# $\omega$ a closed 2-form and $\bigwedge_{i=1}^n \omega \ne 0$ on a compact orientable smooth $2n$-manifold w/o boundary, $M$, then $H^2(M) \ne 0$.

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?

• Fyi, this is telling you something important about symplectic manifolds: their second cohomology can’t vanish. – symplectomorphic Dec 19 '18 at 3:39
• what is $\Lambda_{i=1}^n$ do you mean $\Lambda^n$ ? – Tsemo Aristide Dec 19 '18 at 3:45
• 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. – user10354138 Dec 19 '18 at 13:43
• @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. – user98602 Dec 19 '18 at 21:23
• Please be careful with your notation. You mean to say that $\omega^n$ is nowhere $0$, I believe, not that it is not identically $0$. – 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 $$\omega^n\neq0$$.

• How is $\omega^n= d(\alpha \wedge \omega^{n-1})$? I'm getting $$d(\alpha \wedge \omega^{n-1})=d\alpha \wedge \omega^{n-1} -\alpha \wedge d\omega^{n-1}=\omega^n-\alpha \wedge d\omega^{n-1}.$$ – Al Jebr Dec 19 '18 at 3:45
• @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. – user98602 Dec 19 '18 at 12:02
• Why is $\int_M \omega^n =0$ a contradiction? – Al Jebr Dec 19 '18 at 16:26
• @AlJebr What do you know about $\omega^n$? – user98602 Dec 19 '18 at 21:21
• @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 $\bigwedge_{i=1}^n\omega$. – 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 $$H^2(S^4\times S^4) = 0$$.