\section{Harmonic differential forms and cohomology}
In this chapter, we are going to introduce the theory of harmonic differential forms and review the
concepts of the de Rahm and Dolbeault cohomologies. Our primary objective is the proof of two Hodge
Isomorphism theorems, which establish a fundamental relation between the spaces of harmonic forms
and the cohomologies. These theorems will be crucial for our proof of the Hodge Decomposition
theorem later.

This chapter is based on sections 5.1 to 5.3 and 6.1 in the book \emph{Hodge Theory and Complex
Algebraic Geometry} \cite{Voisin2002} written by Claire Voisin. Some of the statements also
originate from chapter \MakeUppercase{\romannumeral 6} in the book \emph{Complex Analytic and
Differential Geometry} \cite{Demailly1997} written by Jean-Pierre Demailly.

For the remainder of this chapter, we will assume the following setting.
\begin{set}
	Let $X$ be a compact $m$-dimensional Kähler manifold with Riemannian metric $g$, closed 
	fundamental form $\omega$ %\in\mathcal{A}^2_\mathbb{R} \cap \mathcal{A}^{1,1}(X)$
	and Kähler metric $h:= g -i\omega$.
\end{set}
\subsection{Harmonic forms and the Laplacians}\;

Our first goal in this chapter is the definition of a generalization of the Laplace operator
$\Delta$. This operator is already known for $\mathbb{R}$-valued functions on the euclidean space
$\mathbb{R}^n$ and we will generalize it to obtain the Laplacians $\Delta_d, \Delta_\partial$ and
$\Delta_\opartial$ on the smooth differential forms $\mathcal{A}^\bullet_\mathbb{C}(X)$.

Based on the definition of harmonic functions in the euclidean case, we will use these generalized
Laplacians to define different types of harmonic differential forms. After that, we will further
develop this theory and focus on an important theorem that establishes a relation between the three
Laplacians. In order to prove this theorem, we will need the Kähler identities established in the
last chapter.
\begin{defn}[Laplacians]
	For every $k$, we define the \emph{Laplacian of the exterior derivative} $d$ as a linear operator 
	$\Delta_d: \mathcal{A}^k_\mathbb{C}(X)\rightarrow\mathcal{A}^k_\mathbb{C}(X)$ that
	is given as 
	\begin{equation*}
		\Delta_d := d\,d^* + d^*d.
	\end{equation*}
	Similarly, the \emph{Laplacians of the Dolbeault operators} $\partial$ and $\opartial$ are defined as
	\begin{align*}
		\Delta_\partial := \partial\,\partial^* + \partial^*\partial \;\enspace \text{ and } \; \enspace 
		\Delta_{\opartial}:= \opartial\,\opartial^* + \opartial^*\opartial
	\end{align*}
	and it is by definition that these two operators are linear mappings of type 
	$\mathcal A^{p,q}(X) \rightarrow\nolinebreak \mathcal A^{p,q}(X)$.
\end{defn}
\begin{rem}[{\cite[p.\,126]{Voisin2002}}]
	We already know that $d,\partial$ and $\opartial$ are linear differential operators of order 1, and
	the same is true for the formal adjoints $d^*,\partial^*$ and $\opartial^*$. Since the Laplacians 
	are all defined as a concatenation and sum of these linear differential operators, they are also 
	linear differential operators but of order 2.
\end{rem}
\begin{defn}[Harmonic differential forms]
	Let $\alpha \in \mathcal{A}^k_\mathbb{C}(X)$ be a complex differential form. We are going to call 
	$\alpha$ \emph{harmonic} if it is $\alpha \in \ker(\Delta_d)$. Furthermore, if $\alpha$ is of type
	$(p,q)$, we will call it $\mathit{\Delta_\opartial}$\emph{-harmonic} if 
	$\alpha \in \ker(\Delta_\opartial)$.
\end{defn}
At this point, we will prove an essential equality that will be required for subsequent calculations.
\begin{lm}[{\cite[Lemma 5.12]{Voisin2002}}]
	For the Laplacian $\Delta_d$, we have the equality
	\begin{align}
		\label{harmonic-forms:eq:property-laplacian}
		\left(\alpha,\Delta_d\alpha\right)_{L^2} = \left(d\alpha,d\alpha\right)_{L^2} +
		\left(d^*\alpha,d^*\alpha\right)_{L^2}.
	\end{align}
	Also, for the other two Laplacians $\Delta_\partial$ and $\Delta_\opartial$, the analogous 
	equalities are valid as well.
\end{lm}
\begin{proof}This proof was taken from the given lemma in \cite{Voisin2002}.
	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}, 
	we get
	\begin{align*}
		\left(\alpha,\Delta_d \alpha\right)_{L^2} = \left(\alpha, d\,d^* \alpha + d^*d
		\alpha\right)_{L^2}&= \left(\alpha,d\,d^*\alpha\right)_{L^2} + \left(\alpha, d^*d\alpha\right)_{L^2}\\
		%\\&= %\left(d^* \alpha, d^*\alpha\right)_{L^2} + \overline{\left(d^*d\alpha, \alpha\right)}_{L^2}
		%\\&= (d^*\alpha,d^*\alpha)_{L^2} + \overline{(d\alpha,d\alpha)}_{L^2}
		%\overline{\left(d\,d^*\alpha,\alpha\right)}_{L^2} + \left(\alpha,d^*d\alpha\right)_{L^2} \\&=
		%\overline{\left(d^*\alpha,d^*\alpha\right)} + \left(d\alpha,d\alpha\right)_{L^2} 
		&= \left(d^*\alpha,d^*\alpha\right)_{L^2} + \left(d\alpha,d\alpha\right)_{L^2}
	\end{align*}
	and the analogous calculation can be used for the Laplacians $\Delta_\partial$ and 
	$\Delta_\opartial$ too.
\end{proof}
Suppose there exists an $\alpha \in \mathcal A^k_\mathbb{C}(X)$ such that $\Delta_d \alpha = 0$. 
We can use the equality established in \Cref{harmonic-forms:eq:property-laplacian} to obtain
\begin{align*}
	%\label{kaehler-maifolds:eq:laplacian-zero}
	0 = (\alpha,\Delta_d\alpha)_{L^2} =  \left(d^*\alpha,d^*\alpha\right)_{L^2} +
	\left(d\alpha,d\alpha\right)_{L^2}.
\end{align*}
As mentioned in \Cref{kaehler-manifolds:rem:l2-metric-properties}, the hermitian $L^2$-metric is
positive definite, so both of these summands have to be equal to zero, which is only the case if 
$d^*\alpha = 0$ and also $d\alpha =0$. If, on the other hand, $d\alpha = 0$ and $d^*\alpha = 0$, 
then it is by definition of the Laplacian $\Delta_d \alpha = 0$. This already proves the following corollary.
\begin{cor}[{\cite[Corollary 5.13]{Voisin2002}}]
	\label{harmonic-forms:lm:kernel-laplacian}
	For the kernel of the Laplacian $\Delta_d$, we have the relation
	\begin{align*}
		\ker(\Delta_d) = \ker(d) \cap \ker(d^*).
	\end{align*}
	In particular, $\alpha \in \mathcal{A}^k_\mathbb{C}(X)$ is harmonic if and only if \,
	$d\alpha = d^*\alpha = 0$.
\end{cor}
It is also obvious that the same calculation can be repeated for the other two Laplacians
$\Delta_\partial$ and $\Delta_\opartial$. Hence, we also get the same relations for the kernels 
of these Laplacians.
\begin{cor}
	\label{harmonic-forms:rem:kernel-laplacian-partial-opartial}
	For the kernels of the Laplacians $\Delta_\partial$ and $\Delta_\opartial$, we have the relations
	\begin{align*}
		\ker(\Delta_\partial) = \ker(\partial) \cap \ker(\partial^*) \; \enspace \text{ and } \; 
		\enspace \ker(\Delta_\opartial) = \ker(\opartial) \cap \ker(\opartial^*).
	\end{align*}
	In particular, $\alpha \in \mathcal{A}^{p,q}(X)$ is $\Delta_\opartial$-harmonic if and
	only if\, $\opartial\alpha  = 0$ and $\opartial^*\alpha = 0$.
\end{cor}
\begin{prop}[{\cite[p.\,368]{Demailly1997}}]
	\label{harmonic-forms:lm:laplacians-are-self-adjoint}
	The Laplacians $\Delta_d, \Delta_\partial$ and $\Delta_\opartial$ are formally self-adjoint.
\end{prop}
\begin{proof}
	We calculate for all $\alpha,\beta \in \mathcal{A}^k_\mathbb{C}(X)$
	\begin{align*}
		(\Delta_d\alpha,\beta)_{L^2} &= (dd^*\alpha + d^*d\alpha,\beta)_{L^2}\\
		&= (dd^*\alpha,\beta)_{L^2} + (d^*d\alpha,\beta)_{L^2} \\
		&= (d^*\alpha,d^*\beta)_{L^2} + (d\alpha,d\beta)_{L^2}.
	\end{align*}
	Furthermore, we are able to calculate
	\begin{align*}
		(\alpha,\Delta_d\beta)_{L^2} &= (\alpha,dd^* \beta + d^*d\beta)_{L^2}\\
		&= (\alpha,dd^*\beta)_{L^2} + (\alpha, d^*d\beta)_{L^2}\\
		&= (d^*\alpha,d^*\beta)_{L^2} + (d\alpha,d\beta)_{L^2}
	\end{align*}
	This already proves that $\Delta_d$ is formally self-adjoint, and the argument for the other 
	two Laplacians is exactly the same.
\end{proof}
At this point, we have established the necessary theory to prove the existence of a previously 
indicated relation between the different Laplacians. This relation will be fundamental to the proof 
of the Hodge Decomposition theorem.
\begin{thm}[{\cite[Theorem 6.7]{Voisin2002}}]
	For the Laplacians $\Delta_d,\Delta_\partial$ and $\Delta_\opartial$ on our compact Kähler manifold
	$X$, we have the following relation
	\begin{align}
		\label{harmonic-forms:eq:thm-laplacian-realtion}
		\frac{1}{2}\Delta_d = \Delta_\partial = \Delta_\opartial.
	\end{align}
	In particular, any differential form $\alpha \in \mathcal{A}^k_\mathbb{C}(X) \cap
	\mathcal{A}^{p,q}(X)$ is harmonic if and only if it is $\Delta_\opartial$-harmonic.
\end{thm}
\begin{proof} 
	We provide a similar argument as in the proof of \cite[Theorem 6.7]{Voisin2002} but our approach 
	will be slightly different. With the splitting of both, the exterior derivative $d = \partial + \opartial$ 
	and the formal adjoint of the exterior derivative $d^* = \partial^* +\,\opartial^*$ 
	(cf. \Cref{kaehler-manifolds:rem:splitting-of-the-formal-adjoint-of-the-exterior-derivative}), the
	definition of the Laplacian $\Delta_d$ can be rewritten as follows.
	\begin{align*}
		\Delta_d &= d\,d^* + d^*d \\
		&= \big(\partial + \opartial\big)\big(\partial^* + \opartial^*\big) +
			\big(\partial^* + \opartial^*\big) \big(\partial + \opartial\big)\\
		&= \big(\partial\partial^* + \partial \opartial^* + \opartial\partial^* + \opartial\,\opartial^*\big) + 
			\big(\partial^*\partial +\partial^*\opartial +\opartial^* \partial + \opartial^*\opartial\big)\\
		&=\partial\partial^* + \partial^*\partial + \partial\opartial^* + \opartial^*\partial +
			\opartial\partial^* +\partial^*\opartial + \opartial\,\opartial^* + \opartial^*\opartial%\\&=
		%\Delta_\partial + \partial\opartial^* + \opartial^*\partial + \opartial\partial^*
		%+\partial^*\opartial + \opartial\,\opartial^* + \opartial^*\opartial 
	\end{align*}
	Now, we are going to use the Kähler identities established in
	\Cref{kaehler-manifolds:thm:kaehler-identities}, in particular the equality 
	$\partial^* = i[\Lambda,\opartial]$, to get
	\begin{align*}
		\opartial\partial^* + \partial^*\opartial &= \opartial i [\Lambda,\opartial] +
			i[\Lambda,\opartial]\opartial \\
		&=i\big(\opartial(\Lambda\opartial-\opartial\Lambda)+(\Lambda\opartial-\opartial\Lambda)\opartial\big)\\
		&=i\big(\opartial\Lambda\opartial - \opartial^2\Lambda + \Lambda\opartial^2 - \opartial\Lambda\opartial\big)\\
		&=i\big(\opartial\Lambda\opartial - \opartial\Lambda\opartial\big) \\
		&=0.
	\end{align*}
	Applying the other Kähler identity $\opartial^* = -i[\Lambda,\partial]$ from
	\Cref{kaehler-manifolds:thm:kaehler-identities}, we can do a similar calculation to obtain
	\begin{align*}
		\partial\opartial^* + \opartial^*\partial &=  -i\partial[\Lambda,\partial] -i[\Lambda,\partial]\partial \\
		& = -i\big(\partial(\Lambda\partial - \partial\Lambda) + (\Lambda\partial -\partial\Lambda)\partial\big) \\
		&= -i\big(\partial\Lambda\partial -\partial^2\Lambda + \Lambda\partial^2 - \partial\Lambda\partial\big) \\
		&= -i\big(\partial\Lambda\partial - \partial\Lambda\partial\big) \\ 
		&= 0.
	\end{align*}
	Thus, the above equation simplifies to
	\begin{align*}
		\Delta_d =\partial\partial^* + \partial^*\partial +  \opartial\,\opartial^* +
		\opartial^*\opartial= \Delta_\partial +  \Delta_\opartial.
	\end{align*}
	Therefore, it is enough to prove $\Delta_\partial = \Delta_\opartial$. However, we can calculate
	\begin{align*}
		\Delta_\partial - \Delta_\opartial &= \partial\partial^* + \partial^*\partial -
		\big(\opartial\,\opartial^* + \opartial^*\opartial\big)\\
		&=i\big(\partial[\Lambda,\opartial] + [\Lambda,\opartial]\partial + \opartial[\Lambda,\partial] 
			+ [\Lambda,\partial]\opartial\big)\\ 
		&=i\big(\partial\Lambda\opartial -\partial\opartial\Lambda + \Lambda\opartial\partial -
			\opartial\Lambda\partial + \opartial\Lambda\partial -\opartial\partial\Lambda +
			\Lambda\partial\opartial - \partial\Lambda\opartial\big)\\
		&=i\big(\Lambda\opartial\partial-\partial\opartial\Lambda  - \opartial \partial\Lambda +
			\Lambda\partial\opartial\big) \\&= i\big(\Lambda(\opartial\partial + \partial\opartial) -
			(\partial\opartial + \opartial\partial) \Lambda\big)\\ 
		&= 0.
	\end{align*}
	For the last step, we have used 
	$0 = d^2 = \partial^2 + \partial\opartial + \opartial\partial + \opartial^2 = \partial\opartial + \opartial\partial$. 
	Finally, we combine both of these results to obtain
	\begin{align*}
		\Delta_d = 2 \Delta_\partial = 2 \Delta_\opartial\,,
	\end{align*}
	which completes the proof of this theorem.
\end{proof}
%\begin{rem}
%	%\TODO{Counter example for this failing on arbitrary hermitian manifolds.}
%\end{rem}
The relation established in \Cref{harmonic-forms:eq:thm-laplacian-realtion} 
has some significant implications for the operator $\Delta_d$ and the property 
of a differential form being harmonic.
\begin{cor}
	\label{harmonic-forms:cor:harmonic-if-and-only-if-delta-opartial-harmonic}
	A complex differential form $\alpha \in \mathcal A^k_\mathbb{C}(X) \cap \mathcal A^{p,q}(X)$ 
	is harmonic if and only if it is $\Delta_\opartial$-harmonic.
\end{cor}
\begin{proof}
	It is $\Delta_d\alpha = 2\Delta_\opartial\,\alpha$, so either both operators map $\alpha$ to zero 
	or neither of them.
\end{proof}
When dealing with Kähler manifolds, the last corollary allows us not to distinguish between harmonic 
and $\Delta_\opartial$-harmonic forms anymore. Therefore, we will just refer to harmonic forms 
from now on.
\begin{cor}[{\cite[Corollary. 6.9]{Voisin2002}}] 
	\label{harmonic-forms:cor:harmonic-if-and-only-if-components-are-harmonic}
	A complex differential form $\alpha \in \mathcal{A}^k_\mathbb{C}(X)$ is harmonic if and only if its
	components $\alpha^{p,q}$ of type $(p,q)$ are harmonic.
\end{cor}
\begin{proof}
	With the previously used decomposition 
	\begin{align}
		\label{harmonic-forms:eq:decomposition-of-kforms}
	\mathcal{A}^k_\mathbb{C}(X) = \bigoplus_{p+q = k} \mathcal{A}^{p,q}(X),
	\end{align} (cf. \cite[Corollary 2.6.8]{Huybrechts2004})
	we can uniquely write  $\alpha = \sum_{p+q=k} \alpha^{p,q}$. As $\Delta_d$ is a linear operator, 
	the sum of harmonic forms is harmonic again. Therefore, it suffices to prove that $\alpha$ 
	being harmonic already implies that the components $\alpha^{p,q}$ are harmonic. 
	
	However, if $\alpha$ is harmonic we can use \Cref{harmonic-forms:eq:thm-laplacian-realtion} to obtain
	\begin{align*}
		0 = \Delta_d \alpha = \sum_{p+q = k} \Delta_d \alpha^{p,q} = 
			\sum_{p+q = k} 2\,\Delta_\opartial\,\alpha^{p,q} = 2\sum_{p+q = k} \Delta_\opartial\, \alpha^{p,q}.
	\end{align*}
	Since $\Delta_\opartial $ maps forms of type $(p,q)$ to forms of type $(p,q)$ and we also have 
	decomposition \Cref{harmonic-forms:eq:decomposition-of-kforms}, we conclude
	that $\Delta_\opartial\,\alpha^{p,q} = 0$ for each $\alpha^{p,q}$. %$p,q \in \mathbb{N}$ with $p+q = k$. 
	Hence with \Cref{harmonic-forms:cor:harmonic-if-and-only-if-delta-opartial-harmonic}, the 
	components $\alpha^{p,q}$ are each harmonic.
\end{proof}
\begin{nota}
	Let  $\mathcal H^k(X) \subset \mathcal{A}_\mathbb{C}^k(X)$ denote the space of complex harmonic
	differential $k$-forms. Let also $\mathcal H^{p,q}(X) \subset \mathcal{A}^{p,q}(X)$ denote the space
	of complex harmonic differential forms of type $(p,q)$, which is also the space of
	$\Delta_\opartial$-harmonic differential forms of type $(p,q)$ 
	(cf. \Cref{harmonic-forms:cor:harmonic-if-and-only-if-delta-opartial-harmonic}).
\end{nota}
With this newly introduced notation, we get the following corollary, which is just a rewritten version
of \Cref{harmonic-forms:cor:harmonic-if-and-only-if-components-are-harmonic} combined with the direct 
sum decomposition in \Cref{harmonic-forms:eq:decomposition-of-kforms}.
\begin{cor}[{\cite[Corollary 6.10]{Voisin2002}}]
	\label{harmonic-forms:cor:final-harmonic-decomposition}
	The harmonic differential $k$-forms decompose as 
	\begin{align*}
		\mathcal H^k(X) = \bigoplus_{p+q=k}\mathcal H^{p,q}(X).
	\end{align*}
\end{cor}
\subsection{Elliptic differential operators}\;

Before we can proceed with the harmonic differential forms theory, we need to extend our
understanding of linear differential operators. Therefore, we will introduce elliptic operators and
address their implications on the previously established theory of harmonic differential forms.

Note, however, that we will only provide a very brief overview and refer to other sources because a
more detailed discussion would be beyond the scope of this thesis. For an in-depth discussion of
elliptic differential operators and their properties, see §1. and §2. of chapter VI in
\cite{Demailly1997}.

At first, we have the following definition, motivated by the discussion in \cite[Section 15]{Schnell2012} 
and the explanations at the beginning of Section 5.2.1 in \cite{Voisin2002}.
\begin{defn}
	In the same setting as in \Cref{kaehler-manifolds:defn:differential-opperators}, for every
	$D_{U_j}$, we define a function 
	$P_j: U_j \times \mathbb{C}^m \rightarrow \mathbb{C}^{r_1\times r_2}$ as 
	\begin{align*}
		P_j(p,\xi) := \sum_{S} \big(P_{r,S,t}(p)\big)_{rt} \xi^S
	\end{align*}
	with $\big(P_{r,S,t}(p)\big)_{rt} \in \mathbb{C}^{r_1\times r_2}$ being the matrix defined by the
	coefficients $P_{r,S,t}(p) \in \mathbb{C}$, $S := (s_1,\dots,s_m)$ and 
	$\xi^S:= \xi_1^{s_1} \cdots \xi_m^{s_m}$. The functions $P_j$ define the \emph{symbol} of the 
	operator $D$.
\end{defn}
\begin{defn}[Elliptic differential operator]
	In the setting of \Cref{kaehler-manifolds:defn:differential-opperators}, the linear differential 
	operator $D$ is said to be \emph{elliptic} if $r_1 = r_2$ and for any local symbol $P_j$, the 
	image $P_j(p,\xi)$ is an invertible matrix for all $p \in U_j$ and $\xi \in \mathbb{C}^n\setminus\{0\}$.
\end{defn}
\begin{rem}
	The usual definition of an elliptic operator only requires the matrices to define injective
	homomorphisms (cf. \cite[Ch.\,VI §1. Definition 1.8]{Demailly1997}). However, the ranks $r_1$ and
	$r_2$ will always be equal in our setting. Therefore, if we additionally require these ranks to be
	equal, these definitions are equivalent for our purposes.
\end{rem}
Through an extensive calculation (cf. \cite[Lemmas 5.18 and 5.19]{Voisin2002}), it is possible to 
calculate the symbols of the three Laplacians and to conclude the following proposition.
\begin{prop}[{\cite[Corollary 5.20]{Voisin2002}}]
	\label{harmonic-forms:lm:laplacians-are-elliptic}The Laplacians $\Delta_d,\Delta_\partial$ and 
	$\Delta_\opartial$ are each elliptic differential operators.
\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}
