the proof of Theorem B in [6], where the map
Hctakes the place of the semiconjugacy H, with Proposition 3.2 and Lemma 3.3 replaced
by Lemma 4.10 and Lemma 4.11 of the current paper.
5 Maps with periodic centre annuli
With the preceding section having proven a classification in the absence of a periodic centre
annulus, we now classify in the presence of such annuli. That is, we prove Theorems Dand
E.Let f:T2→T2be partially hyperbolic surface endomorphism which admits a periodic
centre annulus.
Central to our classification is the following proposition, which states that the dynamics
on a periodic centre annulus take the form of a skew product.
Proposition 5.1 Suppose f is a partially hyperbolic endomorphism of a closed, oriented
surface M and that there is an invariant annulus M0with the following properties:
1. f (M0)=M0and f restricted to M0is a covering map;
2. the boundary components of M0are circles tangent to the centre direction.
3. no circle tangent to the centre direction intersects the interior U0of M0.
Then, there is an embedding h :U0→S1×Rsuch that the covering map h ◦fk◦h−1
from h(U0)to itself is of the form
hfkh−1(v , s)=(A(v), φ(v , s))
where A : S1→S1is an expanding linear map and φ:h(U0)→Ris continuous. Moreover,
if v∈S1,thenh
−1(v ×R)is a curve tangent to E c.
This result is an analogue of the classification of dynamics of diffeomorphisms with
centre-stable tori. In fact, the preceding theorem may be proved by adjusting each of the
arguments used in [11] to the current setting, where one replaces the notion of a region
between centre-stable tori with a periodic centre annulus, both of which take the role of
M0in the statement above. Since this is long process that does not use any new ideas, it is
completed in an auxiliary document [7].
With the dynamics inside a periodic centre annulus understood, we turn to outside of the
annulus, and in particular the preimages of these annuli. Let Fc
bran be an invariant branching
foliation which contains all the periodic centre annuli, as in Lemma 2.3.Let A:T2→T2
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Geometriae Dedicata (2022) 216:29 Page 17 of 19 29
be the linearisation of f. Then the lift of Fc
bran to R2lies close to the lift of some A-invariant
linear foliation A.Letλ1be the eigenvalue of Aassociated to the foliation A,andλ2the
other (not necessarily distinct) eigenvalue of A. Recall from Sect. 3that we may assume that
λ1,λ
2are both real and positive, and that one of these eigenvalues is greater than 1.
Lemma 5.2 Let S∈Fc
bran be a centre circle. Then the preimage f −1(S)consists of λ2
disjoint circles in Fc
bran.
Proof Since fis a covering map, f−1(S)consists of circles, each of which is mapped onto S.
These circles cannot be null-homotopic since they are tangent to the centre. Let S⊂f−1(S)
be one of these circles. Then since f(S)=Shas the same slope as Aand fis a finite distance
from A,thenSis also has the same slope as A. Since the induced homomorphism of fon
the fundamental group of T2is the same as that of A,then f|S:S→Sis a map of degree
λ1.Thenas fhas degree λ1λ2,theset f−1(S)must consist of λ2distinct circles.
The remaining ingredient for our two classification theorems is that the complement of
the orbits of periodic centre annuli consists of circles.
Proposition 5.3 Suppose that f admits a periodic centre annulus. Let the open sets
X1,...,Xn⊂T2be the distinct, disjoint periodic centre annuli of f . The connected com-
ponents of the set
V=T2\
i
k∈Z
fk(Xi)
are circles.
Proof Let Fc
bran be an invariant centre branching foliation which contains the boundaries of
all the periodic centre annuli as leaves. As Vis the complement of the preimages of these
annuli, then Vis saturated by leaves of Fc
bran. A property of rationally sloped foliations on T2
is that if a leaf is not a circle, then it lies inside a tannulus or Reeb component. By Lemma 2.5,
the non-circle leaves then lie only in periodic centre annuli or their preimages. Moreover,
by a Haefliger compact leaf argument, the limits of accumulating boundary circles in the
preimages of periodic centre annuli is also a circle. This ensures that Vwill be both a set
saturated by leaves, and these leaves must be circles.
Let Y⊂Vbe a connected component, and suppose that Yhas non-empty interior. Since
Yis saturated by leaves of Fc
bran,thenYcan a priori be either an annulus or a ‘pinched
annulus’, i.e., a deformation retract of an annulus. Let the width of Ybe the Hausdorff
distance between its boundary circles, as we used in Sect. 2. Then arguing similarly to
Lemma 2.5, for all sufficiently large k,fk(Y)has width bounded from below. Then fn(Y)
must be pre-periodic, so that Vcontains a periodic centre annulus, contradicting the definition
of V.
We consider now the setting of a non-hyperbolic linearisation and aim to prove Theorem
D. This relies primarily on a length versus volume argument.
Lemma 5.4 If A is non-hyperbolic, a periodic centre annulus has the slope associated to the
eigenvalue λof A for which λ>1.
Proof Let X0⊂T2be a periodic centre annulus. Assume without loss of generality that X0
is invariant and lift X0to a strip X⊂R2. By making an affine change of coordinates on R2,
we can assume that the linearisation takes the form A:(x,y)→ (λ1x,λ
2y)where λ1and λ2
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29 Page 18 of 19 Geometriae Dedicata (2022) 216:29
are the eigenvalues of A,andthat X=˜
f(X)⊂R×[−C,C]for a constant C>0. As Ais
non-hyperbolic, as defined in the introduction, we either have λ1>λ
2=1or1=λ1<λ
2.
We now assume that λ1=1 holds and derive a contradiction.
Since ˜
fis a finite distance from A, it follows that if Jis a bounded subset of R2,then
the diameter of ˜
fn(J)∩(R×[−C,C])grows linearly with respect to n. In particular, if J
is an unstable segment inside X, then the diameter of ˜
fn(J)grows linearly, contradicting
Proposition 2.1.
This allows us to conclude the classification in the case of a non-hyperbolic linearisation.
Proof of Theorem DBy Lemmas 5.2 and 5.4, the preimage of the boundary circle of a periodic
centre annulus is a single circle. In turn, the preimage of a single periodic centre annulus
is a single annulus, which by finiteness of the periodic centre must annuli also be periodic.
By Lemma 5.3, the finitely many periodic centre annuli together with their boundary circles
cover T2.
Now we complete the classification in the case of an expanding linearisation.
Proof of Theorem EFirst we consider the orbits of annuli. As both eigenvalues of Aare larger
than 1, Lemma 5.2 shows that the preimage of a periodic centre annulus is necessarily multiple
disjoint annuli. By continually iterating backwards, we see that the orbit of a periodic centre
annulus consists of infinitely many annuli. By Lemma 5.3, the union of these orbits, ,must
be dense, as otherwise its complement would have non-empty interior and not consist of
disjoint circles.
Next, we turn to how orbits of the annuli correspond under the semiconjugacy. As stated
in the introduction, since Ais expanding, we have a unique semiconjugacy H:T2→T2
from fto A. Observe that points within a given periodic centre annulus remain a bounded
distance apart when iterated on the universal cover, implying that the image of such an annulus
under the semiconjugacy His necessarily a circle in an A-invariant linear foliation A.By
the semiconjugacy property, this circle will periodic under A. Thus, a connected component
of is mapped to a circle which is either periodic or preperiodic under A. Similarly, a circle
S∈T2\will be mapped by Hto a circle in A. This establishes the first two properties of
the theorem.
Finally, suppose that fadmits an invariant unstable direction Euon T2.Let X⊂T2be a
periodic centre annulus. By Lemma 5.2, preimages of Xcontain annuli of arbitrarily small
width. In particular, there is n>0 and an annulus Y⊂f−n(X)with a width small enough
to ensure that any unstable curve contained in Ymust have length less than 1. However,
since Xis periodic, it contains unstable curves of arbitrarily large length. Then there exists
an unstable curve Ju⊂Xsuch that any component of f−n(Ju)has length greater than 1, a
contradiction. Here, we are using that Euis invariant, so that the preimage of a curve tangent
to Euis also tangent to Eu. This property does not hold for curves tangent to a cone family.
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