Example 11.2.23ACA94D5D202C

Proof. Denote the two factor forms by $\alpha _ i,\rho _ i$, and put $\mu _ i=\alpha _ i^2/2$ and $\nu _ i=\rho _ i\wedge \bar\rho _ i$; we suppress pullbacks from the notation. Every holomorphic form on a complex torus has constant coefficients. On each factor, the $L^2$ pairing of $\rho _ i=\partial v_ i$ with a holomorphic two-form is a multiple of their wedge pairing, which vanishes by Stokes' theorem. On the product, pure and mixed two-forms are pointwise orthogonal, and the pairing with a pure-factor form is the vanishing pairing on that factor times the volume of the other factor. Thus $\rho =\rho _1+\rho _2$ is the canonical completion of $\omega $. Since $\mu _ i=\nu _ i$ is half the pullback of $\mu _0$, expansion gives $F(\omega )=\int _ X(\nu _1\wedge \mu _2+\mu _1\wedge \nu _2)=V_0^2/2$.

Put $\Theta _ i=\alpha _ i+\rho _ i+\bar\rho _ i$ and let $\Theta =\Theta _1+\Theta _2$, which is closed. For an arbitrary real $(1,1)$-form $\psi $ and $(2,0)$-form $\sigma $ on $X$, expansion on the two factors gives

Indeed, the mixed components of $\psi $ and $\sigma $ have odd degree on both factors and contribute zero. The pure components on the left give $(\psi +\sigma +\bar\sigma )\wedge (\Theta _1\wedge \mu _2+\Theta _2\wedge \mu _1)$, and since $\Theta _ i^2=4\mu _ i$, the cube on the right is twelve times the same complementary form. Apply lemma 11.2.15 with $\psi =\partial \bar\tau +\bar\partial \tau $ and $\sigma =\partial \tau $. The derivative is $\frac1{12}\int _ X\mathrm d(\tau +\bar\tau )\wedge \Theta ^3=0$, since $\Theta $ is closed. This proves criticality under every Aeppli variation.

Each $\partial \alpha _ i=-\bar\partial \rho _ i$ is nonzero, so $\omega $ is not Kähler. In $\partial \omega \wedge \bar\partial \omega $, the terms coming from a single factor vanish for dimension reasons, while the two cross terms are nonzero and have different bidegrees on the factors, so they cannot cancel. Hence $\mathrm d\omega \wedge \mathrm d^{\mathrm c}\omega \ne 0$, and since $\mathrm{dd}^{\mathrm{c}}\omega =0$, lemma 11.1.3 gives $\mathrm{dd}^{\mathrm{c}}(\omega ^2)\ne 0$. It remains to apply theorem 11.2.19. $\square$