PomeLemma 24.4.17.2066C3EEC6F6 There is a linear projection $S:\Omega ^\bullet (X,\mathbb R)\to E_ D^\bullet $ with $S\mathrm d=\mathrm dS$. Its complexification preserves bidegree and commutes with $\partial $ and $\bar\partial $. It preserves strict positivity in bidegrees $(1,1)$ and $(n-1,n-1)$.
24.4.16 Balanced and pluriclosed metricsB8BA04F9415A
The balanced–pluriclosed conjecture asks whether a compact complex manifold carrying both a balanced metric $\beta $ and a pluriclosed metric $\omega $ must be Kähler. The two metrics may differ; a single metric satisfying both conditions is already Kähler by lemma 24.4.19. The source of the conjecture and its formulation through a flow are given in section 25.22.
A manifold in Fujiki's class $\mathcal C$ that carries a pluriclosed metric is Kähler, by the result recalled there. It would therefore suffice to construct a closed Kähler current from the two metrics. The scalar potential equation alone cannot produce the required closedness.
Indeed, for every distributional potential $u$,
\[ \mathrm{d}(\omega +\mathrm{dd}^{\mathrm{c}}u)=\mathrm{d}\omega . \]
Thus if the pluriclosed metric is not closed, no potential in its background makes it Kähler; the finite-volume equation, including its smooth solution, does not change this differential datum. Similarly, the Gauduchon metric in a conformal class need not be balanced. Positivity and a scalar normalization cannot replace either missing closedness condition.
The closed form $\beta ^{n-1}$ pairs trivially with exact real two-forms, so it can give a conserved trace along a flow. When $c_1(X)=0$ in real cohomology, the flow conjecture recalled in section 25.22 proposes that, in the presence of a balanced metric, the pluriclosed flow exists for all time and converges to a Kähler metric. The invariant examples cited there use the conserved trace. For general metrics, the analytic difficulty is to turn this integral into a pointwise bound.
Parallel torsion gives a setting in which the integral constraint of a balanced trace becomes pointwise. Suppose $n\geq 2$, write $J$ for the complex structure of $X$, let $g$ be a Hermitian metric on $X$, and let $D$ be a real connection on $TX$ that preserves $g$ and $J$ and has parallel torsion $T^ D$:
\[ Dg=0,\qquad DJ=0,\qquad DT^ D=0, \qquad T^ D(U,V)=D_ UV-D_ VU-[U,V]. \]
The balanced and pluriclosed metrics considered below need not equal $g$. Write
\[ E_ D^ k=\{ \alpha \in \Omega ^ k(X,\mathbb R):D\alpha =0\} \]
for the space of $D$-parallel real $k$-forms. Since $X$ is connected, evaluation at one point is injective on this space, so it is finite-dimensional. Its complexification inherits the decomposition by bidegree because $DJ=0$.
Proof. Step 1: the projection. For a parallel form $\alpha $, the formula for exterior differentiation in terms of a connection expresses $\mathrm d\alpha $ as an algebraic contraction of $T^ D$ with $\alpha $. Since $DT^ D=0$, it follows that $\mathrm dE_ D^ k\subseteq E_ D^{k+1}$. The Hodge star of $g$ is also parallel and maps $E_ D^ k$ onto $E_ D^{2n-k}$. Consequently the pairing
\[ E_ D^ k\times E_ D^{2n-k}\longrightarrow \mathbb R, \qquad (\alpha ,\beta )\longmapsto \int _ X\alpha \wedge \beta \]
is nondegenerate: pairing a nonzero $\alpha $ with $*_ g\alpha $ gives its squared $L^2$ norm, which is positive. Hence every smooth real $k$-form $\alpha $ determines a unique $S_ k\alpha \in E_ D^ k$ such that
3EA53204CDA3
24.4.18
\begin{equation} \int _ X S_ k\alpha \wedge \beta =\int _ X\alpha \wedge \beta \qquad (\beta \in E_ D^{2n-k}). \end{equation}
This defines a projection onto $E_ D^ k$. For $\gamma \in E_ D^{2n-k-1}$ we have $\mathrm d\gamma \in E_ D^{2n-k}$, so Stokes' theorem gives
\[ \int _ X\mathrm dS_ k\alpha \wedge \gamma =(-1)^{k+1}\int _ X\alpha \wedge \mathrm d\gamma =\int _ X S_{k+1}(\mathrm d\alpha )\wedge \gamma . \]
Both forms being compared are parallel, so nondegeneracy proves $\mathrm dS_ k\alpha =S_{k+1}\mathrm d\alpha $.
Step 2: positivity. Fix $p\in X$. Let $H$ be the closure of the full $D$-holonomy group at $p$, a compact subgroup of the unitary group, with probability Haar measure $m_ H$. For a path from $p$ to $x$, write $P_ x:T_ pX\to T_ xX$ for parallel transport along it. The average
\[ R_\alpha (x)=\int _ H h^*(P_ x^*\alpha _ x)\, \mathrm dm_ H(h) \]
is independent of the path, since a change of path acts by an element of $H$. It is an $H$-invariant tensor at $p$ depending smoothly on $x$, because paths can be chosen smoothly on each coordinate neighborhood. Average it over $X$:
\[ a_ p=\frac1{\operatorname {Vol}_ g(X)} \int _ X R_\alpha (x)\, \mathrm dV_ g(x). \]
Being $H$-invariant, the tensor $a_ p$ extends uniquely to a global parallel form $\widetilde S_ k\alpha $. For parallel $\beta $, write $\alpha \wedge \beta =f\, \mathrm dV_ g$. Since parallel transport is unitary and $\beta $ is invariant under holonomy, averaging gives
\[ R_\alpha (x)\wedge \beta _ p=f(x)\, \mathrm dV_{g,p}. \]
Integrating this equality shows that $\widetilde S_ k\alpha $ satisfies (24.4.18); hence $\widetilde S_ k=S_ k$.
Every parallel transport and every holonomy element preserves reality, bidegree and the positive cones. A smooth strictly positive $(q,q)$-form, for $q=1$ or $n-1$, is bounded below by a positive multiple of the $q$th power of the fundamental form of $g$, uniformly on the compact manifold, and both probability averages preserve that lower bound. Therefore $S$ preserves strict positivity and bidegree. Comparing bidegrees in $S\mathrm d=\mathrm dS$ proves that $S$ commutes with $\partial $ and $\bar\partial $. $\square$
Lemma 24.4.19.4865DD012C9C If a Hermitian form $\zeta $ satisfies $\mathrm d\zeta ^{n-1}=0$ and $\mathrm{dd}^{\mathrm{c}}\zeta =0$, then $\mathrm d\zeta =0$.
Proof. For $n=2$, the balanced condition is the assertion $\mathrm d\zeta =0$. Suppose $n\geq 3$ and set $\alpha =\partial \zeta $. Since $\zeta $ is balanced, $\alpha \wedge \zeta ^{n-2}=0$, so $\alpha $ is a primitive $(2,1)$-form. Applying $\partial \bar\partial $ to the balanced power gives
\[ 0=\partial \bar\partial \zeta ^{n-1} =(n-1)\partial \bar\partial \zeta \wedge \zeta ^{n-2} +(n-1)(n-2)\alpha \wedge \bar\alpha \wedge \zeta ^{n-3}. \]
The first term vanishes because $\zeta $ is pluriclosed. For a primitive $(2,1)$-form, pointwise Hermitian linear algebra gives
\[ -\mathrm i\, \alpha \wedge \bar\alpha \wedge \frac{\zeta ^{n-3}}{(n-3)!} =|\alpha |_\zeta ^2\frac{\zeta ^ n}{n!}. \]
Thus $\alpha =0$, and since $\zeta $ is real, also $\bar\partial \zeta =0$, so $\mathrm d\zeta =0$. $\square$
Theorem 24.4.20.781DAEC1B4CE Let $D$ be a connection on $TX$ as above, preserving $g$ and $J$ and with parallel torsion. If $X$ admits a balanced metric and a pluriclosed metric, then it admits a $D$-parallel Kähler metric.
Proof. Let $\beta $ be a balanced metric and $\omega $ a pluriclosed metric. By lemma 24.4.17, $S(\beta ^{n-1})$ is strictly positive, parallel and closed. It has a unique smooth positive $(n-1)$-st root $B$. Indeed, after the fixed normalization of wedge powers, the corresponding map on matrices is $G\mapsto C=(\det G)G^{-1}$, whose inverse is
\[ G=(\det C)^{1/(n-1)}C^{-1}. \]
By uniqueness the root is invariant under parallel transport. Thus $DB=0$ and $\mathrm dB^{n-1}=0$. Similarly, $P=S\omega $ is a parallel positive pluriclosed form.
Consider the finite-dimensional space of real parallel directions of exact type
\[ \mathcal A_ D=\{ (\mathrm d\xi )^{1,1}:\xi \in E_ D^1\} . \]
Every such direction is $\partial \bar\partial $-closed, since
\[ (\mathrm d\xi )^{1,1}=\partial \xi ^{0,1}+\bar\partial \xi ^{1,0}. \]
For a parallel positive form $Q$, the volume functional has derivative
B755FCE3BC15
24.4.21
\begin{equation} \left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0} \frac1{n!}\int _ X(Q+t(\mathrm d\xi )^{1,1})^ n =\frac1{(n-1)!}\int _ X\xi \wedge \mathrm dQ^{n-1}, \end{equation}
because the $(2,0)$ and $(0,2)$ parts of $\mathrm d\xi $ pair to zero with $Q^{n-1}$. Since $\mathrm dQ^{n-1}$ is parallel, nondegeneracy of the parallel pairing shows that all these derivatives vanish exactly when $Q$ is balanced.
Apply this to the balanced form $B$. For $a\in \mathcal A_ D$, the function $\operatorname {tr}_ B a$ is parallel and hence constant. Its integral against $B^ n$ vanishes by (24.4.21), so it vanishes identically. Therefore every form in the positive affine slice
\[ \mathcal P=\{ Q\in P+\mathcal A_ D:Q>0\} \]
has the same pointwise trace $\operatorname {tr}_ B Q=\operatorname {tr}_ B P=:C>0$.
Evaluate the parallel forms at one point. The positive eigenvalues of $Q\in \mathcal P$ relative to $B$ sum to $C$, so they are bounded above by $C$. Thus $\mathcal P$ is bounded in its finite-dimensional affine space. Its closure is
\[ \overline{\mathcal P}=\{ Q\in P+\mathcal A_ D:Q\geq 0\} , \]
because $(1-t)Q+tP$ is positive for $0<t\leq 1$ whenever $Q$ is semipositive; in particular, this closure is compact. Since both forms are parallel, the relative eigenvalues and the relative determinant are constant on $X$. Hence
\[ \frac1{n!}\int _ XQ^ n =\det (B^{-1}Q)\frac1{n!}\int _ XB^ n. \]
At a relative boundary point of $\overline{\mathcal P}$ the semipositive matrix is singular, so this volume is zero, whereas it is positive at $P$. A maximum of the volume on the compact set $\overline{\mathcal P}$ therefore lies in $\mathcal P$.
At this interior maximum all the derivatives in (24.4.21) vanish, so the maximizing form is balanced; it is pluriclosed because it lies in $P+\mathcal A_ D$. By lemma 24.4.19 it is closed. Since it is positive and $D$-parallel, this proves the theorem. $\square$
The trace constraint in this argument is pointwise because the metrics and the directions of variation are parallel. This additional structure makes the positive slice compact and the volume vanish on its boundary; for a general family, the conserved balanced trace gives no such control.
Problem 24.4.22.58C2B4B7107E Suppose that a compact complex manifold carries a balanced metric and a pluriclosed metric. Use the two metrics to construct closed positive data, such as a closed Kähler current, or to control a geometric flow that can change the torsion. Determine which estimate, beyond the finite balanced trace, prevents degeneration of the metrics. Can determinant estimates for Gauduchon currents contribute to that estimate?
The complete determinant profile also need not control ellipticity. Under the lower-power $\mathrm{dd}^{\mathrm{c}}$-closed condition, theorem 12.18.8 computes its exact value on every $L^ n$ density root. The nonclosed pluriclosed family of theorem 12.18.15 keeps this entire profile, the bounded-potential mass extrema and a balanced trace fixed while its least relative eigenvalue tends to zero. Its torsion is fixed and nonzero. The underlying torus is already Kähler; the example shows which scalar data an additional estimate must go beyond, without constructing a pluriclosed-flow trajectory.
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