Some virtually abelian subgroups of the group of analytic symplectic diffeomorphisms of $S^2$

We show that if $M$ is a compact oriented surface of genus 0 and $G$ is a subgroup of $\Symp^\omega_\mu(M)$ which has an infinite normal solvable subgroup, then $G$ is virtually abelian. In particular the centralizer of an infinite order $f \in \Symp^\omega_\mu(M)$ is virtually abelian. Another immediate corollary is that if $G$ is a solvable subgroup of $\Symp^\omega_\mu(M)$ then $G$ is virtually abelian. We also prove a special case of the Tits Alternative for subgroups of $\Symp^\omega_\mu(S^2).$


Introduction
If M is a compact connected oriented surface we let Diff 0 r (M ) denote the C r diffeomorphisms isotopic to the identity. In particular if r = ω this denotes the subgroup of those which are real analytic. Let Symp r µ (M ) denote the subgroup of Diff 0 r (M ) which preserve a smooth volume form µ. We will denote by Cent r (f ) and Cent r µ (f ) the subgroups of Diff 0 r (M ) and Symp r µ (M ), respectively, whose elements commute with f. If G is a subgroup of Symp r µ (M ) then Cent r µ (f, G) will denote the the subgroup of G whose elements commute with f. In this article we wish to address the algebraic structure of subgroups of Symp ∞ µ (M ) and our results are largely limited to the case of analytic diffeomorphisms when M has genus 0. Our approach is to understand the possible dynamic behavior of such diffeomorphisms and use the structure exhibited to conclude information about subgroups. Definition 1.1. If N is a connected manifold, an element f ∈ Diff r (N ) will be said to have full support provided N \ Fix(f ) is dense in N . We will say that f has support of finite type if both Fix(f ) and N \ Fix(f ) have finitely many components. We will say that a subgroup G of Diff r (N ) has full support of finite type if every non-trivial element of G has full support of finite type. Question 1.7. Suppose M is a compact surface and G is a subgroup of Diff 0 2 (M ) with full support of finite type and f ∈ G has infinite order. Is the centralizer of f in G always virtually abelian?
The following result gives a partial answer to this question. Proof. If we apply Theorem 1.5 to the group G = Cent ∞ µ (f, H) and observe that the cyclic group generated by f is a normal abelian subgroup, then we conclude that G is virtually abelian. Question 1.9 of [6] asks if any finite index subgroup of MCG(Σ g ) can act faithfully by area preserving diffeomorphisms on a closed surface. The following corollary of Proposition 1.8 and Proposition 1. 2 gives a partial answer in the special case of analytic diffeomorphisms and M of genus 0. Corollary 1.9. Suppose H is a finite index subgroup of MCG(Σ g ) with g ≥ 2 or Out(F n ) with n ≥ 3, and G is a group with the property that every element of infinite order has a virtually abelian centralizer. Then any homomorphism ν : H → G has non-trivial kernel. In particular this holds if G = Symp ω µ (S 2 ).
Proof. Suppose at first that H is a finite index subgroup of MCG(Σ g ) with g ≥ 2.
Let α 1 , α 2 and α 3 be simple closed curves in M such that α 1 and α 2 are disjoint from α 3 and intersect each other transversely in exactly one point. Let T i be a Dehn twist around α i with degree chosen so that T i ∈ H. Then T 1 and T 2 commute with T 3 but no finite index subgroup of the group generated by T 1 and T 2 is abelian. It follows that µ cannot be injective. Suppose now that H is a finite index subgroup of Out(F n ) with n ≥ 3. Let A, B and C be three of the generators of F n . Define Φ ∈ Aut(F n ) by B → BA, define Ψ ∈ Aut(F n ) by C → CBAB −1 A and define Θ ∈ Aut(F n ) by C → CABAB −1 , where Φ fixes all generators other than B and where Ψ and Θ fix all generators other than C. Let φ, ψ and θ be the elements of Out(F n ) determined by Φ, Ψ and Θ respectively. It is straightforward to check that Φ commutes with Ψ and Θ and that for all k > 0, Ψ k (which is defined by C → C(BAB −1 A) k ) does not commute with Θ k (which is defined by C → C(ABAB −1 ) k ). Moreover, [Θ k , Ψ k ] is not a non-trivial inner automorphism because it fixes both A and B. It follows that [θ k , ψ k ] is non-trivial for all k and hence that no finite index subgroup of the group generated by ψ and θ, and hence no finite index subgroup of Cent(φ), the centralizer of φ in Out(F n ), is abelian. The proof now concludes as above.
Another interesting consequence of Theorem 1.5 is the following result. Proposition 1. 10. Suppose M is a compact surface of genus 0 and G is a solvable subgroup of Symp ∞ µ (M ) with full support of finite type, then G is virtually abelian. The Tits alternative (see J. Tits [16]) is satisfied by a group G if every subgroup (or by some definitions every finitely generated subgroup) of G is either virtually solvable or contains a non-abelian free group. This is a deep property known for finitely generated linear groups, mapping class groups [9], [12], and the outer automorphism group of a free group [1] [2]. It is an important open question for Diff ω (S 1 ) (see [5]). It is known that this property is not satisfied for Diff ∞ (S 1 ) (again see [5]).
This naturally raises the question for surfaces.
Conjecture 1.11 (Tits alternative). If M is a compact surface then every finitely generated subgroup of Symp ω µ (M ) is either virtually solvable or contains a non-abelian free group.
In the final section of this paper we are able to use the techniques developed to prove an important special case of this conjecture. Theorem 1.12. Suppose that M is a compact oriented genus zero surface, that G is a subgroup of Symp ω µ (M ) and that G contains an infinite order element with entropy zero and at least three periodic points. Then either G contains a subgroup isomorphic to F 2 , the free group on two generators, or G has an abelian subgroup of finite index.
We observe that one cannot expect the virtually abelian (as opposed to solvable) conclusion to hold more generally as the subgroup of Symp ω µ (T 2 ) generated by an ergodic translation and the automorphism with matrix 1 1 0 1 is solvable, but not virtually abelian.
Definition 2.1. Suppose M is a compact oriented surface. The annular compactification of an open annulus U ⊂ M is obtained by blowup on an end whose frontier is a single point and by the prime end compactification otherwise. We will denote it by U c (see 2.7 [6] for details).
For any diffeomorphism h of M which leaves U invariant there is a canonical extension to a homeomorphism h c : U c → U c which is functorial in the sense that (hg) c = h c g c .
Definition 2.2. Suppose M is a compact genus zero surface, f ∈ Symp ∞ µ (M ), and that the number of periodic points of f is greater than the Euler characteristic of M . If f has infinite order and entropy 0, we will call it a multi-rotational diffeomorphism. This set of diffeomorphisms will be denoted Z(M ).
The rationale for the terminology multi-rotational comes from the following result which is a distillation of several results from [6] (see Theorems 1.2, 1.4 and 1.5 from that paper). In particular, if f ∈ Z(M ) then every point of int(M ) \ Fix(f ) has a well defined rotation number (with respect to any pair of components from ∂M ∪ Fix(f ) and there are non-trivial intervals of rotation numbers. (1) The elements of A are pairwise disjoint.
(4) For each U ∈ A, the rotation number ρ f : U c → S 1 is continuous and nonconstant. Each component of the level set of ρ f which is disjoint from ∂U c is essential in U , i.e. separates U c into two components, each containing a component of ∂U c .
We will make repeated use of the properties in the following straightforward corollary.
Corollary 2.4. Suppose M is a compact genus zero surface, f ∈ Z(M ) and Cent ∞ µ (f ) denotes its centralizer in Symp ∞ µ (M ).
(1) If g ∈ Cent ∞ µ (f ) and U ∈ A f then g(U ) ∈ A f and the Cent ∞ µ (f )-orbit of U is finite.
(2) If U ∈ A f then any g ∈ Cent ∞ µ (f ) which satisfies g(U ) = U must preserve the components of each level set of ρ f : U c → S 1 .
(3) If f has support of finite type then the set A f of maximal annuli for f is finite.
consists of pairwise disjoint annuli each with the same positive measure. There can only be finitely many of them in M. This proves (1).
To show (2) we observe that g is a topological conjugacy from f to itself and rotation number is a conjugacy invariant. Hence g must permute the components of the level sets of ρ f . Clearly those level sets which contain a component of ∂U c must be preserved. Since any other such component separates U , if it were not g-invariant the fact that g is area preserving would be contradicted.
Since the elements of A f are maximal f -invariant annuli in int(M ) \ Fix(f ) , they are mutually non-parallel in int(M ) \ Fix(f ) . Let E be the union of a set of core curves for some given finite subset of A f . Theorem 2.3-(3) implies that each component of M \ E either contains a component of Fix(f ) ∪ ∂M or has negative Euler characteristic. The dual graph for E is a tree that has one edge for each element of E and has at most as many valence one and valence two vertices as there are components of Fix(f ) ∪ ∂M . Since the Euler characteristic of the tree is 1 there is a uniform bound on the number of its edges. It follows that the cardinality of E, and hence the number of elements of A f , is uniformly bounded. This proves (3).
3 The positive entropy case. Lemma 3.1. Suppose that M is a closed surface, that f ∈ Diff 2 (M ) and that g commutes with f . Suppose further that f has a hyperbolic fixed point p of saddle type, and that g fixes p and preserves the branches of W s (p, f ). Then there is a C 1 coordinate function t on W s (p, f ) and a unique number α > 0 such that in these coordinates g(t) = αt. In particular α is an eigenvalue of Dg p .
Proof. By Sternberg linearization (see Theorem 2 of [15]) there is a C 1 coordinate function t(x) on W s (p, f ) in which the restriction of f (also denoted f ) satisfies f (t) = λt where λ ∈ (0, 1) is the eigenvalue of Df p corresponding to the eigenvector tangent to W s (p, f ).
In these coordinates g(t) = λ −n g(λ n t). Applying the chain rule gives g (t) = g (λ n t). Letting n tend to infinity we get g (t) = g (0) so g (x) is constant and g(t) = αt where α = g (0).

Definition 3.2.
Suppose that M is a compact surface and that f ∈ Diff r (M ) has a hyperbolic fixed point p of saddle type. We define Cent r p (f ) to be the subgroup of elements of Diff r (M ) which commute with f , fix the point p, and preserve the branches of W s (f, p). Let R + denote the multiplicative group of positive elements of R. The expansion factor homomorphism is defined by φ(g) = α where α is the unique number given by Lemma (3.1) for which g(x) = αx in C 1 coordinates on W s (f, p). It is immediate that φ is actually a homomorphism. We also observe that φ(g) is just the eigenvalue of Dg p whose eigenvector is tangent to W s (f, p).
The following result is due to Katok [11] who stated it only in the analytic case, but gave a proof very similar to the one below. [11]]). Suppose G is a subgroup of Diff 2 (M ) which has full support of finite type and f ∈ Diff 2 (M ) has positive topological entropy. Then the centralizer of f in G, Cent 2 (f, G), is virtually cyclic. Moreover, every infinite order element of Cent 2 (f, G) has positive topological entropy.
Proof. By a result of Katok [10] there is a hyperbolic periodic point p for f of saddle type with a transverse homoclinic point q. Let φ : Cent 2 p (f ) → R + be the expansion factor homomorphism of Definition 3.2 and let H = Cent 2 p (f ) ∩ G. Then H has finite index in Cent 2 (f, G) because otherwise the Cent 2 (f, G) orbit of p is infinite and consists of hyperbolic fixed points of f all with the same eigenvalues. This is impossible because a limit point would be a non-isolated hyperbolic fixed point. For the main statement of the proposition, it suffices to show that H is cyclic and for this it suffices to show that the restriction φ| H : H → R + is injective and has a discrete image.
To show that φ| H is injective it suffices to show that if g ∈ H and φ(g) = 1 then g = id. But φ(g) = 1 implies W s (p, f ) ⊂ Fix(g). Note that if φ : Cent 2 p (f −1 ) → R + is the expansion homomorphism for f −1 then φ(g)φ (g) = det(Dg p ) = 1 so we also know that W u (p, f ) ⊂ Fix(g). Hence we need only show that this implies g = id. Let J s be the interval in W s (p, f ) joining p to q and define J u ⊂ W u (p, f ) analogously.
. Then the number of components of the complement of K n tends to +∞ with n. Since g has full support of finite type, each component V of the complement of K n contains a component of M \ Fix(g), in contradiction to the fact that M \ Fix(g) has only finitely many components. We conclude that φ is injective.
To show that φ| H has discrete image we assume that there is a sequence {g n } of elements of H such that lim φ(g n ) = 1, and show that φ(g n ) = 1 for n sufficiently large. Let I s = f −1 (J s ) and I u = f (J u ). Since I s and I u are compact intervals and the point q is a point in the interior of each where they intersect transversely, there is a neighborhood U of q such that U ∩ I s ∩ I u = {q}. But for n sufficiently large g n (q) ∈ U and g n (q) ∈ (I s ∩ I u ). It follows that g n (q) = q for n sufficiently large and hence φ(g n ) = 1. This proves that the image of φ H is discrete and hence that H is cyclic.
Each non-trivial element h ∈ H has φ(h) = 1 and φ (h) = 1. Hence p is a hyperbolic fixed point of h and q is a transverse homoclinic point for h. It follows that h has positive entropy (see Theorem 5.5 of [14]). Proposition 3.3. Suppose G is a subgroup of Diff 2 (M 2 ) which has full support of finite type and A is an abelian normal subgroup of G. If there is an element f ∈ A with positive topological entropy then G is virtually cyclic.
Proof. It follows from Proposition 1.3 that the group A is virtually cyclic. Since f ∈ has positive entropy it is infinite order and generates a finite index subgroup A f of A. Hence there exists a positive integer k such that a k ∈ A f for all a ∈ A. In particular, for each g ∈ G, we have gf k g −1 = (gf g −1 ) k = f m for some m ∈ Z . Since f and gf g −1 are conjugate, the topological entropy ent(f ) = ent(gf g −1 ). Hence ent(gf g −1 ) k = |k| ent(gf g −1 ) = |k| ent(f ), and ent(f m ) = |m| ent(f ).
We conclude that m = ±k and hence that gf k g −1 = f ±k for all g ∈ G. Let G 0 be the subgroup of index at most two of G such that gf k g −1 = f k for all g ∈ G 0 . Then G 0 ⊂ Cent 2 (f k , G) and Proposition 1.3 completes the proof.

Mean Rotation Numbers
In this section we record some facts (mostly well known) about rotation numbers for homeomorphisms of area preserving homeomorphisms of the closed annulus. In subsequent sections we will want to apply these results when M is a surface, f ∈ Symp ∞ µ (M ), and U is an f -invariant open annulus. We will do this by considering the extension of f to the closed annulus U c which is the annular compactification of U . When the annulus U is understood, we will use ρ f (x) to mean the rotation number of x ∈ U with respect to the homeomorphism f c : U c → U c of the closed annulus U c .
wherex is a lift of x and p 1 :Ã =R × [0, 1] → R is projection onto the first factor. As reflected in the notation, ∆f (x) is independent of the choice of liftx and hence may be considered as being defined on A.
If X ⊂ A is an f -invariant µ-measurable set then the mean translation number relative to X andf is defined to be We define the mean rotation number relative to X, ρ µ (f, X) to be the coset of T µ (f , X) mod Z thought of as an element of T = R/Z.
The mean rotation number is independent of the choice of liftf since different lifts give values of T (f , X) differing by an element of Z.
We define the translation number of x with respect tof (see e.g. Definition 2.1 of [6]), by A straightforward application of the Birkhoff ergodic theorem implies that τf (x) is well defined for almost all x and If X is also g-invariant and g preserves µ then and hence integrating we obtain Proof. Replacing X with X ∩ int(A) we may assume X ⊂ int(A). If τf (x) = 0 on a positive measure subset of X then the fixed point exists by Proposition 2.4 of [6]. Otherwise for some > 0 there is a positive measure f -invariant subset X + on which ρ f > and another X − on which ρ f < − . If x ∈ X + is recurrent and not fixed then it is contained in a positively recurring disk which is disjoint from its f -image.
Similarly if y ∈ X − is recurrent and not fixed then it is contained in a negatively recurring disk which is disjoint from its f -image. Theorem 2.1 of [7]   Proof.
Replacing f with f m it will suffice to show that f (V ) = V implies ρ f = 0 on a full measure subset of V . Since V does not contain a simple closed curve that separates the components of ∂A, both components of ∂A belong to the same component X of A \ V by Lemma 3.1 of [6]. Note that D := A \ X contains V , is contained in int(A) and is f -invariant. Lemma 3.2 of [6] implies that D is connected and it is simply connected since its complement X is connected. Hence D is an open disk. By the Brouwer plane translation theorem D contains a point of Fix(f ). LetÃ be the universal cover of A and letD ⊂Ã be an open disk which is a lift of D. Since f has a fixed point in D there is a liftf :Ã →Ã which has a fixed point inD. Thereforef (D) =D. If p 1 (D) is bounded then it is obvious that ρ f is constant and 0 on D. Otherwise, note that by Poincaré recurrence and the Birkhoff ergodic theorem there is a full measure subsetW ofD consisting of points which are recurrent and have a well defined translation number. Calculating the translation number of a point x ∈W on a subsequence of iterates which converges to a point ofÃ shows that it must be 0. Hence W , the projection ofW to D, has the desired properties.
, is a subgroup G with the property that every nontrivial element of G has exactly χ(M ) fixed points. An element f ∈ Symp r µ (M ) will be called a pseudo-rotation provided the cyclic group it generates is a pseudo-rotation group.
Our definition may be slightly non-standard in that we consider finite order elements of Symp r µ (M ) to be pseudo-rotations. We observe that if χ(M ) < 0 then any pseudo-rotation group is trivial. Since we assume M is oriented we have only the cases that M is A, D 2 , or S 2 for which any pseudo-rotations must have precisely 0, 1, or 2 fixed points respectively. By far the most interesting case is M = S 2 , since we will show in Lemma 5.7 that when M = D 2 or A 2 any pseudo-rotation group is abelian. This is not the case when M = S 2 since SO(3) acting on the unit sphere in R 3 is a pseudo-rotation group.
There are three immediate but very useful facts about pseudo-rotations which we summarize in the following Lemma: (1) Either f has finite order or Fix(f ) = Per(f ).
(2) If Fix(f ) contains more than χ(M ) points then f = id. In particular if f, g are elements of a pseudo-rotation group which agree at more than χ(M ) points then f g −1 = id so f = g.
(3) If M = D 2 , then the one fixed point of f is in the interior of D 2 .
Proof. Parts (1) and (2) are immediate from the definition of pseudo-rotation and part (3) holds because otherwise the restriction of f to the interior would have recurrent points but no fixed point, contradicting the Brouwer plane translation theorem.
We will make repeated use of these properties.
is an abelian pseudo-rotation group, with Fix(g) = Fix(f ) for all g ∈ G, then the assignment g → ρ g , for g ∈ G, is an injective homomorphism, so, in particular, if g has infinite order then the constant ρ g is irrational.
Proof. The only possibilities for M are S 2 , A or D 2 . If M = D 2 , as remarked above, its one fixed point must be in the interior of M . Hence in all cases U is an open annulus. If ρ f is not constant on U c then the Poincaré-Birkhoff Theorem (see Theorem (2.2) of [6] for example) implies there are periodic points with arbitrarily large period. We may therefore conclude that ρ f is constant and equal to ρ µ (f ). Suppose now that G is an abelian pseudo-rotation subgroup containing f and Fix(f ) = Fix(g) for all g ∈ G, so U is G-invariant. Since ρ µ : G → T is a homomorphism ρ µ (g) being rational implies ρ µ (g n ) = 0 for some n. Then Lemma 4.2 implies the existence of a fixed point in U for g n . Hence g n = id. We conclude that if g has infinite order then We observed above that if f is a non-trivial pseudo-rotation then either f has finite order or Per(f ) = Fix(f ). We now prove the converse to this statement.
with the property that every non-trivial element g of G either has finite order or satisfies Fix(g) = Per(g). Then G is a pseudo-rotation group.
Proof. No element g ∈ G can have positive entropy since that would imply g has points of arbitrarily high period (see Katok [10]), a contradiction. If g ∈ G, then it must have at least χ(M ) fixed points because the Lefschetz index of a fixed point of g is at most 1 (see [13], for example). If g is non-trivial and has more than χ(M ) fixed points it follows from part (4) of Theorem 2.3 that there is an g-invariant annulus U ⊂ int(M ) for which the rotation number function in continuous and non-constant. It then follows from Corollary 2.4 of [8] that there are periodic points with arbitrarily high period, contradicting our hypothesis. We conclude that either g has finite order or card(Per(g)) = χ(M ).
Lemma 5.5. Suppose U ⊂ M is an open annulus and that C 1 and C 2 are disjoint closed connected sets in U each of which separates the ends of U and such that both U \ C 1 and U \ C 2 have two components. Then M \ (C 1 ∪ C 2 ) has three components: one with frontier contained in C 1 , one with frontier contained in C 2 and an open annulus with frontier contained in (C 1 ∪ C 2 ).
Then X 1 and Y 2 are components of M \ (C 1 ∪ C 2 ) with frontiers contained in C 1 and C 2 respectively and it suffices to show that V = X 2 ∩ Y 1 is an open annulus. We use the fact that an open connected subset of S 2 whose complement has two components is an annulus. Of course the same is true for M which may be considered as a subsurface of S 2 . But X 1 ∪ C 1 = cl(X 1 ) ∪ C 1 is connected and similarly so is So V is an annulus.
Our next result includes a very special case of Proposition (1.8) namely we show that the centralizer of an infinite order pseudo-rotation is virtually abelian. In fact, for later use, we need to consider a slightly more general setting, namely, not just the centralizer of a single pseudo-rotation, but the centralizer of an abelian pseudorotation group all of whose non-trivial elements have the same fixed point set.
Lemma 5.6. Suppose A is an abelian pseudo-rotation subgroup of Symp ∞ µ (M ) containing elements of arbitrarily large order. Suppose also that there is a set F with consisting of those elements which pointwise fix F , is abelian and has index at most 2 in Cent ∞ µ (A).
Proof. Elements of A must have entropy 0 since positive entropy implies the existence of infinitely many periodic points by a result of Katok, [10]. Let U be the open annulus , whose elements fix the ends of U , has index at most 2. We claim that the elements of C 0 all have entropy 0. Clearly we need only consider elements of infinite order since all finite order homeomorphisms have entropy 0. Hence the claim follows from Corollary (1.4) if there is an element of infinite order in A. Otherwise if C 0 contains an element with positive entropy then it contains an element g with a hyperbolic fixed point p ∈ U . Each point in the A-orbit of p is in Fix(g) and has the same set of eigenvalues for the derivative Dg. It follows that the A-orbit of p has finite cardinality, say m, and hence that p ∈ Fix(f m ) for all f ∈ A. Part (2) of Lemma 5.2 then implies that f m = id for all f ∈ A, if m > 2. This contradicts the fact that A contains elements of arbitrarily high order. This completes the proof that all elements of C 0 have entropy zero.
To prove that C 0 is abelian, we will show that each commutator h of two elements in C 0 is the identity by assuming that h is non-trivial and arguing to a contradiction. Since h is a commutator the map h c : U c → U c has mean rotation number ρ µ (h c ) = 0 and hence, by Lemma 4.2, h has a fixed point in U . Therefore Fix(h) contains more than χ(M ) points. If h has finite order then in a suitable averaged metric it is an isometry of M . But then each fixed point must have Lefschetz number +1 and hence by the Lefschetz theorem h has χ(M ) fixed points, a contradiction. We conclude therefore that h has infinite order and hence satisfies the hypothesis of Theorem (2.3). By this theorem there exists an element V ∈ A h and it must be the case that V ⊂ U since V cannot contain a point of Fix(h) ⊃ F . We will show this leads to a contradiction, either by showing that h has a fixed point in V or by showing that some non-trivial f ∈ A has a fixed point in U . We may then conclude h = id and hence that C 0 is abelian.
Suppose first that V is inessential in U . Elements of A commute with h and hence permute the elements of A h by Corollary 2.4. Since there can be only finitely many elements of A h of any fixed area, the A-orbit of V is finite. Hence there is m such that f m (V ) = V for all f ∈ A. The union of V with the component of its complement which is disjoint from F is an open disk D ⊂ U satisfying f m (D) = D for all f ∈ A. It follows from the Brouwer plane translation theorem that f m has a fixed point in D. If the order of f is greater than m then f m is a non-trivial element of A with fixed points not in F contradicting the assumption that Fix(f ) = F for all non-trivial f ∈ A. So the possibility that V is inessential in U is contradicted.
We are now reduced to the case that V is essential in U . It follows from part (1) of Corollary 2.4 that each element of A preserves V or maps it to a disjoint element of A h . Since the annulus U is A-invariant and V is essential in U it follows that V is A-invariant as the alternative would contradict the fact that A preserves area. We want to replace V with a slightly smaller essential V 0 ⊂ V which has the property that its frontier (in M ) lies in V and has measure 0. To do this we observe that ρ h is non-constant on V and hence has uncountably many level sets. Hence there must be two of its level sets C 1 , C 2 which have measure 0. Let V 0 be the essential open annulus in V whose frontier lies in C 1 ∪C 2 (see Lemma 5.5).
It follows from part (2) of Corollary 2.4 that each C i is A-invariant and hence V 0 is also.
As a first subcase, suppose that cl Uc (V 0 ) is g-invariant for each g ∈ C 0 . Then h is a commutator of elements that preserve cl Uc (V 0 ), so The last subcase is that there exists g ∈ C 0 such that g(V 0 ) ⊂ cl M (V 0 ) = cl Uc (V 0 ). Choose a component W of g(V 0 ) ∩ (U \ cl Uc (V 0 )). Since g leaves U invariant and preserves area and since g −1 (W ) ∩ W = ∅, W cannot contain a closed curve α that is essential in U . Thus W is inessential in U . As we noted above V 0 is A-invariant.
Since g ∈ C 0 , it commutes with A so g(V 0 ) is also A-invariant. It follows that A permutes the components of g(V 0 ) ∩ (U \ cl Uc (V 0 )). In particular for some m > 0 we have f m (W ) = W for all f ∈ A. Letting D be the union of W with any components of its complement which do not contain ends of U we conclude that D is an open disk and f m (D) = D for all f ∈ A. By the Brouwer plane translation theorem there is a point of Fix(f m ) in D. Since f m is a pseudo-rotation and has more than χ(M ) fixed points, f m = id. Since this holds for any f ∈ A we have contradicted the hypothesis that A contains elements of arbitrarily high order.
Lemma 5.7. Suppose that G is a pseudo-rotation subgroup of Symp r µ (M ) where M = A or D 2 and r ≥ 2. Then G is abelian and Fix(g) is the same for all non-trivial g ∈ G.
Proof. If M = A then Fix(g) = ∅ for all non-trivial g ∈ G. Since ρ µ ([f, g]) = 0 for each f, g ∈ G, Lemma 4.2 implies that each [f, g] has a fixed point in the interior of A and hence is the identity. Thus each f and g commute and we are done.
We assume for the remainder of the proof that M = D 2 . For each f ∈ G we considerf , the restriction of f to ∂D 2 . LetĜ be the image in Diff r (S 1 ) of G under the homomorphism f →f .
Lemma 5.2 (3) implies that f fixes a point in the interior of D 2 . If f k fixes a point in ∂D 2 for some k ≥ 1 then f k has more than χ(M ) fixed points and so is the identity. We conclude that iff has a point of period k then f is periodic of period k. In particular, iff : S 1 → S 1 has a fixed point then f = id. This proves that G acts freely on S 1 . It also proves that the restriction homomorphism f →f is an isomorphism G →Ĝ. SinceĜ acts freely on S 1 it follows from Hölder's Theorem (see, e.g., Theorem 2.3 of [5]) thatĜ (and hence G) is abelian.
Suppose f, g ∈ G and let Fix(f ) = {q}. Since f and g commute, g preserves {q} and so fixes q. This proves that Fix(f ) = Fix(g) and completes the proof.
If f ∈ Diff r (S 2 ) and p ∈ Fix(f ) we can compactify S 2 \ {p} by blowing up p, i.e. adding a boundary component on which we extend f by letting it be the projectivization of Df p . More precisely we definef p : S 1 → S 1 by considering the boundary as the unit circle in R 2 and lettinĝ If Fix(f ) contains two points then we may blow up these points and obtain the annular compactification A of S 2 \ Fix(f ).
Corollary 5.8. Suppose that G is a pseudo-rotation subgroup of Symp ∞ µ (S 2 ) and that G p is the stabilizer of a point p ∈ S 2 . Then G p is abelian and there exists q ∈ S 2 such that Fix(g) = {p, q} for all non-trivial g ∈ G p .
Proof. Suppose that f ∈ G p . Blowing up the point p and extending f we obtain f : D 2 → D 2 . This construction is functorial in the sense that if h = f g thenh =fḡ.
Hence the assignment f →f is a homomorphism from G p to Symp ∞ µ (D 2 ). This homomorphism is injective by part (2) of Lemma 5.2. The result now follows from Lemma 5.7.
It is not quite the case that for abelian pseudo-rotation subgroups of Symp ∞ µ (S 2 ) that the fixed point set for non-trivial elements is independent of the element. There is essentially one exception, namely the group generated by rotations of the sphere through angle π about two perpendicular axes. Lemma 5.9. If G is an abelian pseudo-rotation subgroup of Symp ∞ µ (S 2 ) then either Fix(g) is the same for every non-trivial element g of G or G is isomorphic to Z/2Z ⊕ Z/2Z. In the latter case the fixed point sets of non-trivial elements of G are pairwise disjoint and hence their union contains six points.
Proof. Let g, h be distinct non-trivial elements of G and suppose Fix(g) = Fix(h). Since G is abelian g(Fix(h)) = Fix(h). Since Fix(g) = Fix(h) and each contains two points, these sets are disjoint. Hence g switches the two points of Fix(h). In this case g has two fixed points and a period 2 orbit and hence g 2 has four fixed points and must be the identity. We conclude that g has order two. Switching the roles of g and h we observe that h has order two.
Let G 0 be the abelian group generated by g and h so G 0 ∼ = Z/2Z ⊕ Z/2Z. The possible actions of an element of G 0 on Fix(g) ∪ Fix(h) are: fix all four points, switch the points of one pair and fix the other, or switch both pairs. If f ∈ G then f preserves Fix(g) and Fix(h) so its action on Fix(g) ∪ Fix(h) must be the same as one of the elements of G 0 . Since f agrees with an element of G 0 at four points it must, in fact, be an element of G 0 . Hence any abelian pseudo-rotation group containing non-trivial elements whose fixed point sets do not coincide must be isomorphic to Z/2Z ⊕ Z/2Z.
Suppose now G ∼ = Z/2Z ⊕ Z/2Z is a pseudo-rotation subgroup of Symp ∞ µ (S 2 ). The three non-trivial elements of G must have pairwise disjoint fixed point sets. To see this note that otherwise there would be two distinct elements, say, g and h with a common fixed point. As observed above this implies Fix(g) = Fix(h). But any two non-trivial elements of G generate it and hence Fix(f ) is the same for all elements f ∈ G. Then by Proposition 5.3 there is an injective homomorphism from G to T. This is a contradiction since T contains a unique element of order two. Suppose A is a subgroup of the circle T = R/Z and G = A φ (Z/2Z) is the semidirect product determined by the homomorphism φ : Z/2Z → Aut(A) given by Lemma 5.11. Suppose G is a pseudo-rotation subgroup of Symp ∞ µ (S 2 ) such that either G leaves invariant a non-empty finite set F ⊂ S 2 , or G has a non-trivial normal solvable subgroup. Then either (1) There is a G-invariant set F 0 containing two points in which case G is isomorphic to a subgroup of the generalized dihedral group A φ (Z/2Z) for some subgroup A of the circle T, or (2) Every G-invariant set contains more than two points in which case G is finite.
Moreover, if the set F 0 in (1) is pointwise fixed then G is isomorphic to a subgroup of T.
Proof. Suppose first G has a non-trivial normal solvable subgroup N . We will reduce this case to the case that there is a finite G-invariant set. Let H be the last nontrivial element of the derived series of N so H is an abelian subgroup. Since H is characteristic in N it is normal in G. By Lemma 5.9 the set is finite. If g ∈ G and h ∈ H then ghg −1 ∈ H. Since Fix(ghg −1 ) = g(Fix(h)), it follows that g(F ) = F. Hence we need only prove the conclusion of our result under the assumption that G leaves invariant a finite set F . Let A be the finite index subgroup of G which pointwise fixes F . By Corollary 5.8 A is abelian. If F contains more than two points then A must be trivial and G = G/A is finite since A had finite index. If F is a single point then by Corollary 5.8 there is a set F containing F and one other point which are the common fixed points for every element of A. As above the set F is G-invariant and we replace F by F , so we may assume F contains two points.
In this case A has index at most two and Fix(h) = Fix(h ) for all h, h ∈ A. From Propostion 5.3 it follows that A is isomorphic to a subgroup of T.
If h ∈ G \ A then h interchanges the two points of F and reverses the orientation Also h has two fixed points and h 2 fixes the points of F so it has four fixed points and we conclude that h 2 = id. Hence the map φ : Z/2Z → G given by φ(1) = h is a homomorphism so G ∼ = A φ Z/2Z. Proposition 5.12. If G is a subgroup of Symp ∞ µ (M ) which has an infinite normal abelian subgroup A which is a pseudo-rotation group then G has an abelian subgroup of index at most two.
Proof. Since A is non-trivial, M is A, D 2 , or S 2 . Since A is infinite, Lemma 5.7 and Lemma 5.9 imply there is a set F containing χ(M ) ≤ 2 points such that F = Fix(h) for all h ∈ A. Let U = M \ F . Observe that in all cases U is an annulus. The set F must be invariant under G since F = Fix(ghg −1 ) = g(Fix(h)) = g(F ) for every element g in G. The subgroup G 0 of G that pointwise fixes F has index at most two. Also elements of G 0 leave U invariant and their restrictions to U are isotopic to the identity.
Since the set Fix(h) is the same for all h ∈ A, by Proposition 5.3 the homomorphism ρ µ : A → T 1 given by h → ρ µ (h) is injective, where ρ µ (h) is the mean rotation number on U c . Since conjugating h by an element g ∈ G 0 does not change its mean rotation number we conclude that h = ghg −1 for all h ∈ A. This means that G 0 is contained in Cent ∞ µ (A, G), the centralizer of A in G. If A contains elements of arbitrarily large order then it follows from Lemma 5.6 that G 0 is abelian and we are done. So we may assume the order of elements in A is bounded. Since the order of elements of A is bounded there are only finitely many possible values for ρ µ (f ) with f ∈ A. Hence, since the assignment f → ρ µ (f ) is injective, we may conclude A is finite in contradiction to our hypotheses.
Example 5.13. Let A be the subgroup of all rational rotations around the z-axis and let G be the Z/2Z extension of A obtained by adding an involution g which rotates around the x-axis and reverses the orientation of the z-axis. More precisely let the homomorphism φ : Z/2Z → Aut(A) be given by φ(1) = i g where i g (h) = g −1 hg = h −1 . Every element of A has finite order while A itself has infinite order. Moreover, G ∼ = A φ (Z/2Z) is not abelian even though it has an index two abelian subgroup.
We are now able to classify up to isomorphism all pseudo-rotation subgroups of Symp ∞ µ (S 2 ) which have a non-trivial normal solvable subgroup. We denote by φ the homomorphism φ : Proposition 5.14. If G is a pseudo-rotation subgroup of Symp ∞ µ (S 2 ) which has a normal solvable subgroup then G is isomorphic to either a subgroup of the generalized dihedral group T φ (Z/2Z) or a subgroup of the group O of orientation preserving symmetries of the regular octahedron (or equivalently the orientation preserving symmetries of the cube).
Proof. Suppose first that there is a G-orbit F 0 containing two points. Then by Proposition 5.11, G is isomorphic to a subgroup of the generalized dihedral group T φ (Z/2Z) and we are done. If this is not the case, then by the same proposition G is finite and every G-orbit contains at least three points.
Let A be the last non-trivial element of the derived series of the normal solvable subgroup of G, so A is abelian. Since A is a characteristic subgroup of that normal solvable subgroup it is normal in G. Hence for any g ∈ G, g(Fix(A)) = Fix(gAg −1 ) = Fix(A) and we observe Fix(A) cannot contain only two points since that is the case we already considered. Therefore not all elements of A have the same set of fixed points and we can apply Lemma 5.9 to conclude the group A is isomorphic to Z/2Z ⊕ Z/2Z. We observed in Remark 5.10 that there is an A-invariant set F which is the union of three sets and h i is one of the non-trivial elements of A. We also noted there that the elements of A preserve the pairs F i , fixing the points of one of them and switching the points in each of the other pairs.
We define a homomorphism θ : G → O(3) as follows. Given g ∈ G define a matrix P = P g by P ij = 1 if g(a j ) = a i and g(b j ) = b i , P ij = −1 if g(a j ) = b i and g(b j ) = a i , and P ij = 0 otherwise. Each row and column has one non-zero entry which is ±1 so P g ∈ O(3). It is straightforward to see that θ(g) = P g defines a homomorphism to O(3). It is also clear that it is injective since P g = I implies that g fixes the six points of F . Note that if h i is a non-trivial element of A, then θ(h i ) must be diagonal since h i (F j ) = F j , and θ(h i ) must have two entries equal to −1 since h i switches the points of two of the F j 's. It follows that θ(A) is precisely the diagonal entries of O(3) with an even number of −1's.
We need to show that θ(G) lies in SO (3), i.e., all its elements have determinant 1. Clearly θ(A) ⊂ SO(3). We denote the symmetric group on three elements by S 3 and think of it as the permutation matrices in O(3). We define a homomorphism θ : G → S 3 byθ(g) = Q ∈ O(3) where Q ij = |P ij | and P = θ(g). We observe that A is in the kernel ofθ. In fact A equals the kernel ofθ. To see this note that ifθ(g) = I then P g = θ(g) is a diagonal matrix and hence ker(θ) is abelian. If g ∈ ker(θ) \ A then the abelian group generated by g and A is not Z/2Z⊕Z/2Z contradicting Lemma 5.9. We conclude A = ker(θ).

Proof of the Main Theorem
In this section we prove Theorem (1.5) Suppose M is a compact oriented surface of genus 0 and G is a subgroup of Symp ∞ µ (M ) which has full support of finite type, e.g. a subgroup of Symp ω µ (M ). Suppose further that G has an infinite normal solvable subgroup. Then G is virtually abelian.
Throughout this section M will denote a compact oriented surface of genus 0, perhaps with boundary. We begin with a lemma that allows us to replace the hypothesis that G has an infinite normal solvable subgroup with the hypothesis that G has an infinite normal abelian subgroup.
, which contains an infinite normal solvable subgroup, then G contains a finite index subgroup which has an infinite normal abelian subgroup.
Proof. Let N be the infinite normal solvable subgroup. The proof is by induction on the length k of the derived series of N . If k = 0 then N is abelian and the result holds. Assume the result holds for k ≤ k 0 for some k 0 ≥ 0 and suppose the length of the derived series for N is k = k 0 + 1. Let A be the abelian group which is the last non-trivial term in the derived series of N . The group A is invariant under any automorphism of N and hence is normal in G. If A is infinite we are done.
We may therefore assume A is finite and hence that in a suitable metric A is a group of isometries. No non-trivial orientation preserving isometry of M can have infinitely many fixed points so we let F be the finite set of fixed points of non-trivial elements of A. Since A is normal g(F ) = F for any g ∈ G. Let G 0 be the normal finite index subgroup of G which pointwise fixes F.
Let N 1 = G 0 ∩ N so N 1 is an infinite solvable normal subgroup of G 0 and the k-th term A 1 in the derived series of N 1 is contained in A ∩ G 0 . We claim that A 1 is trivial. If F contains more than χ(M ) points then this follows from the fact that a non-trivial isometry isotopic to the identity fixes exactly χ(M ) points since each fixed point must have Lefschetz index +1.
Otherwise F contains χ(M ) points and we may blow up these points to form an annulus A on which N 1 acts preserving each boundary component. Since each element of A 1 is a commutator of elements in N 1 , it acts on A with mean rotation number zero. Each element of A 1 therefore contains a fixed point in the interior of A by Lemma 4.2. However, a finite order isometry of the annulus which is isotopic to the identity and contains an interior fixed point must be the identity. This is because every fixed point must have Lefschetz index +1 and the Euler characteristic of A is 0. This shows that A 1 acts trivially on A and hence on M . This verifies the claim that A 1 is trivial and hence the length of the derived series for N 1 is at most k 0 . The inductive hypothesis completes the proof.
The next lemma states the condition we use to prove that G is virtually abelian. Proof. We assume that [G 0 , G 0 ] contains a non-trivial element h and argue to a contradiction. For each G 0 -invariant open annulus V , let V c be its annular compactification and let h c : V c → V c be the extension of h| V (see Definition 2.1 or Definition 2.7 of [6] for details). Since h is a commutator of elements of G 0 and since G 0 extends to an action on V c , h c is a commutator and so has mean rotation number zero. Lemma 4.2 therefore implies that Fix(h) ∩ V = ∅.
By assumption, Fix(h) does not contain any open set and so Fix(h| V ) is a proper subset of V . We claim that fr(V ) ∪ Fix(h| V ) is not connected. To see this, let S be the end compactification of V obtained by adding two points, one for each end of V and let h S : S → S be the extension of h| V that fixes the two added points. Brown and Kister [4] asserts W is h S -invariant. However, then the Brouwer plane translation theorem would imply that h has a fixed point in W . This contradiction proves the claim.
By passing to a subfamily of the G 0 -invariant annuli, we may assume that the following is either true for all V or is false for all V : some component of Fix(h) intersects both components of the frontier of V . In the former case, the interior of each V contains a component of Fix(h). In the latter case, no component of Fix(h) intersects more than two of the annuli in our infinite family. In both cases, Fix(h) has infinitely many components in contradiction to the assumptions that h is non-trivial and that G 0 has full support of finite type.
We need an elementary topology result. Lemma 6.3. Suppose that C ⊂ int(M ) is a closed connected set which is nowhere dense and has two complementary components U 1 and U 2 . Then C = fr(U 1 ) ∩ fr(U 2 ) is a closed connected set with two complementary components, U 1 and U 2 , each of which has frontier C and is equal to the interior of its closure. Moreover U i is dense in U i .
Proof. To see that C separates U 1 and U 2 , suppose that σ is a closed path in M \ C with initial endpoint in U 1 and terminal endpoint in U 2 . Then σ ∩ fr(U 2 ) = ∅ and we let σ 0 be the shortest initial segment of σ terminating at a point x ∈ σ ∩ fr(U 2 ). Each y ∈ σ 0 \ x has a neighborhood that is disjoint from U 2 . Since C has empty interior, every neighborhood of y must intersects U 1 . It follows that every neighborhood of x must intersect U 1 and hence that x ∈ C . This contradiction proves that C separates U 1 and U 2 . Since the union of U 1 and U 2 is dense in M , the components V 1 and V 2 of M \ C that contain them are the only components of M \ C . Every neighborhood of a point in C intersects both U 1 and U 2 and so intersects both V 1 and V 2 . Thus C ⊂ fr(V 1 ) ∩ fr(V 2 ). Since fr(V 1 ), fr(V 2 ) ⊂ C we have that C ⊂ fr(V 1 ) ∩ fr(V 2 ) ⊂ fr(V i ) ⊂ C and hence C = f r(V 1 ) = f r(V 2 ) which implies that V 1 and V 2 are the interior of their closures. Since C is nowhere dense in M it follows that U 1 ∪ U 2 is dense in M and hence that U i is dense in U i . Notation 6.4. If C and C are as in Lemma 6.3 then we say that C is obtained from C by trimming. Recall that if f ∈ Z(M ) and U ∈ A f , then the rotation number function ρ = ρ f | U : U → S 1 is well defined and continuous. A component of a point pre-image of ρ is a level set for ρ and is said to be irrational if its ρ-image is irrational and to be interior if it is disjoint from the frontier of U . If C is an interior level set which is nowhere dense and C is obtained from C by trimming then we will call C a trimmed level set. The collection of all trimmed level sets for f will be denoted C(f ).
containing an abelian normal subgroup A, that f ∈ Z(M ) lies in A and that U ∈ A f . Suppose further that C 1 and C 2 are obtained from nowhere dense irrational interior level sets C 1 and C 2 for ρ = ρ f | U by trimming. Letting B i be the component of M \ (C 1 ∪ C 2 ) with frontier C i and V be the component of M \ (C 1 ∪ C 2 ), with frontier (C 1 ∪ C 2 ), assume that g-image is in V and essential in V then there is a proper subsurface of B i whose g image contains B i in contradiction to the fact that g preserves area. It follows that either g(B i ) = B i or there is a component W i of g(B i ) ∩ V which is inessential in V and whose frontier intersects C i . As above, this implies that ρ is constant and rational on W in contradiction to the fact that ρ is continuous on U and takes an irrational value on C i . This contradiction completes the proof.
We are now prepared to complete the proof of our main theorem.
Theorem (1.5) Suppose M is a compact oriented surface of genus 0 and G is a subgroup of Symp ∞ µ (M ) which has full support of finite type, e.g., a subgroup of Symp ω µ (M ). Suppose further that G has an infinite normal solvable subgroup. Then G is virtually abelian.
Proof. By Lemma 6.1 it suffices to prove the result when G has an infinite normal abelian subgroup A. If A contains an element of positive entropy the result follows by Proposition 3.3. If the group A is a pseudo-rotation group the result follows from Proposition 5.12.
We are left with the case that A contains an element f ∈ Z(M ). Let U be an element of A f and let ρ = ρ f | U . Since there are only countably many level sets of ρ with positive measure, all but countably many have empty interior or equivalently are nowhere dense. Choose a nowhere dense interior irrational level set C of ρ. One component of M \ C, say, Y , will be an open subsurface with µ(Y ) ≤ µ(M )/2. Choose a nowhere dense interior irrational level set C ⊂ Y ∩ U and let Y be the open subsurface which is the component of the complement of C satisfying Y ⊂ Y . Finally, choose nowhere dense interior irrational level setsĈ 1 ,Ĉ 2 ⊂ (Y \ cl(Y )) so that the annulusV bounded by the trimmed setsĈ 1 andĈ 2 has measure less than the measure of Y . ThenĈ 1 andĈ 2 satisfy the hypotheses of Lemma 6.5. It follows that the subgroup G 0 of G that preservesV has finite index in G.
Note that any two trimmed irrational level sets C 1 and C 2 inV satisfy the hypotheses of Lemma 6.5. Moreover if V is the essential open subannulus bounded by C 1 and C 2 then g(V ) ∩ V = ∅ for each g ∈ G 0 because each such g preservesV and preserves area. Lemma 6.5 therefore implies that each such V is G 0 -invariant so Lemma 6.2 completes the proof.

The Tits Alternative
We assume throughout this section that M be a compact oriented surface of genus 0, ultimately proving a special case (Theorem 1.12) of the Tits Alternative.
Lemma 7.1. Suppose f ∈ Z(M ) and U ∈ A(f ) is a maximal annulus for f. If G is a subgroup of Symp ω µ (M ) whose elements preserve each element of an infinite family of trimmed level sets for ρ f | U lying in U , then G is abelian.
Proof. Since infinitely many of the trimmed level sets in U are preserved by G so is each open annulus bounded by two such level sets. It is thus clear that one may choose infinitely many disjoint open annuli in U each of which is G-invariant. The result now follows from Lemma 6.2.
Lemma 7.2. Suppose f : A → A is a homeomorphism isotopic to the identity with universal covering liftf :Ã →Ã which has a non-trivial translation interval and suppose α ⊂ A is an embedded arc joining the two boundary components of A. Then there exists m > 0 such that for any γ ⊂ A which is an embedded arc, disjoint from α, joining the two boundary components of A, every lift of f k (γ), |k| > m must intersect more than one lift of α.
Proof. Letα be a lift of α and let T be a generator for the group of covering translations ofÃ. We claim that there is an m > 0 such that for all k ≥ m,f k (α) intersects at least 6 adjacent lifts of α. To show the claim consider the fundamental domain D 0 bounded byα and T (α). If there are arbitrarily large k for whichf k (α) intersects fewer than six translates ofα, then for such k,f k (D 0 ) would lie in an adjacent strip of five adjacent translates of D 0 and the translation interval forf would have length 0. This contradiction verifies the claim.
There is a liftγ of γ that is contained in D 0 . Let X be the region bounded by T −1 (γ) and T (γ) and note thatα ⊂ X. If the lemma fails then we can choose k ≥ m and a lifth of f k such thath(γ) is disjoint from T i (α) for all i = 0. It follows that h(X) is contained in ∪ 2 j=−2 T j (D 0 ) in contradiction to the fact thath(γ) intersects at least six translates ofα. Lemma 7.3. Suppose f, g ∈ Z(M ) and there are trimmed level sets C 1 ∈ C(f ) and C 2 ∈ C(g) such that there exist points x i ∈ M \ (C 1 ∪ C 2 ), 1 ≤ i ≤ 4 with the following properties: (1) x 1 , x 2 lie in the same component of the complement of C 1 and x 3 , x 4 lie in the other component of this complement.
(2) x 1 , x 3 lie in the same component of the complement of C 2 and x 2 , x 4 lie in the other component of this complement.
Then for some n > 0, the diffeomorphisms f n and g n generate a non-abelian free group.
Proof. Let C i be the untrimmed level set whose trimmed version is C i . We claim that by modifying the points {x i } slightly we may assume that the hypotheses (1) and (2) hold with C i replaced by C i . This is because each component of the complement of C i is a dense open subset of a component of the complement of C i . Hence each x j can be perturbed slightly tox j which, for each i, is in the component of complement of x j . Henceforth we will work with C i and refer tox j simply as x j . Let β 1 be a path in M \ C 2 joining x 1 and x 3 ; so β 1 crosses C 1 and is disjoint from C 2 . Likewise let β 2 be a path in M \ C 1 joining x 1 and x 2 ; so β 2 crosses C 2 and is disjoint from C 1 . (See Figure 1.) The level set C 1 is an intersection where each B n is an f -invariant open annulus with cl(B n+1 ) ⊂ B n and the rotation interval ρ(f, B n ) of the annular compactification f c of f | Bn is non-trivial (see section 15 and the proof of Theorem 1.4 in [6]). For n sufficiently large cl(B n ) is disjoint from β 2 and {x i } 4 i=1 . Let A 1 be a choice of B n with this property. We may choose a closed subarc α 1 of β 1 whose interior lies in A 1 and whose endpoints are in different components of the complement of A 1 . We will use intersection number with α 1 with a curve in A 1 to get a lower bound on the number of times that curve "goes around" the annulus A 1 .
Similarly the level set C 2 is an intersection where each B n is a g-invariant open annulus with cl(B n+1 ) ⊂ B n and the rotation interval ρ(g, B n ) of the annular compactification g c of f | B n is non-trivial. We construct A 2 and the arc α 2 in a fashion analogous to the construction of A 1 and α 1 . By construction α 1 is disjoint from A 2 and crosses A 1 while α 2 is disjoint from A 1 and crosses A 2 . (See Figure 2.) Note that any essential closed curve in A 1 must intersect α 1 and must contain points of both components of the complement of C 2 . To see this latter fact we note that we constructed α 1 to lie in one component of the complement of C 2 but we could as well have constructed α 1 in the other component of this complement. Any essential curve in A 1 must intersect both α 1 and α 1 . Similarly any essential curve in A 2 must intersect α 2 and must contain points of both components of the complement of C 1 .
There is a key consequence of these facts which we now explore. Let γ 0 be an arc with interior in A 1 , disjoint from A 2 ∪ α 1 and with endpoints in different components of the complement of A 1 . Replace f and g by f m and g m where m and m are the numbers guaranteed by Lemma 7.2 for f and g respectively. Then we know that for k = 0 the curve f k (γ 0 ) must intersect more than one lift of the arc α 1 in the universal coveringÃ 1 .
It follows that f k (γ 0 ) contains a subarc whose union with a subarc of α 1 is essential in A 1 . Hence we conclude that f k (γ 0 ) contains a subarc crossing A 2 , i.e. a subarc γ 1 whose interior lies in A 1 ∩A 2 (and hence is disjoint from α 2 ), and whose endpoints are in different components of the complement of A 2 . (See Figure 2.) Since we replaced g by g m above we know that for k = 0 the curve g k (γ 1 ) must intersect more than one lift of the arc α 2 in the universal coveringÃ 2 .
We can now construct γ 2 in a similar manner but switching the roles of f and g, α 1 and α 2 , and A 1 and A 2 . More precisely, for any k = 0 the curve g k (γ 1 ) contains a subarc whose union with a subarc of α 2 is essential in A 2 . It follows that g k (γ 1 ) contains a subarc γ 2 whose interior lies in A 1 ∩A 2 and whose endpoints are in different components of the complement of A 1 . Note that γ 2 is a subarc of g m f k (γ 0 ).
We can repeat this construction indefinitely, each time switching f and g. Hence if we are given h = g m 1 f k 1 . . . g mn f kn and m i = 0, k i = 0 we can obtain a non-trivial arc γ 2n which is a subarc of h(γ 0 ). Since γ 2n ⊂ A 1 ∩ A 2 and γ 0 ∩ A 2 = ∅ it is not possible that h = id. But every element of the group generated by f and g is either conjugate to a power of f , a power of g, or an element expressible in the form of h. Hence we conclude that the group generated by f and g is a non-abelian free group.
Proof of Theorem 1.12. Suppose f ∈ G ∩ Z(M ) and U ∈ A(f ) is a maximal annulus for f . One possibility is that there is a finite index subgroup G 0 of G which preserves infinitely many of the trimmed rotational level sets for f which lie in U . In this case Lemma 7.1 implies G 0 is abelian and we are done.
If this possibility does not occur, we claim that there exists a trimmed level set C in U and h 0 ∈ G such that h 0 (C) ∩ C = C but h 0 (C) ∩ C = ∅. If this is not the case then for every h ∈ G and every C either h(C) = C or h(C) ∩ C = ∅. It follows that the G-orbit of C consists of pairwise disjoint copies of C. Since elements of G preserve area this orbit must be finite and it follows that the subgroup of G which stabilizes C has finite index. If we now choose C 0 and C 0 , two trimmed level sets in U and let G 0 be the finite index subgroup of G which stabilizes both of them, then the annulus U 0 which they bound is G 0 -invariant. Now if C is any trimmed level set in U 0 then its G 0 -orbit lies in U and we conclude from area preservation that g(C) ∩ C = ∅ for all g ∈ G 0 . If for some g and C we have g(C) ∩ C = C we have demonstrated the claim and otherwise we are in the previous case since G 0 preserves the infinitely many trimmed level sets lying in U 0 .
So we may assume that the claim holds, i.e., that there is h 0 ∈ G and a trimmed level set C 1 for f in U such that h 0 (C 1 ) ∩ C 1 = C 1 and h 0 (C 1 ) ∩ C 1 = ∅. Let C 2 = h 0 (C 1 ) and g = h 0 f h −1 0 . Since C 1 = C 2 and each is the common frontier of its complementary components, it follows that each of the complementary components of C 1 has non-empty intersection with each of the complementary components of C 2 . Hence we may choose points x i ∈ M \ (C 1 ∪ C 2 ), 1 ≤ i ≤ 4 satisfying the hypothesis of Lemma 7.3 which completes the proof.