Harmonic representatives for compact de Rham and Dolbeault complexes

The de Rham Laplacian satisfies the hypotheses of [F1]. Let $D=d$ on real forms, or on their complexification, and $P=D{D}^{\ast }+D^{\ast }D$. The coordinate formula [F2] exhibits $D$ as a smooth first-order differential operator, so [F4] supplies its unique smooth first-order formal adjoint $D^{\ast }$. In an orthonormal basis of the cotangent space at a point ([F7]), the leading coefficient matrices of $D$ multiply, on a covector $\xi$, to exterior multiplication $\epsilon (\xi )$, and those of $D^{\ast }$ are $(-1{)}^1$ times the metric adjoints, so they multiply to $-\epsilon (\xi {)}^{\ast }$. The coefficient of ${\mathrm{\partial }}_i{\mathrm{\partial }}_j$ in $D{D}^{\ast }$ is therefore $\epsilon (\xi )(-\epsilon (\xi {)}^{\ast })$ and in $D^{\ast }D$ it is $(-\epsilon (\xi {)}^{\ast })\epsilon (\xi )$; by [F5] the second-order coefficient matrix of $P$ is $-(\epsilon (\xi )\epsilon (\xi {)}^{\ast }+\epsilon (\xi {)}^{\ast }\epsilon (\xi ))$. The metric adjoint of $\epsilon (\xi )$ is the contraction that deletes each factor of a wedge monomial pairing with $\xi$, with alternating signs, and deleting and reinserting the same factor on both sides gives the identity $\epsilon (\xi )\epsilon (\xi {)}^{\ast }+\epsilon (\xi {)}^{\ast }\epsilon (\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{i}\mathrm{d}$; hence the second-order coefficient of $P$ is $-\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{i}\mathrm{d}$ and its principal symbol in the sense of [F5] is ${\sigma }_2(P)(x,\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{i}\mathrm{d}$, a smooth positive definite scalar form. Finally, for compactly supported smooth forms, $(Pu,v{)}_{L^2}=(D^{\ast }u,D^{\ast }v{)}_{L^2}+(Du,Dv{)}_{L^2}$ by [F4], which is conjugate-symmetric in $u,v$ and equals $\mathrm{\parallel }Du{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{D}^{\ast }u{\mathrm{\parallel }}_2^2\ge 0$ when $u=v$; so $P$ is formally symmetric and nonnegative. Thus [F1] applies to $P$ with domain $H^2$.

The Dolbeault Laplacian satisfies the hypotheses of [F1]. Fix $p$ and let $D={\bar{\mathrm{\partial }}}_E$ on the sum over $q$, $P=D{D}^{\ast }+D^{\ast }D$, with the operator and its adjoint as in [F3]; by that item $D$ is a smooth first-order operator and the Dolbeault Laplacian is the formally symmetric Laplace-type operator with ${\sigma }_2(P)(x,\xi )=\mathrm{\mid }{\xi }^{0,1}{\mathrm{\mid }}^2\mathrm{i}\mathrm{d}$, notation as in [F5]. The same computation as in step 1.1, now with $\eta ={\xi }^{0,1}$ and unitary frames ([F7]), gives the deletion identity $\epsilon (\eta )\epsilon (\eta {)}^{\ast }+\epsilon (\eta {)}^{\ast }\epsilon (\eta )=\mathrm{\mid }\eta {\mathrm{\mid }}^2\mathrm{i}\mathrm{d}$ including the holomorphic $p$-factors and the bundle factor, so the second-order coefficient is $-\mathrm{\mid }\eta {\mathrm{\mid }}^2\mathrm{i}\mathrm{d}$ and the principal symbol is $\mathrm{\mid }{\xi }^{0,1}{\mathrm{\mid }}^2\mathrm{i}\mathrm{d}$; for a nonzero real covector the $(0,1)$-part is nonzero, so the form is positive definite. The energy identity and formal symmetry are those established in [F3], and no K"ahler hypothesis is used. Hence [F1] applies to this $P$ as well.

Consequences of $D^2=0$. Squaring $D^2=0$ and transposing the adjoint identity gives $(D^{\ast }{)}^2=0$, and expanding the two products gives $PD=DP$ and $P{D}^{\ast }=D^{\ast }P$. On smooth forms the energy identity of steps 1.1 and 2.1 shows $Pu=0$ exactly when $Du=0$ and $D^{\ast }u=0$; [F1] makes every $L^2$-harmonic element smooth, so this describes $\ker P$ on $H^2$ as well. Harmonic forms are orthogonal to the images of $D$ and $D^{\ast }$: for harmonic $h$, $(Dv,h{)}_{L^2}=(v,D^{\ast }h{)}_{L^2}=0$ and $(D^{\ast }v,h{)}_{L^2}=(v,Dh{)}_{L^2}=0$. Since $D$ raises the degree by one in the de Rham grading and the second bidegree by one in the Dolbeault bigrading, while $D^{\ast }$ lowers it, the operator $P$ preserves degrees, and the unique orthogonal solution defining $G$ in [F1] preserves degree: its other degree components solve the homogeneous equation and lie in the orthogonal complement of the kernel, hence vanish; the harmonic projection preserves degree as well.

Commutation identities. For smooth $u$, $P(DGu)=D(PGu)=D(u-\mathrm{\Pi }u)=Du$ by step 3.1 and [F1]. The section $DGu$ is smooth and orthogonal to $\ker P$ by step 3.1, and $Du$ is orthogonal to $\ker P$ as well; by the uniqueness of the orthogonal solution in [F1], $DGu=GDu$. The same reasoning applied to $D^{\ast }$ gives $D^{\ast }G=G{D}^{\ast }$. Since $\mathrm{\Pi }$ projects onto $\ker P$ and the images of $D$ and $D^{\ast }$ are orthogonal to it, $D\mathrm{\Pi }=D^{\ast }\mathrm{\Pi }=0$; taking adjoints and using that $\mathrm{\Pi }$ is an orthogonal projection gives $\mathrm{\Pi }D=\mathrm{\Pi }{D}^{\ast }=0$. All identities are asserted on smooth forms, so no composition of unbounded Hilbert-space domains has been formed.

Decomposition and harmonic representatives. For smooth $u$, [F1] gives $u=\mathrm{\Pi }u+(I-\mathrm{\Pi })u=\mathrm{\Pi }u+D{D}^{\ast }Gu+D^{\ast }DGu$, and the three summands are pairwise orthogonal: the harmonic summand is orthogonal to both images by step 3.1, and $(Da,D^{\ast }b{)}_{L^2}=(D^{2}a,b{)}_{L^2}=0$ since $D^2=0$. Conversely a harmonic form or a form in the image of $D$ or $D^{\ast }$ lies in the displayed ambient degree, so this is the orthogonal smooth decomposition. If $u$ is closed, $Du=0$, then step 4.1 gives $DGu=GDu=0$ and the coexact term vanishes: $u=\mathrm{\Pi }u+D(D^{\ast }Gu)$, a harmonic representative of the class. If a harmonic $h$ is exact, $h=Dv$, then $\mathrm{\parallel }h{\mathrm{\parallel }}_2^2=(Dv,h{)}_{L^2}=(v,D^{\ast }h{)}_{L^2}=0$ because $D^{\ast }h=0$ by step 3.1, so $h=0$; hence $h\mapsto [h]$ is injective, and the preceding formula makes it surjective. It is linear as the composition of a linear projection with a linear quotient map, and its domain is finite dimensional by [F1]; the same argument applies over $\mathbb{R}$ or $\mathbb{C}$.

Conclusion and endpoints. Steps 1.1 and 2.1 verified the exact scalar-positive-symbol, formal-symmetry and nonnegativity hypotheses for the de Rham and Dolbeault Laplacians, so [F1] gives genuine nonnegative selfadjoint realizations with domain $H^2$, finite smooth kernels, smoothing projections and smooth Green inverses; steps 3.1 through 5.1 proved the degree, commutation, orthogonal-decomposition and harmonic-representative assertions for both complexes. At the endpoint degrees the operators mapping outside the displayed range are zero by the degree convention of [F3], so the identities hold there as well. The trivial line bundle case gives the usual Dolbeault cohomology of complex-valued forms, and no sheaf comparison has been used or asserted.