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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6$\begingroup$ If $K$ is the Kaehler form then $K \wedge K$ is not zero, as $K$ has even degree. $\endgroup$– Paul ReynoldsCommented Mar 28, 2012 at 14:23
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6$\begingroup$ 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. $\endgroup$– Gunnar Þór MagnússonCommented Mar 28, 2012 at 14:23
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3$\begingroup$ 12 seconds apart. Well played, Paul, well played. ;) $\endgroup$– Gunnar Þór MagnússonCommented Mar 28, 2012 at 14:25
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$\begingroup$ 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. $\endgroup$– Ago SzekeresCommented Mar 28, 2012 at 14:29
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3$\begingroup$ 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. $\endgroup$– Deane YangCommented Mar 28, 2012 at 15:34
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1 Answer
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(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 two-form, 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.
answered Nov 13, 2014 at 5:06
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1$\begingroup$ Actually we can say more about the exterior product of $K$ with itself. $K^n$ is the volume form and therefore is nonzero everywhere. $\endgroup$ Commented Nov 13, 2014 at 5:21
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$\begingroup$ @DeaneYang: Good point. $\endgroup$ Commented Nov 13, 2014 at 14:10