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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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












    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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      Tsein32

      562




      562




















          1 Answer
          1






          active

          oldest

          votes

















          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






















            Your Answer




            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "504"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f310808%2fare-mapping-class-groups-of-orientable-surfaces-good-in-the-sense-of-serre%23new-answer', 'question_page');

            );

            Post as a guest

















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            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

























                answered 1 hour ago









                Andy Putman

                30.1k5129207




                30.1k5129207



























                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f310808%2fare-mapping-class-groups-of-orientable-surfaces-good-in-the-sense-of-serre%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    Long meetings (6-7 hours a day): Being “babysat” by supervisor

                    Image
                    Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 31 down vote favorite 4 I hold a PhD in mechanical engineering with a 7 year background in self-guided, self-paced research including 2 years as a graduate level course instructor and a subject matter expert for various engineering projects in a university setting in the United States. For the last 10-11 months, I am working in a research industry in France where my task is to debug C++ spaghetti code written over a period of several years by my current supervisor. It is a team of 2: just I and my supervisor. Diurnally, I am posed with a "bug of the day" to get out of this C++ code. My supervisor generally gives me about 15-30 minutes to solve it and then accosts me about whether or not the task is completed and then rinses and repeats this until the end of the day. Off-late, my supervisor has been holding 6-7 hour long "meetin
                    Read more

                    Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Many fantasy worlds have multiple humanoid species (elves, orcs, humans, etc) living in the same world, in addition to that, they often have the ability to interbreed. And I can't help but question their believability... fantasy-races reproduction interspecies-relations share | improve this question edited Aug 9 at 14:46 Michael Kjörling ♦ 21.1k 10 85 161 asked Aug 9 at 13:13 Chlodio 118 6 closed as unclear what you're asking by MichaelK, Gryphon, John, Vincent, Frostfyre Aug 9 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. 3 Stack Exchange exist to
                    Read more

                    Confectionery

                    Image
                    Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
                    Read more