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Let $M$ be a Kahler manifold, with metric $g$, fundamental form $\omega$, and dual Lefschetz operator $\Lambda$. Now $\Lambda$, and contraction with $\omega$, both map the two forms $\Omega^2(M)$ to $0$-forms, ie smooth functions. Are they equal? I think this is almost certainly true, but I can't see a clean argument.

Do I need Kahler here? I would guess this works for all complex manifolds.

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The Kahler form lets you turn a 2-form into a linear endomorphism of the tangent bundle. Taking the trace of that endomorphism gives the value of the dual Lefschetz operator on the 2-form. I don't know of a clean proof, but some calculations in a basis (this is just linear algebra, no Kahler condition is needed) should do the trick. –  Gunnar Magnusson Dec 17 '11 at 11:34
    
Hi Gunnar. Thanks for your answer, but I don't see how Kahler a Kahler form turns a 2-form into an endomorphism of the tangent bundle. –  Ago Szekeres Dec 17 '11 at 16:22
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Dear Ago, I should perhaps have been more precise. Consider a $(1,1)$-form $u$ on $X$. Such a form may be viewed as a sesquilinear form on the holomorphic tangent bundle $T_X$, or equivalently, a $\mathbb C$-linear morphism $T_X \to T_X^*$. The metric $\omega$ is another such morphism, but $\omega$ is invertible. Thus we can consider the endomorphism $A := \omega^{-1} \circ u$ of the holomorphic tangent bundle $T_X$. We now find that $\Lambda u = tr(A) = tr(\omega^{-1} u)$. –  Gunnar Magnusson Dec 24 '11 at 11:00
    
Ah mince, viewing $u$ as a morphism we should have $u : T_X \to \overline T_X^*$. I forgot the conjugation. –  Gunnar Magnusson Dec 24 '11 at 11:01
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