Lemma 11.2.28A9C3139B5B5

Proof. By lemma 11.1.2, $B(v,w)=\int _ Xv\mathrm{dd}^{\mathrm{c}}(w\Omega )$. Since $v$ is arbitrary, symmetry is therefore equivalent to the pointwise identity

for every smooth real $w$. Taking $w=1$ gives $\mathrm{dd}^{\mathrm{c}}\Omega =0$. The remaining expression is linear in the first jet of $w$, so we may complexify the test functions. At a chosen point, take a holomorphic linear function multiplied by a cutoff function, so that $\partial w=\lambda $ is an arbitrary $(1,0)$-covector and $\bar\partial w=0$ at that point. There the first-order expression equals

It follows that $\lambda \wedge \bar\partial \Omega =0$ for every $(1,0)$-covector $\lambda $. Since the wedge pairing between $(1,0)$-covectors and forms of bidegree $(n-1,n)$ is nondegenerate, $\bar\partial \Omega =0$, and $\partial \Omega =0$ because $\Omega $ is real. Conversely, if $\mathrm{d}\Omega =0$, then $\partial \Omega =\bar\partial \Omega =0$ by type, so $\mathrm d^{\mathrm c}\Omega =0$, and the same product formula proves symmetry. $\square$