Smooth vector fields on a surface modulo diffeomorphisms












7












$begingroup$


Let $Sigma$ be a two-dimensional connected smooth manifold without boundary. (Feel free to assume it is compact and orientable.)



Let $mathcal{X}(Sigma)$ denote the smooth vector fields on $Sigma$ and let $operatorname{Diff}(Sigma)$ denote the group of diffeomorphisms of $Sigma$. Clearly $operatorname{Diff}(Sigma)$ acts on $mathcal{X}(Sigma)$.



I would expect naively that the space of orbits $mathcal{M}:=mathcal{X}(Sigma)/operatorname{Diff}(Sigma)$ would be finite-dimensional. Is this actually the case?



Question



What can one say about $mathcal{M}$ in general?



Any references where I could read about this question would be appreciated.






dg.differential-geometry differential-topology smooth-manifolds infinite-dimensional-manifolds







share|cite|improve this question











$endgroup$

















    7












    $begingroup$


    Let $Sigma$ be a two-dimensional connected smooth manifold without boundary. (Feel free to assume it is compact and orientable.)



    Let $mathcal{X}(Sigma)$ denote the smooth vector fields on $Sigma$ and let $operatorname{Diff}(Sigma)$ denote the group of diffeomorphisms of $Sigma$. Clearly $operatorname{Diff}(Sigma)$ acts on $mathcal{X}(Sigma)$.



    I would expect naively that the space of orbits $mathcal{M}:=mathcal{X}(Sigma)/operatorname{Diff}(Sigma)$ would be finite-dimensional. Is this actually the case?



    Question



    What can one say about $mathcal{M}$ in general?



    Any references where I could read about this question would be appreciated.






    dg.differential-geometry differential-topology smooth-manifolds infinite-dimensional-manifolds







    share|cite|improve this question











    $endgroup$















      7












      7








      7





      $begingroup$


      Let $Sigma$ be a two-dimensional connected smooth manifold without boundary. (Feel free to assume it is compact and orientable.)



      Let $mathcal{X}(Sigma)$ denote the smooth vector fields on $Sigma$ and let $operatorname{Diff}(Sigma)$ denote the group of diffeomorphisms of $Sigma$. Clearly $operatorname{Diff}(Sigma)$ acts on $mathcal{X}(Sigma)$.



      I would expect naively that the space of orbits $mathcal{M}:=mathcal{X}(Sigma)/operatorname{Diff}(Sigma)$ would be finite-dimensional. Is this actually the case?



      Question



      What can one say about $mathcal{M}$ in general?



      Any references where I could read about this question would be appreciated.






      dg.differential-geometry differential-topology smooth-manifolds infinite-dimensional-manifolds







      share|cite|improve this question











      $endgroup$




      Let $Sigma$ be a two-dimensional connected smooth manifold without boundary. (Feel free to assume it is compact and orientable.)



      Let $mathcal{X}(Sigma)$ denote the smooth vector fields on $Sigma$ and let $operatorname{Diff}(Sigma)$ denote the group of diffeomorphisms of $Sigma$. Clearly $operatorname{Diff}(Sigma)$ acts on $mathcal{X}(Sigma)$.



      I would expect naively that the space of orbits $mathcal{M}:=mathcal{X}(Sigma)/operatorname{Diff}(Sigma)$ would be finite-dimensional. Is this actually the case?



      Question



      What can one say about $mathcal{M}$ in general?



      Any references where I could read about this question would be appreciated.






      dg.differential-geometry differential-topology smooth-manifolds infinite-dimensional-manifolds




      dg.differential-geometry differential-topology smooth-manifolds infinite-dimensional-manifolds






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 12 hours ago









      YCor

      28.2k483136




      28.2k483136










      asked 12 hours ago









      José Figueroa-O'FarrillJosé Figueroa-O'Farrill

      25.5k373150




      25.5k373150






















          1 Answer
          1






          active

          oldest

          votes


















          10












          $begingroup$

          This is not finite dimensional. For example, consider non-vanishing vector fields on $T^2=mathbb R^2/mathbb Z^2$, transversal to vertical circles. Any such field defines a self diffeo $S^1to S^1$ on a vertical circle, called the return map. Suppose that such a diffeo $varphi$ has $n$ fixed points $x_i$ (they correspond to closed orbits of the field). Then for each fixed point $x_iin S^1$ we have a linear map on the tangent space $dvarphi: T_{x_i}S^1to T_{x_i}S^1$ given by multiplication by $alpha_iin mathbb R^*$. Such $alpha_i$ is an invariant of a vector field under diffeomorphisms. And since the number of closed orbits can be arbitrary and these $alpha_i$'s are independent, we see that the dimension is not finite.



          If you want something which is finite-dimensional, one can restrict to area-preserving vector fields which are same thing as closed $1$-forms. Now the spaces of minimal closed $1$-forms on a closed genus $g$ surface is indeed finite-dimensional by a theorem of Calabi. Each such form is the real part of a holomorphic $1$-form for a certain complex structure on the surface. (the reference is: E. Calabi, An intrinsic characterization of harmonic 1-forms, Global Analysis, Papers in Honor of K. Kodaira, 1969)






          share|cite|improve this answer











          $endgroup$













            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',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            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%2f325016%2fsmooth-vector-fields-on-a-surface-modulo-diffeomorphisms%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            This is not finite dimensional. For example, consider non-vanishing vector fields on $T^2=mathbb R^2/mathbb Z^2$, transversal to vertical circles. Any such field defines a self diffeo $S^1to S^1$ on a vertical circle, called the return map. Suppose that such a diffeo $varphi$ has $n$ fixed points $x_i$ (they correspond to closed orbits of the field). Then for each fixed point $x_iin S^1$ we have a linear map on the tangent space $dvarphi: T_{x_i}S^1to T_{x_i}S^1$ given by multiplication by $alpha_iin mathbb R^*$. Such $alpha_i$ is an invariant of a vector field under diffeomorphisms. And since the number of closed orbits can be arbitrary and these $alpha_i$'s are independent, we see that the dimension is not finite.



            If you want something which is finite-dimensional, one can restrict to area-preserving vector fields which are same thing as closed $1$-forms. Now the spaces of minimal closed $1$-forms on a closed genus $g$ surface is indeed finite-dimensional by a theorem of Calabi. Each such form is the real part of a holomorphic $1$-form for a certain complex structure on the surface. (the reference is: E. Calabi, An intrinsic characterization of harmonic 1-forms, Global Analysis, Papers in Honor of K. Kodaira, 1969)






            share|cite|improve this answer











            $endgroup$


















              10












              $begingroup$

              This is not finite dimensional. For example, consider non-vanishing vector fields on $T^2=mathbb R^2/mathbb Z^2$, transversal to vertical circles. Any such field defines a self diffeo $S^1to S^1$ on a vertical circle, called the return map. Suppose that such a diffeo $varphi$ has $n$ fixed points $x_i$ (they correspond to closed orbits of the field). Then for each fixed point $x_iin S^1$ we have a linear map on the tangent space $dvarphi: T_{x_i}S^1to T_{x_i}S^1$ given by multiplication by $alpha_iin mathbb R^*$. Such $alpha_i$ is an invariant of a vector field under diffeomorphisms. And since the number of closed orbits can be arbitrary and these $alpha_i$'s are independent, we see that the dimension is not finite.



              If you want something which is finite-dimensional, one can restrict to area-preserving vector fields which are same thing as closed $1$-forms. Now the spaces of minimal closed $1$-forms on a closed genus $g$ surface is indeed finite-dimensional by a theorem of Calabi. Each such form is the real part of a holomorphic $1$-form for a certain complex structure on the surface. (the reference is: E. Calabi, An intrinsic characterization of harmonic 1-forms, Global Analysis, Papers in Honor of K. Kodaira, 1969)






              share|cite|improve this answer











              $endgroup$
















                10












                10








                10





                $begingroup$

                This is not finite dimensional. For example, consider non-vanishing vector fields on $T^2=mathbb R^2/mathbb Z^2$, transversal to vertical circles. Any such field defines a self diffeo $S^1to S^1$ on a vertical circle, called the return map. Suppose that such a diffeo $varphi$ has $n$ fixed points $x_i$ (they correspond to closed orbits of the field). Then for each fixed point $x_iin S^1$ we have a linear map on the tangent space $dvarphi: T_{x_i}S^1to T_{x_i}S^1$ given by multiplication by $alpha_iin mathbb R^*$. Such $alpha_i$ is an invariant of a vector field under diffeomorphisms. And since the number of closed orbits can be arbitrary and these $alpha_i$'s are independent, we see that the dimension is not finite.



                If you want something which is finite-dimensional, one can restrict to area-preserving vector fields which are same thing as closed $1$-forms. Now the spaces of minimal closed $1$-forms on a closed genus $g$ surface is indeed finite-dimensional by a theorem of Calabi. Each such form is the real part of a holomorphic $1$-form for a certain complex structure on the surface. (the reference is: E. Calabi, An intrinsic characterization of harmonic 1-forms, Global Analysis, Papers in Honor of K. Kodaira, 1969)






                share|cite|improve this answer











                $endgroup$



                This is not finite dimensional. For example, consider non-vanishing vector fields on $T^2=mathbb R^2/mathbb Z^2$, transversal to vertical circles. Any such field defines a self diffeo $S^1to S^1$ on a vertical circle, called the return map. Suppose that such a diffeo $varphi$ has $n$ fixed points $x_i$ (they correspond to closed orbits of the field). Then for each fixed point $x_iin S^1$ we have a linear map on the tangent space $dvarphi: T_{x_i}S^1to T_{x_i}S^1$ given by multiplication by $alpha_iin mathbb R^*$. Such $alpha_i$ is an invariant of a vector field under diffeomorphisms. And since the number of closed orbits can be arbitrary and these $alpha_i$'s are independent, we see that the dimension is not finite.



                If you want something which is finite-dimensional, one can restrict to area-preserving vector fields which are same thing as closed $1$-forms. Now the spaces of minimal closed $1$-forms on a closed genus $g$ surface is indeed finite-dimensional by a theorem of Calabi. Each such form is the real part of a holomorphic $1$-form for a certain complex structure on the surface. (the reference is: E. Calabi, An intrinsic characterization of harmonic 1-forms, Global Analysis, Papers in Honor of K. Kodaira, 1969)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 11 hours ago

























                answered 11 hours ago









                Dmitri PanovDmitri Panov

                19.5k460121




                19.5k460121






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to MathOverflow!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325016%2fsmooth-vector-fields-on-a-surface-modulo-diffeomorphisms%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Loup dans la culture

                    Image
                    Loups sur une enluminure du Bestiaire d'Aberdeen Le loup est l'animal le plus emblématique de l'histoire de l'Eurasie, il était à l'honneur durant l'Antiquité chez la totalité des anciens peuples européens [ 1 ] . Sommaire 1 Le loup dans la mythologie 2 Les loups dans le folklore 2.1 L'évolution des mentalités 2.2 Le loup dans la psychanalyse 3 Le loup dans les œuvres culturelles 3.1 Le loup dans la littérature 3.1.1 Fables 3.1.2 Poésie 3.1.3 Romans 3.1.4 Littérature jeunesse 3.1.4.1 Quelques titres d'albums contemporains 3.1.5 Bande dessinée 3.2 Le loup à l'écran 3.2.1 Le cinéma et la télévision 3.2.2 Documentaires 3.2.3 Dessins animés 3.2.4 Jeux vidéo 3.3 Le loup en peinture et en sculpture 3.4 Le loup dans les œuvres musicales 4 Le loup et les noms propres 5 Bibliographie 6 Notes et références 7 Voir aussi 7.1 Articles connexes
                    Read more

                    How to solve the problem of ntp “Unable to contact time server” from KDE?

                    Image
                    1 0 I want to set date and time automatically but this error appears when I try to .. I'm using openSUSE, I tried different servers but it didn't work either, help please! It doesn't seem to be a network problem: $ ping pool.ntp.org 64 bytes from 41.78.128.17: icmp_seq=1 ttl=41 time=260 ms To the question of whether ntpd is installed and running: $ ps -C ntpd PID TTY TIME CMD i.e. there's no ntpd process running. $ rpm -qa | grep ntp yast2-ntp-client-3.1.12-1.7.noarch ntp-4.2.6p5-25.2.1.i586 Further information from the comments: $ sudo ntpdate pool.ntp.org 25 Sep 19:41:51 ntpdate[3162]: no server suitable for synchronization found networking kde opensuse ntp ntpd sh
                    Read more

                    Connection limited (no internet access)

                    Image
                    -1 My laptop is running on Windows 10 ، My wifi adapter is tenda V7TW311M ,My router is D-Link , When I connect to the internet I get connection limited with yellow triangle , and no ipv4 address in ipconfig ,Ihave tried manually setting the ipv4 address When dhcp works there is no ipv4 , When I select ipv4 manually dhcp is turned off , I tried an update my wifi to the Latest version But the same problem remained . networking wifi internet net share | improve this question asked 13 mins ago Souchi Souchi 1 1
                    Read more