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 $(n-1,n-1)$.
Lemma 24.4.172066C3EEC6F6
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 $. $\square$
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