Extending a volume form on $S^k$ to $mathbb R^{k+1}$ such that the exterior derivative is zero.












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As we know that $S^k$ is a properly embedded submanifold of $R^{k+1}$. Let $w$ be a volume for of $S^k$(a non vanishing top form). Then $dw = 0$ on $S^k$. Is it possible to extend $w$ to $R^{k+1}$ such that $dw = 0$? We can extend the form for sure by first extending it locally then patch it together. But it does not seem to be the case that the extension should always have exterior differentiation equal to zero.






differential-geometry differential-topology







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    $begingroup$


    As we know that $S^k$ is a properly embedded submanifold of $R^{k+1}$. Let $w$ be a volume for of $S^k$(a non vanishing top form). Then $dw = 0$ on $S^k$. Is it possible to extend $w$ to $R^{k+1}$ such that $dw = 0$? We can extend the form for sure by first extending it locally then patch it together. But it does not seem to be the case that the extension should always have exterior differentiation equal to zero.






    differential-geometry differential-topology







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      As we know that $S^k$ is a properly embedded submanifold of $R^{k+1}$. Let $w$ be a volume for of $S^k$(a non vanishing top form). Then $dw = 0$ on $S^k$. Is it possible to extend $w$ to $R^{k+1}$ such that $dw = 0$? We can extend the form for sure by first extending it locally then patch it together. But it does not seem to be the case that the extension should always have exterior differentiation equal to zero.






      differential-geometry differential-topology







      share|cite|improve this question









      $endgroup$




      As we know that $S^k$ is a properly embedded submanifold of $R^{k+1}$. Let $w$ be a volume for of $S^k$(a non vanishing top form). Then $dw = 0$ on $S^k$. Is it possible to extend $w$ to $R^{k+1}$ such that $dw = 0$? We can extend the form for sure by first extending it locally then patch it together. But it does not seem to be the case that the extension should always have exterior differentiation equal to zero.






      differential-geometry differential-topology




      differential-geometry differential-topology






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      asked Dec 12 '18 at 6:02









      kochkoch

      18418




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          3 Answers
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          $begingroup$

          If that were possible, then by Stokes,
          $$int_{B^{k+1}}dw=int_{S^{k}}w$$
          where $B^{k+1}$ is the unit ball in $Bbb R^{k+1}$. But the LHS is zero, and the RHS is positive.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Let $dtheta$ be the $1-$form corresponding to the polar coordinate on $S^1$, why dont we have $dtheta$ on the whole $mathbb R^2$?
            $endgroup$
            – koch
            Dec 12 '18 at 13:52










          • $begingroup$
            $dtheta$ is not defined at the origin. @koch
            $endgroup$
            – Lord Shark the Unknown
            Dec 12 '18 at 18:19



















          1












          $begingroup$

          I would say no, because then, from Stokes’ theorem,
          $$0 = int_{B^{k+1}}{dw}=int_{S^k}{w},$$ whoch is clearly false.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Using Stokes' theorem is probably the best argument. But here is another perspective, using de Rham cohomology: The inclusion $iota: S^krightarrow mathbb{R}^{k+1}$ induces a map $$
            iota^*: H^k_{dR}(mathbb{R}^{k+1}) rightarrow H_{dR}^k(S^k) .
            $$
            Since the space on the right hand side is spanned by the cohomology class $[w]$, being able to find a closed extension of $w$ would imply surjectivity of $iota^*$. But this is impossible since $H^k_{dR}(mathbb{R}^{k+1}) = 0$.






            share|cite|improve this answer









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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              If that were possible, then by Stokes,
              $$int_{B^{k+1}}dw=int_{S^{k}}w$$
              where $B^{k+1}$ is the unit ball in $Bbb R^{k+1}$. But the LHS is zero, and the RHS is positive.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Let $dtheta$ be the $1-$form corresponding to the polar coordinate on $S^1$, why dont we have $dtheta$ on the whole $mathbb R^2$?
                $endgroup$
                – koch
                Dec 12 '18 at 13:52










              • $begingroup$
                $dtheta$ is not defined at the origin. @koch
                $endgroup$
                – Lord Shark the Unknown
                Dec 12 '18 at 18:19
















              2












              $begingroup$

              If that were possible, then by Stokes,
              $$int_{B^{k+1}}dw=int_{S^{k}}w$$
              where $B^{k+1}$ is the unit ball in $Bbb R^{k+1}$. But the LHS is zero, and the RHS is positive.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Let $dtheta$ be the $1-$form corresponding to the polar coordinate on $S^1$, why dont we have $dtheta$ on the whole $mathbb R^2$?
                $endgroup$
                – koch
                Dec 12 '18 at 13:52










              • $begingroup$
                $dtheta$ is not defined at the origin. @koch
                $endgroup$
                – Lord Shark the Unknown
                Dec 12 '18 at 18:19














              2












              2








              2





              $begingroup$

              If that were possible, then by Stokes,
              $$int_{B^{k+1}}dw=int_{S^{k}}w$$
              where $B^{k+1}$ is the unit ball in $Bbb R^{k+1}$. But the LHS is zero, and the RHS is positive.






              share|cite|improve this answer









              $endgroup$



              If that were possible, then by Stokes,
              $$int_{B^{k+1}}dw=int_{S^{k}}w$$
              where $B^{k+1}$ is the unit ball in $Bbb R^{k+1}$. But the LHS is zero, and the RHS is positive.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 12 '18 at 7:02









              Lord Shark the UnknownLord Shark the Unknown

              103k1160132




              103k1160132












              • $begingroup$
                Let $dtheta$ be the $1-$form corresponding to the polar coordinate on $S^1$, why dont we have $dtheta$ on the whole $mathbb R^2$?
                $endgroup$
                – koch
                Dec 12 '18 at 13:52










              • $begingroup$
                $dtheta$ is not defined at the origin. @koch
                $endgroup$
                – Lord Shark the Unknown
                Dec 12 '18 at 18:19


















              • $begingroup$
                Let $dtheta$ be the $1-$form corresponding to the polar coordinate on $S^1$, why dont we have $dtheta$ on the whole $mathbb R^2$?
                $endgroup$
                – koch
                Dec 12 '18 at 13:52










              • $begingroup$
                $dtheta$ is not defined at the origin. @koch
                $endgroup$
                – Lord Shark the Unknown
                Dec 12 '18 at 18:19
















              $begingroup$
              Let $dtheta$ be the $1-$form corresponding to the polar coordinate on $S^1$, why dont we have $dtheta$ on the whole $mathbb R^2$?
              $endgroup$
              – koch
              Dec 12 '18 at 13:52




              $begingroup$
              Let $dtheta$ be the $1-$form corresponding to the polar coordinate on $S^1$, why dont we have $dtheta$ on the whole $mathbb R^2$?
              $endgroup$
              – koch
              Dec 12 '18 at 13:52












              $begingroup$
              $dtheta$ is not defined at the origin. @koch
              $endgroup$
              – Lord Shark the Unknown
              Dec 12 '18 at 18:19




              $begingroup$
              $dtheta$ is not defined at the origin. @koch
              $endgroup$
              – Lord Shark the Unknown
              Dec 12 '18 at 18:19











              1












              $begingroup$

              I would say no, because then, from Stokes’ theorem,
              $$0 = int_{B^{k+1}}{dw}=int_{S^k}{w},$$ whoch is clearly false.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                I would say no, because then, from Stokes’ theorem,
                $$0 = int_{B^{k+1}}{dw}=int_{S^k}{w},$$ whoch is clearly false.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I would say no, because then, from Stokes’ theorem,
                  $$0 = int_{B^{k+1}}{dw}=int_{S^k}{w},$$ whoch is clearly false.






                  share|cite|improve this answer









                  $endgroup$



                  I would say no, because then, from Stokes’ theorem,
                  $$0 = int_{B^{k+1}}{dw}=int_{S^k}{w},$$ whoch is clearly false.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 7:03









                  MindlackMindlack

                  3,53717




                  3,53717























                      1












                      $begingroup$

                      Using Stokes' theorem is probably the best argument. But here is another perspective, using de Rham cohomology: The inclusion $iota: S^krightarrow mathbb{R}^{k+1}$ induces a map $$
                      iota^*: H^k_{dR}(mathbb{R}^{k+1}) rightarrow H_{dR}^k(S^k) .
                      $$
                      Since the space on the right hand side is spanned by the cohomology class $[w]$, being able to find a closed extension of $w$ would imply surjectivity of $iota^*$. But this is impossible since $H^k_{dR}(mathbb{R}^{k+1}) = 0$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Using Stokes' theorem is probably the best argument. But here is another perspective, using de Rham cohomology: The inclusion $iota: S^krightarrow mathbb{R}^{k+1}$ induces a map $$
                        iota^*: H^k_{dR}(mathbb{R}^{k+1}) rightarrow H_{dR}^k(S^k) .
                        $$
                        Since the space on the right hand side is spanned by the cohomology class $[w]$, being able to find a closed extension of $w$ would imply surjectivity of $iota^*$. But this is impossible since $H^k_{dR}(mathbb{R}^{k+1}) = 0$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Using Stokes' theorem is probably the best argument. But here is another perspective, using de Rham cohomology: The inclusion $iota: S^krightarrow mathbb{R}^{k+1}$ induces a map $$
                          iota^*: H^k_{dR}(mathbb{R}^{k+1}) rightarrow H_{dR}^k(S^k) .
                          $$
                          Since the space on the right hand side is spanned by the cohomology class $[w]$, being able to find a closed extension of $w$ would imply surjectivity of $iota^*$. But this is impossible since $H^k_{dR}(mathbb{R}^{k+1}) = 0$.






                          share|cite|improve this answer









                          $endgroup$



                          Using Stokes' theorem is probably the best argument. But here is another perspective, using de Rham cohomology: The inclusion $iota: S^krightarrow mathbb{R}^{k+1}$ induces a map $$
                          iota^*: H^k_{dR}(mathbb{R}^{k+1}) rightarrow H_{dR}^k(S^k) .
                          $$
                          Since the space on the right hand side is spanned by the cohomology class $[w]$, being able to find a closed extension of $w$ would imply surjectivity of $iota^*$. But this is impossible since $H^k_{dR}(mathbb{R}^{k+1}) = 0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 12 '18 at 8:25









                          Jan BohrJan Bohr

                          3,3071421




                          3,3071421






























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