In this note we show that many subgroups of mapping class groups of
infinite-type surfaces without boundary have trivial centers, including all
normal subgroups. Using similar techniques, we show that every nontrivial
normal subgroup of a big mapping class group contains a nonabelian free group.
In contrast, we show that no big mapping class group satisfies the strong Tits
alternative enjoyed by finite-type mapping class groups. We also give examples
of big mapping class groups that fail to satisfy even the classical Tits
alternative and give a proof that every countable group appears as a subgroup
of some big mapping class group.Comment: 6 pages, 1 figur

Lanier, Justin

Loving, Marissa

English

arXiv

In this note we show that many subgroups of mapping class groups of infinite-type surfaces without boundary have trivial centers, including all normal subgroups. Using similar techniques, we show that every nontrivial normal subgroup of a big mapping class group contains a nonabelian free group. In contrast, we show that no big mapping class group satisfies the strong Tits alternative enjoyed by finite-type mapping class groups. We also give examples of big mapping class groups that fail to satisfy even the classical Tits alternative; consequently, these examples are not linear

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GLASNIK MATEMATIČKIVol. 55(75)(2020), 85 – 91CENTERS OF SUBGROUPS OF BIG MAPPING CLASSGROUPS AND THE TITS ALTERNATIVEJustin Lanier and Marissa LovingGeorgia Institute of Technology, USA and University of Illinois, USAAbstract. In this note we show that many subgroups of mappingclass groups of infinite-type surfaces without boundary have trivial centers,including all normal subgroups. Using similar techniques, we show thatevery nontrivial normal subgroup of a big mapping class group containsa nonabelian free group. In contrast, we show that no big mapping classgroup satisfies the strong Tits alternative enjoyed by finite-type mappingclass groups. We also give examples of big mapping class groups that failto satisfy even the classical Tits alternative; consequently, these examplesare not linear.Let S be a connected orientable surface without boundary that is of in-finite type, so that π1(S) is infinitely generated. The mapping class groupMap(S) is the group of homotopy classes of orientation-preserving homeo-morphisms of S. The mapping class group of an infinite-type surface is oftencalled a big mapping class group. Similarly, let Sg be the compact connectedorientable surface of genus g and let Map(Sg) be its mapping class group.In this note we address several questions about subgroups of Map(S). Wefirst prove some results about the triviality of centers of subgroups of Map(S).We then show that Map(S) never satisfies the strong Tits alternative enjoyedby Map(Sg), as well as some related results.Centers. The center of Map(Sg) is trivial for g ≥ 3, while its center is acyclic group generated by the hyperelliptic involution when g = 1 or 2 ([4,Section 3.4]). The centers of mapping class groups of finite-type surfaces withpunctures and boundary components were computed by Paris–Rolfsen ([12,Theorem 5.6]). Although Dehn twists about boundary components are always2020 Mathematics Subject Classification. 20F34, 57M60, 37E30.Key words and phrases. Mapping class groups, normal subgroups, free subgroups, Titsalternative, centers.8586 J. LANIER AND M. LOVINGcentral, for finite-type surfaces without boundary there are only finitely manyexceptional cases of low complexity where the center is nontrivial.These results where the center is trivial follow from the existence of a(stable) Alexander system—a collection Γ of simple closed curves and arcs inthe surface such that no nontrivial mapping class fixes Γ up to isotopy ([4,Section 2.3]). Hernández–Morales–Valdez recently proved that the Alexandermethod carries over to the infinite-type setting ([11]).Theorem 1 (Hernández–Morales–Valdez). Let S be an orientable surfaceof infinite topological type, with possibly non-empty boundary. There existsa collection of essential arcs and simple closed curves Γ on S such that anyorientation-preserving homeomorphism fixing pointwise the boundary of S thatpreserves the isotopy classes of the elements of Γ, is isotopic to the identity.With this result in hand, it is straightforward to compute the centers ofbig mapping class groups, just as in the finite-type case. Throughout thisnote, we restrict our attention to surfaces without boundary. When S hasboundary components, Dehn twists about boundary components are alwayscentral, and the center of Map(S) can be analyzed by applying a cappinghomomorphism to each boundary component ([4, Proposition 3.19]).Proposition 2. If S is an infinite-type surface without boundary, thecenter of Map(S) is trivial.Proof. Suppose f ∈ Map(S) is nontrivial; we show that f is not central.By Theorem 1, there exists a curve c such that f(c) 6= c. Consider fTcf−1,which equals Tf(c). If f were central, the product fTcf−1 would also equalTc. But Tf(c) 6= Tc, since curves are compact and f(c) 6= c.If we now consider subgroups of Map(S), a similar proof goes through forany subgroup that contains every Dehn twist, or even some nonzero powerof every Dehn twist. Examples of such subgroups include: the pure mappingclass group PMod(S), which is the subgroup that acts trivially on the spaceof ends of S; the compactly supported mapping class group Mapc(S); andthe level m subgroup Map(S)[m], which is the subgroup that acts trivially onH1(S;Z/mZ). Thus all of these subgroups of Map(S) have trivial center.Patel–Vlamis showed that the center of PMod(S) is trivial in the casewhere S is an infinite-type surface with finite genus and without boundary([13, Lemma 3.3]). Their proof uses the same standard trick of conjugatingDehn twists, but leverages the compact-open topology on Map(S) as a stand-in for the Alexander method.Note that PMod(S), Mapc(S), and Map(S)[m] are all normal in Map(S).In fact, we can extend the result of Proposition 2 to the centers of all normalsubgroups of Map(S).Theorem 3. If S is an infinite-type surface without boundary, every nor-mal subgroup of Map(S) has trivial center.CENTERS OF SUBGROUPS OF BIG MAPPING CLASS GROUPS 87Proof. Let N be any normal subgroup of Map(S) and let f be anynontrivial element of N . As f is nontrivial, by Theorem 1 there exists a curvec such that f(c) 6= c. SinceN is normal, the product f(Tcf−1T−1c ) = Tf(c)T−1cis also in N . We will show that f and Tf(c)T−1c do not commute. ConjugatingTf(c)T−1c by f yields Tf2(c)T−1f(c) ∈ N . If Tf(c)T−1c and f were to commute, wewould have Tf2(c)T−1f(c) = Tf(c)T−1c . But this is not possible. First, considerthe case when i(c, f(c)) = 0. Since c and f(c) are distinct, so are f(c) andf2(c). But then the two products are clearly distinct multitwists, even if fswaps c and f(c). Then consider the case when i(c, f(c)) ≥ 1. Rearranging thesupposed equality yields Tf2(c) = Tf(c)T−1c Tf(c). The product on the right-hand side is a partial pseudo-Anosov on the surface filled by c and f(c), sinceits conjugate T 2f(c)T−1c is pseudo-Anosov by Penner’s construction ([14]); seealso [4, p. 396]. However, the left-hand side of the equation is a Dehn twist,a contradiction. We therefore have that f does not commute with Tf(c)T−1cand so conclude that the center of N is trivial.We note that this argument also holds for finite-type surfaces with stableAlexander systems and recovers the corresponding fact about normal sub-groups in that setting. This is a well-known fact for finite-type mapping classgroups. A proof can be given by considering the action of the normal sub-group on the space of projective measured foliations; see, for instance, [2, p.52]. Theorem 3 can be also be proved by appealing to the finite-type result,in a similar way to how we proceed in our proof of Proposition 7.The Tits alternative. Of course, for any S there do exist subgroups ofMap(S) that have nontrivial center. For instance, Map(S) has abelian sub-groups, such as any cyclic subgroup or more generally any subgroup generatedby mapping classes supported on disjoint subsurfaces. It is a theorem, provedindependently by Ivanov and McCarthy, that Map(Sg) satisfies a strong ver-sion of the Tits alternative: every subgroup is either virtually abelian orcontains a nonabelian free group ([7, 10]). In contrast, big mapping classgroups do not satisfy this strong version of the Tits alternative.Theorem 4. For every surface S of infinite type without boundary,Map(S) contains the restricted wreath product Z ≀ Z as a subgroup and sodoes not satisfy the strong Tits alternative.Proof. By the classification theorem of Kerékjártó and Richards ([8,15]), every surface S of infinite type without boundary contains at least oneof the following: infinite genus, a countable collection of isolated punctures,or a Cantor set of punctures where none of these punctures is accumulatedby isolated punctures. We consider these possibilities in turn.Let S be a surface with infinite genus. Then S has at least one endaccumulated by genus. Consider the subgroup of Map(S) generated by a88 J. LANIER AND M. LOVINGhandle shift h (possibly one-ended) and a mapping class g supported on oneof the handles in the support of h (which is necessarily of infinite order). (See[13] for a definition of handle shifts and [9] for their classification.) Then 〈g, h〉is isomorphic to the restricted wreath product Z ≀ Z, which is not virtuallyabelian and does not contain a nonabelian free group.Similarly, let S be a surface with a countable set of isolated punctures.We will construct a (one-ended) puncture shift, in analogy with a handle shift.Take a subset P of the countable set of isolated punctures that accumulate toan end e. We may index P by Z as {. . . p−1, p0, p1, . . . }. Take a bi-infinite stripΣ containing P that has two boundary components that are arcs connecting eto itself, so that Σ is homeomorphic to R× [−1, 1] with punctures at Z×{0}.This is depicted in Figure 1. Let H be a homeomorphism that fixes theboundary arcs as well as S \Σ and that pushes each pi to pi+1. Let h be themapping class of H and let g be a nontrivial mapping class supported on adisk containing only p0 and p1, such that the disk and its translates underpowers of h2 are all pairwise disjoint. Then the subgroup 〈g, h2〉 is isomorphicto the restricted wreath product Z ≀ Z.Finally, let S be a surface with a Cantor set of ends and no infinite setof isolated ends. Then none of the ends in the Cantor set C is accumulatedby isolated ends. We will construct a (one-ended) Cantor disk shift. Pick oneend e ∈ C. Partition C \ e into a countable number of Cantor sets and indexthese by Z: {. . . , C−1, C0, C1, . . . }. Let Di be a disk that supports Ci and noother ends. Let h be a mapping class that takes each Di to Di+1. Let g beany nontrivial mapping class supported in D0. Then the subgroup 〈g, h〉 isisomorphic to the restricted wreath product Z ≀ Z.Figure 1. A puncture shift.The classical Tits alternative replaces virtually abelian with virtually solv-able. Since the Z ≀ Z subgroup of Map(S) from Proposition 4 is solvable, itCENTERS OF SUBGROUPS OF BIG MAPPING CLASS GROUPS 89is not a counterexample to Map(S) satisfying the classical Tits alternative.We now give an example of a big surface whose mapping class group does notsatisfy the classical Tits alternative.Example 5. Thompson’s group F does not satisfy the classical Tits al-ternative, as it is not virtually solvable and does not have a nonabelian freesubgroup ([3, Corollary 3.3]). Let S be R2 \ C, where C is a Cantor set embed-ded in [0, 1]×{0}. Let F act by piecewise-linear homeomorphisms on [0, 1]×Rso that it also acts faithfully by homeomorphsisms on C. All of these homeo-morphisms are orientation preserving and so F is a subgroup of Homeo+(S).Since F acts faithfully on C, the quotient map Homeo+(S) → Mod(S) is in-jective on this F subgroup. It follows that F is a subgroup of Map(S), whichtherefore does not satisfy the classical Tits alternative.The construction in Example 5 can be extended to other examples ofbig surfaces that have a Cantor set in their space of ends. For instance, theconstruction holds for any surface S where Ends(S) contains a Cantor setsuch that every point in the Cantor set is in the same mapping class grouporbit. Note that no big mapping class group containing Thompson’s group Fas a subgroup is linear, since F , not satisfying the classical Tits alternative,fails to be linear ([16]). (In the finite-type setting, linearity of most mappingclass groups remains an open question.) Still, there are many surfaces wherethe construction of Example 5 does not hold.Question 6. Do there exist infinite-type surfaces whose mapping classgroups satisfy the classical Tits alternative?Another way of framing the Tits alternative is that there are groups thatare not subgroups of any finite-type mapping class group. Thompson’s groupF is of course an example of such a group. On the other hand, consideringfinite subgroups in the finite-type setting yields a very different phenomenon:every finite group is a subgroup of Map(Sg) for some g. See [5] and [4,Theorem 7.12].Allcock and Winkelmann each proved that for every countable group G,there is a complete hyperbolic surface S of infinite type without boundary suchthat G is the isometry group of S ([1, 17]). Like Example 5, their argumentsimply that there are big mapping class groups that do not satisfy the classicalTits alternative, but they are not sufficient to resolve Question 6.Even if Map(S) does not satisfy the classical Tits alternative, it is possiblethat it satisfies some variant. One result in this direction was proved byHurtado–Militon, who showed that a version of the Tits alternative holds forthe smooth mapping class group of S whenever S has finite genus and a spaceof ends homeomorphic to a Cantor set ([6, Theorem 1.3]).We conclude with a result showing that whenever Map(S) does not satisfythe classical Tits alternative, the subgroup given as a counterexample cannot90 J. LANIER AND M. LOVINGbe normal in Map(S). Observe that in the case of a surface S with boundary,a normal subgroup consisting of boundary twists is abelian and thereforesatisfies the other half of the alternative.Proposition 7. If S is an infinite-type surface without boundary, everynontrivial normal subgroup of Map(S) contains a nonabelian free group.Proof. Let N be any normal subgroup of Map(S) and let f be anynontrivial element of N . Then by Theorem 1 there is a curve c such thatf(c) 6= c. We have that g = f(Tcf−1T−1c ) = Tf(c)T−1c is also in N . Theelement g has as support a compact subsurface Σ′. Let Σ be a connectedsubsurface of S containing Σ′ such that no boundary component of Σ′ ishomotopic to any boundary component of Σ. Then g does not act as a Dehntwist about any of the boundary components of Σ.Consider MapΣ(S), the subgroup of Map(S) that sends Σ to itself. Ob-serve that the restrictionMapΣ(S)|Σ is isomorphic to Map(Σ). By conjugatingg by some h ∈ MapΣ(S), we may then produce a nonabelian free group as asubgroup of 〈g, hgh−1〉 by applying the Tits alternative to normal subgroupsin the finite-type setting.Acknowledgements.The authors would like to thank Jim Belk, Kevin Kordek, and Jing Tao forhelpful conversations and Santana Afton, Tyrone Ghaswala, Chris Leininger,Dan Margalit, and Nick Vlamis for their comments on a draft of this note. Wewould also like to thank the anonymous referee for their helpful suggestions.References[1] D. Allcock, Hyperbolic surfaces with prescribed infinite symmetry groups, Proc. Amer.Math. Soc. 134 (2006), 3057–3059.[2] T. Brendle and D. Margalit, Normal subgroups of mapping class groups and the meta-conjecture of Ivanov, J. Amer. Math. Soc. 32 (2019), 1009–1070.[3] M. G. Brin and C. C. Squier, Groups of piecewise linear homeomorphisms of the realline, Invent. Math. 79 (1985), 485–498.[4] B. Farb and D. Margalit, A primer on mapping class groups, Princeton UniversityPress, Princeton, 2012.[5] L. Greenberg, Maximal groups and signatures, Ann. of Math. Studies, No. 79, 1974,207–226.[6] S. Hurtado and E. Militon, Distortion and Tits alternative in smooth mapping classgroups, Trans. Amer. Math. Soc. 371 (2019), 8587–8623.[7] N. V. Ivanov, Algebraic properties of the Teichmüller modular group, Dokl. Akad.Nauk SSSR 275 (1984), 786–789.[8] B. Kerékjártó, Vorlesungen über Topologie. I. Springer, Berlin, 1923.[9] S. Afton, S. Freedman, J. Lanier and L. Yin, Generators, relations, and homomor-phisms of big mapping class groups, in preparation.[10] J. McCarthy, A “Tits-alternative” for subgroups of surface mapping class groups,Trans. Amer. Math. Soc. 291 (1985), 583–612.[11] J. Hernández Hernández, I. Morales and F. Valdez, The Alexander method for infinite-type surfaces, Michigan Math. J. 68 (2019), 743–753.CENTERS OF SUBGROUPS OF BIG MAPPING CLASS GROUPS 91[12] L. Paris and D. Rolfsen, Geometric subgroups of mapping class groups, J. ReineAngew. Math. 521 (2000), 47–83.[13] P. Patel and N. G. Vlamis, Algebraic and topological properties of big mapping classgroups, Algebr. Geom. Topol. 18 (2018), 4109–4142.[14] R. C. Penner, A construction of pseudo-Anosov homeomorphisms, Trans. Amer. Math.Soc. 310 (1988), 179–197.[15] I. Richards, On the classification of noncompact surfaces, Trans. Amer. Math. Soc.106 (1963), 259–269.[16] J. Tits, Free subgroups in linear groups, J. Algebra 20 (1972), 250–270.[17] J. Winkelmann, Realizing countable groups as automorphism groups of Riemann sur-faces, Doc. Math. 6 (2001), 413–417.J. LanierSchool of MathematicsGeorgia Institute of Technology686 Cherry St., Atlanta, GA 30332USAE-mail : jlanier8@gatech.eduM. LovingDepartment of MathematicsUniversity of Illinois1409 W. Green Street, Urbana, IL 61801USAE-mail : mloving2@illinois.eduReceived : 20.7.2019.Revised : 14.8.2019.

Centers of subgroups of big mapping class groups and the Tits alternative

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