Parabolic isometries of visible CAT(0) spaces and metrics on moduli space
aa r X i v : . [ m a t h . G T ] F e b PARABOLIC ISOMETRIES OF VISIBLE CAT(0) SPACES ANDMETRICS ON MODULI SPACE
YUNHUI WU
Abstract.
We show that the translation length of any parabolic isometryon a complete semi-uniformly visible CAT(0) space is always zero. As aconsequence, we will classify the isometries on visible CAT(0) spaces in termsof translation lengths. We will also show that the moduli space M ( S g,n ) ofsurface S g,n of g genus with n punctures admits no complete visible CAT(0)Riemannian metric if 3 g + n ≥
5, which answers the Brock-Farb-McMullenquestion in the visible case. Introduction
CAT(0) spaces are generalizations of nonpositively curved Riemannian man-ifolds. Nonsmooth spaces, which are also called singular spaces, may occur inCAT(0) spaces. A classical example of a singular CAT(0) space is a tree, whichis an one-dimensional graph without loops. The vertices are the singular set. ACAT(0) space could be no proper, that is, not locally compact. The properness ofa tree depends on whether it is locally finite. Actually a locally infinite tree is thesimplest example for CAT(0) spaces which are not locally compact. The first partof this paper will focus on parabolic isometries of complete CAT(0) spaces whichmay be no proper.Let M be a complete CAT(0) space. An isometry γ of M is a map γ : M → M which satisfies dist ( γ · p, γ · q ) = dist ( p, q ), for all p, q ∈ M . The set of isometrieson a metric space is a group. An isometry can be classified as elliptic, hyperbolic,or parabolic. An isometry is called elliptic if it has a fixed point in M . Any finiteorder element in the isometry group of a complete CAT(0) space is always elliptic(see lemma 2.10 in section 2). An isometry γ is called hyperbolic if there existsa geodesic line c : ( −∞ , + ∞ ) → M such that γ acts on c ( R ) by a non-trivialtranslation. The fundamental group of a closed nonpositively curved Riemannianmanifold consists of hyperbolic isometries except the unit. If an isometry is nei-ther elliptic nor hyperbolic, then we call it to be parabolic . Parabolic isometriesmay occur in the fundamental group of an open nonpositively curved Riemannianmanifold.Let γ be an isometry of a complete CAT(0) space M . We define the translationlength | γ | of γ as | γ | := inf p ∈ M dist ( γ · p, p ) . From the definition of translation length, | γ | may be not achieved. If | γ | is achievedin M , | γ | = 0 corresponds to the elliptic case, and | γ | > hyperbolic case. If | γ | can not be achieved in M , γ is parabolic. One can see moredetails in [BH99, BGS85]. Let us look at the following two examples. Let H be theupper half plane endowed with the hyperbolic metric and define γ : H → H to be γ · ( x, y ) = ( x + 1 , y ). It is easy to see that | γ | = 0, which can not be achieved in H .Thus, γ is parabolic. Similarly consider R × H and define γ : R × H → R × H to be γ · ( z, ( x, y )) = ( z + 1 , ( x + 1 , y )). It is easy to see that γ is parabolicand | γ | = 1. So parabolic isometry with positive translation length may occur inCAT(0) spaces.A visible CAT(0) space, introduced by Eberlein and O’Neill in [EO73], needsthe space to be more curved. In some sense it means that for any two differentpoints at “infinity”, they can be viewed from each other along the space, whichcan not happen in R n . We call a complete CAT(0) space M is visible if forany x = y ∈ M ( ∞ ) there exists a geodesic line c : ( −∞ , + ∞ ) → M such that c ( −∞ ) = x and c (+ ∞ ) = y , where M ( ∞ ) is the visual boundary of M consistingof asymptotic geodesic rays (see [BH99] for details). In particular, a completeCAT(0) space, whose visual boundary is empty, is a visible space. When thevisual boundary is not empty, classical examples for visible spaces include trees,which are singular, and complete simply connected Riemannian manifolds whosesectional curvatures are bounded above by negative numbers, which are smooth.For the example R × H above which contains parabolic isometries with positivetranslation lengths, there exists embedding totally geodesical Euclidean planesin it. In particular, R × H is not visible. It was shown in [Wu11] that anyparabolic isometry of a complete proper visible CAT(0) space has zero translationlength, which was also served as a bridge to find a counterexample for the Eberleinconjecture which said that a complete open manifold M with sectional curvature − ≤ K M ≤ and finite volume is visible if the universal covering space ˜ M of M contains no imbedded flat half planes . And the proof in [Wu11] requires that thespace is proper. So it is natural to ask Question 1.1.
Does every parabolic isometry of a complete visible CAT(0) spacehave zero translation length?
In general, the answer to the question above is not true, one can see the examplein remark 3.5. If we just consider complete visible CAT(0) spaces, which may be noproper, we still would like to know what kind of spaces such that parabolic isome-tries have zero translation length. A semi-uniformly visible CAT(0) space, roughlyspeaking, means a visible CAT(0) space which can not contain flat strips with ar-bitrary large widths, one can see definition 2.6 for details. The motivation thatwe introduce semi-uniformly visible CAT(0) spaces is that the counterexample inremark 3.5 for question 1.1 contains flat strips with arbitrary large widths. Propo-sition 2.7 in section 2 tells that both complete proper visible CAT(0) spaces andcomplete Gromov-hyperbolic CAT(0) spaces are semi-uniformly visible CAT(0)spaces. The following theorem gives an affirmative answer to question 1.1 forsemi-uniformly visible CAT(0) spaces.
Theorem 1.2.
Let M be a complete semi-uniformly visible CAT(0) space. Thenany parabolic isometry has zero translation length. i.e., for any parabolic isometry γ on M we have | γ | = 0 . ARABOLIC ISOMETRY 3
Buyalo in [Buy98] proved theorem 1.2 when M is a complete Gromov-hyperbolicCAT(0) space. Since a complete Gromov-hyperbolic CAT(0) space is a completesemi-uniformly visible CAT(0) space (see proposition 2.5 in section 2), theorem1.2 gives a generalization. And the method here is different as what Buyalo didin [Buy98]. For the manifold case, Bishop and O’Neill in [BO69] proved that anyparabolic isometry of a complete simply connected manifold M with sectional cur-vature K M ≤ − Theorem 1.3.
Let M be a complete semi-uniformly visible CAT(0) space, and γ be an isometry of M . Then,(1): γ is elliptic if and only if γ has a fixed point in M .(2): γ is hyperbolic if and only if | γ | > .(3): γ is parabolic if and only if | γ | = 0 and γ does not have any fixed point in M . Let S g,n be a surface of g genus with n punctures, and M ( S g,n ) be the modulispace of S g,n . There are a lot of canonical metrics on M ( S g,n ) like the Teichm¨ullermetric, the Weil-Petersson metric, and so on (see [IT92]). Kravetz in [Kra59]proved the Teichm¨uller metric has negative curvature. However, a mistake inthe proof of Kravetz’s theorem was found by Linch in [Lin71]. Masur in [Mas75]proved that the Teichm¨uller metric is not nonpositively curved except severalcases. Although the Weil-Petersson metric is negatively curved (see [Wol86]), butnot complete (see [Wol75]). The McMullen metric, constructed by McMullen in[McM00], is a complete K¨ahler-hyperbolic metric in the sense of Gromov. Liu, Sunand Yau in [LSY05] perturbed the Weil-Petersson metric to construct the so-calledperturbed Ricci metric, which is complete and whose Ricci curvatures are pinchedby two negative numbers. There is a question in Brock-Farb’s paper [BF06] whichstates Question 1.4 (Brock-Farb-McMullen) . Does M ( S g,n ) admit a complete, nonpos-itively curved Riemannian metric? Since M ( S g,n ) is an orbifold, here a Riemannian metric on M ( S g,n ) means aRiemannian metric on the Teichm¨uller space T ( S g,n ), the universal covering spaceof M ( S g,n ), on which the natural action of the mapping class group Mod S g,n on T ( S g,n ) is an isometric action.We call a Riemannian metric on M ( S g,n ) is called visible CAT(0) Riemann-ian if the sectional curvature of T ( S g,n ) is nonpositive and T ( S g,n ) is visible.If 3 g + n ≤ M ( S g,n ) has positive dimension, then ( g, n ) must be one of { (1 , , (1 , , (0 , } . For these three cases, it is well known that the Teichm¨ullermetric on M ( S g,n ) is a complete hyperbolic metric (see [FW10]). In particular, theTeichm¨uller metric on M ( S g,n ) is complete visible CAT(0) Riemannian. Hence,we can always assume that 3 g + n ≥ YUNHUI WU
Theorem 1.5. If g + n ≥ , then M ( S g,n ) admits no complete visible CAT(0)Riemannian metric. Since a complete Riemannian manifold with sectional curvatures bounded aboveby a negative number is a visible CAT(0) manifold, the following theorem followsimmediately from theorem 1.5.
Theorem 1.6. If g + n ≥ , M ( S g,n ) admits no complete Riemannian metricsuch that the sectional curvature K ( M ( S g,n )) ≤ − . Ivanov in [Iva88] showed that M ( S g,n ) (3 g + n ≥
5) admits no complete, finitevolume Riemannian metric whose sectional curvature is pinched by two negativenumbers. McMullen in [McM00] stated that M ( S g,n ) (3 g + n ≥
5) admits nocomplete Riemannian metric whose sectional curvature is pinched by two negativenumbers, which was proved by Brock and Farb in [BF06]. Moreover, the authorsin [BF06] showed that M ( S g,n ) (3 g + n ≥
5) admits no complete, finite volumeRiemannian metric such that the universal covering space is Gromov-hyperbolic.For the visible and finite volume case, one can refer to [Wu11]. What is new fortheorem 1.5 and 1.6 is that there is no finite volume restriction. For related topics,one can also see [Ji12, KM96, KN04, LSY04, MP99, MW95].Throughout this paper, we always assume that the geodesics use arc-length pa-rameters.
Plan of the paper.
In section 2 we set out necessary backgrounds, and provesome basic properties on CAT(0) spaces and mapping class groups, which will beapplied in subsequent sections. Section 3 establishes theorem 1.2 and theorem 1.3.Theorem 1.5 is proved in section 4.
Acknowledgments.
The author is indebted to Andy Putman for the discussionson the proof of theorem 1.5, in particular for his suggestion on writing this articleand correction on English for the original manuscript. Thank also to Benson Farband Mike Wolf for the discussions on the proof of theorem 1.5. The author alsowould like to thank Tushar Das for the discussions on theorem 1.2 and sharing hisidea in remark 3.5. 2.
Notations and Preliminaries
CAT(0) spaces.
A CAT(0) space is a geodesic metric space in which eachgeodesic triangle is no fatter than a triangle in the Euclidean plane with the sameedge lengths.
Definition 2.1. let M be a geodesic metric space. For any a, b, c ∈ M , threegeodesics [ a, b ] , [ b, c ] , [ c, a ] form a geodesic triangle ∆ . Let ∆( a, b, c ) ⊂ R be atriangle in the Euclidean plane with the same edge lengths as ∆ . Let p, q be pointson [ a, b ] and [ a, c ] respectively, and let p, q be points on [ a, b ] and [ a, c ] respectively,such that dist M ( a, p ) = dist R ( a, p ) , dist M ( a, q ) = dist R ( a, q ) . We call M a CAT(0) space if for all ∆ the inequality dist M ( p, q ) ≤ dist R ( p, q ) holds. ARABOLIC ISOMETRY 5
Let M be a complete CAT(0) space. The ideal boundary, denoted by M ( ∞ ),consists of asymptotic rays. For each point p ∈ M and x ∈ M ( ∞ ), since thedistance function between geodesics is convex, it is not hard to see that thereexists a unique geodesic ray c which represents x and starts from p (see [BH99]).We write c (+ ∞ ) = x . Although a complete CAT(0) space may be singular, thedefinition of CAT(0) spaces can guarantee that the notation of the angle, like thesmooth case, still make sense (see [BH99]). Given two points x, y in M ( ∞ ). Let ∠ p ( x, y ) denote the angle at p between the unique geodesics rays which issue from p and lie in the classes x and y respectively. The angular metric is defined to be ∠ ( x, y ) := sup p ∈ M ∠ p ( x, y ). It is easy to see that ∠ ( x, y ) = 0 if and only if x = y .On a complete visible CAT(0) space M , for any x = y ∈ M ( ∞ ), ∠ ( x, y ) = π . Sothe angular metric gives a discrete topology on the ideal boundary of a completevisible CAT(0) space.The following lemma will be used in next section, which gives us a way tocompute the angular metric. Lemma 2.2.
Let M be a complete CAT(0) space with a basepoint p . Let x, y ∈ M ( ∞ ) and c, c ′ be two geodesic rays with c (0) = c ′ (0) = p, c (+ ∞ ) = x and c ′ (+ ∞ ) = y . Then, ∠ ( x, y )2 ) = lim t → + ∞ dist ( c ( t ) , c ′ ( t )) t . Proof.
See proposition 9.8 on page 281 in chapter II.9 of [BH99]. (cid:3)
Product.
Let X and X be two metric spaces. The product X = X × X has a natural metric which is called the product metric. Let γ i be an isometry of X i ( i = 1 , γ = ( γ , γ ) is an isometry of X under the naturalaction. The following lemma tells when the converse is true. Lemma 2.3.
Let X = X × X . Then an isometry γ on X decomposes as ( γ , γ ) ,with γ i be an isometry of X i ( i = 1 , , if and only if, for every x ∈ X , thereexists a point denoted γ · x ∈ X such that γ · ( { x } × X ) = { γ · x } × X .Proof. See proposition 5.3 on page 56 in chapter I.5 of [BH99]. (cid:3)
The following product decomposition theorem will be applied for several timesin this article.
Proposition 2.4.
Assume that M is a complete CAT(0) space. Let c : R → M be a geodesic line and P c be the set of geodesic lines which are parallel to c . Then, P c is isometric to the product P ′ c × R where P ′ c is a closed convex subset in M .Proof. See theorem 2.14 on page 183 in chapter II.2 of [BH99]. (cid:3)
Semi-uniformly visible CAT(0) spaces.
Recall a metric space M is called Gromov-hyperbolic if there exists a δ > δ -thin. Where a δ -thin geodesic triangle means that each of its sides is containedin the δ -neighborhood of the union of the other two sides. A R -tree is a Gromov-hyperbolic space which holds for any δ >
0. For more details one can see [BH99,Gro87]. The following proposition tells that Gromov-hyperbolic CAT(0) spacesare stronger than visible CAT(0) spaces.
YUNHUI WU
Proposition 2.5.
Every complete Gromov-hyperbolic CAT(0) space is a visibleCAT(0) space.Proof.
See proposition 10.1 in [Buy98]. For the proper case, one can also seeproposition 1.4 in chapter III.H of [BH99]. (cid:3)
Before we define semi-uniformly visible CAT(0) spaces, let us consider the fol-lowing two examples.
Example . For every positive integer n , let I n be the vertical segment in R asfollows I n := { ( x, y ) ∈ R ; x = 1 n , ≤ y ≤ n } and I be the horizontal segment with unit length as follows I := { ( x, y ) ∈ R ; 0 ≤ x ≤ , y ≡ } . Consider the space M := (( + ∞ [ n =1 I n ) [ I, ds )where ds is the induced metric from R .It is easy to see that M is a complete tree. So M is a complete Gromov-hyperbolic space. By proposition 2.5, M is also a complete CAT(0) visible space.It is not hard to see that M is not locally compact (around the point ( x, y ) = (0 , M is empty.Let M be a complete CAT(0) space. We call M has an infinite-flat-strip ifthere exists a totally geodesical convex subset U × R ⊆ M where U is unbounded.From proposition 2.4 we can always assume that U is closed convex. The followingexample tells that the convex hull of an infinite-flat-strip may not contain a halfflat plane. Example . Let M be the metric space in example 1. Consider the product space N := M × R which is endowed with the product metric.It is obvious that N is a complete CAT(0) space, but not locally compact.Actually the visual boundary N ( ∞ ) of N consists of two points, which are thepositive and negative directions of the R component. In particular they can bejoined by a geodesic line in N . Thus N is a complete visible CAT(0) space. Itis easy to see that N contains an infinite-flat-strip. And N is not a Gromov-hyperbolic CAT(0) space.Example 2 tells us that a complete visible CAT(0) space may contain an infinite-flat-strip. The following definition will exclude complete CAT(0) spaces withinfinite-flat-strip. Definition 2.6.
Let M be a complete CAT(0) space. M is called semi-uniformlyvisible if M is visible and M can not contain an infinite-flat-strip. ARABOLIC ISOMETRY 7
It is obvious that any complete simply connected Riemannian manifold whosesectional curvatures are bounded by a negative number is a semi-uniformly visibleCAT(0) space. The following proposition tells that semi-uniformly visible CAT(0)spaces contain a lot of standard visible spaces.
Proposition 2.7. (1): Every complete proper visible CAT(0) space is a semi-uniformly visible CAT(0) space.(2): Every complete Gromov-hyperbolic CAT(0) space is a semi-uniformly visibleCAT(0) space.Proof. Proof of part (1):
If not. Then we can assume that M is a complete propervisible CAT(0) space which contains an infinite-flat-strip U × R where U is un-bounded. Let c : ( −∞ , + ∞ ) → M be the geodesic line x × R where x ∈ U .And let P c be the set of geodesic lines which are parallel to c . By proposition2.4, P c is isometric to the product P ′ c × R where P ′ c is a closed convex subset in M . Hence U ⊆ P ′ c . Since U is unbounded, P ′ c is also unbounded. Thus P ′ c acomplete, unbounded, proper space. The Arzel`a Ascoli theorem would guaranteethat there exists a geodesic ray d : [0 , + ∞ ) → P ′ c . Hence M contains a flat halfplane [0 , + ∞ ) × R in M which is impossible because M is visible. Proof of part (2):
By proposition 2.5 it suffices to show that every completeGromov-hyperbolic CAT(0) space is semi-uniform. Assume not. Then we canassume that M is a complete Gromov-hyperbolic CAT(0) space which contains aninfinite-flat-strip U × R where U is unbounded. Let δ > M is δ -thin. Let c : ( −∞ , + ∞ ) → M be the geodesicline x × R where x ∈ U . And let P c be the set of geodesic lines which are parallelto c . By proposition 2.4, P c is isometric to the product P ′ c × R where P ′ c is a closedconvex subset in M . Hence U ⊆ P ′ c . Since U is unbounded, we can find a flat strip[0 , k ] × R with width k where k is an arbitrary positive number. If we choose k tobe greater enough than δ , it is not hard to find a geodesic triangle ∆ in [0 , k ] × R such that ∆ is not δ -thin, which is a contradiction. (cid:3) Remark . From part (1) of the proposition above, if the space M is complete,locally compact, CAT(0) space, then semi-uniform visibility is equvilent to vis-ibility. Hence, the interesting aspect for semi-uniform visibility is for completeCAT(0) spaces which are not locally compact.2.4. Horospheres in CAT(0) spaces.
Let M be a complete CAT(0) space and x ∈ M ( ∞ ). c : [0 , + ∞ ) → M is a geodesic ray with c (+ ∞ ) = x , which determinesa Busemann function f = f c at x given by f ( p ) = lim t → + ∞ ( dist ( p, c ( t )) − t ). f isa convex function. For each t ∈ R , the set B ( t ) := { q ∈ M, f ( q ) ≤ t } is a horoball at x , whose boundary, a level set of f , is a horosphere at x . Every horoball isconvex because of the convexity of f . If x is fixed by some parabolic isometry γ of M , then each horoball at x is γ -invariant. In particular each horosphere at x ,the boundary of some horoball, is also γ -invariant. One can see more details in[BH99]. The following lemma will be applied to prove theorem 1.5. YUNHUI WU
Proposition 2.8.
Let M be a complete, simply connected, nonpositively curvedRiemannian manifold and c : [0 , + ∞ ) → M be a geodesic ray. Then every horo-sphere at c (+ ∞ ) is diffeomorphic to R k where k = dim ( M ) − .Proof. Let H be a horosphere at c (+ ∞ ) and B be the horoball whose boundary is H . Choose a point q in the complement of B and consider the distance function f : H → (0 , + ∞ ), given by f ( p ) = dist ( q, p ). The idea is prove that f only hasone critical point on H , which is the intersection point of H and the geodesic ray,starting at q , which goes to c (+ ∞ ). Then the conclusion follows from standardMorse theory. One can see lemma 3-4 on page 511 in [Fuk84] for details. (cid:3) Isometries on CAT(0) spaces.
Let M be a complete CAT(0) space and γ be an isometry of M . If the translation length | γ | is achieved in M , γ is eitherelliptic or hyperbolic. The following lemma tells us any isometry on R n can achieveits translation length, which will be applied in subsequent sections. Lemma 2.9.
Let γ be an isometry of the Euclidean space R n where n is a positiveinteger. Then γ is either elliptic or hyperbolic.Proof. One possible way is to argue it by contradiction. Assume that γ is para-bolic. Then there exists a horosphere H such that γ acts on H as an isometry.Since H is a horosphere of R n , H is isometric to R n − which is totally geodesicin R n . The conclusion follows by induction on the dimensions. For the details weleave it as an exercise to the reader. (cid:3) From definition an elliptic isometry γ of M has at least one fixed point. Thefollowing is one of the basic fixed point theorems in CAT(0) geometry. Lemma 2.10.
An isometry γ of a complete CAT(0) space M has a fixed point ifand only if there exists a γ -invariant bounded subset in M .Proof. The idea is to prove that, for any bounded subset in M there exists a uniquecentre point for this bounded subset, and then prove this point is fixed by γ . Onecan see the proof of proposition 6.7 on page 231 in chapter II.6 of [BH99]. (cid:3) Recall that the translation length γ of an isometry of M is defined as the infi-mum of translation of γ over M . The following lemma gives us another viewpointfor the translation length of a single isometry. Lemma 2.11.
Let M be complete CAT(0) space and γ be an isometry of M .Then, for all p ∈ M , we have | γ | = lim n → + ∞ dist ( γ n · p, p ) n . Proof.
See lemma 6.6 on page 83 in [BGS85] or lemma 3.4 in [Wu11]. (cid:3)
As a direct corollary,
Lemma 2.12.
Let M be complete CAT(0) space and γ be an isometry of M .Then | γ | = 2 | γ | . ARABOLIC ISOMETRY 9
Proof.
Let p ∈ M . By lemma 2.11, | γ | = lim n → + ∞ dist ( γ n · p, p ) n = 2 lim n → + ∞ dist ( γ n · p, p )2 n = 2 | γ | . (cid:3) A group G acting on a metric space X is called properly if for each compactsubset K ⊂ X , the set K ∩ gK is nonempty for only finitely many g in G . Lemma 2.13.
Let M be a complete visible CAT(0) Riemannian manifold and G = Z ⊕ Z , a free abelian group of rank 2, act properly on M by isometries. Thenfor any nontrivial g ∈ G , g is parabolic.Proof. Since G acts properly on M , for any nontrivial g ∈ G , g is either parabolicor hyperbolic. Assume that g is hyperbolic. Let M in ( g ) := { p ∈ M, dist ( g · p, p ) = | g |} . From theorem 6.8 on page 231 in chapter II.6 of [BH99] we know that M in ( g )is isometric to the product Y × R on which g acts trivially on the Y component,where Y is a closed convex subset in M . Since M is a visible CAT(0) manifold, Y is bounded, otherwise there exists a flat half plane [0 , + ∞ ) × R in X whichis impossible in visible CAT(0) spaces (this step needs that the space is proper).Let h ∈ G such that the group < g, h > , generated by g and h , is a free abeliangroup of rank 2. Since gh = hg , M in ( g ) is h -invariant. Since h is an isometry, itmust send a geodesic line to another geodesic line. Thus, by lemma 2.3, h splitsas ( h , h ) where h is an isometry on Y and h is an isometry on R . Firstly Y isa complete CAT(0) space because Y is closed convex in M . Since Y is bounded,by lemma 2.10, there exists x ∈ Y such that h · x = x . Hence x × R is < g, h > -invariant. Since G acts properly on M and < g, h > is a subgroup of G , < g, h > also acts properly on x × R , which is impossible because the rank of < g, h > is 2. (cid:3) Mapping class groups.
Let S g,n be a Riemann surface of genus g with n punctures, and Mod S g,n be the mapping class group of S g,n , i.e., the group ofisotopy classes of self-homeomorphisms of S g,n which preserve the orientation andthe punctures. The following proposition lists the basic properties of Mod S g,n ,which will be used. Proposition 2.14.
Let
Mod S g,n be the mapping class group of S g,n . Then(1): Mod S g,n acts properly on the Teichm¨uller space.(2): Mod S g,n is finitely generated by Dehn-twists along simple closed curves.(3): There exist torsion-free subgroups of finite index in Mod S g,n .Proof. See the details in [FM12]. (cid:3)
Bestvina, Kapovich and Kleiner in [BKK02] defined the action dimension of agroup G , denoted by actdim ( G ), to be the minimum dimension of a contractiblemanifold on which G properly acts. For examples, the action dimension of Z is 1.The action dimension of the fundamental group of a closed hyperbolic surface is2. In [BKK02] there is a notation, which is called the obstructor dimension of agroup G , denoted by obdim ( G ) (see [BKK02]). Proposition 2.15 (BKK) . (1): For any group G , actdim ( G ) ≥ obdim ( G ) .(2): The obstructor dimension is a quasi-isometric invariance.Proof. See theorem 1 in [BKK02] for part (1) and remark 11 in [BKK02] for part(2). (cid:3)
In [Des06] Z. Despotovic proved that the action dimension of the mapping classgroup actdim (Mod S g,n ) = 6 g − n if 3 g + n ≥ obdim (Mod S g,n ) = 6 g − n and thenused part (1) of proposition 2.15 to conclude actdim (Mod S g,n ) = 6 g − n .The following result tells that the action dimension is preserved by subgroups ofmapping class groups, up to finite index. Proposition 2.16 (Despotovic) . If g + n ≥ , then for any finite index subgroup G of Mod S g,n , actdim ( G ) = 6 g − n .Proof. Let G be a subgroup of Mod S g,n with finite index. So G endowed witha word metric is quasi-isometric to Mod S g,n . Since the obstructor dimension ofMod S g,n obdim (Mod S g,n ) = 6 g − n (see [Des06]), by part (2) of proposition2.15 we have obdim ( G ) = 6 g − n . Thus from part (1) of proposition 2.15 weknow that actdim ( G ) ≥ g − n .On the other hand it is obvious that actdim ( G ) ≤ g − n because G acts properly on the Teichm¨uller space which is contractible. Hence actdim ( G ) =6 g − n . (cid:3) Proofs of theorem 1.2 and theorem 1.3
Before we go to prove theorem 1.2, let us control the size of the fixed points ofparabolic isometries.
Proposition 3.1.
Let M be a complete semi-uniformly visible CAT(0) space and γ be a parabolic isometry on M . Then, the number of the fixed points of γ satisfy { x ∈ M ( ∞ ) : γ · x = x } ≤ . Proof.
Assume not. That is | F ix ( γ ) | ≥
2, where
F ix ( γ ) denotes the set of thefixed points of γ in M ( ∞ ). Let x = y ∈ F ix ( γ ). Since M is visible, there existsa geodesic line c : R → M such that c ( −∞ ) = x and c (+ ∞ ) = y . Let P c be theset of geodesic lines which are parallel to c . By proposition 2.4, P c is isometric tothe product P ′ c × R where P ′ c is a convex subset in M . Since M is semi-uniformlyvisible, P ′ c is bounded, otherwise there exists an infinite-flat-strips in M , whichcontradicts the definition of semi-uniformly CAT(0) spaces.Since c ( −∞ ) , c (+ ∞ ) ∈ F ix ( γ ), γ · ( c ( R )) is also geodesic line which is parallelto c . In particular, P c = P ′ c × R is a γ -invariant subset in M . From lemma 2.3, γ splits as ( γ , γ ) where γ is an isometry on P ′ c and γ is an isometry on R .Firstly since P ′ c is convex in M , P ′ c is also a CAT(0) space. Since P ′ c is bounded,by lemma 2.10, there exists x ∈ P ′ c such that γ · x = x . By lemma 2.9, γ iseither elliptic or hyperbolic. Case 1: γ is elliptic. ARABOLIC ISOMETRY 11
There exists x ∈ R such that γ · x = x . In particular γ = ( γ , γ ) fixes thepoint ( x , x ), which means that γ is elliptic, which contradicts the assumptionthat γ is parabolic. Case 2: γ is hyperbolic. Since γ · x = x , γ = ( γ , γ ) acts on the line x × R as a translation. Fromthe definition of hyperbolic isometries, x × R is an axis for γ . In particular, γ ishyperbolic, which also contradicts our assumption that γ is parabolic. (cid:3) Remark . If M is proper, i.e., locally compact, it is not hard to prove theexistence of fixed points for parabolic isometry (see proposition 8.25 on page 275in chapter II.8 of [BH99]). By part (1) of proposition 2.7, proposition 3.1 tellsthat that the fixed point of any parabolic isometry of a complete proper visibleCAT(0) space is unique. One can also see [FNS06]. Remark . If M is a complete Gromov-hyperbolic CAT(0) space (may be noproper!), the existence of fixed points of any parabolic isometry is also guaranteed(see theorem 0.3 in [Buy98] or one can also see [GdlH90]). I am grateful to TusharDas for this point. Proposition 3.2.
Let M be a complete CAT(0) space and γ be an isometry on M . If | γ | > , then { x ∈ M ( ∞ ) : γ · x = x } ≥ . Proof.
Since | γ | > γ is either hyperbolic or parabolic.If γ is hyperbolic, let c : ( −∞ , + ∞ ) → M be an axis for γ . The conclusionfollows from the fact that { c ( −∞ ) , c (+ ∞ ) } belongs to { x ∈ M ( ∞ ) : γ · x = x } .If γ is parabolic. Since | γ | >
0, a special case of a result of Karlsson andMargulis [KM99] shows that γ has a unique fixed point x ∈ M ( ∞ ) such thatfor every p ∈ M and every geodesic ray c : [0 , + ∞ ) → M with c (0) = p and c (+ ∞ ) = x we have lim n → + ∞ dist ( γ n · p, c ( n | γ | )) n = 0 . Similarly | γ − | = | γ | > γ − has a unique fixed point y ∈ M ( ∞ ) such thatfor every p ∈ M and every geodesic ray c ′ : [0 , + ∞ ) → M with c ′ (0) = p and c ′ (+ ∞ ) = y we have lim n → + ∞ dist ( γ − n · p, c ′ ( n | γ | )) n = 0 . By triangle inequality, we havelim n → + ∞ dist ( c ( n | γ | ) , c ′ ( n | γ | )) n = lim n → + ∞ dist ( γ n · p, γ − n · p ) n . Since γ is an isometry, by lemma 2.2 and lemma 2.11, we have2 | γ | sin( ∠ ( c (+ ∞ ) , c ′ (+ ∞ ))2 ) = | γ | . From lemma 2.12, we know that2 | γ | sin( ∠ ( c (+ ∞ ) , c ′ (+ ∞ ))2 ) = 2 | γ | . Since | γ | 6 = 0, ∠ ( c (+ ∞ ) , c ′ (+ ∞ )) = π = 0. That is ∠ ( x, y ) = 0. Since x, y ∈ F ix ( γ ), we have { x ∈ M ( ∞ ) : γ · x = x } ≥ . (cid:3) Remark . The result of Karlsson and Margulis above was also used by M.Bridson in [Bri] to show that any Dehn-twist has zero translation length whena mapping class group of a closed surface with genus g ≥ Proof of theorem 1.2.
Suppose not, we assume that | γ | >
0. From proposition 3.2we know that { x ∈ M ( ∞ ) : γ · x = x } ≥ . On the other hand, since M is complete semi-uniformly visible and γ is parabolic,by proposition 3.1, we have { x ∈ M ( ∞ ) : γ · x = x } ≤ (cid:3) Since a complete proper visible CAT(0) space is semi-uniformly visible (see part(1) of proposition 2.7), theorem 1.2 implies
Theorem 3.3. ( [Wu11] ) Let M be a complete proper visible CAT(0) space. Thenany parabolic isometry of M has zero translation length. i.e., for any parabolicisometry γ of M we have | γ | = 0 . Since a complete Gromov-hyperbolic CAT(0) space is semi-uniformly visible(see part (2) of proposition 2.7), by theorem 1.2 we immediately obtain
Theorem 3.4 (Buyalo) . Let M be a complete Gromov-hyperbolic CAT(0) space.Then any parabolic isometry of M has zero translation length. i.e., for any para-bolic isometry γ of M we have | γ | = 0 .Remark . We call a manifold M is tame if M is the interior of some compactmanifold M with boundary. Phan conjectures in [Tam11] that let M be a tame,finite volume, negatively curved manifold, then M is not visible if the fundamentalgroup of M contains a parabolic isometry of ˜ M with positive translation length.This conjecture is confirmed in [Wu11]. Since the proof in this paper is differentas the one in [Wu11], theorem 1.2 gives a new proof to confirm this conjecture. Remark . A complete visible CAT(0) space may contain parabolic isometrieswith positive translation lengths. One possible example is the following:
Firstlyfind a complete CAT(0) space M such that there exists a parabolic isometry γ onit and the visual boundary is empty (one may construct M by taking the convexhull of the orbits of zero under α on page 6 in section 1.3.3 in [Val07] , which ismodified by Edelstein’s example in [Ede64] ). Then take the product of M with R we get a space N := M × R . Since the visual boundary of N consists of two pointswhich can be joined by a geodesic line, N is a complete visible CAT(0) space. ARABOLIC ISOMETRY 13
Consider the isometry γ : N → N which is defined as γ · ( m, t ) = ( γ · m, t + 1) . Itis easy to see that γ is a parabolic isometry on N whose translation length | γ | > . I am greatly indebted to Tushar Das to share his idea on this example.Theorem 1.3 is an easy application of theorem 1.2.
Proof of theorem 1.3.
Proof of (1) : By definition.
Proof of (2) : If γ is hyperbolic, by the definition we know that | γ | > | γ | >
0, assume that γ was not hyperbolic, so γ is parabolic. From theorem1.2 we have | γ | = 0 which contradicts with our assumption. Proof of (3) : If γ is parabolic, it is obvious that γ does not have fixed points. | γ | = 0 follows from theorem 1.2.If γ does not have fixed points and | γ | = 0, the conclusion that γ is parabolicfollows from part (2). (cid:3) Proof of theorem 1.5
Before we go to prove theorem 1.5, let us make some preparations. The followingproposition has been proven in different literature (see [BF06, KN04, MP99]). Forcompleteness we still give the proof here.
Proposition 4.1.
Let M be a complete, visible CAT(0) Riemannian manifold.Assume that the mapping class group Mod S g,n acts properly on M . If g + n ≥ ,then there exists a point x ∈ M ( ∞ ) such that γ · x = x for all γ ∈ Mod S g,n .Proof. Let α be a simple closed curve on S g,n and τ α be the Dehn twist along α (see the definition of Dehn twist in [FM12]). Since 3 g + n ≥
5, there exists asimple closed curve β which is disjoint with α . Let τ β be the Dehn twist along β .Since < τ α , τ β > is a free abelian group of rank 2, by lemma 2.13, τ α is parabolic.Since M is proper visible, by remark 3.1 there exists a unique x ∈ M ( ∞ ) suchthat F ix ( τ α ) = { x } . Claim 1: If τ α · τ β = τ β · τ α , then F ix ( τ β ) = F ix ( τ α ) = { x } .Proof claim 1: Since τ α · τ β = τ β · τ α and τ α · x = x , τ α · ( τ β · x ) = τ β · x . F ix ( τ β ) = { x } follows from the fact that F ix ( τ α ) = { x } . Claim 2: For any simple closed curve β on S g,n , F ix ( τ β ) = { x } .Proof claim 2: Let β be a simple closed curve. Since 3 g + n ≥
5, the curvecomplex is connected (see theorem 4.3 in [FM12]). In particular, there exists asequence of simple closed curves { α i } ki =1 such that α = α, α k = β and α i T α i +1 = ∅ . Hence, by claim 1, F ix ( τ β ) = F ix ( τ α k − ) = F ix ( τ α k − ) = · · · = F ix ( τ α ) = { x } . The conclusion follows from part (2) of proposition 2.14 and claim 2. (cid:3)
Proposition 4.2.
Let M be a complete, visible CAT(0) Riemannian manifoldand the mapping class group Mod S g,n acts properly on M . If g + n ≥ , then anyinfinite ordered element φ ∈ Mod S g,n acts as a parabolic isometry.Proof. Suppose that there exists an element φ ∈ Mod S g,n with infinite order whichacts on M as a hyperbolic isometry. Then there exists a geodesic line γ : R → M ,an axis for φ , such that φ · γ ( t ) = γ ( | φ | + t ) for all t ∈ R . Since M is a visible CAT(0) space, it is not hard to see that F ix ( φ ) = { γ (+ ∞ ) , γ ( −∞ ) } . From proposition4.1 we assume that γ (+ ∞ ) is fixed by Mod S g,n . Let σ ∈ Mod S g,n , since σ fixes γ (+ ∞ ) there exists some a C > dist ( σ · γ ( n · | φ | ) , γ ( n · | φ | )) ≤ C forall n >
0. Hence dist (( φ − n · σ · φ n ) · γ (0) , γ (0)) ≤ C . Since the action is proper,there exists a subsequence { n i } such that φ − n i · σ · φ n i ≡ φ − n · σ · φ n , hence φ n − n i · σ = σ · φ n − n i . Since σ is arbitrary and φ has infinite order in Mod S g,n ,we can choose σ to be pseudo-Anosov such that { σ, φ } generates a free group ofrank 2 (see [Iva92]). Since φ n − n i · σ = σ · φ n − n i , the group < σ, φ > contains afree abelian subgroup of rank 2, which is a contradiction since < σ, φ > is a freegroup. (cid:3) Proposition 4.3.
Let M be a complete, visible CAT(0) Riemannian manifoldand the mapping class group Mod S g,n acts properly on M . If g + n ≥ , thenthere exists a horosphere H such that every torsion free subgroup of Mod S g,n actsproperly on H .Proof. From proposition 4.1, there exists a point x ∈ M ( ∞ ) such that Mod S g,n fixes x . Let H be a horosphere at x and G be a torsion free subgroup of Mod S g,n .By proposition 4.2 we know that G consists of parabolic isometries except the unit.Hence, by proposition 8.25 on page 275 in chapter II.8 of [BH99], H is G -invariantbecause G fixes x . Let d be the metric of M and d H be the induced metric on H . It is obvious that d H ( p, q ) ≥ d ( p, q ) for all p, q ∈ H . The conclusion that G acts properly on H follows easily from the facts that G acts properly on M and d H ( p, q ) ≥ d ( p, q ). (cid:3) Now we are ready to prove theorem 1.5.
Proof of theorem 1.5.
We argue it by contradiction. Assume that M ( S g,n ) admitsa complete, visible CAT(0) Riemannian metric ds . Let T ( S g,n ) be the universalcovering space of ( M ( S g,n ) , ds ), which is the Teichm¨uller space endowed with thepull back metric ds . By part (3) of proposition 2.14 we can assume that G isa torsion free subgroup of Mod S g,n with finite index. By proposition 4.3, thereexists a horosphere H such that G acts properly on H . Proposition 2.8 tells usthat H is homeomorphic to R g − n , in particular H is a contractible manifold.By the definition of action dimension we know that actdim ( G ) ≤ g − n. On the other hand, since 3 g + n ≥ G is a finite index subgroup of Mod S g,n ,by proposition 2.16, we have actdim ( G ) = 6 g − n which is a contradiction. (cid:3) Remark . If we carefully check the proof of theorem 1.5, we can conclude asfollows:
The Teichm¨uller space T ( S g,n ) (3 g + n ≥ admits no complete Mod S g,n -invariant CAT(0) Riemannian metric such that every Dehn twist has only onefixed point in the visual boundary of T ( S g,n ) . I am very grateful to Benson Farbfor this point.
ARABOLIC ISOMETRY 15
Remark . If we consider the case that n = 0 and g ≥
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