611 lines
34 KiB
TeX
611 lines
34 KiB
TeX
\section{Harmonic differential forms and cohomology}
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In this chapter, we are going to introduce the theory of harmonic differential forms and review the
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concepts of the de Rahm and Dolbeault cohomologies. Our primary objective is the proof of two Hodge
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Isomorphism theorems, which establish a fundamental relation between the spaces of harmonic forms
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and the cohomologies. These theorems will be crucial for our proof of the Hodge Decomposition
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theorem later.
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This chapter is based on sections 5.1 to 5.3 and 6.1 in the book \emph{Hodge Theory and Complex
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Algebraic Geometry} \cite{Voisin2002} written by Claire Voisin. Some of the statements also
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originate from chapter \MakeUppercase{\romannumeral 6} in the book \emph{Complex Analytic and
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Differential Geometry} \cite{Demailly1997} written by Jean-Pierre Demailly.
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For the remainder of this chapter, we will assume the following setting.
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\begin{set}
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Let $X$ be a compact $m$-dimensional Kähler manifold with Riemannian metric $g$, closed
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fundamental form $\omega$ %\in\mathcal{A}^2_\mathbb{R} \cap \mathcal{A}^{1,1}(X)$
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and Kähler metric $h:= g -i\omega$.
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\end{set}
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\subsection{Harmonic forms and the Laplacians}\;
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Our first goal in this chapter is the definition of a generalization of the Laplace operator
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$\Delta$. This operator is already known for $\mathbb{R}$-valued functions on the euclidean space
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$\mathbb{R}^n$ and we will generalize it to obtain the Laplacians $\Delta_d, \Delta_\partial$ and
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$\Delta_\opartial$ on the smooth differential forms $\mathcal{A}^\bullet_\mathbb{C}(X)$.
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Based on the definition of harmonic functions in the euclidean case, we will use these generalized
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Laplacians to define different types of harmonic differential forms. After that, we will further
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develop this theory and focus on an important theorem that establishes a relation between the three
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Laplacians. In order to prove this theorem, we will need the Kähler identities established in the
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last chapter.
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\begin{defn}[Laplacians]
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For every $k$, we define the \emph{Laplacian of the exterior derivative} $d$ as a linear operator
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$\Delta_d: \mathcal{A}^k_\mathbb{C}(X)\rightarrow\mathcal{A}^k_\mathbb{C}(X)$ that
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is given as
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\begin{equation*}
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\Delta_d := d\,d^* + d^*d.
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\end{equation*}
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Similarly, the \emph{Laplacians of the Dolbeault operators} $\partial$ and $\opartial$ are defined as
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\begin{align*}
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\Delta_\partial := \partial\,\partial^* + \partial^*\partial \;\enspace \text{ and } \; \enspace
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\Delta_{\opartial}:= \opartial\,\opartial^* + \opartial^*\opartial
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\end{align*}
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and it is by definition that these two operators are linear mappings of type
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$\mathcal A^{p,q}(X) \rightarrow\nolinebreak \mathcal A^{p,q}(X)$.
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\end{defn}
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\begin{rem}[{\cite[p.\,126]{Voisin2002}}]
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We already know that $d,\partial$ and $\opartial$ are linear differential operators of order 1, and
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the same is true for the formal adjoints $d^*,\partial^*$ and $\opartial^*$. Since the Laplacians
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are all defined as a concatenation and sum of these linear differential operators, they are also
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linear differential operators but of order 2.
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\end{rem}
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\begin{defn}[Harmonic differential forms]
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Let $\alpha \in \mathcal{A}^k_\mathbb{C}(X)$ be a complex differential form. We are going to call
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$\alpha$ \emph{harmonic} if it is $\alpha \in \ker(\Delta_d)$. Furthermore, if $\alpha$ is of type
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$(p,q)$, we will call it $\mathit{\Delta_\opartial}$\emph{-harmonic} if
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$\alpha \in \ker(\Delta_\opartial)$.
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\end{defn}
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At this point, we will prove an essential equality that will be required for subsequent calculations.
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\begin{lm}[{\cite[Lemma 5.12]{Voisin2002}}]
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For the Laplacian $\Delta_d$, we have the equality
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\begin{align}
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\label{harmonic-forms:eq:property-laplacian}
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\left(\alpha,\Delta_d\alpha\right)_{L^2} = \left(d\alpha,d\alpha\right)_{L^2} +
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\left(d^*\alpha,d^*\alpha\right)_{L^2}.
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\end{align}
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Also, for the other two Laplacians $\Delta_\partial$ and $\Delta_\opartial$, the analogous
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equalities are valid as well.
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\end{lm}
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\begin{proof}This proof was taken from the given lemma in \cite{Voisin2002}.
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With the definition of the Laplacian $\Delta_d$ and the formal adjunction properties \Cref{kaehler-manifolds:eq:formal-adjunction-property,kaehler-manifolds:eq:formal-adjunction-property-2},
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we get
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\begin{align*}
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\left(\alpha,\Delta_d \alpha\right)_{L^2} = \left(\alpha, d\,d^* \alpha + d^*d
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\alpha\right)_{L^2}&= \left(\alpha,d\,d^*\alpha\right)_{L^2} + \left(\alpha, d^*d\alpha\right)_{L^2}\\
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%\\&= %\left(d^* \alpha, d^*\alpha\right)_{L^2} + \overline{\left(d^*d\alpha, \alpha\right)}_{L^2}
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%\\&= (d^*\alpha,d^*\alpha)_{L^2} + \overline{(d\alpha,d\alpha)}_{L^2}
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%\overline{\left(d\,d^*\alpha,\alpha\right)}_{L^2} + \left(\alpha,d^*d\alpha\right)_{L^2} \\&=
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%\overline{\left(d^*\alpha,d^*\alpha\right)} + \left(d\alpha,d\alpha\right)_{L^2}
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&= \left(d^*\alpha,d^*\alpha\right)_{L^2} + \left(d\alpha,d\alpha\right)_{L^2}
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\end{align*}
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and the analogous calculation can be used for the Laplacians $\Delta_\partial$ and
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$\Delta_\opartial$ too.
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\end{proof}
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Suppose there exists an $\alpha \in \mathcal A^k_\mathbb{C}(X)$ such that $\Delta_d \alpha = 0$.
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We can use the equality established in \Cref{harmonic-forms:eq:property-laplacian} to obtain
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\begin{align*}
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%\label{kaehler-maifolds:eq:laplacian-zero}
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0 = (\alpha,\Delta_d\alpha)_{L^2} = \left(d^*\alpha,d^*\alpha\right)_{L^2} +
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\left(d\alpha,d\alpha\right)_{L^2}.
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\end{align*}
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As mentioned in \Cref{kaehler-manifolds:rem:l2-metric-properties}, the hermitian $L^2$-metric is
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positive definite, so both of these summands have to be equal to zero, which is only the case if
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$d^*\alpha = 0$ and also $d\alpha =0$. If, on the other hand, $d\alpha = 0$ and $d^*\alpha = 0$,
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then it is by definition of the Laplacian $\Delta_d \alpha = 0$. This already proves the following corollary.
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\begin{cor}[{\cite[Corollary 5.13]{Voisin2002}}]
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\label{harmonic-forms:lm:kernel-laplacian}
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For the kernel of the Laplacian $\Delta_d$, we have the relation
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\begin{align*}
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\ker(\Delta_d) = \ker(d) \cap \ker(d^*).
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\end{align*}
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In particular, $\alpha \in \mathcal{A}^k_\mathbb{C}(X)$ is harmonic if and only if \,
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$d\alpha = d^*\alpha = 0$.
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\end{cor}
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It is also obvious that the same calculation can be repeated for the other two Laplacians
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$\Delta_\partial$ and $\Delta_\opartial$. Hence, we also get the same relations for the kernels
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of these Laplacians.
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\begin{cor}
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\label{harmonic-forms:rem:kernel-laplacian-partial-opartial}
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For the kernels of the Laplacians $\Delta_\partial$ and $\Delta_\opartial$, we have the relations
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\begin{align*}
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\ker(\Delta_\partial) = \ker(\partial) \cap \ker(\partial^*) \; \enspace \text{ and } \;
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\enspace \ker(\Delta_\opartial) = \ker(\opartial) \cap \ker(\opartial^*).
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\end{align*}
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In particular, $\alpha \in \mathcal{A}^{p,q}(X)$ is $\Delta_\opartial$-harmonic if and
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only if\, $\opartial\alpha = 0$ and $\opartial^*\alpha = 0$.
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\end{cor}
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\begin{prop}[{\cite[p.\,368]{Demailly1997}}]
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\label{harmonic-forms:lm:laplacians-are-self-adjoint}
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The Laplacians $\Delta_d, \Delta_\partial$ and $\Delta_\opartial$ are formally self-adjoint.
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\end{prop}
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\begin{proof}
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We calculate for all $\alpha,\beta \in \mathcal{A}^k_\mathbb{C}(X)$
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\begin{align*}
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(\Delta_d\alpha,\beta)_{L^2} &= (dd^*\alpha + d^*d\alpha,\beta)_{L^2}\\
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&= (dd^*\alpha,\beta)_{L^2} + (d^*d\alpha,\beta)_{L^2} \\
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&= (d^*\alpha,d^*\beta)_{L^2} + (d\alpha,d\beta)_{L^2}.
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\end{align*}
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Furthermore, we are able to calculate
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\begin{align*}
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(\alpha,\Delta_d\beta)_{L^2} &= (\alpha,dd^* \beta + d^*d\beta)_{L^2}\\
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&= (\alpha,dd^*\beta)_{L^2} + (\alpha, d^*d\beta)_{L^2}\\
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&= (d^*\alpha,d^*\beta)_{L^2} + (d\alpha,d\beta)_{L^2}
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\end{align*}
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This already proves that $\Delta_d$ is formally self-adjoint, and the argument for the other
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two Laplacians is exactly the same.
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\end{proof}
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At this point, we have established the necessary theory to prove the existence of a previously
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indicated relation between the different Laplacians. This relation will be fundamental to the proof
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of the Hodge Decomposition theorem.
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\begin{thm}[{\cite[Theorem 6.7]{Voisin2002}}]
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For the Laplacians $\Delta_d,\Delta_\partial$ and $\Delta_\opartial$ on our compact Kähler manifold
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$X$, we have the following relation
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\begin{align}
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\label{harmonic-forms:eq:thm-laplacian-realtion}
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\frac{1}{2}\Delta_d = \Delta_\partial = \Delta_\opartial.
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\end{align}
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In particular, any differential form $\alpha \in \mathcal{A}^k_\mathbb{C}(X) \cap
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\mathcal{A}^{p,q}(X)$ is harmonic if and only if it is $\Delta_\opartial$-harmonic.
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\end{thm}
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\begin{proof}
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We provide a similar argument as in the proof of \cite[Theorem 6.7]{Voisin2002} but our approach
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will be slightly different. With the splitting of both, the exterior derivative $d = \partial + \opartial$
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and the formal adjoint of the exterior derivative $d^* = \partial^* +\,\opartial^*$
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(cf. \Cref{kaehler-manifolds:rem:splitting-of-the-formal-adjoint-of-the-exterior-derivative}), the
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definition of the Laplacian $\Delta_d$ can be rewritten as follows.
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\begin{align*}
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\Delta_d &= d\,d^* + d^*d \\
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&= \big(\partial + \opartial\big)\big(\partial^* + \opartial^*\big) +
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\big(\partial^* + \opartial^*\big) \big(\partial + \opartial\big)\\
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&= \big(\partial\partial^* + \partial \opartial^* + \opartial\partial^* + \opartial\,\opartial^*\big) +
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\big(\partial^*\partial +\partial^*\opartial +\opartial^* \partial + \opartial^*\opartial\big)\\
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&=\partial\partial^* + \partial^*\partial + \partial\opartial^* + \opartial^*\partial +
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\opartial\partial^* +\partial^*\opartial + \opartial\,\opartial^* + \opartial^*\opartial%\\&=
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%\Delta_\partial + \partial\opartial^* + \opartial^*\partial + \opartial\partial^*
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%+\partial^*\opartial + \opartial\,\opartial^* + \opartial^*\opartial
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\end{align*}
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Now, we are going to use the Kähler identities established in
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\Cref{kaehler-manifolds:thm:kaehler-identities}, in particular the equality
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$\partial^* = i[\Lambda,\opartial]$, to get
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\begin{align*}
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\opartial\partial^* + \partial^*\opartial &= \opartial i [\Lambda,\opartial] +
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i[\Lambda,\opartial]\opartial \\
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&=i\big(\opartial(\Lambda\opartial-\opartial\Lambda)+(\Lambda\opartial-\opartial\Lambda)\opartial\big)\\
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&=i\big(\opartial\Lambda\opartial - \opartial^2\Lambda + \Lambda\opartial^2 - \opartial\Lambda\opartial\big)\\
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&=i\big(\opartial\Lambda\opartial - \opartial\Lambda\opartial\big) \\
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&=0.
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\end{align*}
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Applying the other Kähler identity $\opartial^* = -i[\Lambda,\partial]$ from
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\Cref{kaehler-manifolds:thm:kaehler-identities}, we can do a similar calculation to obtain
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\begin{align*}
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\partial\opartial^* + \opartial^*\partial &= -i\partial[\Lambda,\partial] -i[\Lambda,\partial]\partial \\
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& = -i\big(\partial(\Lambda\partial - \partial\Lambda) + (\Lambda\partial -\partial\Lambda)\partial\big) \\
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&= -i\big(\partial\Lambda\partial -\partial^2\Lambda + \Lambda\partial^2 - \partial\Lambda\partial\big) \\
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&= -i\big(\partial\Lambda\partial - \partial\Lambda\partial\big) \\
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&= 0.
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\end{align*}
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Thus, the above equation simplifies to
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\begin{align*}
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\Delta_d =\partial\partial^* + \partial^*\partial + \opartial\,\opartial^* +
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\opartial^*\opartial= \Delta_\partial + \Delta_\opartial.
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\end{align*}
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Therefore, it is enough to prove $\Delta_\partial = \Delta_\opartial$. However, we can calculate
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\begin{align*}
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\Delta_\partial - \Delta_\opartial &= \partial\partial^* + \partial^*\partial -
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\big(\opartial\,\opartial^* + \opartial^*\opartial\big)\\
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&=i\big(\partial[\Lambda,\opartial] + [\Lambda,\opartial]\partial + \opartial[\Lambda,\partial]
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+ [\Lambda,\partial]\opartial\big)\\
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&=i\big(\partial\Lambda\opartial -\partial\opartial\Lambda + \Lambda\opartial\partial -
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\opartial\Lambda\partial + \opartial\Lambda\partial -\opartial\partial\Lambda +
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\Lambda\partial\opartial - \partial\Lambda\opartial\big)\\
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&=i\big(\Lambda\opartial\partial-\partial\opartial\Lambda - \opartial \partial\Lambda +
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\Lambda\partial\opartial\big) \\&= i\big(\Lambda(\opartial\partial + \partial\opartial) -
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(\partial\opartial + \opartial\partial) \Lambda\big)\\
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&= 0.
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\end{align*}
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For the last step, we have used
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$0 = d^2 = \partial^2 + \partial\opartial + \opartial\partial + \opartial^2 = \partial\opartial + \opartial\partial$.
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Finally, we combine both of these results to obtain
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\begin{align*}
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\Delta_d = 2 \Delta_\partial = 2 \Delta_\opartial\,,
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\end{align*}
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which completes the proof of this theorem.
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\end{proof}
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%\begin{rem}
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% %\TODO{Counter example for this failing on arbitrary hermitian manifolds.}
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%\end{rem}
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The relation established in \Cref{harmonic-forms:eq:thm-laplacian-realtion}
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has some significant implications for the operator $\Delta_d$ and the property
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of a differential form being harmonic.
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\begin{cor}
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\label{harmonic-forms:cor:harmonic-if-and-only-if-delta-opartial-harmonic}
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A complex differential form $\alpha \in \mathcal A^k_\mathbb{C}(X) \cap \mathcal A^{p,q}(X)$
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is harmonic if and only if it is $\Delta_\opartial$-harmonic.
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\end{cor}
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\begin{proof}
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It is $\Delta_d\alpha = 2\Delta_\opartial\,\alpha$, so either both operators map $\alpha$ to zero
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or neither of them.
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\end{proof}
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When dealing with Kähler manifolds, the last corollary allows us not to distinguish between harmonic
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and $\Delta_\opartial$-harmonic forms anymore. Therefore, we will just refer to harmonic forms
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from now on.
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\begin{cor}[{\cite[Corollary. 6.9]{Voisin2002}}]
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\label{harmonic-forms:cor:harmonic-if-and-only-if-components-are-harmonic}
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A complex differential form $\alpha \in \mathcal{A}^k_\mathbb{C}(X)$ is harmonic if and only if its
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components $\alpha^{p,q}$ of type $(p,q)$ are harmonic.
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\end{cor}
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\begin{proof}
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With the previously used decomposition
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\begin{align}
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\label{harmonic-forms:eq:decomposition-of-kforms}
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\mathcal{A}^k_\mathbb{C}(X) = \bigoplus_{p+q = k} \mathcal{A}^{p,q}(X),
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\end{align} (cf. \cite[Corollary 2.6.8]{Huybrechts2004})
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we can uniquely write $\alpha = \sum_{p+q=k} \alpha^{p,q}$. As $\Delta_d$ is a linear operator,
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the sum of harmonic forms is harmonic again. Therefore, it suffices to prove that $\alpha$
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being harmonic already implies that the components $\alpha^{p,q}$ are harmonic.
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However, if $\alpha$ is harmonic we can use \Cref{harmonic-forms:eq:thm-laplacian-realtion} to obtain
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\begin{align*}
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0 = \Delta_d \alpha = \sum_{p+q = k} \Delta_d \alpha^{p,q} =
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\sum_{p+q = k} 2\,\Delta_\opartial\,\alpha^{p,q} = 2\sum_{p+q = k} \Delta_\opartial\, \alpha^{p,q}.
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\end{align*}
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Since $\Delta_\opartial $ maps forms of type $(p,q)$ to forms of type $(p,q)$ and we also have
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decomposition \Cref{harmonic-forms:eq:decomposition-of-kforms}, we conclude
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that $\Delta_\opartial\,\alpha^{p,q} = 0$ for each $\alpha^{p,q}$. %$p,q \in \mathbb{N}$ with $p+q = k$.
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Hence with \Cref{harmonic-forms:cor:harmonic-if-and-only-if-delta-opartial-harmonic}, the
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components $\alpha^{p,q}$ are each harmonic.
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\end{proof}
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\begin{nota}
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Let $\mathcal H^k(X) \subset \mathcal{A}_\mathbb{C}^k(X)$ denote the space of complex harmonic
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differential $k$-forms. Let also $\mathcal H^{p,q}(X) \subset \mathcal{A}^{p,q}(X)$ denote the space
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of complex harmonic differential forms of type $(p,q)$, which is also the space of
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$\Delta_\opartial$-harmonic differential forms of type $(p,q)$
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(cf. \Cref{harmonic-forms:cor:harmonic-if-and-only-if-delta-opartial-harmonic}).
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\end{nota}
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With this newly introduced notation, we get the following corollary, which is just a rewritten version
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of \Cref{harmonic-forms:cor:harmonic-if-and-only-if-components-are-harmonic} combined with the direct
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sum decomposition in \Cref{harmonic-forms:eq:decomposition-of-kforms}.
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\begin{cor}[{\cite[Corollary 6.10]{Voisin2002}}]
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\label{harmonic-forms:cor:final-harmonic-decomposition}
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The harmonic differential $k$-forms decompose as
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\begin{align*}
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\mathcal H^k(X) = \bigoplus_{p+q=k}\mathcal H^{p,q}(X).
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\end{align*}
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\end{cor}
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\subsection{Elliptic differential operators}\;
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Before we can proceed with the harmonic differential forms theory, we need to extend our
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understanding of linear differential operators. Therefore, we will introduce elliptic operators and
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address their implications on the previously established theory of harmonic differential forms.
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Note, however, that we will only provide a very brief overview and refer to other sources because a
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more detailed discussion would be beyond the scope of this thesis. For an in-depth discussion of
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elliptic differential operators and their properties, see §1. and §2. of chapter VI in
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\cite{Demailly1997}.
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At first, we have the following definition, motivated by the discussion in \cite[Section 15]{Schnell2012}
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and the explanations at the beginning of Section 5.2.1 in \cite{Voisin2002}.
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\begin{defn}
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In the same setting as in \Cref{kaehler-manifolds:defn:differential-opperators}, for every
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$D_{U_j}$, we define a function
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$P_j: U_j \times \mathbb{C}^m \rightarrow \mathbb{C}^{r_1\times r_2}$ as
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\begin{align*}
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P_j(p,\xi) := \sum_{S} \big(P_{r,S,t}(p)\big)_{rt} \xi^S
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\end{align*}
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with $\big(P_{r,S,t}(p)\big)_{rt} \in \mathbb{C}^{r_1\times r_2}$ being the matrix defined by the
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coefficients $P_{r,S,t}(p) \in \mathbb{C}$, $S := (s_1,\dots,s_m)$ and
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$\xi^S:= \xi_1^{s_1} \cdots \xi_m^{s_m}$. The functions $P_j$ define the \emph{symbol} of the
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operator $D$.
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\end{defn}
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\begin{defn}[Elliptic differential operator]
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In the setting of \Cref{kaehler-manifolds:defn:differential-opperators}, the linear differential
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operator $D$ is said to be \emph{elliptic} if $r_1 = r_2$ and for any local symbol $P_j$, the
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image $P_j(p,\xi)$ is an invertible matrix for all $p \in U_j$ and $\xi \in \mathbb{C}^n\setminus\{0\}$.
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\end{defn}
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\begin{rem}
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The usual definition of an elliptic operator only requires the matrices to define injective
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homomorphisms (cf. \cite[Ch.\,VI §1. Definition 1.8]{Demailly1997}). However, the ranks $r_1$ and
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$r_2$ will always be equal in our setting. Therefore, if we additionally require these ranks to be
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equal, these definitions are equivalent for our purposes.
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\end{rem}
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Through an extensive calculation (cf. \cite[Lemmas 5.18 and 5.19]{Voisin2002}), it is possible to
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calculate the symbols of the three Laplacians and to conclude the following proposition.
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\begin{prop}[{\cite[Corollary 5.20]{Voisin2002}}]
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\label{harmonic-forms:lm:laplacians-are-elliptic}The Laplacians $\Delta_d,\Delta_\partial$ and
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$\Delta_\opartial$ are each elliptic differential operators.
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\end{prop}
|
||
After establishing this property, we are able to apply a fundamental result from the theory of
|
||
Sobolev spaces. There is a more general version with proof in {\cite[Ch.\,VI §2. Corollary 2.4]{Demailly1997}}
|
||
but in our setting, this theorem states the following.
|
||
\begin{thm}
|
||
\label{harmonic-forms:thm:fundamental-theorem}
|
||
Let $D: \mathcal{A}^k_\mathbb{C}(X) \rightarrow \mathcal{A}^k_\mathbb{C}(X)$ be an elliptic linear
|
||
differential operator. Also let
|
||
$D^*: \mathcal{A}^k_\mathbb{C}(X)\rightarrow \mathcal{A}^k_\mathbb{C}(X)$
|
||
be its formal adjoint with respect to the hermitian $L^2$-metric. Then, $D$ has the following
|
||
properties:
|
||
\begin{enumerate}
|
||
\item $\mathit{ker(D)}$ is of finite dimension and
|
||
$\mathit{Im(D)}\subset\mathcal{A}^k_\mathbb{C}(X)$ is closed and of finite codimension.
|
||
\label{harmonic-forms:prop:fundamental-theorem-1}
|
||
\item There is a decomposition
|
||
\begin{align*}
|
||
\mathcal{A}^k_\mathbb{C}(X) = Im(D) \oplus \ker(D^*),
|
||
\end{align*}
|
||
which is an orthogonal direct sum decomposition with respect to the $L^2$-metric.
|
||
\end{enumerate}
|
||
\end{thm}
|
||
\begin{rem}
|
||
\label{harmonic-forms:rem:elliptic-theorem}
|
||
For $D: \mathcal{A}^{p,q}(X) \rightarrow \mathcal{A}^{p,q}(X)$ an elliptic operator with formal
|
||
adjoint $D^*: \mathcal{A}^{p,q}(X) \rightarrow \mathcal{A}^{p,q}(X)$, we get a similar theorem,
|
||
but it is instead $\img(D) \subset \nolinebreak \mathcal{A}^{p,q}(X)$ closed and the
|
||
decomposition is given as
|
||
\begin{align*}
|
||
\mathcal{A}^{p,q}(X) = \img(D) \oplus \ker(D^*).
|
||
\end{align*}
|
||
\end{rem}
|
||
At this point, we have two corollaries resulting from this theorem together with the previously
|
||
established fact that the Laplacians are elliptic operators.
|
||
\begin{cor}[{\cite[p.\,58]{Schnell2012}}]
|
||
\label{harmonic-forms:lm:finite-dimensional}
|
||
If regarded as complex vector spaces, it is $\dim_\mathbb{C} \mathcal{H}^k(X) < \infty$ and
|
||
$\dim_\mathbb{C} \mathcal{H}^{p,q}(X) < \infty$.
|
||
\end{cor}
|
||
\begin{proof}
|
||
This is a direct consequence of \Cref{harmonic-forms:lm:laplacians-are-elliptic}
|
||
and \creflmpart{harmonic-forms:thm:fundamental-theorem}{harmonic-forms:prop:fundamental-theorem-1}.
|
||
% Note custom command as defined in main_thesis
|
||
\end{proof}
|
||
\begin{cor}[{\cite[p.\,129]{Voisin2002}}]
|
||
For the differential $k$-forms, we have the orthogonal decomposition
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:decomposition-of-the-k-forms}
|
||
\mathcal{A}^k_\mathbb{C}(X) = \Delta_d(\mathcal{A}^k_\mathbb{C}(X)) \oplus \mathcal H^k(X).
|
||
\end{align}
|
||
At the same time, the differential forms of type $(p,q)$ decompose orthogonally as
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:decomposition-of-the-pq-forms}
|
||
\mathcal{A}^{p,q}(X) = \Delta_\opartial(\mathcal{A}^{p,q}(X)) \oplus \mathcal H^{p,q}(X).
|
||
\end{align}
|
||
\end{cor}
|
||
\begin{proof}
|
||
If we combine \Cref{harmonic-forms:lm:laplacians-are-elliptic},
|
||
\Cref{harmonic-forms:thm:fundamental-theorem} and the fact that $\Delta_d$ is formally self-adjoint
|
||
(cf. \Cref{harmonic-forms:lm:laplacians-are-self-adjoint}), we get
|
||
\begin{align*}
|
||
\mathcal{A}^k_\mathbb{C}(X) = \img(\Delta_d) \oplus \ker(\Delta_d^*) =
|
||
\img(\Delta_d) \oplus \ker(\Delta_d) =
|
||
\Delta_d(\mathcal{A}^k_\mathbb{C}(X)) \oplus \mathcal H^k(X).
|
||
\end{align*}
|
||
Using \Cref{harmonic-forms:rem:elliptic-theorem}, the same argument yields
|
||
\begin{align*}
|
||
\mathcal{A}^{p,q}(X) = \img(\Delta_\opartial) \oplus \ker(\Delta_\opartial^*) =
|
||
\img(\Delta_\opartial) \oplus \ker(\Delta_\opartial) =
|
||
\Delta_\opartial(\mathcal{A}^{p,q}(X)) \oplus \mathcal H^{p,q}(X).
|
||
\end{align*}
|
||
\end{proof}
|
||
\subsection{De Rahm and Dolbeault cohomologies}\;
|
||
|
||
In the next section, we aim to prove the Hodge Isomorphism theorems, which state that every class of
|
||
closed differential forms (of type $(p,q)$) possesses a unique harmonic representative (of type
|
||
$(p,q)$). Although we will not conduct an in-depth discussion of the theory behind cohomology, we
|
||
will briefly revisit the fundamental definitions of the de Rahm and Dolbeault cohomologies to
|
||
provide the necessary context. For a more detailed insight, see the chapter
|
||
\emph{Sheaves and Cohomology} in \cite{Voisin2002}.
|
||
|
||
\begin{defn}[de Rahm complex]
|
||
The \emph{de Rahm complex} is defined as
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:derahm}
|
||
0 \longrightarrow C^\infty(X,\mathbb{C})\cong\mathcal{A}^0_\mathbb{C}(X) \xlongrightarrow{d_0}
|
||
\mathcal{A}^1_\mathbb{C}(X) \xlongrightarrow{d_1} \dots \xlongrightarrow{d_{2m-1}}
|
||
\mathcal{A}^{2m}_\mathbb{C}(X) \xlongrightarrow{d_{2m}} 0,
|
||
\end{align}
|
||
with $d_k: \mathcal{A}^k_\mathbb{C}(X) \rightarrow \mathcal{A}^{k+1}_\mathbb{C}(X)$ being the
|
||
restriction of the exterior derivative to the differential $k$-forms $\mathcal{A}^k_\mathbb{C}(X)$.
|
||
\end{defn}
|
||
Note that the property $d^2 = 0$ ensures that this is indeed a complex.
|
||
Using this property, we are able to define the de Rahm cohomology.
|
||
\begin{defn}[de Rahm cohomology]
|
||
Let $Z^k(X,\mathbb{C}):= \ker(d_k)$ be the vector space of \emph{closed} differential $k$-forms and
|
||
$B^k(X,\mathbb{C}):= \img(d_{k-1})$ be the vector space of \emph{exact} differential $k$-forms. The
|
||
\emph{$k$-th complex de Rahm cohomology group} is defined as
|
||
\begin{align*}
|
||
H^k_{dR}(X,\mathbb{C}) := Z^k(X,\mathbb{C}) / B^k(X,\mathbb{C}).
|
||
\end{align*}
|
||
It is $B^k(X,\mathbb{C}) \subset Z^k(X,\mathbb{C})$ because \Cref{harmonic-forms:eq:derahm} is
|
||
a complex, so this is well-defined.
|
||
\end{defn}
|
||
\begin{defn}[Dolbeault complex]
|
||
Similar to \Cref{harmonic-forms:eq:derahm}, we define the \emph{$p$-th Dolbeault complex} for all
|
||
$0 \leq p \leq m$ as
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:dolbeault}
|
||
0 \longrightarrow \mathcal{A}^{p,0}(X) \xlongrightarrow{\opartial_0}\mathcal{A}^{p,1}(X)
|
||
\xlongrightarrow{\opartial_1}\dots\xlongrightarrow{\opartial_{m-1}}\mathcal{A}^{p,m}(X)
|
||
\xlongrightarrow{} 0
|
||
\end{align}
|
||
with $\opartial_k: \mathcal{A}^{p,k}(X) \rightarrow \mathcal{A}^{p,k+1}(X)$ being the restriction
|
||
of $\opartial$ to the differential forms of type $(p,k)$.
|
||
\end{defn}
|
||
It is again obvious that this is indeed a complex because we have $\opartial^2 = 0$. Using this
|
||
property, we can define the Dolbeault cohomology.
|
||
\begin{defn}[Dolbeault cohomology]
|
||
Let $Z_p^k(X,\mathbb{C}):= \ker(\opartial_k)$ be the vector space of $\opartial$-closed
|
||
differential forms and $B_p^k(X,\mathbb{C}):= \img(\opartial_{k-1})$ be the vector space of
|
||
$\opartial$-exact differential forms of type $(p,k)$. The
|
||
\emph{$k$-th Dolbeault cohomology group} of the $p$-th Dolbeault complex is defined as
|
||
\begin{align*}
|
||
H^{p,k}_\opartial(X,\mathbb{C}) := Z_p^k(X,\mathbb{C}) / B_p^k(X,\mathbb{C}).
|
||
\end{align*}
|
||
It is $B_p^k(X,\mathbb{C}) \subset Z_p^k(X,\mathbb{C})$ because \Cref{harmonic-forms:eq:dolbeault}
|
||
is a complex, so this is well-defined.
|
||
\end{defn}
|
||
Before we are able to prove the Hodge Isomorphism theorems, we have to show the following propositions.
|
||
\begin{prop}
|
||
\label{harmonic-forms:lm:harmonic-forms-are-closed}
|
||
Let $\alpha\in\mathcal{H}^k(X)$ be harmonic, then $\alpha$ is also closed.
|
||
\end{prop}
|
||
\begin{proof}
|
||
In \Cref{harmonic-forms:lm:kernel-laplacian}, it was shown that $\ker(\Delta_d) = \ker(d) \cap
|
||
\ker(d^*)$. Therefore, it is $\mathcal{H}^k(X) \subset \ker(d)$ and the statement is proven.
|
||
\end{proof}
|
||
\begin{prop}[{\cite[Ch.\,VI §3. Theorem 3.16]{Demailly1997}}]
|
||
\label{harmonic-forms:lm:improved-decomposition-of-the-k-forms}
|
||
For the differential $k$-forms, there is another orthogonal decomposition given as
|
||
\begin{align}
|
||
\mathcal{A}^k_\mathbb{C}(X) = \mathcal{H}^k(X) \oplus d(\mathcal A^{k-1}_\mathbb{C}(X)) \oplus
|
||
d^*(\mathcal A^{k+1}_\mathbb{C}(X)).
|
||
\end{align}
|
||
\end{prop}
|
||
\begin{proof}
|
||
This is a restated version of the proof of the given theorem in \cite{Demailly1997}.
|
||
With the definition of the Laplacian $\Delta_d$, it is obvious that
|
||
$\Delta_d(\mathcal{A}^k_\mathbb{C}(X))\subset d(\mathcal A^{k-1}_\mathbb{C}(X)) \oplus d^*(\mathcal A^{k+1}_\mathbb{C}(X))$.
|
||
Thus, because of \Cref{harmonic-forms:eq:decomposition-of-the-k-forms}, it is enough to prove the
|
||
orthogonality of the decomposition. Therefore, let $\alpha\in \mathcal{A}^{k+1}_\mathbb{C}(X)$
|
||
and $\beta \in \mathcal{A}_\mathbb{C}^{k-1}(X)$. It \nolinebreak is
|
||
\begin{align*}
|
||
(d^*\alpha,d\beta)_{L^2} = (\alpha,d^2\beta)_{L^2} = 0.
|
||
\end{align*}
|
||
Hence, $d(\mathcal A^{k-1}_\mathbb{C}(X))$ and $d^*(\mathcal A^{k+1}_\mathbb{C}(X))$ are orthogonal.
|
||
With the previously established equality $\ker(\Delta_d) = \ker(d) \cap \ker(d^*)$
|
||
(cf. \Cref{harmonic-forms:lm:kernel-laplacian}), it is also for all $\gamma \in \mathcal{H}^k(X)$
|
||
\begin{align*}
|
||
(\gamma,d\beta)_{L^2} = (d^*\gamma,\beta)_{L^2} = (0,\beta)_{L^2} = 0 \; \text{ and } \;
|
||
(d^*\alpha,\gamma)_{L^2} = (\alpha,d\gamma)_{L^2} = (\alpha, 0)_{L^2} = 0.
|
||
\end{align*} Therefore, the statement is proven.
|
||
\end{proof}
|
||
\begin{cor}
|
||
The kernel of the exterior derivative $d$ decomposes orthogonally as
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:kernel-of-the-exterior-derivative}
|
||
\ker(d) = \mathcal{H}^k(X) \oplus d(\mathcal{A}^{k-1}_\mathbb{C}(X)).
|
||
\end{align}
|
||
\end{cor}
|
||
\begin{proof}
|
||
With \Cref{harmonic-forms:lm:kernel-laplacian}, it is obvious that
|
||
$\mathcal{H}^k(X) \oplus d(\mathcal{A}^{k-1}) \subset \ker(d)$.
|
||
Also for all $\alpha \in \mathcal{A}^{k+1}_\mathbb{C}(X)$, it is $d^*\alpha \in \ker(d^*)$
|
||
because $(d^*)^2 = 0$. Therefore, if it would be $d^*\alpha \in \ker(d)$, it would already be
|
||
$d^*\alpha \in \mathcal{H}^k(X)$. Thus, this statement is a consequence of the last proposition.
|
||
\end{proof}
|
||
The same argument yields a similar statement for the formal adjoint of the exterior derivative.
|
||
\begin{cor}
|
||
The kernel of the formal adjoint $d^*$\! decomposes orthogonally as
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:kernel-of-the-formal-adjoint-of-the-exterior-derivative}
|
||
\ker(d^*) = \mathcal{H}^k(X) \oplus d^*(\mathcal{A}_\mathbb{C}^{k+1}(X)).
|
||
\end{align}
|
||
\end{cor}
|
||
\begin{rem}
|
||
It has to be mentioned that both of these corollaries have been motivated by an attempt to
|
||
separate the proof of the Hodge Isomorphism theorem found in \cite[Theorem 5.23]{Voisin2002}
|
||
into multiple components to make it more accessible to the reader.
|
||
\end{rem}
|
||
At this time, we are able to prove the first of the above-mentioned theorems, which is going to be
|
||
essential for the proof of the Hodge Decomposition.
|
||
\begin{thm}[Hodge Isomorphism theorem \MakeUppercase{\romannumeral 1} {\cite[Theorem
|
||
5.23]{Voisin2002}}]
|
||
\label{harmonic-forms:thm:hodge-isomorphism-1}
|
||
The natural mapping
|
||
\begin{align*}
|
||
\mathcal{H}^k(X) \rightarrow H^k_{dR}(X,\mathbb{C}), \;\enspace
|
||
\alpha \mapsto [\alpha]
|
||
\end{align*}
|
||
is an isomorphism. In particular, any class of closed forms in $H^k_{dR}(X,\mathbb{C})$ has a
|
||
unique harmonic representative.
|
||
\end{thm}
|
||
\begin{proof}
|
||
We use the same proof as in the given theorem in \cite{Voisin2002} and provide links to our
|
||
previously provided statements.
|
||
Note that the given mapping is meaningful because harmonic forms are closed
|
||
(cf. \Cref{harmonic-forms:lm:harmonic-forms-are-closed}).
|
||
Let now $\alpha \in \mathcal{A}^k_\mathbb{C}(X)$ be a closed differential form. With the
|
||
decomposition in \Cref{harmonic-forms:eq:decomposition-of-the-k-forms} we can uniquely rewrite
|
||
$\alpha =\beta + \Delta_d\gamma = \beta + dd^*\gamma + d^*d\gamma$ with
|
||
$\beta, \gamma \in \mathcal{A}^k_\mathbb{C}(X)$ and $\beta$ harmonic. Since $\alpha$ is a closed
|
||
form, it is
|
||
\begin{align*}
|
||
0 = d\alpha = d\beta + d^2d^*\gamma + dd^*d\gamma= d\beta + dd^*d\gamma.
|
||
\end{align*}
|
||
As harmonic forms are closed, this simplifies to $0 = dd^*d\gamma$. Hence, with the decomposition in
|
||
\Cref{harmonic-forms:eq:kernel-of-the-exterior-derivative}, we know that $d^*d\gamma = 0$. Thus, it is
|
||
$\alpha = \beta + dd^*\gamma$ and therefore we have $[\alpha] = [\beta]$ because $dd^* \gamma$ is
|
||
exact. This already proves the surjectivity.
|
||
In order to prove the injectivity, we chose $\alpha' \in \mathcal{A}^k_\mathbb{C}(X)$ to be harmonic and
|
||
exact. We can rewrite $\alpha' = d\beta'$ for $\beta'\in \mathcal{A}_\mathbb{C}^{k-1}(X)$.
|
||
It is $d\beta' \in \ker(d^*)$ because $d\beta'$ is harmonic.
|
||
However, with \Cref{harmonic-forms:eq:kernel-of-the-formal-adjoint-of-the-exterior-derivative}, we
|
||
know that $\ker(d^*) \cap d(\mathcal{A}^{k-1}_\mathbb{C}(X)) = \{0\}$. Thus, $\alpha' = d\beta' = 0$, and
|
||
therefore, the mapping is injective because $\alpha'$ only maps to an exact differential form if $\alpha' = 0$.
|
||
\end{proof}
|
||
As a direct result, we get the following corollary.
|
||
\begin{cor}[{\cite[Ch.\,VI §3. Theorem 3.17]{Demailly1997}}]
|
||
The de Rahm cohomology groups $H_{dR}^k(X,\mathbb{C})$ have finite dimensions.
|
||
\end{cor}
|
||
\begin{proof}
|
||
In \Cref{harmonic-forms:lm:finite-dimensional}, we have already established that $\mathcal{H}^k(X)$
|
||
has finite dimension. Thus, the first Hodge Isomorphism theorem immediately proves this result.
|
||
\end{proof}
|
||
Next, we will prove the same statement but this time for the Dolbeault cohomology groups. It should
|
||
not be surprising that the proof will use the same arguments.
|
||
\begin{prop}
|
||
\label{harmonic-forms:lm:opartial-harmonic-is-opartial-closed}
|
||
Let $\alpha \in \mathcal{H}^{p,q}(X)$ be a harmonic differential form, then
|
||
$\alpha$ is also $\opartial$-closed.
|
||
\end{prop}
|
||
\begin{proof} The differential form $\alpha$ is harmonic if and only if it is $\Delta_\opartial$-harmonic
|
||
(cf. \Cref{harmonic-forms:cor:harmonic-if-and-only-if-delta-opartial-harmonic}). Therefore, this
|
||
statement is a direct consequence of \Cref{harmonic-forms:rem:kernel-laplacian-partial-opartial}.
|
||
\end{proof}
|
||
\begin{prop}[{\cite[Ch.\,VI §7. Theorem 7.1]{Demailly1997}}]
|
||
\label{harmonic-forms:lm:refined-decomposition-of-the-pq-forms}
|
||
For the differential forms of type $(p,q)$, there is another orthogonal decomposition given as
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:another-decomposition-for-k-forms-with-opartial}
|
||
\mathcal{A}^{p,q}(X) = \mathcal{H}^{p,q}(X) \oplus \opartial\,(\mathcal{A}^{p,q-1}(X)) \oplus
|
||
\opartial^*\!(\mathcal{A}^{p,q+1}(X))
|
||
\end{align}
|
||
\end{prop}
|
||
\begin{proof}
|
||
Just repeat the proof of \Cref{harmonic-forms:lm:improved-decomposition-of-the-k-forms} using the
|
||
decomposition in \Cref{harmonic-forms:eq:decomposition-of-the-pq-forms} and the properties for the
|
||
kernel of $\Delta_\opartial$ established in
|
||
\Cref{harmonic-forms:rem:kernel-laplacian-partial-opartial}.
|
||
\end{proof}
|
||
With this decomposition, we are able to find a description for the kernel of the Dolbeault operators
|
||
$\opartial$ and $\opartial^*$. It is obvious that
|
||
$\mathcal{H}^{p,q}(X) \oplus \opartial\,(\mathcal{A}^{p,q-1}(X)) \subset \ker(\opartial)$. It is
|
||
also for all $\alpha \in \mathcal{A}^{p,q+1}(X)$ the image $\opartial^*\!\alpha \in \ker(\opartial^*)$.
|
||
Therefore, if it would be $\opartial^*\!\alpha \in \linebreak\ker(\opartial)$, then it would already
|
||
be $\opartial^*\!\alpha \in \mathcal{H}^{p,q}(X)$. Hence, the last proposition yields
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:kernel-of-opartial}
|
||
\ker(\opartial) = \mathcal{H}^{p,q}(X) \oplus \opartial\,(\mathcal{A}^{p,q-1}(X)).
|
||
\end{align}
|
||
Using the same argument again but for the formal adjoint $\opartial^*\!,$ we obtain
|
||
\begin{align}
|
||
\label{harmonic-forms:eq:kernel-of-the-formal-adjoint-of-opartial}
|
||
\ker(\opartial^*) = \mathcal{H}^{p,q}(X) \oplus \opartial^*(\mathcal{A}^{p,q+1}(X)).
|
||
\end{align}
|
||
This is everything necessary to prove the second Hodge Isomorphism theorem for differential forms
|
||
of type $(p,q)$.
|
||
\begin{thm}[Hodge Isomorphism theorem \MakeUppercase{\romannumeral 2} {\cite[Ch.\,VI §7. Theorem 7.2]{Demailly1997}}]
|
||
\label{harmonic-forms:thm:hodge-iso-2}
|
||
The natural mapping
|
||
\begin{align*}
|
||
\mathcal{H}^{p,q}(X) \rightarrow H^{p,q}_\opartial(X,\mathbb{C}),\; \enspace \alpha \mapsto [\alpha]
|
||
\end{align*}
|
||
is an isomorphism. In particular, any class of \,$\opartial$-closed forms in
|
||
$H^{p,q}_\opartial(X,\mathbb{C})$ has a unique harmonic representative.
|
||
\end{thm}
|
||
\begin{proof}
|
||
Just repeat the proof of \Cref{harmonic-forms:thm:hodge-isomorphism-1} using
|
||
\Cref{harmonic-forms:lm:opartial-harmonic-is-opartial-closed},
|
||
\Cref{harmonic-forms:lm:refined-decomposition-of-the-pq-forms} and the just established
|
||
decompositions \Cref{harmonic-forms:eq:kernel-of-opartial} and
|
||
\Cref{harmonic-forms:eq:kernel-of-the-formal-adjoint-of-opartial}.
|
||
\end{proof}
|