Is there a pattern for closed and co-closed $n$-forms on $mathbb{R}^{2n}$?












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$begingroup$


Consider $mathbb{R}^{2n}$ with its standard Euclidean Riemannian metric. Let $omega in Omega^n(mathbb{R}^{2n})$ be an $n$-form.



I am trying to understand if there is a succinct way to express the equation



$$ domega=0 , , text{ and }, delta omega=0$$



i.e. $omega$ is closed and co-closed. (Here $delta=pm star d star$ is the adjoint of the $d$.)



For $n=1$, every one-form $omega in Omega^1(mathbb{R}^{2})$ can be written as
$$ omega=fdx+gdy,$$ so
$ domega=(g_x-f_y) dxwedge dy $. Thus, $domega=0 iff f_y =g_x$.



Since $star omega=fdy-gdx$, we have
$delta omega=0 iff dstar omega=0 iff f_x=-g_y$, so




$$ domega=0 , , text{ and }, delta omega=0 iff g+if , , text{ is holomorphic. }$$



Question: Is there some "nice" equivalent formulation of $0= delta omega= d omega$ for middle-dimensional forms in $mathbb{R}^{2n}$? Some sort of "generalized holomorphicity"?




Here is an attempt to write the equations for $n=2$:



Write



$omega=f_{12}dx_1 wedge dx_2 + f_{13}dx_1 wedge dx_3+f_{14}dx_1 wedge dx_4+f_{23}dx_2 wedge dx_3+f_{24}dx_2 wedge dx_4+f_{34}dx_3 wedge dx_4$,



for some $f_{ij}:mathbb{R}^4 to mathbb{R}$.



Then $domega=0$ is equivalent to the following system of $4$ equations:



$$ (f_{12})_3-(f_{13})_2+(f_{23})_1=0$$



$$ (f_{12})_4-(f_{14})_2+(f_{24})_1=0$$



$$ (f_{13})_4-(f_{14})_3+(f_{34})_1=0$$



$$ (f_{23})_4-(f_{24})_3+(f_{34})_2=0.$$



If $star omega = (tilde f_{ij})$, then $tilde f_{ij}=text{sgn}(ijkl)f_{kl}$, so (if am not mistaken) $d(star omega)=0$ is equivalent to



$$ (f_{34})_3+(f_{24})_2+(f_{14})_1=0$$



$$ (f_{34})_4-(f_{23})_2-(f_{13})_1=0$$



$$ -(f_{24})_4-(f_{23})_3+(f_{12})_1=0$$



$$ (f_{14})_4+(f_{13})_3+(f_{12})_2=0.$$




Is there any pattern here?




Any relation to holomorphicity? Or is this connection something special for dimension $2$?



We have here $6$ functions $f_{ij}$ which we can "pair naturally" (by $f_{12} iff f_{34}$ etc), and so get $3$ functions $mathbb{R}^4 cong mathbb{C}^2 to mathbb{C}$, but I am not sure that the holomorphicity of these pairs has any relation to the systems above.



Any ideas about this?










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Consider $mathbb{R}^{2n}$ with its standard Euclidean Riemannian metric. Let $omega in Omega^n(mathbb{R}^{2n})$ be an $n$-form.



    I am trying to understand if there is a succinct way to express the equation



    $$ domega=0 , , text{ and }, delta omega=0$$



    i.e. $omega$ is closed and co-closed. (Here $delta=pm star d star$ is the adjoint of the $d$.)



    For $n=1$, every one-form $omega in Omega^1(mathbb{R}^{2})$ can be written as
    $$ omega=fdx+gdy,$$ so
    $ domega=(g_x-f_y) dxwedge dy $. Thus, $domega=0 iff f_y =g_x$.



    Since $star omega=fdy-gdx$, we have
    $delta omega=0 iff dstar omega=0 iff f_x=-g_y$, so




    $$ domega=0 , , text{ and }, delta omega=0 iff g+if , , text{ is holomorphic. }$$



    Question: Is there some "nice" equivalent formulation of $0= delta omega= d omega$ for middle-dimensional forms in $mathbb{R}^{2n}$? Some sort of "generalized holomorphicity"?




    Here is an attempt to write the equations for $n=2$:



    Write



    $omega=f_{12}dx_1 wedge dx_2 + f_{13}dx_1 wedge dx_3+f_{14}dx_1 wedge dx_4+f_{23}dx_2 wedge dx_3+f_{24}dx_2 wedge dx_4+f_{34}dx_3 wedge dx_4$,



    for some $f_{ij}:mathbb{R}^4 to mathbb{R}$.



    Then $domega=0$ is equivalent to the following system of $4$ equations:



    $$ (f_{12})_3-(f_{13})_2+(f_{23})_1=0$$



    $$ (f_{12})_4-(f_{14})_2+(f_{24})_1=0$$



    $$ (f_{13})_4-(f_{14})_3+(f_{34})_1=0$$



    $$ (f_{23})_4-(f_{24})_3+(f_{34})_2=0.$$



    If $star omega = (tilde f_{ij})$, then $tilde f_{ij}=text{sgn}(ijkl)f_{kl}$, so (if am not mistaken) $d(star omega)=0$ is equivalent to



    $$ (f_{34})_3+(f_{24})_2+(f_{14})_1=0$$



    $$ (f_{34})_4-(f_{23})_2-(f_{13})_1=0$$



    $$ -(f_{24})_4-(f_{23})_3+(f_{12})_1=0$$



    $$ (f_{14})_4+(f_{13})_3+(f_{12})_2=0.$$




    Is there any pattern here?




    Any relation to holomorphicity? Or is this connection something special for dimension $2$?



    We have here $6$ functions $f_{ij}$ which we can "pair naturally" (by $f_{12} iff f_{34}$ etc), and so get $3$ functions $mathbb{R}^4 cong mathbb{C}^2 to mathbb{C}$, but I am not sure that the holomorphicity of these pairs has any relation to the systems above.



    Any ideas about this?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Consider $mathbb{R}^{2n}$ with its standard Euclidean Riemannian metric. Let $omega in Omega^n(mathbb{R}^{2n})$ be an $n$-form.



      I am trying to understand if there is a succinct way to express the equation



      $$ domega=0 , , text{ and }, delta omega=0$$



      i.e. $omega$ is closed and co-closed. (Here $delta=pm star d star$ is the adjoint of the $d$.)



      For $n=1$, every one-form $omega in Omega^1(mathbb{R}^{2})$ can be written as
      $$ omega=fdx+gdy,$$ so
      $ domega=(g_x-f_y) dxwedge dy $. Thus, $domega=0 iff f_y =g_x$.



      Since $star omega=fdy-gdx$, we have
      $delta omega=0 iff dstar omega=0 iff f_x=-g_y$, so




      $$ domega=0 , , text{ and }, delta omega=0 iff g+if , , text{ is holomorphic. }$$



      Question: Is there some "nice" equivalent formulation of $0= delta omega= d omega$ for middle-dimensional forms in $mathbb{R}^{2n}$? Some sort of "generalized holomorphicity"?




      Here is an attempt to write the equations for $n=2$:



      Write



      $omega=f_{12}dx_1 wedge dx_2 + f_{13}dx_1 wedge dx_3+f_{14}dx_1 wedge dx_4+f_{23}dx_2 wedge dx_3+f_{24}dx_2 wedge dx_4+f_{34}dx_3 wedge dx_4$,



      for some $f_{ij}:mathbb{R}^4 to mathbb{R}$.



      Then $domega=0$ is equivalent to the following system of $4$ equations:



      $$ (f_{12})_3-(f_{13})_2+(f_{23})_1=0$$



      $$ (f_{12})_4-(f_{14})_2+(f_{24})_1=0$$



      $$ (f_{13})_4-(f_{14})_3+(f_{34})_1=0$$



      $$ (f_{23})_4-(f_{24})_3+(f_{34})_2=0.$$



      If $star omega = (tilde f_{ij})$, then $tilde f_{ij}=text{sgn}(ijkl)f_{kl}$, so (if am not mistaken) $d(star omega)=0$ is equivalent to



      $$ (f_{34})_3+(f_{24})_2+(f_{14})_1=0$$



      $$ (f_{34})_4-(f_{23})_2-(f_{13})_1=0$$



      $$ -(f_{24})_4-(f_{23})_3+(f_{12})_1=0$$



      $$ (f_{14})_4+(f_{13})_3+(f_{12})_2=0.$$




      Is there any pattern here?




      Any relation to holomorphicity? Or is this connection something special for dimension $2$?



      We have here $6$ functions $f_{ij}$ which we can "pair naturally" (by $f_{12} iff f_{34}$ etc), and so get $3$ functions $mathbb{R}^4 cong mathbb{C}^2 to mathbb{C}$, but I am not sure that the holomorphicity of these pairs has any relation to the systems above.



      Any ideas about this?










      share|cite|improve this question











      $endgroup$




      Consider $mathbb{R}^{2n}$ with its standard Euclidean Riemannian metric. Let $omega in Omega^n(mathbb{R}^{2n})$ be an $n$-form.



      I am trying to understand if there is a succinct way to express the equation



      $$ domega=0 , , text{ and }, delta omega=0$$



      i.e. $omega$ is closed and co-closed. (Here $delta=pm star d star$ is the adjoint of the $d$.)



      For $n=1$, every one-form $omega in Omega^1(mathbb{R}^{2})$ can be written as
      $$ omega=fdx+gdy,$$ so
      $ domega=(g_x-f_y) dxwedge dy $. Thus, $domega=0 iff f_y =g_x$.



      Since $star omega=fdy-gdx$, we have
      $delta omega=0 iff dstar omega=0 iff f_x=-g_y$, so




      $$ domega=0 , , text{ and }, delta omega=0 iff g+if , , text{ is holomorphic. }$$



      Question: Is there some "nice" equivalent formulation of $0= delta omega= d omega$ for middle-dimensional forms in $mathbb{R}^{2n}$? Some sort of "generalized holomorphicity"?




      Here is an attempt to write the equations for $n=2$:



      Write



      $omega=f_{12}dx_1 wedge dx_2 + f_{13}dx_1 wedge dx_3+f_{14}dx_1 wedge dx_4+f_{23}dx_2 wedge dx_3+f_{24}dx_2 wedge dx_4+f_{34}dx_3 wedge dx_4$,



      for some $f_{ij}:mathbb{R}^4 to mathbb{R}$.



      Then $domega=0$ is equivalent to the following system of $4$ equations:



      $$ (f_{12})_3-(f_{13})_2+(f_{23})_1=0$$



      $$ (f_{12})_4-(f_{14})_2+(f_{24})_1=0$$



      $$ (f_{13})_4-(f_{14})_3+(f_{34})_1=0$$



      $$ (f_{23})_4-(f_{24})_3+(f_{34})_2=0.$$



      If $star omega = (tilde f_{ij})$, then $tilde f_{ij}=text{sgn}(ijkl)f_{kl}$, so (if am not mistaken) $d(star omega)=0$ is equivalent to



      $$ (f_{34})_3+(f_{24})_2+(f_{14})_1=0$$



      $$ (f_{34})_4-(f_{23})_2-(f_{13})_1=0$$



      $$ -(f_{24})_4-(f_{23})_3+(f_{12})_1=0$$



      $$ (f_{14})_4+(f_{13})_3+(f_{12})_2=0.$$




      Is there any pattern here?




      Any relation to holomorphicity? Or is this connection something special for dimension $2$?



      We have here $6$ functions $f_{ij}$ which we can "pair naturally" (by $f_{12} iff f_{34}$ etc), and so get $3$ functions $mathbb{R}^4 cong mathbb{C}^2 to mathbb{C}$, but I am not sure that the holomorphicity of these pairs has any relation to the systems above.



      Any ideas about this?







      complex-analysis riemannian-geometry differential-forms complex-geometry harmonic-functions






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 12:59







      Asaf Shachar

















      asked Dec 13 '18 at 11:13









      Asaf ShacharAsaf Shachar

      5,66131141




      5,66131141






















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