= Solution
Let $\omega$ be the Kähler form associated with the metric. The <Lefschetz operator of a Kähler manifold> and <adjoint Lefschetz operator> are
$$
\boxed{L\alpha=\omega\wedge\alpha,\qquad\Lambda=L^*.}
$$
Here the adjoint is taken with respect to the pointwise <Hermitian inner product> on complexified forms, or equivalently the global $L^2$ inner product. It lowers degree by two and bidegree by $(1,1)$. In terms of the <Hodge star operator>, $\Lambda=*^{-1}L*$.
Since $d\omega=0$, $dL=Ld$, so $L$ takes <closed differential forms> to <closed differential forms> and <exact differential forms> to <exact differential forms>. This already gives its action on <de Rham cohomology>. For both operators together, use the <Kähler identities>: $L$ commutes with the <Hodge Laplacian>, and taking adjoints shows that $\Lambda$ does too. Thus both preserve <harmonic differential forms>.
When $M$ is compact, the <Hodge decomposition theorem> identifies $H^k(M,\mathbb C)$ with its space of harmonic $k$-forms. If $h_\alpha$ is the unique <harmonic differential form> representing $[\alpha]$, define
$$
\boxed{L[\alpha]=[Lh_\alpha],\qquad\Lambda[\alpha]=[\Lambda h_\alpha].}
$$
The resulting forms are harmonic and therefore closed, and uniqueness of $h_\alpha$ makes these definitions independent of the initial representative. The cohomological definition of $\Lambda$ uses <harmonic differential forms> as representatives: $\Lambda$ need not take an arbitrary closed form to a closed form.
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