Fubini's Theorem, but $dxdy = -dydx$












1












$begingroup$


As differential $2$-forms, clearly $dxdy=-dydx$ by alternation. Yet just as clearly, $int_0^1 int_0^1 dxdy neq -int_0^1 int_0^1 dydx$. Where's the abuse of notation here?






integration differential-forms







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$endgroup$












  • $begingroup$
    You've written the same on both sides of the second equation. And you haven't explained where and why you see a problem.
    $endgroup$
    – Henrik
    Nov 6 '18 at 6:43










  • $begingroup$
    Sorry, fat fingers.
    $endgroup$
    – Feryll
    Nov 6 '18 at 6:44










  • $begingroup$
    You need to match up the orientation when integrating differential forms by definition, to avoid exactly this problem.
    $endgroup$
    – user10354138
    Nov 6 '18 at 7:08












  • $begingroup$
    Check carefully how the integral of a $2$-form over an oriented $2$-dimensional manifold is defined. In particular, how is the integral of $f,dxwedge dy$ over a region in $Bbb R^2$ (with the usual orientation) defined?
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 23:14
















1












$begingroup$


As differential $2$-forms, clearly $dxdy=-dydx$ by alternation. Yet just as clearly, $int_0^1 int_0^1 dxdy neq -int_0^1 int_0^1 dydx$. Where's the abuse of notation here?






integration differential-forms







share|cite|improve this question











$endgroup$












  • $begingroup$
    You've written the same on both sides of the second equation. And you haven't explained where and why you see a problem.
    $endgroup$
    – Henrik
    Nov 6 '18 at 6:43










  • $begingroup$
    Sorry, fat fingers.
    $endgroup$
    – Feryll
    Nov 6 '18 at 6:44










  • $begingroup$
    You need to match up the orientation when integrating differential forms by definition, to avoid exactly this problem.
    $endgroup$
    – user10354138
    Nov 6 '18 at 7:08












  • $begingroup$
    Check carefully how the integral of a $2$-form over an oriented $2$-dimensional manifold is defined. In particular, how is the integral of $f,dxwedge dy$ over a region in $Bbb R^2$ (with the usual orientation) defined?
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 23:14














1












1








1


3



$begingroup$


As differential $2$-forms, clearly $dxdy=-dydx$ by alternation. Yet just as clearly, $int_0^1 int_0^1 dxdy neq -int_0^1 int_0^1 dydx$. Where's the abuse of notation here?






integration differential-forms







share|cite|improve this question











$endgroup$




As differential $2$-forms, clearly $dxdy=-dydx$ by alternation. Yet just as clearly, $int_0^1 int_0^1 dxdy neq -int_0^1 int_0^1 dydx$. Where's the abuse of notation here?






integration differential-forms




integration differential-forms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 6 '18 at 6:44







Feryll

















asked Nov 6 '18 at 6:41









FeryllFeryll

544212




544212












  • $begingroup$
    You've written the same on both sides of the second equation. And you haven't explained where and why you see a problem.
    $endgroup$
    – Henrik
    Nov 6 '18 at 6:43










  • $begingroup$
    Sorry, fat fingers.
    $endgroup$
    – Feryll
    Nov 6 '18 at 6:44










  • $begingroup$
    You need to match up the orientation when integrating differential forms by definition, to avoid exactly this problem.
    $endgroup$
    – user10354138
    Nov 6 '18 at 7:08












  • $begingroup$
    Check carefully how the integral of a $2$-form over an oriented $2$-dimensional manifold is defined. In particular, how is the integral of $f,dxwedge dy$ over a region in $Bbb R^2$ (with the usual orientation) defined?
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 23:14


















  • $begingroup$
    You've written the same on both sides of the second equation. And you haven't explained where and why you see a problem.
    $endgroup$
    – Henrik
    Nov 6 '18 at 6:43










  • $begingroup$
    Sorry, fat fingers.
    $endgroup$
    – Feryll
    Nov 6 '18 at 6:44










  • $begingroup$
    You need to match up the orientation when integrating differential forms by definition, to avoid exactly this problem.
    $endgroup$
    – user10354138
    Nov 6 '18 at 7:08












  • $begingroup$
    Check carefully how the integral of a $2$-form over an oriented $2$-dimensional manifold is defined. In particular, how is the integral of $f,dxwedge dy$ over a region in $Bbb R^2$ (with the usual orientation) defined?
    $endgroup$
    – Ted Shifrin
    Nov 7 '18 at 23:14
















$begingroup$
You've written the same on both sides of the second equation. And you haven't explained where and why you see a problem.
$endgroup$
– Henrik
Nov 6 '18 at 6:43




$begingroup$
You've written the same on both sides of the second equation. And you haven't explained where and why you see a problem.
$endgroup$
– Henrik
Nov 6 '18 at 6:43












$begingroup$
Sorry, fat fingers.
$endgroup$
– Feryll
Nov 6 '18 at 6:44




$begingroup$
Sorry, fat fingers.
$endgroup$
– Feryll
Nov 6 '18 at 6:44












$begingroup$
You need to match up the orientation when integrating differential forms by definition, to avoid exactly this problem.
$endgroup$
– user10354138
Nov 6 '18 at 7:08






$begingroup$
You need to match up the orientation when integrating differential forms by definition, to avoid exactly this problem.
$endgroup$
– user10354138
Nov 6 '18 at 7:08














$begingroup$
Check carefully how the integral of a $2$-form over an oriented $2$-dimensional manifold is defined. In particular, how is the integral of $f,dxwedge dy$ over a region in $Bbb R^2$ (with the usual orientation) defined?
$endgroup$
– Ted Shifrin
Nov 7 '18 at 23:14




$begingroup$
Check carefully how the integral of a $2$-form over an oriented $2$-dimensional manifold is defined. In particular, how is the integral of $f,dxwedge dy$ over a region in $Bbb R^2$ (with the usual orientation) defined?
$endgroup$
– Ted Shifrin
Nov 7 '18 at 23:14










2 Answers
2






active

oldest

votes


















3












$begingroup$

$(x,y)mapsto(y,x)$ is an orientation-reversing map of the plane to itself.
For orientation-reversing maps one has to introduce a minus sign factor in the
change of variables:
$$int_0^1int_0^1,dxwedge,dy=-int_0^1int_0^1-,dywedge,dx
=int_0^1int_0^1,dywedge,dx.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Expanding upon Lord Shark's answer: remember that when changing coordinate systems, you need to introduce the determinant of the Jacobian of the coordinate transformation. In your case, the Jacobian is



    $$
    J(x,y) = begin{bmatrix}
    frac{partial y}{partial x} & frac{partial x}{partial x} \
    frac{partial y}{partial y} & frac{partial x}{partial y}
    end{bmatrix}
    =begin{bmatrix} 0 & 1 \ 1 & 0 end{bmatrix} , ,
    $$
    whose determinant is $-1$. Therefore,
    $$
    int_0^1 int_0^1 dx wedge dy = int_0^1 int_0^1 det(J(x,y)) , (-dy wedge dx) = int_0^1 int_0^1 dy wedge dx
    $$



    As Lord Shark pointed out, any orientation-reversing transformation will have a Jacobian whose determinant is $-1$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      $(x,y)mapsto(y,x)$ is an orientation-reversing map of the plane to itself.
      For orientation-reversing maps one has to introduce a minus sign factor in the
      change of variables:
      $$int_0^1int_0^1,dxwedge,dy=-int_0^1int_0^1-,dywedge,dx
      =int_0^1int_0^1,dywedge,dx.$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        $(x,y)mapsto(y,x)$ is an orientation-reversing map of the plane to itself.
        For orientation-reversing maps one has to introduce a minus sign factor in the
        change of variables:
        $$int_0^1int_0^1,dxwedge,dy=-int_0^1int_0^1-,dywedge,dx
        =int_0^1int_0^1,dywedge,dx.$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          $(x,y)mapsto(y,x)$ is an orientation-reversing map of the plane to itself.
          For orientation-reversing maps one has to introduce a minus sign factor in the
          change of variables:
          $$int_0^1int_0^1,dxwedge,dy=-int_0^1int_0^1-,dywedge,dx
          =int_0^1int_0^1,dywedge,dx.$$






          share|cite|improve this answer









          $endgroup$



          $(x,y)mapsto(y,x)$ is an orientation-reversing map of the plane to itself.
          For orientation-reversing maps one has to introduce a minus sign factor in the
          change of variables:
          $$int_0^1int_0^1,dxwedge,dy=-int_0^1int_0^1-,dywedge,dx
          =int_0^1int_0^1,dywedge,dx.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 6 '18 at 7:16









          Lord Shark the UnknownLord Shark the Unknown

          102k1159132




          102k1159132























              0












              $begingroup$

              Expanding upon Lord Shark's answer: remember that when changing coordinate systems, you need to introduce the determinant of the Jacobian of the coordinate transformation. In your case, the Jacobian is



              $$
              J(x,y) = begin{bmatrix}
              frac{partial y}{partial x} & frac{partial x}{partial x} \
              frac{partial y}{partial y} & frac{partial x}{partial y}
              end{bmatrix}
              =begin{bmatrix} 0 & 1 \ 1 & 0 end{bmatrix} , ,
              $$
              whose determinant is $-1$. Therefore,
              $$
              int_0^1 int_0^1 dx wedge dy = int_0^1 int_0^1 det(J(x,y)) , (-dy wedge dx) = int_0^1 int_0^1 dy wedge dx
              $$



              As Lord Shark pointed out, any orientation-reversing transformation will have a Jacobian whose determinant is $-1$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Expanding upon Lord Shark's answer: remember that when changing coordinate systems, you need to introduce the determinant of the Jacobian of the coordinate transformation. In your case, the Jacobian is



                $$
                J(x,y) = begin{bmatrix}
                frac{partial y}{partial x} & frac{partial x}{partial x} \
                frac{partial y}{partial y} & frac{partial x}{partial y}
                end{bmatrix}
                =begin{bmatrix} 0 & 1 \ 1 & 0 end{bmatrix} , ,
                $$
                whose determinant is $-1$. Therefore,
                $$
                int_0^1 int_0^1 dx wedge dy = int_0^1 int_0^1 det(J(x,y)) , (-dy wedge dx) = int_0^1 int_0^1 dy wedge dx
                $$



                As Lord Shark pointed out, any orientation-reversing transformation will have a Jacobian whose determinant is $-1$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Expanding upon Lord Shark's answer: remember that when changing coordinate systems, you need to introduce the determinant of the Jacobian of the coordinate transformation. In your case, the Jacobian is



                  $$
                  J(x,y) = begin{bmatrix}
                  frac{partial y}{partial x} & frac{partial x}{partial x} \
                  frac{partial y}{partial y} & frac{partial x}{partial y}
                  end{bmatrix}
                  =begin{bmatrix} 0 & 1 \ 1 & 0 end{bmatrix} , ,
                  $$
                  whose determinant is $-1$. Therefore,
                  $$
                  int_0^1 int_0^1 dx wedge dy = int_0^1 int_0^1 det(J(x,y)) , (-dy wedge dx) = int_0^1 int_0^1 dy wedge dx
                  $$



                  As Lord Shark pointed out, any orientation-reversing transformation will have a Jacobian whose determinant is $-1$.






                  share|cite|improve this answer









                  $endgroup$



                  Expanding upon Lord Shark's answer: remember that when changing coordinate systems, you need to introduce the determinant of the Jacobian of the coordinate transformation. In your case, the Jacobian is



                  $$
                  J(x,y) = begin{bmatrix}
                  frac{partial y}{partial x} & frac{partial x}{partial x} \
                  frac{partial y}{partial y} & frac{partial x}{partial y}
                  end{bmatrix}
                  =begin{bmatrix} 0 & 1 \ 1 & 0 end{bmatrix} , ,
                  $$
                  whose determinant is $-1$. Therefore,
                  $$
                  int_0^1 int_0^1 dx wedge dy = int_0^1 int_0^1 det(J(x,y)) , (-dy wedge dx) = int_0^1 int_0^1 dy wedge dx
                  $$



                  As Lord Shark pointed out, any orientation-reversing transformation will have a Jacobian whose determinant is $-1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 22:45









                  mcatmcat

                  1087




                  1087






























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