nth-power of the dual Lefshetz operator

Question

Let $(X,\omega)$ be a Kahler manifold, denote by $\Lambda$ the dual of the Lefshetz operator $\omega\wedge$ (see e.g. Dual Lefschetz Operator and Contraction with the Fundamental Form). Let $\zeta\in\Omega^{1,1}(X)$ be a (1,1)-form . Do we have the identity $$\Lambda^n\zeta^n=(\Lambda\zeta)^n\in C^{\infty}(X)$$ where $\Lambda^n:\Omega^{n,n}(X)\to \Omega^0(X)$ is just the nth power of $\Lambda$?

Answer 1

No.

For $X =\mathbb P^1 \times \mathbb P^1$, $n=2$, $\alpha$ and $\beta$ the pullbacks of the standard $(1,1)$-forms from the two copies of $\mathbb P^1$, and $\zeta = c \alpha + d\beta$, the left side is proportional to $cd$ while the right side is the square of a linear form. These can never be equal.