Lemma 24.4.2490818FEADA4E

Proof. Step 1: an elliptic operator. Fix $t$ and suppress it from the notation; the auxiliary metric defines the adjoints. Put $C=\partial \bar\partial $ and $E=\bar\partial \partial ^*$. The nonnegative self-adjoint operator

is elliptic of order four. Indeed, at a nonzero real covector $\xi $, the principal symbols of $\partial $ and $\bar\partial $ are, up to the common factor $\mathrm i$, exterior multiplication by the $(1,0)$ and $(0,1)$ parts of $\xi $. These parts are orthogonal and have the same norm $r>0$. Split the exterior algebra according to which of these two directions are present. The four terms of order four act on the four resulting summands respectively, and the sum of their symbols is $r^4\operatorname {Id}$. The terms of lower order do not affect ellipticity.

The quadratic form of $\mathcal L$ is a sum of squared norms, which gives the inclusion $\subseteq $ in

For the reverse inclusion, note that $C^*=\bar\partial ^*\partial ^*$, $E=\bar\partial \partial ^*$ and $E^*=\partial \bar\partial ^*$ all vanish under these three conditions. Elliptic theory on the compact manifold gives a finite-dimensional space of smooth forms $\ker \mathcal L$, the orthogonal projection $\mathcal H$ onto it and the Green operator $\mathcal G$, with $\mathcal L\mathcal G=\operatorname {Id}-\mathcal H$.

Step 2: the projection. Set $R=\mathcal L-C^*C$. Each term of $R$ has image in $\operatorname {im}\partial +\operatorname {im}\bar\partial $. Thus every smooth $(1,1)$-form $\alpha $ has the decomposition

The first two terms lie in $\ker C$, since $C\partial =C\bar\partial =0$, and the last is orthogonal to $\ker C$. Consequently

is the orthogonal projection onto $\ker C$. If $C\alpha =0$, the last term in (24.4.26) vanishes, so the Aeppli class of $\alpha $ is represented by $\mathcal H\alpha $. A harmonic form in $\operatorname {im}\partial +\operatorname {im}\bar\partial $ is zero, by orthogonality. This identifies $H_ A^{1,1}$ with $\ker \mathcal L$ and proves finiteness.

Step 3: dependence on $t$. Now identify the varying type bundles and their $L^2$ spaces smoothly with those at $t=0$, using a unitary identification for the auxiliary metrics. The coefficients of $\mathcal L_ t$ then vary smoothly. Choose a circle $|z|=\varepsilon $ enclosing no nonzero eigenvalue of $\mathcal L_0$. The resolvents of $\mathcal L_0$ on this circle are bounded from $H^ s$ to $H^{s+4}$. The Neumann series for their perturbations gives smooth resolvents on the circle for small $t$, and the contour projections have constant rank. Constancy of $h_ A^{1,1}$ makes that rank equal to $\dim \ker \mathcal L_ t$. Hence the contour projections are exactly $\mathcal H_ t$, and $\mathcal L_ t$ has no additional small eigenvalues.

The operators

vary smoothly on every Sobolev scale, by the same inverse argument and elliptic regularity. Formula (24.4.27) therefore gives the asserted smooth family $\mathcal P_ t$. Complex conjugation sends $C_ t$ to $-C_ t$, exchanges $E_ t$ with $E_ t^*$, and exchanges the last two terms of (24.4.25). Thus $\mathcal L_ t$, $\mathcal H_ t$, $\mathcal G_ t$ and $\mathcal P_ t$ commute with conjugation. $\square$