differential 1 - form of $omega$












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Let $omega$ be a differential form.
$$omega = dfrac{xdywedge dz + ydzwedge dx + zdxwedge dy}{(x^2 + y^2+ z^2)^{3/2}}$$
1)Find $domega$.



2)show that exist 1-form $theta$ such that $dtheta = omega$ in half-space $z>0$.



3) Show that doesn't exist 1-form $theta$ such that $dtheta = omega$ in $ mathbb R^3 setminus{0} $



I found $domega$, it equals zero, but i have no idea how to show in 2) and 3). Can you help me with this?






differential-geometry differential-forms







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    $begingroup$


    Let $omega$ be a differential form.
    $$omega = dfrac{xdywedge dz + ydzwedge dx + zdxwedge dy}{(x^2 + y^2+ z^2)^{3/2}}$$
    1)Find $domega$.



    2)show that exist 1-form $theta$ such that $dtheta = omega$ in half-space $z>0$.



    3) Show that doesn't exist 1-form $theta$ such that $dtheta = omega$ in $ mathbb R^3 setminus{0} $



    I found $domega$, it equals zero, but i have no idea how to show in 2) and 3). Can you help me with this?






    differential-geometry differential-forms







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $omega$ be a differential form.
      $$omega = dfrac{xdywedge dz + ydzwedge dx + zdxwedge dy}{(x^2 + y^2+ z^2)^{3/2}}$$
      1)Find $domega$.



      2)show that exist 1-form $theta$ such that $dtheta = omega$ in half-space $z>0$.



      3) Show that doesn't exist 1-form $theta$ such that $dtheta = omega$ in $ mathbb R^3 setminus{0} $



      I found $domega$, it equals zero, but i have no idea how to show in 2) and 3). Can you help me with this?






      differential-geometry differential-forms







      share|cite|improve this question









      $endgroup$




      Let $omega$ be a differential form.
      $$omega = dfrac{xdywedge dz + ydzwedge dx + zdxwedge dy}{(x^2 + y^2+ z^2)^{3/2}}$$
      1)Find $domega$.



      2)show that exist 1-form $theta$ such that $dtheta = omega$ in half-space $z>0$.



      3) Show that doesn't exist 1-form $theta$ such that $dtheta = omega$ in $ mathbb R^3 setminus{0} $



      I found $domega$, it equals zero, but i have no idea how to show in 2) and 3). Can you help me with this?






      differential-geometry differential-forms




      differential-geometry differential-forms






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 28 '18 at 23:12









      Gera SlanovaGera Slanova

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          2) You have shown in the first question that $omega$ is closed, whence using Poincaré's lemma, it is exact on each contractible subsets of $mathbb{R}^3$ on which $omega$ is defined, such that the half-space $z>0$, since it is convex.



          3) If $omega$ were exact on $mathbb{R}^3setminus{0}$, using Stokes' theorem, its integral on $S^2$ would be zero, since $S^2$ has no boundary, but it is equal to $4pi$.



          I feel a bit guilty about not computing this integral, so let's do it:
          $$begin{align}int_{S^2}omega&=int_{S^2}x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedge mathrm{d}y,tag{1}\tag{2}&=int_{B^3}mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y),\tag{3}&=3int_{B^3}mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z,\&=4pi.tag{4}end{align}$$
          Here are the steps behing each equality:





          • $(1)$: $omega_{vert S^2}=x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y$, since $x^2+y^2+z^2=1$ for $(x,y,z)in S^2$,


          • $(2)$: Stokes' theorem on $B^3$, noticing that $partial B^3=S^2$,


          • $(3)$: $mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y)=3,mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$, using standard properties of the wedge product and the exterior derivative,


          • $(4)$: $mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$ is the Lebesgue measure on $mathbb{R}^3$ and the Lebesgue measure of $B^3$ is $4pi/3$.


          I hope that everything is crystal clear.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            2) You have shown in the first question that $omega$ is closed, whence using Poincaré's lemma, it is exact on each contractible subsets of $mathbb{R}^3$ on which $omega$ is defined, such that the half-space $z>0$, since it is convex.



            3) If $omega$ were exact on $mathbb{R}^3setminus{0}$, using Stokes' theorem, its integral on $S^2$ would be zero, since $S^2$ has no boundary, but it is equal to $4pi$.



            I feel a bit guilty about not computing this integral, so let's do it:
            $$begin{align}int_{S^2}omega&=int_{S^2}x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedge mathrm{d}y,tag{1}\tag{2}&=int_{B^3}mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y),\tag{3}&=3int_{B^3}mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z,\&=4pi.tag{4}end{align}$$
            Here are the steps behing each equality:





            • $(1)$: $omega_{vert S^2}=x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y$, since $x^2+y^2+z^2=1$ for $(x,y,z)in S^2$,


            • $(2)$: Stokes' theorem on $B^3$, noticing that $partial B^3=S^2$,


            • $(3)$: $mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y)=3,mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$, using standard properties of the wedge product and the exterior derivative,


            • $(4)$: $mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$ is the Lebesgue measure on $mathbb{R}^3$ and the Lebesgue measure of $B^3$ is $4pi/3$.


            I hope that everything is crystal clear.






            share|cite|improve this answer











            $endgroup$


















              6












              $begingroup$

              2) You have shown in the first question that $omega$ is closed, whence using Poincaré's lemma, it is exact on each contractible subsets of $mathbb{R}^3$ on which $omega$ is defined, such that the half-space $z>0$, since it is convex.



              3) If $omega$ were exact on $mathbb{R}^3setminus{0}$, using Stokes' theorem, its integral on $S^2$ would be zero, since $S^2$ has no boundary, but it is equal to $4pi$.



              I feel a bit guilty about not computing this integral, so let's do it:
              $$begin{align}int_{S^2}omega&=int_{S^2}x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedge mathrm{d}y,tag{1}\tag{2}&=int_{B^3}mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y),\tag{3}&=3int_{B^3}mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z,\&=4pi.tag{4}end{align}$$
              Here are the steps behing each equality:





              • $(1)$: $omega_{vert S^2}=x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y$, since $x^2+y^2+z^2=1$ for $(x,y,z)in S^2$,


              • $(2)$: Stokes' theorem on $B^3$, noticing that $partial B^3=S^2$,


              • $(3)$: $mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y)=3,mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$, using standard properties of the wedge product and the exterior derivative,


              • $(4)$: $mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$ is the Lebesgue measure on $mathbb{R}^3$ and the Lebesgue measure of $B^3$ is $4pi/3$.


              I hope that everything is crystal clear.






              share|cite|improve this answer











              $endgroup$
















                6












                6








                6





                $begingroup$

                2) You have shown in the first question that $omega$ is closed, whence using Poincaré's lemma, it is exact on each contractible subsets of $mathbb{R}^3$ on which $omega$ is defined, such that the half-space $z>0$, since it is convex.



                3) If $omega$ were exact on $mathbb{R}^3setminus{0}$, using Stokes' theorem, its integral on $S^2$ would be zero, since $S^2$ has no boundary, but it is equal to $4pi$.



                I feel a bit guilty about not computing this integral, so let's do it:
                $$begin{align}int_{S^2}omega&=int_{S^2}x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedge mathrm{d}y,tag{1}\tag{2}&=int_{B^3}mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y),\tag{3}&=3int_{B^3}mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z,\&=4pi.tag{4}end{align}$$
                Here are the steps behing each equality:





                • $(1)$: $omega_{vert S^2}=x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y$, since $x^2+y^2+z^2=1$ for $(x,y,z)in S^2$,


                • $(2)$: Stokes' theorem on $B^3$, noticing that $partial B^3=S^2$,


                • $(3)$: $mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y)=3,mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$, using standard properties of the wedge product and the exterior derivative,


                • $(4)$: $mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$ is the Lebesgue measure on $mathbb{R}^3$ and the Lebesgue measure of $B^3$ is $4pi/3$.


                I hope that everything is crystal clear.






                share|cite|improve this answer











                $endgroup$



                2) You have shown in the first question that $omega$ is closed, whence using Poincaré's lemma, it is exact on each contractible subsets of $mathbb{R}^3$ on which $omega$ is defined, such that the half-space $z>0$, since it is convex.



                3) If $omega$ were exact on $mathbb{R}^3setminus{0}$, using Stokes' theorem, its integral on $S^2$ would be zero, since $S^2$ has no boundary, but it is equal to $4pi$.



                I feel a bit guilty about not computing this integral, so let's do it:
                $$begin{align}int_{S^2}omega&=int_{S^2}x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedge mathrm{d}y,tag{1}\tag{2}&=int_{B^3}mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y),\tag{3}&=3int_{B^3}mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z,\&=4pi.tag{4}end{align}$$
                Here are the steps behing each equality:





                • $(1)$: $omega_{vert S^2}=x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y$, since $x^2+y^2+z^2=1$ for $(x,y,z)in S^2$,


                • $(2)$: Stokes' theorem on $B^3$, noticing that $partial B^3=S^2$,


                • $(3)$: $mathrm{d}(x,mathrm{d}ywedgemathrm{d}z+y,mathrm{d}zwedgemathrm{d}x+z,mathrm{d}xwedgemathrm{d}y)=3,mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$, using standard properties of the wedge product and the exterior derivative,


                • $(4)$: $mathrm{d}xwedgemathrm{d}ywedgemathrm{d}z$ is the Lebesgue measure on $mathbb{R}^3$ and the Lebesgue measure of $B^3$ is $4pi/3$.


                I hope that everything is crystal clear.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 29 '18 at 0:20

























                answered Dec 28 '18 at 23:20









                C. FalconC. Falcon

                15.2k41950




                15.2k41950






























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