# Why does the Lefschetz Operator not Square to Zero?

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?

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If $K$ is the Kaehler form then $K \wedge K$ is not zero, as $K$ has even degree. –  Paul Reynolds Mar 28 '12 at 14:23
If $K$ is a form of degree $p$, then we have $K \wedge K = (-1)^p K \wedge K$ by anticommutativity. When $p$ is odd this entails that $K \wedge K = 0$, but gives no information when $p$ is even, which is for example the case for the fundamental form of a hermitian metric. –  Gunnar Magnusson Mar 28 '12 at 14:23
12 seconds apart. Well played, Paul, well played. ;) –  Gunnar Magnusson Mar 28 '12 at 14:25
Yes, of course!! If the form is a simple product $f$d$g \wedge$d$h$, then it's square is zero, but when it's a sum of such products the same is not true. Thanks for your help, and sorry for the stupid question. –  Ago Szekeres Mar 28 '12 at 14:29
Well, it's not really a stupid question. Many of us had the same question, when we first learned about Kahler metrics. So in that sense it is an elementary question, and maybe more appropriate for math.stackexchange.com than here. But never mind. –  Deane Yang Mar 28 '12 at 15:34