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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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  • $\begingroup$ 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. $\endgroup$
    – Gunnar Þór Magnússon
    Commented Dec 17, 2011 at 11:34
  • $\begingroup$ 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. $\endgroup$
    – Ago Szekeres
    Commented Dec 17, 2011 at 16:22
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    $\begingroup$ 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 $\Lambda u = tr(A) = tr(\omega^{-1} u)$. $\endgroup$
    – Gunnar Þór Magnússon
    Commented Dec 24, 2011 at 11:00
  • $\begingroup$ Ah mince, viewing $u$ as a morphism we should have $u : T_X \to \overline T_X^*$. I forgot the conjugation. $\endgroup$
    – Gunnar Þór Magnússon
    Commented Dec 24, 2011 at 11:01

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The question asks whether $\boxed{ \Lambda \alpha = g(\omega,\alpha)}$ for any $\alpha \in \Omega^2(X)$.

This is indeed true, as explained by Gunnar, but here is a simple way of seeing it. Firstly, since we're just using linear operators we can work fiber-wise (hence indeed no closedness of $\omega$ is needed, and whether we work in the context of $TM$ or general vector bundles is irrelevant). Remember that by definition $g(\Lambda \eta, \mu) = g(\eta, L \mu)$ where $L$ is the wedge product with $\omega$.

Using this in our case: let $\alpha \in \Lambda^2 E_x$, then we can rewrite $\boxed{\Lambda \alpha}$ as $g(\Lambda \alpha, 1) = g(\alpha, L \; 1) = \boxed{ g(\alpha, \omega) }$.

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  • $\begingroup$ can this proof be used also for arbitrary $k$-forms? It doesn't seem to require modifications $\endgroup$
    – jj_p
    Commented May 21, 2014 at 17:17
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    $\begingroup$ No, since I used that $\Lambda \alpha$ is a number to write $\Lambda \alpha = g(\Lambda \alpha,1)$. (Also, it would not be clear what is meant by $g(\alpha, \omega)$ with $\alpha$ a $k$-form and $\omega$ a $2$-form.) $\endgroup$
    – Ruben Verresen
    Commented May 22, 2014 at 21:07
  • $\begingroup$ Still, the statement should hold true also for $k$-forms, right? $\endgroup$
    – jj_p
    Commented May 23, 2014 at 6:32
  • $\begingroup$ Not the statement that I put in the box, no. All we know (at least without doing any more effort) is that $g(\Lambda \alpha, \beta ) = g(\alpha, L\beta) = g(\alpha, \omega \wedge \beta)$ (which are sadly nothing but the definitions!) The equation $\Lambda \alpha = g(\alpha, \omega)$ doesn't make sense in general, since if $\alpha$ is a $k$-form, what would the RHS mean? $\endgroup$
    – Ruben Verresen
    Commented May 23, 2014 at 6:48
  • $\begingroup$ I was a bit imprecise, by statement I meant 'contraction with $\omega$ is the same as operating with $L$ on a $k$-form': can one try to see this by using the first equation in your last post? $\endgroup$
    – jj_p
    Commented May 23, 2014 at 7:55

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