How does this example from Spivak that $H_c^n(mathbb R^n) ne 0$?












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I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $mathbb R^n$ is not trivial.



How does the fact that the integral of $omega$ over $mathbb R^n$ is zero show that $H_c^n(mathbb R^n) ne 0$?




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differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology







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    0












    $begingroup$


    I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $mathbb R^n$ is not trivial.



    How does the fact that the integral of $omega$ over $mathbb R^n$ is zero show that $H_c^n(mathbb R^n) ne 0$?




    enter image description here






    differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $mathbb R^n$ is not trivial.



      How does the fact that the integral of $omega$ over $mathbb R^n$ is zero show that $H_c^n(mathbb R^n) ne 0$?




      enter image description here






      differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology







      share|cite|improve this question









      $endgroup$




      I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $mathbb R^n$ is not trivial.



      How does the fact that the integral of $omega$ over $mathbb R^n$ is zero show that $H_c^n(mathbb R^n) ne 0$?




      enter image description here






      differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology




      differential-geometry proof-explanation differential-topology homology-cohomology de-rham-cohomology






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      asked Dec 10 '18 at 23:28









      Al JebrAl Jebr

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          $begingroup$

          Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.






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          • $begingroup$
            Why is $int_{partial mathbb R^n} eta =0$?
            $endgroup$
            – Al Jebr
            Dec 11 '18 at 0:34






          • 1




            $begingroup$
            @AlJebr It's because $partial mathbf R^n = emptyset$.
            $endgroup$
            – Zircht
            Dec 11 '18 at 0:41











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          $begingroup$

          Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why is $int_{partial mathbb R^n} eta =0$?
            $endgroup$
            – Al Jebr
            Dec 11 '18 at 0:34






          • 1




            $begingroup$
            @AlJebr It's because $partial mathbf R^n = emptyset$.
            $endgroup$
            – Zircht
            Dec 11 '18 at 0:41
















          2












          $begingroup$

          Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why is $int_{partial mathbb R^n} eta =0$?
            $endgroup$
            – Al Jebr
            Dec 11 '18 at 0:34






          • 1




            $begingroup$
            @AlJebr It's because $partial mathbf R^n = emptyset$.
            $endgroup$
            – Zircht
            Dec 11 '18 at 0:41














          2












          2








          2





          $begingroup$

          Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.






          share|cite|improve this answer









          $endgroup$



          Since $mathbf R^n$ has dimension $n$ and $omega$ is an $n$-form, we have that $omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $mathbf R^n$). Hence $omegain Z_c^n(mathbf R^n)$. But $omeganotin B_c^n(mathbf R^n)$ because we'd would have $eta$ with compact support such that $omega=deta$, and that would imply that $int_{mathbf R^n} omega =0$ as Spivak claims. This is a contradiction because $omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $omega$ is a nonzero element of $H_c^n(mathbf R^n)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 23:40









          ZirchtZircht

          1,623913




          1,623913












          • $begingroup$
            Why is $int_{partial mathbb R^n} eta =0$?
            $endgroup$
            – Al Jebr
            Dec 11 '18 at 0:34






          • 1




            $begingroup$
            @AlJebr It's because $partial mathbf R^n = emptyset$.
            $endgroup$
            – Zircht
            Dec 11 '18 at 0:41


















          • $begingroup$
            Why is $int_{partial mathbb R^n} eta =0$?
            $endgroup$
            – Al Jebr
            Dec 11 '18 at 0:34






          • 1




            $begingroup$
            @AlJebr It's because $partial mathbf R^n = emptyset$.
            $endgroup$
            – Zircht
            Dec 11 '18 at 0:41
















          $begingroup$
          Why is $int_{partial mathbb R^n} eta =0$?
          $endgroup$
          – Al Jebr
          Dec 11 '18 at 0:34




          $begingroup$
          Why is $int_{partial mathbb R^n} eta =0$?
          $endgroup$
          – Al Jebr
          Dec 11 '18 at 0:34




          1




          1




          $begingroup$
          @AlJebr It's because $partial mathbf R^n = emptyset$.
          $endgroup$
          – Zircht
          Dec 11 '18 at 0:41




          $begingroup$
          @AlJebr It's because $partial mathbf R^n = emptyset$.
          $endgroup$
          – Zircht
          Dec 11 '18 at 0:41


















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