PomeLemma 11.2.2.8A9C3139B5B5 For a smooth real $(n-1,n-1)$-form $\Omega $, the pairing $B(v,w)=\int _ Xw\mathrm{dd}^{\mathrm{c}}v\wedge \Omega $ is symmetric on smooth real functions if and only if $\mathrm{d}\Omega =0$.
11.2 Monge–Ampère energy and torsion1FC29EAFAA97
1FC29EAFAA97
Throughout this chapter, $\omega $ is a Hermitian form on $X$, and $\omega _ u=\omega +\mathrm{dd}^{\mathrm{c}}u$ and the space $\mathcal H_\omega $ of smooth strictly $\omega $-plurisubharmonic functions are as in section 11.1.
For a Kähler form, the Monge–Ampère operator is the gradient of an energy: there is a functional on $\mathcal H_\omega $ whose derivative at $u$ in every smooth real direction $v$ is $\int _ Xv\, \omega _ u^ n$. For a Hermitian form, such a primitive exists exactly when $\omega $ is closed. Symmetry of its mixed second derivatives first forces $\mathrm d(\omega _ u^{n-1})=0$, and varying the potential then forces $\mathrm d\omega =0$; this is stronger than the mass invariance of section 11.1. For a nonclosed background, integrating the Monge–Ampère one-form along radial segments still defines a functional, and its first variation shows which torsion terms enter the corresponding Euler equation. A different energy is available for Hermitian symplectic forms on fourfolds: along potential variations, this completion energy changes by minus the change of the volume. It is therefore constant along potential variations exactly when the mass is invariant, although even then the Monge–Ampère one-form has no primitive unless $\omega $ is closed.
11.2.1 Primitives of the Monge–Ampère one-form49D0776D0F6B
On $\mathcal H_\omega $ consider the Monge–Ampère one-form
\[ \alpha _ u(v)=\int _ Xv\, \omega _ u^ n,\qquad v\in C^\infty (X,\mathbb R). \]
Let $U\subseteq \mathcal H_\omega $ be open in the $C^2$ topology on smooth functions. A primitive of $\alpha $ on $U$ is a functional $E$ on $U$ such that, for every $u\in U$ and every smooth real direction $v$, the directional derivative
\[ DE_ u(v)=\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}E(u+tv) \]
exists and equals $\alpha _ u(v)$. Restrict $E$ to the intersection of $U$ with a finite-dimensional affine plane. Its first derivatives are values of $\alpha $, which depend smoothly on the parameters, so its mixed second derivatives agree. The following integration-by-parts criterion identifies the differential condition imposed by this symmetry.
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
\[ \mathrm{dd}^{\mathrm{c}}(w\Omega )-\mathrm{dd}^{\mathrm{c}}w\wedge \Omega =w\mathrm{dd}^{\mathrm{c}}\Omega +\mathrm{d}w\wedge \mathrm d^{\mathrm c}\Omega -\mathrm d^{\mathrm c}w\wedge \mathrm{d}\Omega =0 \]
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
\[ \frac{\mathrm i}{2\pi }\bigl (\partial w\wedge \bar\partial \Omega -\bar\partial w\wedge \partial \Omega \bigr ). \]
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. $\square$
Lemma 11.2.3.8535B6F80ED4 Let $T$ be a form of type $(2,1)+(1,2)$ on a complex vector space of dimension $n\geq 2$. If $T\wedge A^{n-2}=0$ for every real $(1,1)$-form $A$ in an open neighborhood of a positive form, then $T=0$.
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. $\square$
Theorem 11.2.4.7D069B347672 For $n\geq 2$, the one-form $\alpha $ has a primitive on $\mathcal H_\omega $ if and only if $\mathrm{d}\omega =0$. In fact, a primitive on a $C^2$-open neighborhood of $0$ already forces $\mathrm{d}\omega =0$.
Proof. Suppose that a primitive $E$ exists on such a neighborhood $U$. For $u\in U$ and smooth real directions $v,w$, the mixed second derivatives of $(s,t)\mapsto E(u+sv+tw)$ at the origin agree. After division by $n$, this equality reads
B4B9D0AD937E
11.2.5
\begin{equation} \int _ Xw\mathrm{dd}^{\mathrm{c}}v\wedge \omega _ u^{n-1} =\int _ Xv\mathrm{dd}^{\mathrm{c}}w\wedge \omega _ u^{n-1}. \end{equation}
By lemma 11.2.2,
\[ 0=\mathrm{d}(\omega _ u^{n-1}) =(n-1)\mathrm{d}\omega \wedge (\omega +\mathrm{dd}^{\mathrm{c}}u)^{n-2}. \]
Fix a point $x$. A real quadratic polynomial in local coordinates, multiplied by a cutoff function equal to one near $x$, realizes any prescribed real $(1,1)$-form $H$ as $\mathrm{dd}^{\mathrm{c}}u(x)$. These functions can be chosen to depend linearly on $H$, with global $C^2$ norms bounded by a fixed constant times the coefficient norm of $H$. Thus every sufficiently small $H$ is realized by a potential in $U$, and the preceding identity holds at $x$ with $\omega (x)+H$ ranging over an open neighborhood of the positive form $\omega (x)$. Applying lemma 11.2.3 with $T=\mathrm{d}\omega (x)$, we conclude that $\mathrm{d}\omega (x)=0$.
Since $x$ was arbitrary, $\mathrm{d}\omega =0$. Conversely, proposition 11.2.8 below constructs a primitive for a closed background. $\square$
In dimension one every real $(1,1)$-form is closed, and proposition 11.2.8 always gives a primitive. For $n\geq 2$, the symmetry condition at a single potential $u$ is $\mathrm d(\omega _ u^{n-1})=0$; the theorem uses it on an open set of potentials. Mass invariance alone does not suffice. Indeed, let $X$ be a compact complex surface admitting no Kähler metric, and let $\omega $ be a Gauduchon form on $X$, that is, $\mathrm{dd}^{\mathrm{c}}\omega =0$ (definition 11.4.1); such forms exist by theorem 11.4.2. The mass of $\omega $ is constant by corollary 11.1.10, but $\omega $ is not closed, so its Monge–Ampère one-form has no primitive.
11.2.6 Radial energy and closed backgroundsB9357DC44463
Integrating the Monge–Ampère one-form along the segment from $0$ to $u$ defines a functional for an arbitrary background. Let $\theta $ be a smooth real $(1,1)$-form on $X$. For $u\in C^\infty (X,\mathbb R)$, put $\theta _ u=\theta +\mathrm{dd}^{\mathrm{c}}u$ and define the radial energy
1CE478F66A34
11.2.7
\begin{equation} \begin{aligned} E_\theta (u) & =\int _0^1\int _ Xu(\theta +t\mathrm{dd}^{\mathrm{c}}u)^ n\, \mathrm{d}t\\ & =\frac1{n+1}\sum _{j=0}^ n \int _ Xu\, \theta _ u^ j\wedge \theta ^{n-j}. \end{aligned} \end{equation}
To obtain the second expression, write $\theta +t\mathrm{dd}^{\mathrm{c}}u=(1-t)\theta +t\theta _ u$, expand, and use
\[ \binom nj\int _0^1t^ j(1-t)^{n-j}\, \mathrm dt=\frac1{n+1}. \]
We write $DE_{\theta ,u}(v)$ for the derivative of $E_\theta $ at $u$ in the direction $v$. When $\theta $ is closed, integration by parts makes the derivatives of the mixed terms telescope.
Proposition 11.2.8.E345C8B2A48C If $\mathrm d\theta =0$, then $DE_{\theta ,u}(v)=\int _ Xv\theta _ u^ n$ for every pair of smooth real functions $u,v$.
Proof. Differentiate the finite sum in (11.2.7). The term indexed by $j$ contributes
\[ \int _ Xv\theta _ u^ j\wedge \theta ^{n-j} +j\int _ Xu\mathrm{dd}^{\mathrm{c}}v\wedge \theta _ u^{j-1}\wedge \theta ^{n-j}, \]
the second summand being absent for $j=0$. For $j\geq 1$, the form $\theta _ u^{j-1}\wedge \theta ^{n-j}$ is closed, so by lemma 11.2.2 we may interchange $u$ and $v$ in the second summand. Using $\mathrm{dd}^{\mathrm{c}}u=\theta _ u-\theta $, we can then write the contribution of the term with $j\geq 1$ as
\[ (j+1)\int _ Xv\theta _ u^ j\wedge \theta ^{n-j} -j\int _ Xv\theta _ u^{j-1}\wedge \theta ^{n-j+1}. \]
Together with the contribution $\int _ Xv\theta ^ n$ of the term $j=0$, these expressions telescope to $(n+1)\int _ Xv\theta _ u^ n$. Dividing by $n+1$ completes the proof. $\square$
Thus, when $\mathrm d\omega =0$, the restriction of $E_\omega $ to $\mathcal H_\omega $ is the primitive required in theorem 11.2.4.
11.2.9 First variation in a nonclosed backgroundAEF022344DD6
For the Hermitian background $\theta =\omega $, the functional (11.2.7) integrates $\alpha $ along the segment from $0$ to $u\in \mathcal H_\omega $. When $\omega $ is not closed, the derivative of $E_\omega $ differs from $\alpha $ by the terms produced when integration by parts differentiates the background powers. Computing this difference gives the Euler equation of $E_\omega $ even when no primitive exists.
Proposition 11.2.10.4793125897B2 For $u\in \mathcal H_\omega $ and $v\in C^\infty (X,\mathbb R)$, put $\Omega _ t=(\omega +t\mathrm{dd}^{\mathrm{c}}u)^{n-1}$ for $0\leq t\leq 1$. Then
91AE93E20D38
11.2.11
\begin{equation} \begin{split} DE_{\omega ,u}(v)-\int _ Xv\omega _ u^ n =n\int _0^1t\int _ Xv\bigl (& u\mathrm{dd}^{\mathrm{c}}\Omega _ t +\mathrm{d}u\wedge \mathrm d^{\mathrm c}\Omega _ t\\ & -\mathrm d^{\mathrm c}u\wedge \mathrm{d}\Omega _ t\bigr )\, \mathrm{d}t. \end{split} \end{equation}
Proof. Differentiating the radial integral in (11.2.7) in the direction $v$ gives
\[ DE_{\omega ,u}(v) =\int _0^1\left(\int _ Xv(\omega +t\mathrm{dd}^{\mathrm{c}}u)^ n +nt\int _ Xu\mathrm{dd}^{\mathrm{c}}v\wedge \Omega _ t\right)\mathrm dt. \]
On the other hand,
\[ \begin{aligned} \int _ Xv\omega _ u^ n & =\int _0^1\frac{\mathrm d}{\mathrm dt} \left(t\int _ Xv(\omega +t\mathrm{dd}^{\mathrm{c}}u)^ n\right)\mathrm dt\\ & =\int _0^1\left(\int _ Xv(\omega +t\mathrm{dd}^{\mathrm{c}}u)^ n +nt\int _ Xv\mathrm{dd}^{\mathrm{c}}u\wedge \Omega _ t\right)\mathrm dt. \end{aligned} \]
The terms without a factor $nt$ cancel in the difference, leaving
\[ \begin{aligned} DE_{\omega ,u}(v)-\int _ Xv\omega _ u^ n & =n\int _0^1t\int _ X (u\mathrm{dd}^{\mathrm{c}}v-v\mathrm{dd}^{\mathrm{c}}u)\wedge \Omega _ t\, \mathrm dt\\ & =n\int _0^1t\int _ Xv \bigl (\mathrm{dd}^{\mathrm{c}}(u\Omega _ t)-\mathrm{dd}^{\mathrm{c}}u\wedge \Omega _ t\bigr )\, \mathrm dt. \end{aligned} \]
The second equality is lemma 11.1.2. Expanding the last parenthesis by the product rule in the proof of lemma 11.2.2 gives (11.2.11). When $n=1$, one has $\Omega _ t=1$ and the right-hand side vanishes. $\square$
Proposition 11.2.12.0009683E4137 If $n=2$, then every $u\in \mathcal H_\omega $ satisfies
\[ E_\omega (u)=\int _ Xu\omega ^2 +\int _ Xu\omega \wedge \mathrm{dd}^{\mathrm{c}}u +\frac13\int _ Xu(\mathrm{dd}^{\mathrm{c}}u)^2. \]
For every smooth real direction $v$, its differential satisfies
173E9E78941E
11.2.13
\begin{equation} DE_{\omega ,u}(v)-\int _ Xv\omega _ u^2 =\int _ Xv\bigl (u\mathrm{dd}^{\mathrm{c}}\omega +\mathrm{d}u\wedge \mathrm d^{\mathrm c}\omega -\mathrm d^{\mathrm c}u\wedge \mathrm{d}\omega \bigr ). \end{equation}
Proof. Expanding the radial integral gives the first identity. For $n=2$, one has $\Omega _ t=\omega +t\mathrm{dd}^{\mathrm{c}}u$, and since $\mathrm d$ and $\mathrm d^{\mathrm c}$ annihilate $\mathrm{dd}^{\mathrm{c}}u$, the derivatives $\mathrm d\Omega _ t$, $\mathrm d^{\mathrm c}\Omega _ t$ and $\mathrm{dd}^{\mathrm{c}}\Omega _ t$ are those of $\omega $. Formula (11.2.11) and $2\int _0^1t\, \mathrm dt=1$ give the second identity. $\square$
Consequently, if $n=2$ and $\mu $ is a fixed smooth real $(2,2)$-form on $X$, a critical point $u\in \mathcal H_\omega $ of the functional $E_\omega (u)-\int _ Xu\, \mu $ satisfies
\[ \omega _ u^2+u\mathrm{dd}^{\mathrm{c}}\omega +\mathrm{d}u\wedge \mathrm d^{\mathrm c}\omega -\mathrm d^{\mathrm c}u\wedge \mathrm{d}\omega =\mu \]
rather than the Monge–Ampère equation $\omega _ u^2=\mu $. On a Gauduchon surface the term $u\mathrm{dd}^{\mathrm{c}}\omega $ vanishes, but the first-order torsion terms can persist. The radial Euler equation can therefore differ from the Monge–Ampère equation even when the total mass is invariant.
11.2.14 Symplectic completions and potential variationsA10584C44FB2
The radial energy is built from the Monge–Ampère one-form, and its first variation exhibits the torsion of the background. When a Hermitian form is the $(1,1)$-part of a closed real two-form, there is another natural energy: the squared $L^2$ norm of the smallest $(2,0)$-part needed to close it. In complex dimension four, adding a complex Hessian to the form changes this energy by minus the change of the volume. Its potential variations are therefore governed by the volume, and not by a primitive of the Monge–Ampère one-form.
Throughout this section we assume $n=4$. A Hermitian form $\omega $ is Hermitian symplectic if there is a smooth $(2,0)$-form $\rho $, called a completion of $\omega $, such that
\[ \partial \rho =0,\qquad \bar\partial \rho =-\partial \omega . \]
Equivalently, the real two-form $\omega +\rho +\bar\rho $ is closed. Applying $\bar\partial $ to the second equation shows that $\omega $ is pluriclosed, meaning $\mathrm{dd}^{\mathrm{c}}\omega =0$. Two completions differ by a closed holomorphic $(2,0)$-form, so the completions of $\omega $ form an affine space over the finite-dimensional space of closed holomorphic $(2,0)$-forms. The quadratic norm
\[ \| \rho \| _{L^2(\omega )}^2 =\frac12\int _ X\rho \wedge \bar\rho \wedge \omega ^2 \]
is positive definite on $(2,0)$-forms, and minimizing it over this affine space gives a unique canonical completion $\rho _\omega $. It is characterized by being orthogonal to every closed holomorphic $(2,0)$-form. Define the completion energy and the volume of $\omega $ by
\[ F(\omega )=\| \rho _\omega \| _{L^2(\omega )}^2, \qquad V(\omega )=\frac1{4!}\int _ X\omega ^4. \]
The completion energy is nonnegative and vanishes exactly when $\omega $ is Kähler: a zero canonical completion forces $\partial \omega =0$, while a closed $\omega $ has the zero completion.
Since $\omega $ is pluriclosed, it defines an Aeppli cohomology class, whose real $(1,1)$-representatives are the forms $\omega +\partial \bar\tau +\bar\partial \tau $, with $\tau $ a smooth $(1,0)$-form. We call a real form $\psi =\partial \bar\tau +\bar\partial \tau $ an Aeppli variation. If $\rho $ completes $\omega $ and $\omega +t\psi $ is positive for a real $t$, then $\omega +t\psi $ has the completion $\rho +t\partial \tau $. In particular, the positive representatives $\omega +\partial \bar\tau +\bar\partial \tau >0$ of the Aeppli class of $\omega $ are again Hermitian symplectic. Every positive potential variation $\omega +\mathrm{dd}^{\mathrm{c}}\varphi $ belongs to this class and has the same exterior derivative as $\omega $. We write $DF_\omega (\psi )$ for the derivative of $t\mapsto F(\omega +t\psi )$ at $t=0$.
Lemma 11.2.15.19D4882CC9D5 Let $\omega $ be Hermitian symplectic, with canonical completion $\rho =\rho _\omega $. Every Aeppli variation $\psi =\partial \bar\tau +\bar\partial \tau $ satisfies
7C73F0133C9D
11.2.16
\begin{equation} DF_\omega (\psi )=\int _ X\left( \operatorname {Re}(\partial \tau \wedge \bar\rho \wedge \omega ^2) +\rho \wedge \bar\rho \wedge \psi \wedge \omega \right). \end{equation}
Proof. The canonical completion of $\omega +t\psi $ is $\rho +t\partial \tau -h(t)$, where $h(t)$ is a closed holomorphic $(2,0)$-form and $h(0)=0$. In a fixed basis of these holomorphic forms, the orthogonal projection defining $h(t)$ solves a positive definite Gram system whose coefficients are smooth in $t$. Since $\rho $ is orthogonal to closed holomorphic $(2,0)$-forms, the contribution of $h'(0)$ to the first derivative of the squared norm vanishes. We may therefore differentiate using the completion $\rho +t\partial \tau $. Differentiating its two factors and the factor $(\omega +t\psi )^2/2$ gives the formula. $\square$
Proposition 11.2.17.AE6CE389C569 Let $\omega $ be Hermitian symplectic. For every smooth real function $\varphi $ with $\omega _\varphi =\omega +\mathrm{dd}^{\mathrm{c}}\varphi >0$, one has $\rho _{\omega _\varphi }=\rho _\omega $ and
BAFFEA611460
11.2.18
\begin{equation} F(\omega _\varphi )+V(\omega _\varphi )=F(\omega )+V(\omega ). \end{equation}
Moreover, $DF_\omega (\mathrm{dd}^{\mathrm{c}}\varphi )=0$ for every smooth real $\varphi $ if and only if $\mathrm{dd}^{\mathrm{c}}(\omega ^3)=0$. Thus a critical point of $F$ in its Aeppli class is Gauduchon.
Proof. Fix any completion $\zeta $ of $\omega $ and put $\Theta _\zeta =\omega +\zeta +\bar\zeta $. Expansion by type gives
\[ \frac1{4!}\int _ X\Theta _\zeta ^4 =V(\omega )+\frac12\int _ X\zeta \wedge \bar\zeta \wedge \omega ^2 +\frac14\int _ X\zeta ^2\wedge \bar\zeta ^2. \]
Since $\partial \omega _\varphi =\partial \omega $, the forms $\omega _\varphi $ and $\omega $ have the same completions. With $\zeta $ fixed, replacing $\omega $ by $\omega _\varphi $ changes the closed form $\Theta _\zeta $ by an exact form, so the integral on the left is unchanged; so is the last integral on the right, which does not involve $\omega $. Hence
\[ \| \zeta \| _{L^2(\omega _\varphi )}^2 -\| \zeta \| _{L^2(\omega )}^2=V(\omega )-V(\omega _\varphi ). \]
This difference is independent of $\zeta $, so the two norms have the same minimizer, which gives $\rho _{\omega _\varphi }=\rho _\omega $ and (11.2.18). Differentiating at $t=0$ along $\omega +t\mathrm{dd}^{\mathrm{c}}\varphi $ and using lemma 11.1.2, we obtain
\[ DF_\omega (\mathrm{dd}^{\mathrm{c}}\varphi )=-\frac16\int _ X\varphi \mathrm{dd}^{\mathrm{c}}(\omega ^3). \]
Since $\varphi $ is arbitrary, this proves the remaining assertions. $\square$
The positive index of a real quadratic form on a real vector space is the supremum of the dimensions of the finite-dimensional subspaces on which it is positive definite; the negative index is defined in the same way. Thus the positive index is infinite when the form is positive definite on subspaces of arbitrarily large finite dimension.
Theorem 11.2.19.04BD04D6C676 Let $\omega $ be Hermitian symplectic, and let $H_\omega $ be the Hessian of $F$ at $\omega $ along the affine space of real representatives of the Aeppli class of $\omega $, so that $H_\omega (\psi ,\psi )$ is the second derivative of $t\mapsto F(\omega +t\psi )$ at $t=0$. If $\mathrm{dd}^{\mathrm{c}}(\omega ^2)\ne 0$, the restriction of $H_\omega $ to potential directions $\psi =\mathrm{dd}^{\mathrm{c}}\varphi $ has infinite positive and negative index. If $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$, then $F(\omega +\mathrm{dd}^{\mathrm{c}}\varphi )=F(\omega )$ whenever $\omega +\mathrm{dd}^{\mathrm{c}}\varphi >0$. Consequently, every local minimum of $F$ in an Aeppli class satisfies $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$.
Proof. Step 1: the Hessian formula. Pluriclosedness and Stokes' theorem give $\int _ X\omega \wedge (\mathrm{dd}^{\mathrm{c}}\varphi )^3=\int _ X(\mathrm{dd}^{\mathrm{c}}\varphi )^4=0$. Thus the volume and, by (11.2.18), the energy along a potential line $t\mapsto \omega +t\mathrm{dd}^{\mathrm{c}}\varphi $ are polynomials of degree at most two in $t$. Differentiating twice and integrating by parts, we obtain
8B6CFC8EA37B
11.2.20
\begin{equation} \begin{aligned} H_\omega (\mathrm{dd}^{\mathrm{c}}\varphi ,\mathrm{dd}^{\mathrm{c}}\varphi ) & =-\frac12\int _ X\omega ^2\wedge (\mathrm{dd}^{\mathrm{c}}\varphi )^2\\ & =\frac12\int _ X\mathrm{dd}^{\mathrm{c}}(\omega ^2)\wedge \mathrm d\varphi \wedge \mathrm d^{\mathrm c}\varphi . \end{aligned} \end{equation}
For the second equality, apply lemma 11.1.2 once and then use the identity $\mathrm{dd}^{\mathrm{c}}(\varphi ^2/2)=\varphi \mathrm{dd}^{\mathrm{c}}\varphi +\mathrm d\varphi \wedge \mathrm d^{\mathrm c}\varphi $ against the closed form $\mathrm{dd}^{\mathrm{c}}(\omega ^2)$.
Step 2: both indices are infinite if $\mathrm{dd}^{\mathrm{c}}(\omega ^2)\ne 0$. Put $\Psi =\mathrm{dd}^{\mathrm{c}}(\omega ^2)$. At each point, $\xi \mapsto \Psi \wedge \mathrm i\xi \wedge \bar\xi $, divided by a positive volume form, is a Hermitian quadratic form in the $(1,0)$-covector $\xi $, and these quadratic forms determine $\Psi $. Their traces with respect to $\omega $ are positive multiples of the density $\Psi \wedge \omega $, whose integral vanishes by Stokes' theorem and $\mathrm{dd}^{\mathrm{c}}\omega =0$. If the quadratic forms took no strictly positive value, they would be negative semidefinite everywhere, with nonpositive traces of integral zero; the traces, and hence $\Psi $, would vanish identically. The same argument applies with the signs reversed. Therefore, if $\Psi \ne 0$, the quadratic forms take a strictly positive value somewhere and a strictly negative value somewhere.
Fix one of the two signs. Choose holomorphic coordinates on a ball such that, with $x=\operatorname {Re}z_1$, the density $\Psi \wedge \mathrm dx\wedge \mathrm d^{\mathrm c}x$ has that sign on the ball; this is possible because $\mathrm dx\wedge \mathrm d^{\mathrm c}x$ is a positive multiple of $\mathrm i\, \mathrm dz_1\wedge \mathrm d\bar z_1$. For a nonzero real cutoff function $a$ supported in the ball, set $\varphi _ N=a\cos (Nx)$. Formula (11.2.20) gives
\[ H_\omega (\mathrm{dd}^{\mathrm{c}}\varphi _ N,\mathrm{dd}^{\mathrm{c}}\varphi _ N) =\frac{N^2}{2}\int _ Xa^2\sin ^2(Nx)\Psi \wedge \mathrm dx\wedge \mathrm d^{\mathrm c}x+O(N), \]
where the constant in $O(N)$ depends on $\Psi $, the chart and the cutoff, but not on $N$. By integration by parts in $x$, the integral tends to half the same integral without the factor $\sin ^2(Nx)$. Hence the Hessian has the chosen sign for large $N$. For a positive integer $k$, apply this construction on $k$ disjoint smaller balls in that region. By the local integral formula, the resulting nonzero directions are linearly independent and pairwise orthogonal for the Hessian, so their span is a $k$-dimensional subspace on which $H_\omega $ is definite of the chosen sign. Repeating the argument with the opposite sign proves that both indices are infinite. Each finite-dimensional span admits sufficiently small variations preserving positivity.
Step 3: flatness if $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$. If $\Psi =0$, lemma 11.1.3 and pluriclosedness give $\mathrm{dd}^{\mathrm{c}}(\omega ^3)=3\omega \wedge \Psi =0$. Both derivatives at $t=0$ of the quadratic polynomial $t\mapsto F(\omega +t\mathrm{dd}^{\mathrm{c}}\varphi )$ vanish, so it is constant on the interval of $t$ for which $\omega +t\mathrm{dd}^{\mathrm{c}}\varphi >0$. This interval contains $0$ and $1$ whenever $\omega +\mathrm{dd}^{\mathrm{c}}\varphi >0$. Finally, a local minimum cannot have a potential direction along which the Hessian is negative, so it satisfies $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$ by Step 2. $\square$
Corollary 11.2.21.67DF1C6C0509 For a Hermitian symplectic form $\omega $ on $X$, the following conditions are equivalent:
$F(\omega +\mathrm{dd}^{\mathrm{c}}\varphi )$ is independent of the smooth real function $\varphi $ with $\omega +\mathrm{dd}^{\mathrm{c}}\varphi >0$;
$\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$;
$\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}u)^4$ is independent of the bounded $\omega $-plurisubharmonic function $u$.
The common mass in the third condition is $24V(\omega )$. Every local minimum of $F$ in an Aeppli class has these properties.
Proof. By (11.2.18), the energy is constant along smooth positive potential variations if and only if their volume is. Since $\mathrm{dd}^{\mathrm{c}}\omega =0$, lemma 11.1.3, theorem 11.1.7 show that this smooth mass invariance is equivalent to $\mathrm{dd}^{\mathrm{c}}(\omega ^2)=0$ and to mass invariance for bounded potentials. The zero potential gives the common mass. The last assertion follows from theorem 11.2.19. $\square$
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 $\mu _0=\eta ^2/2$,
\[ \frac{\alpha _ s^2}{2}=(1-s^2)\mu _0, \qquad \rho _ s\wedge \bar\rho _ s=s^2\mu _0. \]
Example 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 $.
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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