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.

  • $\begingroup$ 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 Dec 17 '11 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 Dec 17 '11 at 16:22
  • 1
    $\begingroup$ 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 Dec 24 '11 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 Dec 24 '11 at 11:01

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 $g(\Lambda \alpha, 1) = g(\alpha, L \; 1) = \boxed{ g(\alpha, \omega) }$.

  • $\begingroup$ can this proof be used also for arbitrary $k$-forms? It doesn't seem to require modifications $\endgroup$ – jj_p May 21 '14 at 17:17
  • 1
    $\begingroup$ 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 May 22 '14 at 21:07
  • $\begingroup$ Still, the statement should hold true also for $k$-forms, right? $\endgroup$ – jj_p May 23 '14 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 May 23 '14 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 May 23 '14 at 7:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.