Theorem 10.4.3 (The mixed Hodge index theorem)2AE132953DC7

Proof. Let $\gamma $ be a Kähler class with a Kähler form $h\in \gamma $, and let $\alpha $ satisfy $Q(\alpha ,\gamma )=0$. Choose a smooth closed real form $a\in \alpha $; then $\int _ Xa\wedge h\wedge \Omega =0$. Apply lemma 10.2.9 with $A=h\wedge \Omega $, a product of $n-1$ Kähler forms, and $\eta =-a\wedge h\wedge \Omega $. The resulting $f$ makes $a'=a+\mathrm{dd}^{\mathrm{c}}f\in \alpha $ satisfy $a'\wedge h\wedge \Omega =0$ at every point. At a point, in holomorphic coordinates, the wedge products of real $(1,1)$-forms are expressed by mixed discriminants of their Hermitian matrices, so lemma 5.3.4 gives $a'\wedge a'\wedge \Omega \leq 0$, with equality only if $a'$ vanishes at that point. By Stokes' formula,

If this integral vanishes, the smooth nonpositive density $a'\wedge a'\wedge \Omega $ vanishes everywhere, so $a'=0$ and $\alpha =0$. Thus $Q$ is negative definite on the hyperplane $\{ Q(\cdot ,\gamma )=0\} $.

Since $Q(\gamma ,\gamma )=\int _ Xh^2\wedge \Omega >0$, every class splits uniquely as a multiple of $\gamma $ plus a class in that hyperplane. The space $H^{1,1}(X,\mathbb R)$ is finite dimensional by theorem 10.2.15, and $Q$ is positive on the line spanned by $\gamma $ and negative definite on its $Q$-orthogonal complement; this gives the signature. Finally, let $\alpha ,\beta $ be Kähler classes and write $\beta =c\alpha +\delta $ with $Q(\delta ,\alpha )=0$. Negative definiteness gives $Q(\beta ,\beta )=c^2Q(\alpha ,\alpha )+Q(\delta ,\delta )\leq c^2Q(\alpha ,\alpha )$, while $Q(\alpha ,\beta )=cQ(\alpha ,\alpha )$. Since $Q(\alpha ,\alpha )>0$, we obtain $Q(\alpha ,\beta )^2=c^2Q(\alpha ,\alpha )^2\geq Q(\alpha ,\alpha )Q(\beta ,\beta )$. $\square$