Are mapping class groups of orientable surfaces good in the sense of Serre?

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A group G is called ‘good’ if the canonical map $GtohatG$ to the profinite completion induces isomorphisms $H^i(hatG,M)to H^i(G,M)$ for any finite $G$-module $M$. I’ve had multiple academics in my department claim to me that the orientation-preserving mapping class group of an orientable surface of genus $g$ with $n$ boundary components satisfies this property, though none have been able to give me a reference.



I think the easiest way to prove this would be to exhibit a finite index solvable subgroup, which for mapping class groups is equivalent to a finite index abelian subgroup (A theorem in Sullivan’s ‘genetics of homotopy theory’ says this implies that G is good). I was thinking of the subgroup of homeomorphisms which act trivially on cohomology with $mathbbZ/nmathbbZ$ coefficients, but I have very little experience with solvable groups and couldn’t even figure out if this was a reasonable guess.



So, does anyone know if there exists a finite index solvable subgroup of the mapping class groups? Or is there a different proof that the mapping class groups are good?



Thanks for any help.






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    up vote
    3
    down vote

    favorite












    A group G is called ‘good’ if the canonical map $GtohatG$ to the profinite completion induces isomorphisms $H^i(hatG,M)to H^i(G,M)$ for any finite $G$-module $M$. I’ve had multiple academics in my department claim to me that the orientation-preserving mapping class group of an orientable surface of genus $g$ with $n$ boundary components satisfies this property, though none have been able to give me a reference.



    I think the easiest way to prove this would be to exhibit a finite index solvable subgroup, which for mapping class groups is equivalent to a finite index abelian subgroup (A theorem in Sullivan’s ‘genetics of homotopy theory’ says this implies that G is good). I was thinking of the subgroup of homeomorphisms which act trivially on cohomology with $mathbbZ/nmathbbZ$ coefficients, but I have very little experience with solvable groups and couldn’t even figure out if this was a reasonable guess.



    So, does anyone know if there exists a finite index solvable subgroup of the mapping class groups? Or is there a different proof that the mapping class groups are good?



    Thanks for any help.






    at.algebraic-topology riemann-surfaces







    share|cite|improve this question























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      A group G is called ‘good’ if the canonical map $GtohatG$ to the profinite completion induces isomorphisms $H^i(hatG,M)to H^i(G,M)$ for any finite $G$-module $M$. I’ve had multiple academics in my department claim to me that the orientation-preserving mapping class group of an orientable surface of genus $g$ with $n$ boundary components satisfies this property, though none have been able to give me a reference.



      I think the easiest way to prove this would be to exhibit a finite index solvable subgroup, which for mapping class groups is equivalent to a finite index abelian subgroup (A theorem in Sullivan’s ‘genetics of homotopy theory’ says this implies that G is good). I was thinking of the subgroup of homeomorphisms which act trivially on cohomology with $mathbbZ/nmathbbZ$ coefficients, but I have very little experience with solvable groups and couldn’t even figure out if this was a reasonable guess.



      So, does anyone know if there exists a finite index solvable subgroup of the mapping class groups? Or is there a different proof that the mapping class groups are good?



      Thanks for any help.






      at.algebraic-topology riemann-surfaces







      share|cite|improve this question













      A group G is called ‘good’ if the canonical map $GtohatG$ to the profinite completion induces isomorphisms $H^i(hatG,M)to H^i(G,M)$ for any finite $G$-module $M$. I’ve had multiple academics in my department claim to me that the orientation-preserving mapping class group of an orientable surface of genus $g$ with $n$ boundary components satisfies this property, though none have been able to give me a reference.



      I think the easiest way to prove this would be to exhibit a finite index solvable subgroup, which for mapping class groups is equivalent to a finite index abelian subgroup (A theorem in Sullivan’s ‘genetics of homotopy theory’ says this implies that G is good). I was thinking of the subgroup of homeomorphisms which act trivially on cohomology with $mathbbZ/nmathbbZ$ coefficients, but I have very little experience with solvable groups and couldn’t even figure out if this was a reasonable guess.



      So, does anyone know if there exists a finite index solvable subgroup of the mapping class groups? Or is there a different proof that the mapping class groups are good?



      Thanks for any help.






      at.algebraic-topology riemann-surfaces




      at.algebraic-topology riemann-surfaces






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      asked 2 hours ago









      Tsein32

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          This is an open and probably very difficult question. There have been purported proofs (for instance, this one), but they have all had fatal flaws.



          The mapping class group is definitely not virtually abelian (or virtually solvable; for other readers, the relevant fact [alluded to by the OP] is that all solvable subgroups of the mapping class group are virtually abelian). For instance, it contains huge numbers of nonabelian free subgroups. One easy-to-state example due to Ishida is that the subgroup generated by two Dehn twists about curves that intersect at least twice is free. See



          Ishida, Atsushi,
          The structure of subgroup of mapping class groups generated by two Dehn twists.
          Proc. Japan Acad. Ser. A Math. Sci. 72 (1996), no. 10, 240-241.






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            1 Answer
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            1 Answer
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            active

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            active

            oldest

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            up vote
            4
            down vote













            This is an open and probably very difficult question. There have been purported proofs (for instance, this one), but they have all had fatal flaws.



            The mapping class group is definitely not virtually abelian (or virtually solvable; for other readers, the relevant fact [alluded to by the OP] is that all solvable subgroups of the mapping class group are virtually abelian). For instance, it contains huge numbers of nonabelian free subgroups. One easy-to-state example due to Ishida is that the subgroup generated by two Dehn twists about curves that intersect at least twice is free. See



            Ishida, Atsushi,
            The structure of subgroup of mapping class groups generated by two Dehn twists.
            Proc. Japan Acad. Ser. A Math. Sci. 72 (1996), no. 10, 240-241.






            share|cite|improve this answer


























              up vote
              4
              down vote













              This is an open and probably very difficult question. There have been purported proofs (for instance, this one), but they have all had fatal flaws.



              The mapping class group is definitely not virtually abelian (or virtually solvable; for other readers, the relevant fact [alluded to by the OP] is that all solvable subgroups of the mapping class group are virtually abelian). For instance, it contains huge numbers of nonabelian free subgroups. One easy-to-state example due to Ishida is that the subgroup generated by two Dehn twists about curves that intersect at least twice is free. See



              Ishida, Atsushi,
              The structure of subgroup of mapping class groups generated by two Dehn twists.
              Proc. Japan Acad. Ser. A Math. Sci. 72 (1996), no. 10, 240-241.






              share|cite|improve this answer
























                up vote
                4
                down vote










                up vote
                4
                down vote









                This is an open and probably very difficult question. There have been purported proofs (for instance, this one), but they have all had fatal flaws.



                The mapping class group is definitely not virtually abelian (or virtually solvable; for other readers, the relevant fact [alluded to by the OP] is that all solvable subgroups of the mapping class group are virtually abelian). For instance, it contains huge numbers of nonabelian free subgroups. One easy-to-state example due to Ishida is that the subgroup generated by two Dehn twists about curves that intersect at least twice is free. See



                Ishida, Atsushi,
                The structure of subgroup of mapping class groups generated by two Dehn twists.
                Proc. Japan Acad. Ser. A Math. Sci. 72 (1996), no. 10, 240-241.






                share|cite|improve this answer














                This is an open and probably very difficult question. There have been purported proofs (for instance, this one), but they have all had fatal flaws.



                The mapping class group is definitely not virtually abelian (or virtually solvable; for other readers, the relevant fact [alluded to by the OP] is that all solvable subgroups of the mapping class group are virtually abelian). For instance, it contains huge numbers of nonabelian free subgroups. One easy-to-state example due to Ishida is that the subgroup generated by two Dehn twists about curves that intersect at least twice is free. See



                Ishida, Atsushi,
                The structure of subgroup of mapping class groups generated by two Dehn twists.
                Proc. Japan Acad. Ser. A Math. Sci. 72 (1996), no. 10, 240-241.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 42 mins ago

























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                Andy Putman

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