= Solution
Represent the middle-degree class by its unique <harmonic differential form> $h$. The <Lefschetz commutator> on degree $k$ is $[\Lambda,L]=(n-k)\operatorname{id}$, so at $k=n$ the operators commute. Since the <adjoint Lefschetz operator> is the adjoint of the <Lefschetz operator of a Kähler manifold>,
$$
\|Lh\|^2=\langle\Lambda Lh,h\rangle
=\langle L\Lambda h,h\rangle=\|\Lambda h\|^2.
$$
Both $Lh$ and $\Lambda h$ are harmonic. A harmonic form represents zero in <de Rham cohomology> exactly when the form itself is zero. Consequently $L[\alpha]=0$ if and only if $Lh=0$, which by the norm identity is equivalent to $\Lambda h=0$, hence to $\Lambda[\alpha]=0$. Therefore
$$
\boxed{\alpha\in H^n(M,\mathbb C)\text{ is primitive}\ \Longleftrightarrow\ \Lambda\alpha=0.}
$$
This identifies the two middle-degree descriptions of a <primitive differential form on a Kähler manifold> at the cohomology level.
Back to article page
