\section{Hodge Decomposition}
In this final chapter, we are left with the proof of the Hodge Decomposition theorem, which is the
primary goal of this thesis. \!With the already established theory from the preceding chapters, the
proof will be straightforward. Since it is apriori unclear whether the Hodge Decomposition depends
on the choice of the Kähler metric, we will also show that this decomposition does not depend on the
metric.

In the second section of this chapter, we are also going to provide a topological application of the
Hodge Decomposition theorem that illustrates only one of the many important consequences in complex
and algebraic geometry.

For the remainder of this chapter, we are going to assume the following setting.
\begin{set}
	Let $X$ be a compact Kähler manifold with Riemannian metric $g$ and closed fundamental form
	$\omega$. Also, let $h:= g -i\omega$ denote the Kähler metric.
\end{set}
\subsection{Proof of the Hodge Decomposition theorem}\;
\begin{thm}[Hodge Decomposition {\cite[Ch.\,VI §8. Theorem 8.5]{Demailly1997}}]
	For the compact Kähler manifold $X$, there does exist an isomorphism 
	\begin{align}
		\label{hodge-decomposition:eq:main-theorem}
		H^k_{dR}(X,\mathbb{C}) \cong \bigoplus_{p+q=k} H^{p,q}_{\opartial}(X,\mathbb{C}).
	\end{align}
\end{thm}
\begin{proof}
	The existence of the decomposition
	\begin{align*}
		\mathcal{H}^k(X) = \bigoplus_{p+q=k} \mathcal{H}^{p,q}(X)
	\end{align*} has been established in \Cref{harmonic-forms:cor:final-harmonic-decomposition}. Using
	the two Hodge Isomorphism
	\Cref{harmonic-forms:thm:hodge-isomorphism-1,harmonic-forms:thm:hodge-iso-2}, we get
	\begin{align*}
		H^k_{dR}(X,\mathbb{C}) \cong\mathcal{H}^k(X) = \bigoplus_{p+q=k} \mathcal{H}^{p,q}(X)\cong
		\bigoplus_{p+q=k}H^{p,q}_\opartial(X,\mathbb{C}).
	\end{align*}
\end{proof}
Note that the definition of the harmonic forms depends on the definition of the Laplacians
$\Delta_d,\Delta_\partial$ and $\Delta_\opartial$. These Laplacians are defined using the formal
adjoint operators $d^*, \partial^*$ and $\opartial^*$ that depend on the $L^2$-metric induced by the
metric $h$. Hence, it is unclear whether the Hodge Decomposition is independent of the choice of
this metric.
\begin{prop}[{\cite[Proposition 6.11]{Voisin2002}}]
	\label{hodge-decomposition:prop:independence-of-metric}
	The Hodge Decomposition \Cref{hodge-decomposition:eq:main-theorem} does not depend on the choice 
	of the Kähler metric.
\end{prop}
\begin{proof}
	This proof was taken from the stated proposition in \cite{Voisin2002}.
	Let $K^{p,q} \subset H^k_{dR}(X,\mathbb{C})$ denote the subspace of cohomology classes that are
	representable by a closed form of type $(p,q)$. In \Cref{harmonic-forms:thm:hodge-iso-2}, we have
	shown that every class in $H_\opartial^{p,q}(X,\mathbb{C})$ has a harmonic representative, which is in
	particular also closed. Therefore, it is already $H_\opartial^{p,q}(X,\mathbb{C}) \subset K^{p,q}$.\\
	Now, we are going to show that the other inclusion is also true. Let $\eta \in K^{p,q}$ be a
	closed form of type $(p,q)$. With \Cref{harmonic-forms:eq:decomposition-of-the-k-forms}, we can
	uniquely rewrite $\eta = \alpha + \Delta_d \beta$ with $\alpha \in \mathcal{H}^k(X)$ and $\beta
	\in \mathcal{A}_\mathbb{C}^k(X)$. Since $\Delta_\opartial$ maps forms of type $(p,q)$ to forms of type
	$(p,q)$, we know with \Cref{harmonic-forms:eq:thm-laplacian-realtion} that this also holds for
	$\Delta_d$. Therefore, we can only consider the components of type $(p,q)$ to get the again unique
	equality 
	\begin{align*}
		\eta = \alpha^{p,q} + \Delta_d\beta^{p,q} = \alpha^{p,q} + (dd^*\beta^{p,q} + d^*d\beta^{p,q}).
	\end{align*}
	With \Cref{harmonic-forms:cor:harmonic-if-and-only-if-components-are-harmonic}, we know that
	the component $\alpha^{p,q}$ is also harmonic and particularly closed. As $\eta$ is closed too, we get
	\begin{align*}
		0 = d\eta = d\alpha^{p,q} + d(dd^* \beta^{p,q} + d^*d\beta^{p,q}) = dd^*d\beta^{p,q}.
	\end{align*}
	With \Cref{harmonic-forms:eq:kernel-of-the-exterior-derivative}, we already know that
	$d^*d\beta^{p,q} = 0$. Thus, we obtain $\eta = \alpha^{p,q} + dd^* \beta^{p,q}$.
	Since $\opartial(dd^*\beta^{p,q}) = 0$, we know that 
	$[\eta] = [\alpha^{p,q}] \in H^{p,q}_\opartial(X,\mathbb{C})$. This proofs that 
	$H^{p,q}_\opartial(X,\mathbb{C}) = K^{p,q}$. Furthermore, since $K^{p,q}$ and the inclusion of 
	$K^{p,q}$ into $H^k_{dR}(X,\mathbb{C})$ do not depend on the choice of the metric, this proves 
	the statement.
\end{proof}
\begin{rem}
	After the independence of the Kähler metric has been proven, it is justified to write an equality in 
	\Cref{hodge-decomposition:eq:main-theorem}.
\end{rem}

\subsection{Application of the Hodge Decomposition}\;

The Hodge Decomposition theorem has numerous applications in complex and algebraic geometry and also
in topology. Since most of these applications need some additional theory that would be beyond the
scope of this thesis, we will only provide one of them, which has historically provided the first
example of a complex manifold that can not be equipped with a Kähler metric.

In the proof of \Cref{hodge-decomposition:prop:independence-of-metric}, we have defined $K^{p,q}
\subset H^k_{dR}(X,\mathbb{C})$ to be the subspace of cohomology classes that are representable by a
closed form of type $(p,q)$. Since the exterior derivative is compatible with complex conjugation
and the conjugation of a closed differential form of type $(p,q)$ is a closed differential form of
type $(q,p)$, we obtain $\overline{K^{p,q}} = K^{q,p}$. Hence, we have the following proposition.
\begin{prop}[{\cite[Corollary 6.12]{Voisin2002}}]
	\label{hodge-decomposition:prop:conjugation-dolbeault-cohomologies}
	For the Dolbeault cohomology groups, we have the equality \vspace*{-0.1cm}
	\begin{align*}
		\overline{H^{p,q}_\opartial(X,\mathbb{C})}=H^{q,p}_\opartial(X,\mathbb{C}).
	\end{align*}
\end{prop}
\begin{proof}
	This proof was taken from the given corollary in \cite{Voisin2002}.
	In the proof of \Cref{hodge-decomposition:prop:independence-of-metric}, we have shown that 
	$K^{p,q} = H_\opartial^{p,q}(X,\mathbb{C})$. Therefore, the statement is a consequence of the 
	just established similar equality $\overline{K^{p,q}} = K^{q,p}$.
\end{proof}

This proposition can be used to obtain an interesting corollary that uses the Hodge Decomposition
theorem and gives a strong constraint to the topological properties of a Kähler manifold. However,
before we focus on this corollary, we will introduce a new notation inspired by 
\cite[Ch. VI §8. Equation 8.11]{Demailly1997} to improve the readability.
\begin{nota}
	The $k$-th \emph{Betti number} and the $(p,q)$-th \emph{Hodge number} of the Kähler manifold $X$ 
	are respectively defined as
	\begin{align*}
		b_k(X) := \dim_\mathbb{C} H^{k}(X,\mathbb{C}), \qquad h_{p,q}(X) := \dim_\mathbb{C}H_\opartial^{p,q}(X,\mathbb{C}).
	\end{align*}
\end{nota}
Using this notation, the Hodge Decomposition \Cref{hodge-decomposition:eq:main-theorem} immediately 
yields the equality
\begin{align}
	\label{hodge-decomposition:eq:betti-and-hodge-numbers}
	b_k(X) = \sum_{p+q=k}h_{p,q}(X)
\end{align}
(cf. \cite[Ch.\,VI §8. Equation 8.12]{Demailly1997}).
This equality can now be used to obtain the following corollary.
\begin{cor}[{\cite[Corollary 6.13]{Voisin2002}}]
	\label{hodge-decomposition:cor:odd-betti-numbers}
	The odd Betti numbers $b_{2k+1}(X)$ are even.
\end{cor}
\begin{proof}
	With \Cref{hodge-decomposition:prop:conjugation-dolbeault-cohomologies}, we get 
	$h_{p,q}(X) = h_{q,p}(X)$. Thus, \Cref{hodge-decomposition:eq:betti-and-hodge-numbers} yields
	\begin{align*}
		b_{2k+1}(X) = \sum_{p=0}^{2k+1} h_{p,2k+1-p}(X) 
		= \sum_{p=0}^{k} 2\,h_{p,2k+1-p}(X) 
		= 2 \sum_{p=0}^{k} h_{p,2k+1-p}(X).
	\end{align*}
	Therefore, the odd Betti number $b_{2k+1}(X)$ is even.
\end{proof}
This corollary allows us to rule out the existence of a Kähler metric for certain compact complex
manifolds by calculating the odd Betti numbers. One example of such a type of manifold that cannot
be endowed with a Kähler metric are the Hopf surfaces.
\begin{defn}[Hopf surfaces]
	Let $\lambda_1,\lambda_2 \in \mathbb{C}$ with $0 < |\lambda_1| \leq |\lambda_2| < 1$. Let 
	$\mathbb{Z}$ act freely and properly discontinuously on $\mathbb{C}^2\setminus\{0\}$ by the 
	biholomorphic transformations
	\begin{align*}
		\mathbb{Z} \times \mathbb{C}^2\!\setminus\!\{0\} \rightarrow \mathbb{C}^2\!\setminus\!\{0\},
		\;\quad(l,z_1,z_2) \mapsto (\lambda_1^lz_1,\lambda_2^lz_2).
	\end{align*}
	The resulting compact complex quotient manifold $(\mathbb{C}^2\setminus\{0\}) / \mathbb{Z}$ is 
	called a \emph{Hopf surface}.
\end{defn}
\begin{rem}
	It has to be mentioned that the definition of the Hopf surfaces differs substantially among 
	different authors. The provided definition is a combination of the two found in 
	\cite[p.\,143]{Voisin2002} and \cite[p.\,61]{Huybrechts2004}
\end{rem}
Let now $Y$ be such a Hopf surface. Using the approach mentioned in \cite[p.\,143]{Voisin2002}, 
one can calculate that $b_1(Y) = 1$. As shown in \Cref{hodge-decomposition:cor:odd-betti-numbers}, 
the odd Betti numbers must be even for a Kähler manifold. Thus, this calculation proves that $Y$ 
cannot be a Kähler manifold.
\begin{rem}
	According to \cite[p.\,172]{Barth1984}, one of these Hopf surfaces has been the first example of 
	a compact complex manifold that is not Kähler. It was discovered by Heinz Hopf in 1948. Note, 
	however, that his definition of the Hopf surfaces has been slightly different than the one 
	presented in this thesis.
\end{rem}