Lemma 11.2.38535B6F80ED4

Proof. For $n=2$ the assertion is immediate. Otherwise the expression $T\wedge A^{n-2}$ is polynomial in the real coefficients of $A$, so it vanishes for every real $(1,1)$-form $A$. Polarization and then complexification give $T\wedge A_1\wedge \cdots \wedge A_{n-2}=0$ for arbitrary complex $(1,1)$-forms $A_ j$, and such products span all forms of bidegree $(n-2,n-2)$.

Since these wedges preserve type, the identity holds separately for the two type components of $T$. Wedging the identity for $T^{2,1}$ with arbitrary $(0,1)$-covectors, we obtain test forms spanning bidegree $(n-2,n-1)$, which pairs perfectly with bidegree $(2,1)$ into top degree. Thus $T^{2,1}=0$. The component $T^{1,2}$ is treated in the same way with $(1,0)$-covectors. $\square$