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






















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          $begingroup$

          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)






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            $begingroup$

            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)






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              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






























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