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Consider two smooth manifolds $M_1, M_2$ with symplectic forms $\omega_1, \omega_2$. The following excerpt is from Cannas da Silva, Lectures on Symplectic Geometry:

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Concerning the last line. For notational convenience, write $\omega = \theta + \eta$. For $n=1$, we get $$ \omega^2 = (\theta + \eta) \wedge (\theta + \eta) = \theta^2 + 2 (\theta \wedge \eta) + \eta^2, $$ since $\theta \wedge \eta = \eta \wedge \theta$. But what happens with the $\eta^2$ and $\theta^2$, why do they vanish? Similarly, one finds, if I am not mistaken, for $n = 2$, $$ \omega^{4} = \theta^4 + 4 \theta^3 \wedge \eta + 6(\theta^2 \wedge \eta^2) + 4 \eta^3 \wedge \theta + \eta^4. $$ (The resemblance with the binomial formula is obvious. But again, the middle term is what we need to get, but what happens with others?)

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    $\begingroup$ The others are $0$, since a $k$-form on an $n$-manifold is $0$ automatically when $k>n$ $\endgroup$ Commented Oct 24, 2022 at 17:38
  • $\begingroup$ Aren't all summands of top degree? For $n=1$, the manifold is of dimension 4, and the degree of $\theta^2$, $\theta \wedge \eta$ and $\eta^2$ is all 2+2. $\endgroup$
    – ferhenk
    Commented Oct 24, 2022 at 19:03
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    $\begingroup$ In $\sum_{j=1}^k \binom{k}{j} \eta^j\wedge \theta^{k-j}$, with $k<n$, there is no reason for any term to vanish. But for $k=2n$, it only remains $\binom{2n}{n} \eta^n\wedge \theta^n$, since any other term has $\eta^{n+1}$ or $\theta^{n+1}$ as a factor, which is zero $\endgroup$
    – Didier
    Commented Oct 24, 2022 at 19:33
  • $\begingroup$ Presumably, $M_1$ and $M_2$ are $n$-dimensional manifolds! $\endgroup$
    – Didier
    Commented Oct 24, 2022 at 20:02
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    $\begingroup$ But $\theta^2$ is the pullback of the square of a form on $M_1$, and that makes it the pullback of $0$. $\endgroup$ Commented Oct 25, 2022 at 2:44

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