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Involving symmetries of Riemann surfaces to a study of the mapping class group

Gromadzki, Grzegorz; Stukow, Michal

Abstract

A pair of symmetries (σ, τ ) of a Riemann surface X is said to be perfect if their product belongs to the derived subgroup of the group Aut+(X) of orientation preserving automorphisms. We show that given g 6= 2, 3, 5, 7 there exists a Riemann surface X of genus g admitting a perfect pair of symmetries of certain topological type. On the other hand we show that a twist can be written as a product of two symmetries of the same type which leads to a decomposition of a twist as a product of two commutators: one from M0 which entirely lives on a Riemann surface and one from M±0 . As a result we obtain the perfectness of the mapping class group Mg for such g relying only on results of Birman [1] but not on influential paper of Powell [6] nor on Johnson's rediscovery of Dehn lantern relation [3] and nor on recent results of Korkmaz-Ozbagci [4] who found explicit presentation of a twist as a product of two commutators.

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Publ. Mat. 48 (2004), 103–106 INVOLVING SYMMETRIES OF RIEMANN SURFACES TO A STUDY OF THE MAPPING CLASS GROUP Grzegorz Gromadzki and Micha l Stukow Abstract A pair of symmetries (σ, τ ) of a Riemann surface Xis said to be perfect if their product belongs to the derived subgroup of the group Aut+(X) of orientation preserving automorphisms. We show that given g6= 2,3,5,7 there exists a Riemann surface Xof genus gadmitting a perfect pair of symmetries of certain topological type. On the other hand we show that a twist can be written as a product of two symmetries of the same type which leads to a decomposition of a twist as a product of two commutators: one from M0which entirely lives on a Riemann surface and one from M±0. As a result we obtain the perfectness of the mapping class group Mgfor such grelying only on results of Birman [1] but not on influential paper of Powell [6] nor on Johnson’s rediscovery of Dehn lantern relation [3] and nor on recent results of Korkmaz-Ozbagci [4] who found explicit presentation of a twist as a product of two commutators. Let Xbe a compact surface of genus g≥2 and let M±=M± gbe the extended mapping class group which is the group of isotopy classes of homeomorphisms of X. The mapping class group Mis the subgroup of M±consisting of the isotopy classes of orientation preserving homeomorphisms. It is well known [5] that Mis generated by 3g−1 canonical twists and together with a symmetry (by which we mean an orientation reversing involution) they generate M±. Classical results of Harnack and Weichhold assert that the conjugacy class of a symmetry σ in M±is determined by its topological type εk, where kis the number of connected components of the set Fix(σ) of fixed points of σ(ovals in Hilbert’s terminology) and ε=ε(σ) is the separability index, which is equal to −1 or +1 according to whether X\Fix(σ) is connected or 2000 Mathematics Subject Classification. Primary: 57A05, 30F10; Secondary: 57M50, 20F05. Key words. Mapping class group, symmetries of Riemann surfaces. Both authors supported by BW5100-5-0080-3. 104 G. Gromadzki, M. Stukow not. Let σbe a symmetry with kovals c1,...,ckand let hcibe a twist about ci. Then σhcibelongs to the class of a non-separating symmetry with k−1 ovals c1,...,ˆci,...,ck. Iterating this construction for a separating symmetry with g+ 1 ovals we obtain Theorem 1. Given g≥1and kin range 0≤k≤g−1there is a pair of non-separating symmetries on a surface of genus gwith kand k+ 1 ovals whose product is a twist. In particular the extended mapping class group is generated by involutions. A pair of symmetries of a Riemann surface Xis said to be perfect if their product belong to the derived subgroup of the group of automorphisms of X. Lemma 2. Given an even integer g≥4there exists a Riemann surface Xof genus gadmitting a perfect pair of non-separating symmetries with 1and 2ovals. For a given odd integer g≥9, there exists a Riemann surface of genus gadmitting a perfect pair of non-separating symmetries with 2and 3ovals. Proof: We shall prove the lemma using theory of Fuchsian and NEC groups; we send the reader to [2] where he can find necessary background. Let Λ be an NEC group with signature (0; +; [−]; {(2,s . . ., 2,4)}), where s= (g+ 4)/2 and let G= D4⊕Z2=ha, b |a2, b2,(ab)4i ⊕ hx|x2i. Let θ: Λ →Gbe an epimorphism induced by the assignment: θ(e) = 1, θ(c0) = θ(cs+1) = a,θ(c1) = x,θ(c2) = b(ab)2(s+1),θ(c3) = x(ab)2and θ(ci) = b(ab)2(i+s). Neither a reflection nor an elliptic elements belong Γ = ker θand so it is a surface Fuchsian group. Thus X=H/Γ is a Riemann surface on which Gacts as a group of automorphisms. By the Hurwitz Riemann formula Xhas genus g. Now we shall show that σ=xis a non-separating symmetry of Xwith 1 oval. For, observe first that σis central in Gand consider induced epimorphism ˜ θ: Λ →G/hσi. Then Γσ= ker ˜ θis a surface NEC group which by [2, 2.3.3] has 1 empty period-cycle. In addition by [2, 2.1.3] Γσhas the sign −and so σis a non-separating symmetry with 1 oval indeed. In the same way one can show that τ=x(ab)2is a non-separating symmetry with 2 ovals. Finally, στ = [ab, aσ] belongs to the derived group of Aut+(X). The case of odd gis similar. As before G= D4⊕Z2but now let Λ be an NEC group with signature (0; +; [−]; {(2,s . . ., 2,4,4)}), where s= (g+ 1)/2. We define an epimorphism θ: Λ →Gby the assignment: θ(e) = 1, θ(c0) = θ(cs+2) = a,θ(c1) = θ(c4) = x,θ(c2) = θ(cs+1) = b, θ(c3) = (ab)2xand θ(ci) = (ab)2(i+s)afor the remaining reflections ci. Symmetries in Mapping Class Group 105 Now Xis a Riemann surface of genus gand τ=x(ab)2,σ=xrepresent two non-separating symmetries of Xwith 2 and 3 ovals respectively. Corollary 3. The mapping class group Mgof a compact Riemann surface of genus g6= 2,3,5,7is perfect. Proof: Take k= 1 or 2 if gis even or odd respectively. By Theorem 1 there is a pair (σ0, τ0) of non-separating symmetries with kand k+1 ovals whose product σ0τ0is a twist and by Lemma 2 there is a perfect pair (σ, τ) of such symmetries. By the mentioned above theorem of Weichhold, σ0=ασα−1, and τ0=βτβ−1for some α, β ∈ M. So h=α−1σ0τ0α= στ[τ, α−1β] is a twist and thus M ⊆ M±0. Now M±/M= Z2and by [1], |M/M0| ≤ 2. So M±/M0is abelian as a group of order ≤4 and therefore M±0⊆ M0. Remark. Observe that the proof of the second part of Lemma 2 works only for s≥5 which forces g≥9. We guess that probably also for g=3,5,7, Riemann surfaces which admits perfect pairs of non-separating symmetries with kand k+ 1 ovals exists. However our aim was rather to show that the extended mapping class group is generated by classes of symmetries and to show how one can use symmetries surfaces to the study of the mapping class involving methods of Riemann surfaces than in the proof of its perfectness for itself and so we have skipped this question. References [1] J. S. Birman, Abelian quotients of the mapping class group of a 2-manifold, Bull. Amer. Math. Soc. 76 (1970), 147–150. [2] E. Bujalance, J. J. Etayo, J. M. Gamboa and G. Gromadzki, “Automorphism groups of compact bordered Klein surfaces. A combinatorial approach”, Lecture Notes in Mathematics 1439, SpringerVerlag, Berlin, 1990. [3] D. L. Johnson, Homeomorphisms of a surface which act trivially on homology, Proc. Amer. Math. Soc. 75(1) (1979), 119–125. [4] M. Korkmaz and B. Ozbagci, Minimal number of singular fibers in a Lefschetz fibration, Proc. Amer. Math. Soc. 129(5) (2001), 1545–1549. [5] W. B. R. Lickorish, A finite set of generators for the homeotopy group of a 2-manifold, Proc. Cambridge Philos. Soc. 60 (1964), 769–778. [6] J. Powell, Two theorems on the mapping class group of a surface, Proc. Amer. Math. Soc. 68(3) (1978), 347–350. 106 G. Gromadzki, M. Stukow Institute of Mathematics University of Gda´nsk Wita Stwosza 57 80-952 Gda´nsk Poland E-mail address:[email protected]l E-mail address:[email protected] Rebut el 24 de febrer de 2003.