$omega$ a closed 2-form and $bigwedge_{i=1}^n omega ne 0$ on a compact orientable smooth $2n$-manifold w/o...












0












$begingroup$



Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $omega$ be a closed $2$-form such that $bigwedge_{i=1}^n omega_p ne 0$ at every point $p$. Show that $H^2_{dR}(M) ne 0$.




That is, $(M, omega)$ is a symplectic manifold.



I'm not sure what to try with this. Any ideas on how to show this?






differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology







share|cite|improve this question











$endgroup$












  • $begingroup$
    Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
    $endgroup$
    – symplectomorphic
    Dec 19 '18 at 3:39










  • $begingroup$
    what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
    $endgroup$
    – Tsemo Aristide
    Dec 19 '18 at 3:45








  • 1




    $begingroup$
    Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
    $endgroup$
    – user10354138
    Dec 19 '18 at 13:43








  • 1




    $begingroup$
    @user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
    $endgroup$
    – user98602
    Dec 19 '18 at 21:23






  • 2




    $begingroup$
    Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
    $endgroup$
    – Ted Shifrin
    Dec 19 '18 at 22:57
















0












$begingroup$



Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $omega$ be a closed $2$-form such that $bigwedge_{i=1}^n omega_p ne 0$ at every point $p$. Show that $H^2_{dR}(M) ne 0$.




That is, $(M, omega)$ is a symplectic manifold.



I'm not sure what to try with this. Any ideas on how to show this?






differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology







share|cite|improve this question











$endgroup$












  • $begingroup$
    Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
    $endgroup$
    – symplectomorphic
    Dec 19 '18 at 3:39










  • $begingroup$
    what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
    $endgroup$
    – Tsemo Aristide
    Dec 19 '18 at 3:45








  • 1




    $begingroup$
    Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
    $endgroup$
    – user10354138
    Dec 19 '18 at 13:43








  • 1




    $begingroup$
    @user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
    $endgroup$
    – user98602
    Dec 19 '18 at 21:23






  • 2




    $begingroup$
    Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
    $endgroup$
    – Ted Shifrin
    Dec 19 '18 at 22:57














0












0








0





$begingroup$



Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $omega$ be a closed $2$-form such that $bigwedge_{i=1}^n omega_p ne 0$ at every point $p$. Show that $H^2_{dR}(M) ne 0$.




That is, $(M, omega)$ is a symplectic manifold.



I'm not sure what to try with this. Any ideas on how to show this?






differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology







share|cite|improve this question











$endgroup$





Suppose $M$ is a compact orientable smooth $2n$-manifold without boundary, and let $omega$ be a closed $2$-form such that $bigwedge_{i=1}^n omega_p ne 0$ at every point $p$. Show that $H^2_{dR}(M) ne 0$.




That is, $(M, omega)$ is a symplectic manifold.



I'm not sure what to try with this. Any ideas on how to show this?






differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology




differential-geometry differential-topology smooth-manifolds exterior-algebra de-rham-cohomology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 15:44







user98602

















asked Dec 19 '18 at 3:17









Al JebrAl Jebr

4,38943378




4,38943378












  • $begingroup$
    Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
    $endgroup$
    – symplectomorphic
    Dec 19 '18 at 3:39










  • $begingroup$
    what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
    $endgroup$
    – Tsemo Aristide
    Dec 19 '18 at 3:45








  • 1




    $begingroup$
    Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
    $endgroup$
    – user10354138
    Dec 19 '18 at 13:43








  • 1




    $begingroup$
    @user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
    $endgroup$
    – user98602
    Dec 19 '18 at 21:23






  • 2




    $begingroup$
    Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
    $endgroup$
    – Ted Shifrin
    Dec 19 '18 at 22:57


















  • $begingroup$
    Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
    $endgroup$
    – symplectomorphic
    Dec 19 '18 at 3:39










  • $begingroup$
    what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
    $endgroup$
    – Tsemo Aristide
    Dec 19 '18 at 3:45








  • 1




    $begingroup$
    Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
    $endgroup$
    – user10354138
    Dec 19 '18 at 13:43








  • 1




    $begingroup$
    @user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
    $endgroup$
    – user98602
    Dec 19 '18 at 21:23






  • 2




    $begingroup$
    Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
    $endgroup$
    – Ted Shifrin
    Dec 19 '18 at 22:57
















$begingroup$
Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
$endgroup$
– symplectomorphic
Dec 19 '18 at 3:39




$begingroup$
Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish.
$endgroup$
– symplectomorphic
Dec 19 '18 at 3:39












$begingroup$
what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
$endgroup$
– Tsemo Aristide
Dec 19 '18 at 3:45






$begingroup$
what is $Lambda_{i=1}^n$ do you mean $Lambda^n$ ?
$endgroup$
– Tsemo Aristide
Dec 19 '18 at 3:45






1




1




$begingroup$
Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
$endgroup$
– user10354138
Dec 19 '18 at 13:43






$begingroup$
Be careful! Do you want $omega^nneq 0$ happening in $Omega^{2n}(M)$ or in $H^{2n}(M)$? The latter is easy, the former is not true.
$endgroup$
– user10354138
Dec 19 '18 at 13:43






1




1




$begingroup$
@user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
$endgroup$
– user98602
Dec 19 '18 at 21:23




$begingroup$
@user10354138 I think your comment is incorrect; the definition of a symplectic form is a closed 2-form so that $omega^n$ is pointwise nonzero. Such 2-forms are popular, both at conferences and at parties.
$endgroup$
– user98602
Dec 19 '18 at 21:23




2




2




$begingroup$
Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:57




$begingroup$
Please be careful with your notation. You mean to say that $omega^n$ is nowhere $0$, I believe, not that it is not identically $0$.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 22:57










2 Answers
2






active

oldest

votes


















3












$begingroup$

If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.



Then use the non-boundary condition and Stokes via



$$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$



to derive a contradiction with $omega^nneq0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
    $endgroup$
    – Al Jebr
    Dec 19 '18 at 3:45












  • $begingroup$
    @AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
    $endgroup$
    – user98602
    Dec 19 '18 at 12:02












  • $begingroup$
    Why is $int_M omega^n =0$ a contradiction?
    $endgroup$
    – Al Jebr
    Dec 19 '18 at 16:26










  • $begingroup$
    @AlJebr What do you know about $omega^n$?
    $endgroup$
    – user98602
    Dec 19 '18 at 21:21










  • $begingroup$
    @MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:22





















1












$begingroup$

This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.



    Then use the non-boundary condition and Stokes via



    $$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$



    to derive a contradiction with $omega^nneq0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
      $endgroup$
      – Al Jebr
      Dec 19 '18 at 3:45












    • $begingroup$
      @AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
      $endgroup$
      – user98602
      Dec 19 '18 at 12:02












    • $begingroup$
      Why is $int_M omega^n =0$ a contradiction?
      $endgroup$
      – Al Jebr
      Dec 19 '18 at 16:26










    • $begingroup$
      @AlJebr What do you know about $omega^n$?
      $endgroup$
      – user98602
      Dec 19 '18 at 21:21










    • $begingroup$
      @MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
      $endgroup$
      – user10354138
      Dec 20 '18 at 8:22


















    3












    $begingroup$

    If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.



    Then use the non-boundary condition and Stokes via



    $$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$



    to derive a contradiction with $omega^nneq0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
      $endgroup$
      – Al Jebr
      Dec 19 '18 at 3:45












    • $begingroup$
      @AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
      $endgroup$
      – user98602
      Dec 19 '18 at 12:02












    • $begingroup$
      Why is $int_M omega^n =0$ a contradiction?
      $endgroup$
      – Al Jebr
      Dec 19 '18 at 16:26










    • $begingroup$
      @AlJebr What do you know about $omega^n$?
      $endgroup$
      – user98602
      Dec 19 '18 at 21:21










    • $begingroup$
      @MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
      $endgroup$
      – user10354138
      Dec 20 '18 at 8:22
















    3












    3








    3





    $begingroup$

    If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.



    Then use the non-boundary condition and Stokes via



    $$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$



    to derive a contradiction with $omega^nneq0$.






    share|cite|improve this answer









    $endgroup$



    If the second de Rham cohomology were trivial, $omega$ would be $dalpha$ for some 1-form $alpha$.



    Then use the non-boundary condition and Stokes via



    $$int_Momega^n=int_Md(alphawedgeomega^{n-1})=int_{partial M}cdots=cdots$$



    to derive a contradiction with $omega^nneq0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 19 '18 at 3:34









    symplectomorphicsymplectomorphic

    12.6k22041




    12.6k22041












    • $begingroup$
      How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
      $endgroup$
      – Al Jebr
      Dec 19 '18 at 3:45












    • $begingroup$
      @AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
      $endgroup$
      – user98602
      Dec 19 '18 at 12:02












    • $begingroup$
      Why is $int_M omega^n =0$ a contradiction?
      $endgroup$
      – Al Jebr
      Dec 19 '18 at 16:26










    • $begingroup$
      @AlJebr What do you know about $omega^n$?
      $endgroup$
      – user98602
      Dec 19 '18 at 21:21










    • $begingroup$
      @MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
      $endgroup$
      – user10354138
      Dec 20 '18 at 8:22




















    • $begingroup$
      How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
      $endgroup$
      – Al Jebr
      Dec 19 '18 at 3:45












    • $begingroup$
      @AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
      $endgroup$
      – user98602
      Dec 19 '18 at 12:02












    • $begingroup$
      Why is $int_M omega^n =0$ a contradiction?
      $endgroup$
      – Al Jebr
      Dec 19 '18 at 16:26










    • $begingroup$
      @AlJebr What do you know about $omega^n$?
      $endgroup$
      – user98602
      Dec 19 '18 at 21:21










    • $begingroup$
      @MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
      $endgroup$
      – user10354138
      Dec 20 '18 at 8:22


















    $begingroup$
    How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
    $endgroup$
    – Al Jebr
    Dec 19 '18 at 3:45






    $begingroup$
    How is $omega^n= d(alpha wedge omega^{n-1})$? I'm getting $$d(alpha wedge omega^{n-1})=dalpha wedge omega^{n-1} -alpha wedge domega^{n-1}=omega^n-alpha wedge domega^{n-1}.$$
    $endgroup$
    – Al Jebr
    Dec 19 '18 at 3:45














    $begingroup$
    @AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
    $endgroup$
    – user98602
    Dec 19 '18 at 12:02






    $begingroup$
    @AlJebr Show that if a form $omega$ is closed, so is $omega^k$ for all $k$, by showing that the wedge product of any two closed forms is closed.
    $endgroup$
    – user98602
    Dec 19 '18 at 12:02














    $begingroup$
    Why is $int_M omega^n =0$ a contradiction?
    $endgroup$
    – Al Jebr
    Dec 19 '18 at 16:26




    $begingroup$
    Why is $int_M omega^n =0$ a contradiction?
    $endgroup$
    – Al Jebr
    Dec 19 '18 at 16:26












    $begingroup$
    @AlJebr What do you know about $omega^n$?
    $endgroup$
    – user98602
    Dec 19 '18 at 21:21




    $begingroup$
    @AlJebr What do you know about $omega^n$?
    $endgroup$
    – user98602
    Dec 19 '18 at 21:21












    $begingroup$
    @MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:22






    $begingroup$
    @MikeMiller You only know that there is a point $pin M$ such that $omega^n(p)neq 0$. You are not given this for all $pin M$ in the question statement by one of the possible interpretation of $bigwedge_{i=1}^nomega$.
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:22













    1












    $begingroup$

    This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.






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        $begingroup$

        This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.






        share|cite|improve this answer











        $endgroup$



        This is a long comment. To emphasize that you must interpret the hypothesis as being that $omega^n (p) ne 0$ for all $pin M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4times S^4$, with obvious projection maps $pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $eta$ on $S^4$ with the property that $eta^2 = etawedgeeta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $rho_1$ to be a smooth function that is $1$ on the unit ball in $Bbb R^3$ and $0$ outside the ball of radius $2$, and let $rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $eta = d(rho_1,dx-rho_2,dy)$. Note that $eta^2 = drho_1wedge drho_2wedge dxwedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $omega = pi_1^*eta + pi_2^*eta$. Then $omega^4 = 6pi_1^*(eta^2)wedgepi_2^*(eta^2)$ will be not identically zero on $S^4times S^4$, and yet $H^2(S^4times S^4) = 0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 20 '18 at 17:58

























        answered Dec 20 '18 at 17:48









        Ted ShifrinTed Shifrin

        64.7k44692




        64.7k44692






























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