Following Huybrechts book Complex Geometry – An Introduction, the author states in Corollary 3.3.10, that on a compact Kähler manifold $(X,g)$ the Lefschetz operator $L$ and its dual $\Lambda$ give rise to maps in Dolbeault cohomology, i. e.
\begin{eqnarray}
L\colon H^{p,q}(X) &&\rightarrow H^{p+1,q+1}(X)\\
[\alpha] &&\mapsto [\alpha \wedge \omega],
\end{eqnarray}
where $\omega$ denotes the associated fundamental form to $g$; and
\begin{eqnarray}
\Lambda\colon H^{p,q}(X) &&\rightarrow H^{p-1,q-1}(X).
\end{eqnarray}
For sure $L$ only depends on the Kähler class $[\omega] \in H^{1,1}(X)$. My question is:
Why does $\Lambda$ only depend on the Kähler class and not on the choosen metric $g$?
Best Answer
Let $\omega$ and $\omega'$ be two Kahler metrics on a compact Kahler manifold $X$. We denote by $L, *, \Lambda$ and $L', *', \Lambda'$ the operators in cohomology that the harmonic forms with respect to these metrics induce, and write $\omega$ and $\omega'$ for the Kahler classes in abuse of notation. (We won't speak of the actual metrics again.) It is obvious that if $\omega = \omega'$, then $L = L'$.
Claim. If $\omega = \omega'$, then $\operatorname{Ker} \Lambda = \operatorname{Ker} \Lambda'$.
It's enough to prove this for classes of a fixed degree $k$. Recall that for a class of degree $k \leq n$, we have $$ [L^{r+1}, \Lambda] u = r(n-k+1-r) L^r u. $$ Picking $r = n-k+1$ we get $$ L^{n-k+1} \Lambda u = \Lambda L^{n-k+1} u. $$ Recall that $\Lambda$ is injective on classes of degree $> n$, because $L$ is surjective on those classes and $\Lambda$ is its adjoint; and recall that $L^{n-k+1}$ is an isomorphism on classes of degree $k-1$. By comparing degrees on the left- and right-hand side above, we see that $\Lambda u = 0$ if and only if $L^{n-k+1}u = 0$. But if $\omega = \omega'$, then $L = L'$, so we conclude the claim.
Now consider the decompositions $$ u = \sum L^j u_{k-j} = \sum (L')^j u_{k-j}' $$ into products of primitive classes, where $u_{k-j}$ is primitive with respect to $\omega$, and $u_{k-j}'$ is primitive with respect to $\omega'$. By the claim, each $u_{k-j}'$ is actually primitive with respect to $\omega$, and by the unicity of the factors in the decomposition we conclude that $u_{k-j}' = u_{k-j}$ if $\omega = \omega'$. The formula for $*(L^k u_{k-j})$ then shows that if $\omega = \omega'$, then $* = *'$. As $\Lambda = *^{-1}L*$, we then conclude that $\Lambda' = \Lambda$ if $\omega = \omega'$.