# Elementary question on wedging 2-forms

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:

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?)

• The others are $0$, since a $k$-form on an $n$-manifold is $0$ automatically when $k>n$ Commented Oct 24, 2022 at 17:38
• 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. Commented Oct 24, 2022 at 19:03
• 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 Commented Oct 24, 2022 at 19:33
• Presumably, $M_1$ and $M_2$ are $n$-dimensional manifolds! Commented Oct 24, 2022 at 20:02
• But $\theta^2$ is the pullback of the square of a form on $M_1$, and that makes it the pullback of $0$. Commented Oct 25, 2022 at 2:44