Why does the Lefschetz Operator not Square to Zero? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T17:19:13Z http://mathoverflow.net/feeds/question/92458 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92458/why-does-the-lefschetz-operator-not-square-to-zero Why does the Lefschetz Operator not Square to Zero? Ago Szekeres 2012-03-28T14:10:55Z 2012-03-28T14:10:55Z <p>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?</p>