I'm trying to learn about the Lefschetz decomposition but am having a very basic problem: For the fundamental form $K$ of a Kahler metric on a complex manifold $M$, the corresponding Lefschetz operator $L$ is defined by $$ L:\Omega^k(M) \to \Omega^{k+2}(M), ~~~~~~~ \omega \mapsto K \wedge \omega. $$ From basic exterior algebra we must have $K \wedge K = 0$. Thus, to my eyes, we should have $$ L^2(\omega) = L(K \wedge \omega) = K \wedge (K \wedge \omega) = (K \wedge K) \wedge \omega = 0 \wedge \omega = 0. $$ However, the repeated Lefschetz operator is a central feature in Kahler theory. What am I missing?
closed as offtopic by abx, Stefan Waldmann, Peter Michor, Ricardo Andrade, Qfwfq Nov 13 at 16:30This question appears to be offtopic. The users who voted to close gave this specific reason:



(Just so this question has an answer.) If $\alpha \in \Omega^k(M)$ and $\beta \in \Omega^l(M)$, $\alpha\wedge\beta = (1)^{kl}\beta\wedge\alpha$. So if one of $\alpha$ or $\beta$ has even degree (i.e. $k$ or $l$ is even), $\alpha\wedge\beta = \beta\wedge\alpha$; if both $\alpha$ and $\beta$ have odd degree, then $\alpha\wedge\beta = \beta\wedge\alpha$. In particular, if $\alpha$ has odd degree, $\alpha\wedge\alpha = \alpha\wedge\alpha$ so $\alpha\wedge\alpha = 0$. Note, if $\alpha$ has even degree, the above discussion gives the tautology $\alpha\wedge\alpha = \alpha\wedge\alpha$. In the case of the Lefschetz operator, we have $L\circ L : \Omega^k(M) \to \Omega^{k+4}(M)$ given by $$(L\circ L)(\alpha) = L(K\wedge\alpha) = K\wedge(K\wedge\alpha) = (K\wedge K)\wedge\alpha$$ where $K$ is the fundamental form. As $K$ is a twoform, it has even degree, so we cannot deduce that it is zero. However, as Deane Yang points out below, we know that $K\wedge K$ is nowhere zero as $K^n$ is a volume form. 

