A CLASSIFICATION OF BRAID TYPES FOR PERIODIC ORBITS OF DIFFEOMORPHISMS OF SURFACES OF GENUS ONE WITH TOPOLOGICAL ENTROPY ZERO

GUASCHI, J . LLIBRE AND R.S . MACKAY We classify the braid types that can occur for finite unions of periodic orbits of diffeomorphisms of surfaces of genus one with zero topological entropy .


. Introduction
In this paper we classify the braid types that can occur for finite unions of periodic orbits of diffeomorphisms of surfaces of genus one with zero topological entropy.This extends the analysis from the case of genus zero [LMI] .The case of most interest to us is diffeomorphisms of the torus, isotopic to the identity.This is relevant to the behaviour of three coupled oscillators, for example .A good picture of their dynamics is developing [KMG], [LM2], [MZ], [H2], [F], [BGKM] .We hope that our results will helo solve the intriguing problem of understanding the boundary of zero topological entropy in the space of C 1 diffeomorphisms of the torus .
We begin by establishing some notation and recalling the definition of braid type.Let f : X -> X be a diffeomorphism of an oriented manifold X .Write h(f) for the topological entropy of f .We write o -p, o -r for orientationpreserving and reversing, respectively.Given two diffeomorphisms f : X -; X, g : Y ~Y of oriented manifolds, we write f -g if there exists an conjugacy between them.
Let M be a súrface, i.e. a compact connected oriented f : M -> M be a diffeomorphism of M and let P be a finite orbits for f .Then we define f-p : Mp , M-p by removing P from M and recompactifying by replacing the points of P by circles on which fP is the projective action of Df [B] .o-p 2-manifold .Let union of periodic Given two diffeomorphisms f, g : X -> X of a surface X and finite unions P, Q of periodic orbits for f, g respectively, we say the pairs (P, f) and (2, g) have the same braid type if there exists an o-p homeomorphism k : XP -XQ such that kfP k-1 is isotopic to gQ .The equivalence class of (P, f), denoted [P, f], is called its braid type .
To specify the braid type of a pair (P, f) we will use Nielsen-Thurston theory to select a simplest representative of the equivalence class.We refer the reader to [LM1] for the necessary information about Nielsen-Thurston theory of classification of surface homeomorphisms up to isotopy, and the definitions of the classes of diffeomorphisms which we call dise trees, reversing disc trees and reversing annulus trees.
The plan of the paper is as follows.In Section 2, we classify finite order homeomorphisms of the torus, which is an important preliminary result.Then in Section 3, we use Nielsen-Thurston theory to isotopc fP to a standard form, and analyse the possibilities .Our main result is Theorem 4, but because it is rather long to state and requires notions introduced in Section 2, we leave its statement until Section 3. In Section 4 we rederive results of [H1], [H2], and [LM2] as corollaries of those of Section 3.
We thank the referee for a careful reading .The first two authors were partially supported by a SERC studentship and a DGICYT grant respectively.

Classification of finite order homeomorphisms of the torus
In this section, we give a classification of finite order homeomorphisms of the torus up to o -p conjugacy.Let Tx , x E R2, be translation by x on R2, Le.Tx (y) = y + x, y E ff82 .Let R4 w E R/27r7L be rotation about 0 on R2 by angle w.Let r be the reflection { x / = x r y, _ -y on IE82.Define ro to be the group generated by Til ol and Tlo,ll, Fo that generated by T(I,o) and T(2, ~) .If f : R2 _ R2 commutes with the action of a group I' on R2, we define f/I' : R2 /I' , R2 /I' by identifying points of the same orbit under I'.For q E NN, let w be the quotient of Z under the equivalence relation generated by the relations pp + q, p --p.
Theorem 1.If f : T2 -> T2 is a homeomorphism of finite order, then it is o -p conjugate to one of the following: (a) T(. oyq/I'o for some q E NI, p E íq, with p, q having no common factor (order q) .
(c) r o T( p o)1 q/Fo, for some q E N1, p E with p, q having no common factor (order q if q is even, order 2q if q is odd) .
No two of the aboye are o -p conjugate.
The proof of Theorem 1 reduces to the classification of isometries of tori, a well-studied problem (e.g.see [NS]) .However, we have not found a compact treatment in the literature, nor one which considers the question of which isometries are equivalent up to o -p conjugacy.So we give a derivation in the Appendix .
For case (b) of Theorem 1, the numbers of periodie orbits of smaller period than the order are given by the following theorem.
Theorem 2 ( [E]) .The rotation of order 2 has 4 fixed points .The rotations of order 3 have 3 fixed points .The rotations of order 4 have 2 fixed points and one orbit of period 2. The rotations of order 6 have one fixed point, one orbit of period 2, and one orbit of period 3. (See Figure 1) .

Results
We firstly recall some definitions and theorems that we will need.We then go on to state and prove our results .
Let f : M -> M be a homeomorphism of a surface of genus one, then its completion g : T2 _ T2 , is defined by considering M to be T2 minus a disjoint union of equal size dises Di , and extending glaD ; radially into Di [E] .If h(f) = 0, then h(g) = 0. We say a simple closed curve on M is rotational if it is homotopically non-trivial on T2 after filling in the holes.
We recall the following two results .
Theorem A ([LMl]) .Let f : X -> X be a homeomorphism of a surface of genus zero .So in both the o-p and o-r cases, if P is a finite union of periodic orbits for f, and fP does not have a boundary component of period 1 (o -p) or period 1 or 2 (o -r), we can always append a fixed point or period 2 orbit to P in order to achieve this.Theorem B ([LMl]) .Let f : X --> X be a diffeomorphism of a surface of genus zero, with h(f) = 0, and let P be a finite union of periodic orbits .
( Let f : M , M be a diffeomorphism of a surface of genus one with h(f) = 0, and let P be a finite union of periodic orbits for f.Let F be a Thurston canonical form for fP .Then from Thurston's classification of surface homeomorphisms, F is either reducible, or of finite order .First we consider the case that F has a rotational reducing curve.Theorem 3 .Suppose f : M --> M is a diffemorphism of a surface of genus one with h(f) = 0, and P is a finite union of periodic orbits for f .Let F be a Thurston canonical form for fv, such that F has a rotational reducing curve F, of period p. Remove the tubular neighbourhood of F and its images, to obtain a disjoint union of punctured annuli A i .Then (i) If all Aá have period p, then [P, f] has a representative which is p annuli, joined by generalised twists, permuted like a rotation, and if f is o -p or p is even there exists a dise tree d : X' ~X' such that FPI A jd or if f is o -r and p is odd there exists a reversing dise tree d' : X' -+ X' such that FPIAj -d' .(ii) Suppose some Ai has period q :~p (so q = p/2), then p = 2, and there are two such annuli Al, A 2 .If f is o -p, [P, f] has a representative which is two annuli, joined by generalised twists, and there exist dise trees Di : has a representative which is two annuli, joined by generalised twists, and for each of Al, A 2 there exist reversing dise trees D' : X' , X' such that FIAj -D' or reversing annulus trees Si : X --> X, such that FIAj -Si, according as FIAj has an invariant boundary component or fixed point, or not.
Proof.Suppose F has period p. Define A = {Fkr : 0 < k < p} .Then F permutes the elements of A. If we remove the tubular neighbourhood of F and its images, we obtain a disjoint union of annuli A i with holes, which are permuted.Then the Ai either have period p, or if p is even, some of the Ai may have period p/2 (Le. the two boundaries of Ai in A are interchanged by Fp/2).This gives two cases: (i) If the Ai have period p, F permutes the Ai, and there are two subcases : so by Theorem B, there exists a disc tree d : X' , X' such that FP IAj -d.
(b) If f is o -r and p is odd, then FPIAj is o -r, so by Theorem B, there exists a reversing disc tree d' : X' -X' such that FPI Aj -d' .(ii) Suppose p is even, and there exists some A i (without loss of generality i = 0) such that Ai has period p/2.Put q = p/2.Then we claim that q=1 .
To prove the claim, suppose Ao has period q > 1 .Write Ak = Fk(Ao), 1 < k < q.Consider the situation on the torus ; then each Ak has two boundary components rkll , rk2) E A, say, for 0 <_ k < q .Then for j =~k, I'~') ~rkl, i l = 1, 2, Le.no two distinct Ak have a boundary component in common .For suppose FjZI .
hence there are only two distinct elements of A, so q = 1, a contradiction .
Since the Ak have no boundary components in common, there exist precisely q tubular regions Bl, . . ., B9, such that each Bi lies between two Ak, Le. each Bi has two distinct elements of A as its (rotational) boundary components.Consider one such element of {Bi}91 , B, say.Then without loss of generality, it lies between Ao and Ak, for some k, and has boundary components róll, 1'kl) E A (see Figure 3(b)) .B is invariant under F2q .Since r has period 2q, B must have period either q or 2q .Suppose it has period q, then F9 interchanges its boundary components, soFkl) = F9 (rol» .But F9 interchanges the boundary components of Ao so F9 (Fó ') ) = roe) , therefore Ao and Ak have a boundary component in common, which we have shown does not occur.Hence B has period 2q .
However Fk (Ao) = Ak, fór k < q, and considering Fk +q(Ao) = Ak, if necessary, then Fk (r( 1» = Fkl), and Fk (ro21 ) = rk21 .But the elements of A are permuted like a rotation or rotation with reflection by F, since f is a diffeomorphism, hence Fk (Fkl» = ról), therefore Fk (B) = B, a contradiction, as B has period 2q .This proves the claim.So q = 1, and r has period 2, with A = {F, F(F)} .If we remove the tubular neighbourhoods of r and F(I'), we obtain two disjoint annuli A o and A1 .Then there are two subcases to consider: (a) If f is o -p, then FIAj (i = 0, 1) is o -p, so by Theorem B, [P, f] has a representative which is two annuli joined by generalised twists, such that FI Aj is a dise tree.(b) If f is r -o, then FEA¡ (i = 0,1) is o -r, so by Theorem B, ['P, f] has a representative which is two annuli joined by generalised twists, such that FI Aj is a reversing dise tree or a reversing annulus tree, according as f has an invariant boundary component or P contains a fixed point in Ai, or not .
This completes the proof of Theorem 3.
Next we consider the case where there is no rotational reducing curve.Either F is of finite order, or has a non-rotational reducing curve C, in which case we may remove the decomposition componente (of genus zero) of C and its images from M. So in both cases, we can find a unique decomposition component S of genus one, such that FAS is of finite order.Let G : T2 -> T2 be the completion of FI S. Then G is of finite order, so is o -p conjugate to one of the cases of Theorem 1.
This leads to the following theorem, which is our main result : Theorem 4 .Suppose F : M -M is a Thurston canonical form for a homeomorphism of a surface of genus one, with unique decomposition component S of genus one.Let G : T2 -> T2 be the completion of FAS.Then one of the following is true : (a) G -T(P.o)/q/ro, and all boundary componente of S have period q.F~ar- ther, for each orbit al , . . ., áq of boundary componente of S not in aM, there exists a dise tree d : X' , X' such that the component Xi of M\S inside Ói is homeomorphic to X', and F9 IXi -d.
(b) G -R,/I'o, w = f7r/2, 7r (order k = 4, 2), or G -R,/ro, w = for/3, ±27x/3 (order k = 6, 3) .If k = 2, then all boundary components of S have period 1 or 2, with at most 4 of period 1.If k = 3, then all boundary components of S Nave period 1 or 3, with at most 3 of period 1.If k = 4, then all boundary components of S have period 1, 2 or 4, with at most 2 of period 1, and 1 orbit of period 2. If k = 6, then all boundary components of S have period 1, 2, 3 or 6 with at most 1 of period 1, 1 orbit of period 2, and 1 orbit of period 3.In case, for each orbit of boundary components of period p, al, . . ., OP, of S, not in OM, there exists a dise tree d : X' -4 X' such that the component Xi of M\S inside ái is homeomorphic to X', and FPIXi -d.
(e) (i) G -(r o TOP o)/q)/Po .If q is even, all boundary components have period q.If q is odd, all boundary components have period q or 2q .
In both parts of case (c), for each orbit of boundary components of period p, 91 5 . . ., ap, of S not in OM, there exists a disc tree d : X' -X' such that the component Xi of M\S inside Vi is homeomorphic to X' and FPIXi -d if p is even, or there exists a reversing disc tree d' : X' -X' such that the component Xi of M\S inside al is homeomorphic to X' and FP I Xi -d if p is odd.
Case (a) G ^-TOP o)/q/I'o, so all points of T2 have period q under G. Hence all boundary components of S have period q, and if we consider the orbits al , . . . .0, of those not in aM, the corresponding decomposition components Xi of M\S inside al are of genus zero, and we may apply Theorem B, so there exists a disc tree d : X' -> X' such that Xi is homeomorphic to X', and The statements about the orbits of boundary components are an immediate consequence of applying Theorem 2 to the completion G : T2 -> T2 .As in case (a), since G is o-p, we obtain disc trees in each orbit of boundary components of S not in 8M .

Case (c) (i) G -(r o TO P o)/ a )/I'o .
From the proof of Theorem 1, the boundary components of S have period q if q is even, or they have period q or 2q if q is odd .
Again from the proof of Theorem 1, the boundary components of S have period 1 or 2.
In both parts of case (c), each component of T2\S inside a boundary component not in 8M of even period r contains a disc tree since Fr is o -p, whilst those of odd period r' contain a reversing disc tree since Fr' is o-r, by Theorem B.

Two Corollaries
As a corollary of the results of Section 3 we obtain the genus one case of [H1]: Theorem 5. Let f : M -M be an o-r diffeomorphism of a surface M of genus one.If f has periodic orbits, or orbits of boundary components, with 3 distinct odd peros, then h(f) > 0.
To derive this from the above, we require the genus zero result of [BF], [H1], also derived by [LM1] : Theorem C .Let f : X -X be an o -r diffeomorphism of a surface of genus zero .If f has periodic orbits or orbits of boundary components with two distinct odd peros, then h(f) > 0.
Proof of Theorem 5: Let f : M -~M be an o-r diffeomorphism of a surface M of genus one.Let P be the union of three orbits of distinct odd period .Let F be a Thurston canonical form for fP.Suppose h(f) --0.Then there are two cases: Case (a) : Suppose there is a rotational reducing curve, I', say, with period p. Remove its annular neighbourhood and its images, then we obtain a disjoint union of annuli Ai .From Theorem 3, there are two possibilities .In the first case, the Ai have period p, so the Ai are permuted by F, and thus p divides the order of each periodic orbit, hence p is odd.So if we consider any Ai, FP ¡A¡ is o -r, since f is o -r.Ai must contain boundary components corresponding to the three orbits of distinct odd period .So by Theorem C, h(f) > 0, a contradiction .
The other possibility is when p = 2, and there are two invariant annuli Ao, Al .Then one of Ao, Al must contain boundary components corresponding to at least two of the three orbits, and since FIAjj (i = 0,1) is o -r, Theorem C implies that h(f) > 0, a contradiction .all boundary components of S llave period q or 2q.If G -(r o R,/2)/I'o, then all boundary components of S llave period 1 or 2. In particular, all boundary components of S of odd period llave the same period p, and the remaining boundary components corresponding to P must lie within decomposition components Xi of genus zero whose outer boundaries have period p.Since FP I Xi is o-r, and at least one of the Xi contains boundary components of odd order, not p, corresponding to the orbits of P, then Theorem C implies that h(f ) > 0, a contradiction .
As a second corollary we will derive a result of [LM2], [H2] for diffeomorphisms of the torus isotopic to the identity.We refer the reader to [LM2] for definitions of lifts, rotation vectors, etc.We recall that for a continuous map f : T2 , T2 with lift f : R2 , If82, if I is the group of integer translations y, , : x 1--> x + m, x E R2, m E Z2, and if f is homotopic to the identity, then fy = -yf for all "y E F. Theorem 6.Let f : T2 -3 T2 be a homeomorphism of the torus isotopic to the identity, an.d suppose h(f) = 0. Then all rotation vectors associated with the periodic orbits of f are collinear.
Proof. .Suppose that f has a finite union of periodic orbits, with associated distinct rotation vectors pi/gi, i = 1, . . ., N. Then from [LM2], since f is homotopic to the identity, for each i E {1,. . ., N} there exists a periodic orbit Qá of primitive rotation type (pi, qi).Let P =UN 1 Qá, and let F be a Thurston canonical form for fP .
Suppose F is of finite order, then using (*) and Theorem 1, we see that F -T(P o)1e/I'o, for some q E N, p E w, so all points of T2 are periodic with period q and rotation vector (p, 0)/q .The remaining possibility is that F is reducible (with finite order components, though we shall not need this) .There are no non-rotational reducing curves for F. For suppose F were such a curve, then it must surround at least two holes, but they must come from the same orbit, so the rotation type of that orbit cannot be primitive, which is a contradiction .
Suppose there is a rotational reducing curve 1' for F. Let G : T2 -> T2 be the completion of F, and let G : R2 --> R2 be a lift of G. Let m E 71 2 \{0} be the homotopy type of 1', and q be its period .Then F lifts to an infinite set of curves F, each invariant under T,,, which partition the plane hito infinite strips S each within a bounded distance of some straight line of direction m.Furthermore, there exists p E 71 2 such that FqT F = 1', and since F is invertible the same holds for the strips S. Hence the rotation set of F, and in particular the rotation vectors of the chosen periodic orbits, are contained in the straight line {p/q + tm : t E IR} .
This completes the proof.

Appendix : Proof of Theorem 1
To prove Theorem 1, we require the following : Theorem 7 [E] .If f is a finite order homeomorphism of a compact orientable manifold M, then there exists a Riemannian metric R of constant curvature on M such that f is a diffeomorphism preserving R.
Let £ be the Euclidean metric on Rn, and let E' be the group of isometries of (IFB", £).It consists of all transformations y ~--> Ay -I-x, with x E Rn, A E O(n) .In the case n = 2, E2 is generated by Tx , R, and r, where x E R2, w E R/27rZ .Theorem 8 (Killing, Hopf, see [W]) .Let (M n , R) be a Riemannian manifold of dimension n >_ 2 with metric R. Then (Mn,7Z) is complete, connected and of constant curvature K = 0 if and only if it, is isometric to a quotient (Rn, £)/I', where I' is a subgroup of En which acts freely and properly discontinuously.
Theorem 9 ([S]) .Suppose R,, E SO(2), R4, =~Id, commutes with I' on R2, where F is a subgroup of E2 with two generators, and which acts freely and properly discontinuously.Then w = ±7r/3, ±7r/2, f2rr/3, or 7r, and I' is conjugate in the group generated by SO(2) and isotropic scale changes to FE] if w = ±7r/2 or 7r, or I'o if w = f7r/3 or f27r/3 .Proof of Theorem 1: Suppose f is a finite order homeomorphism of T2 .Theorem 7 implies that there exists a Riemannian metric R of constant curvature on T2 , such that f is a diffeomorphism and preserves R. By the Gauss-Bonnet formula (e.g.[DC]), the curvature is zero.By Theorem 8, (T 2 , R) is isometric to (R2, £)/I', for some I' < E2 .Since T2 is compact, I' must be a group generated by two translations Ty Ty2, with yI, y2 linearly independent, and without loss of generality, yI is along the x-axis (by rotating coordinates).To find the finite order isometries f : (T 2 , R) o, it suffices to find all the isometries f : (R2, £) ~, such that f I' = Ff, and f'' E I', for some m.If f is o -r, then r o f is o -p and satisfies the same conditions .So let us take f to be o -p.
If f has no fixed point, then it is a translation Tx , x E R2 .If f" E I then x is a rational combination of yI and y2 .Hence f -TP79 /Fo, for some p E Z2, q E IN.However, many of there are o -p conjugate .Let 9 = PigP2 , which we assume is in lowest terms .Suppose pI and p2 have no common factors.Consider the orbit of TP1g/Fo under SL(2, 7L) .Then for and Qpi e bP2 and ~pi e dP2 have no common factors .For Suppose apl + bp2 = knl, CPI + dp2 = kn2, for some integers nI, n2, k :,A ±1.Then P2] Pi n i = kA_I In2] but we assumed that pl and p2 were coprime, a contradiction .Now pI and p2 are coprime if and only if there exist integers a, b such that apl + bp2 = 1.Let c = -p2 and d = pi, then La dJ Conversely given coprime integers a, c we can find integers b, d satisfying adbc = 1, and Suppose pI and p2 are not coprime but have highest common factor hcf (PI, P2) = k .Then pi9PZ = k 9~, where m and n are coprime .Hence the orbit of Tp/ q/ro under SL(2, 7L) is all Tp , Ig l/ro, with q' = q and hcf (pi, p2) hcf (pl, p2), where p' = (pi,p2) .
Thus every element in this orbit is a translation by v (a, b), with k = hcf (P1, P2), and hcf (a, b) = 1.Thus we may choose a representative T(p,o)Iq/ro, with p E 7L 9, and p, q having no common factors.
If f has a fixed point, we may assume by making a translation if necessary that 0 is fixed.So Px) = Ax, for some A E SO(2) .The case of f = Id is included in part (a) .Otherwise, by Theorem 9, r is conjugate to ro or ro, and w = f7r/2, n in the case ro, ±21r/3, for/3 in the case I'o .
Thus when f is o -p, it is o -p conjugate to one of the cases given in parts (a) and (b) .It can be seen that no two of these cases are o -p conjugate .
If f is o -r, then it is o -p conjugate to the composition of r with one of the cases in parts (a) and (b) .However, some of these are o -p conjugate to each other, so they are not distinct cases .There are two classes to consider .
(i) f -g = (r-Tlp,ol/j/ro (Case (c)) .On T2 , there are two curves invariant under g; they are y = 0 and y =-2 .Under g, the x-coordinate has period q, and the y-coordinate has period 1 if y = 0 or 2, and period 2 otherwise .So if q is even all points in T2 have period q.If q is odd, all points in T2 have period q or 2q, according as they lie on an invariant curve or not .Thé invariant curves have rotation number p/q E i 1 , where iI stands for thé quotient of R under (ii) fg = (r o &J/ro, w = 7r, f7r/2, or f -g = (r o R,,,)/ro, w = tir/3, ±27r/3 .Without loss of generality (by conjugation by R_W ), we take w to be positive.
If w = 7r, then on ff8 2 this corresponds to perpendicular reflection in x = 0, so by a change of coordinates, f -r, which is already included in Case (c) .
If w = 7r/2 (Case (d)), then on R2 this corresponds to perpendicular reflection in the curve x + y = 0.It is the only invariant curve, and, is composed of fixed points .
If w = 7r/3, then on R2 this corresponds to perpendicular reflection in x + Ní3 -y = 0. On T2, there is one curve fixed pointwise by g, and g is a reflection in it .
If w = 27r/3, then on R2 this corresponds to perpendicular reflection in x + ~y = 0. On T2, x + ~y = 0 is the only curve fixed pointwise by g, and g is a reflection in it.
Let hl = rOR,y2/I'o, h2 = roR2,/3/Po .Then hl and h2 are o-p conjugate .To see this, we need to find an o-p homeomorphism k : R2 /I'o --+ R2 /1'o such that khl = h2k, or equivalently an o -p homeomorphism K : U82 -R2 such that KHl = H2K, where hl = Hl/ro and h2 = H2/I'o, and also such that for each -yo E I'o there exists a -yo E I'o such that K-yo = -yáK and vice versa .This is satisfied by The cases hl and r are distinct because the case r has two invariant curves, whilst the case hl has only invariant curve.
Case (b) : Suppose there is no rotational reducing curve.Then there exists a unique decomposition component S of genus one .Let G : T2 -> T2 be the completion of FAS.Then we are in Case (c) of Theorem 4. If G -(r-T(P o)M/I'o, the equivalence relation generated by x -x+ 1, x --x.The rotation number is an invariant of o -p conjugaty.Hence the cases in part (c) are distinct .
to the basis vectors of I'o), which is a shear taking the generators of PC] to FA .Hence hl and h2 are o-p conjugate .Similarly the case roR,y3/I'o is o -p conjugate to hl , by taking K= 0 2