PomeExample 11.2.23.ACA94D5D202C On $X=S\times S$, let $\omega $ be the sum of the pullbacks of $\alpha _{1/\sqrt2}$ from the two factors. This form is Hermitian symplectic but not Kähler, and it is critical for $F$ under every Aeppli variation, including those with mixed factor components. If $V_0=\int _ S\mu _0$, then $F(\omega )=V_0^2/2>0$. The potential Hessian of $F$ at $\omega $ has infinite positive and negative index, even though potential variations preserve $\mathrm d\omega $.
11.2.22 Completion energy on products of complex toriEF1FC602D899
Products of two complex two-tori give explicit Hermitian symplectic forms in both regimes of theorem 11.2.19. Each factor carries a nonclosed form with an explicit completion. If both factors carry such a form, the product is critical for $F$ under every Aeppli variation although its potential Hessian is indefinite. If only one factor does, the energy is constant along potential variations but varies in the Aeppli class.
Consider the square complex two-torus $S=\mathbb C^2/(\mathbb Z+\mathrm i\mathbb Z)^2$. Let $z_ j=x_ j+\mathrm i y_ j$ be the coordinates on its cover and put $\eta =\mathrm i\sum _{j=1}^2\mathrm dz_ j\wedge \mathrm d\bar z_ j$. For $0<s<1$, the periodic forms
\[ v_ s=\frac{s}{\pi }\mathrm e^{2\pi \mathrm i x_1}\mathrm dz_2, \qquad \alpha _ s=\eta +\partial \bar v_ s+\bar\partial v_ s \]
are well defined on $S$. The Hermitian matrix of $\alpha _ s$ with respect to the forms $\mathrm i\, \mathrm dz_ j\wedge \mathrm d\bar z_ k$ is
\[ \begin{pmatrix} 1
& -s\mathrm e^{-2\pi \mathrm i x_1}
\\ -s\mathrm e^{2\pi \mathrm i x_1}
& 1
\end{pmatrix}, \]
with eigenvalues $1-s$ and $1+s$, so $\alpha _ s$ is a Hermitian form. It has the completion $\rho _ s=\partial v_ s=\mathrm i s\mathrm e^{2\pi \mathrm i x_1}\mathrm dz_1\wedge \mathrm dz_2$, and with $\mu _0=\eta ^2/2$,
\[ \frac{\alpha _ s^2}{2}=(1-s^2)\mu _0, \qquad \rho _ s\wedge \bar\rho _ s=s^2\mu _0. \]
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
\[ \operatorname {Re}(\sigma \wedge \bar\rho \wedge \omega ^2) +\rho \wedge \bar\rho \wedge \psi \wedge \omega =\frac1{12}(\psi +\sigma +\bar\sigma )\wedge \Theta ^3. \]
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. $\square$
If only one factor carries a nonclosed form, the energy becomes constant along potential variations, while the same explicit family shows that it still varies in the Aeppli class.
Example 11.2.24.F9D5DB66D92A For $0<s<1$, let $\widehat\omega _ s$ on $X=S\times S$ be the sum of the pullbacks of $\alpha _ s$ from the first factor and of $\eta $ from the second, and set $V_0=\int _ S\mu _0$. Then $\mathrm d\widehat\omega _ s\ne 0$ and
\[ F(\widehat\omega _ s)=s^2V_0^2,\qquad V(\widehat\omega _ s)=(1-s^2)V_0^2. \]
The energy is constant on all positive smooth potential variations of $\widehat\omega _ s$, and every bounded $\widehat\omega _ s$-plurisubharmonic function has mass $24(1-s^2)V_0^2$. The form $\widehat\omega _ s$ is not a critical point of $F$ in its Aeppli class, and its Monge–Ampère one-form has no primitive on any $C^2$-open neighborhood of zero.
Proof. The pullback of $\rho _ s$ from the first factor is a completion of $\widehat\omega _ s$. It is canonical: pure and mixed holomorphic two-forms are orthogonal for the product metric, and its pairing with pure-factor forms vanishes by the Stokes argument in the proof of example 11.2.23. The energy is therefore $\int _ X\rho _ s\wedge \bar\rho _ s\wedge \eta ^2/2=s^2V_0^2$. The volume is the product of the two surface volumes, which gives the second formula.
The exterior derivatives of $\widehat\omega _ s$ come entirely from the first factor. Their wedge product $\mathrm d\widehat\omega _ s\wedge \mathrm d^{\mathrm c}\widehat\omega _ s$ vanishes for degree reasons, while $\mathrm{dd}^{\mathrm{c}}\widehat\omega _ s=0$. Thus $\mathrm{dd}^{\mathrm{c}}(\widehat\omega _ s^2)=0$ by lemma 11.1.3, and corollary 11.2.21 gives the constancy of the energy and the common bounded mass.
The forms $\widehat\omega _{(1+t)s}$ are positive for small $|t|$ and lie in the Aeppli class of $\widehat\omega _ s$, since $v_{(1+t)s}=(1+t)v_ s$. The derivative of their energy at $t=0$ is $2s^2V_0^2>0$, so $\widehat\omega _ s$ is not a critical point in its Aeppli class. Finally, $\partial \alpha _ s=-\bar\partial \rho _ s\ne 0$. Hence $\mathrm d\widehat\omega _ s\ne 0$, and theorem 11.2.4 shows that no primitive exists. $\square$
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