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?

  • $\begingroup$ 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
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    $\begingroup$ Be careful! Do you want $\omega^n\neq 0$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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

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

Then use the non-boundary condition and Stokes via

$$\int_M\omega^n=\int_Md(\alpha\wedge\omega^{n-1})=\int_{\partial M}\cdots=\cdots$$

to derive a contradiction with $\omega^n\neq0$.

  • $\begingroup$ How is $\omega^n= d(\alpha \wedge \omega^{n-1})$? I'm getting $$d(\alpha \wedge \omega^{n-1})=d\alpha \wedge \omega^{n-1} -\alpha \wedge d\omega^{n-1}=\omega^n-\alpha \wedge d\omega^{n-1}.$$ $\endgroup$ – 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 $p\in M$ such that $\omega^n(p)\neq 0$. You are not given this for all $p\in M$ in the question statement by one of the possible interpretation of $\bigwedge_{i=1}^n\omega$. $\endgroup$ – user10354138 Dec 20 '18 at 8:22

This is a long comment. To emphasize that you must interpret the hypothesis as being that $\omega^n (p) \ne 0$ for all $p\in M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4\times S^4$, 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 $\eta^2 = \eta\wedge\eta$ 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 $\eta^2 = d\rho_1\wedge d\rho_2\wedge dx\wedge dy$ 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 $\omega^4 = 6\pi_1^*(\eta^2)\wedge\pi_2^*(\eta^2)$ will be not identically zero on $S^4\times S^4$, and yet $H^2(S^4\times S^4) = 0$.


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