[proofplan] We first identify the sheaf cohomology groups $H^q(X,\Omega_X^p)$ with Dolbeault cohomology groups of type $(p,q)$. The Kähler [Hodge decomposition](/theorems/2745) then expresses each complex de Rham cohomology group $H^k(X;\mathbb C)$ as the direct sum of its bidegree pieces, and taking dimensions gives the Betti number formula. The first symmetry follows from conjugation symmetry of Hodge numbers on compact Kähler manifolds. The second symmetry follows from Serre duality in complex dimension $2$, together with the wedge-pairing identification $(\Omega_X^p)^\vee\otimes \Omega_X^2 \cong \Omega_X^{2-p}$. [/proofplan]

[step:Identify sheaf cohomology with Dolbeault cohomology] For integers $p,q$ with $0 \le p,q \le 2$, let $A^{p,q}(X)$ denote the complex [vector space](/page/Vector%20Space) of smooth $(p,q)$-forms on $X$. For every such pair $(p,q)$, let \begin{align*} \bar\partial_{p,q}:A^{p,q}(X)\to A^{p,q+1}(X) \end{align*} denote the Dolbeault operator in bidegree $(p,q)$, and write $\bar\partial$ for $\bar\partial_{p,q}$ when the bidegree is determined by the source and target. Define $A^{p,-1}(X):=0$ and define the Dolbeault cohomology group by \begin{align*} H^{p,q}_{\bar\partial}(X):=\ker\left(\bar\partial_{p,q}:A^{p,q}(X)\to A^{p,q+1}(X)\right)\big/\operatorname{im}\left(\bar\partial_{p,q-1}:A^{p,q-1}(X)\to A^{p,q}(X)\right). \end{align*} By the Dolbeault theorem for compact complex manifolds, the natural Dolbeault resolution of $\Omega_X^p$ induces a complex-linear isomorphism \begin{align*} H^q(X,\Omega_X^p)\cong H^{p,q}_{\bar\partial}(X). \end{align*} Therefore \begin{align*} h^{p,q}(X)=\dim_{\mathbb C}H^{p,q}_{\bar\partial}(X). \end{align*} In the remaining steps, write $H^{p,q}(X)$ for $H^{p,q}_{\bar\partial}(X)$ under this identification. [/step]

[step:Take dimensions in the Kähler Hodge decomposition]Since $X$ is compact Kähler, [Kähler Cohomology Carries a Pure Hodge Structure]([citetheorem:8066]) applies to $X$. Thus, for every integer $k$ with $0 \le k \le 4$, there is a direct sum decomposition of complex vector spaces \begin{align*} H^k(X;\mathbb C)=\bigoplus_{\substack{0 \le p,q \le 2 \;:\; p+q=k}}H^{p,q}(X). \end{align*} Taking complex dimensions of this finite direct sum gives \begin{align*} \dim_{\mathbb C}H^k(X;\mathbb C)=\sum_{\substack{0 \le p,q \le 2 \;:\; p+q=k}}\dim_{\mathbb C}H^{p,q}(X). \end{align*} Using the definitions of $b_k(X)$ and $h^{p,q}(X)$, this becomes \begin{align*} b_k(X)=\sum_{\substack{0 \le p,q \le 2 \;:\; p+q=k}}h^{p,q}(X). \end{align*}[/step]

[guided]The point of this step is to turn the [Hodge decomposition](/theorems/3941) into a numerical identity. Because $X$ is assumed to be a compact Kähler manifold, the hypotheses of [Kähler Cohomology Carries a Pure Hodge Structure]([citetheorem:8066]) are satisfied. That result gives, for each cohomological degree $k$, a direct sum decomposition \begin{align*} H^k(X;\mathbb C)=\bigoplus_{\substack{0 \le p,q \le 2 \;:\; p+q=k}}H^{p,q}(X). \end{align*} The bounds $0 \le p,q \le 2$ come from the fact that $X$ has complex dimension $2$, so there are no nonzero smooth forms of type $(p,q)$ outside this range. Now use the elementary dimension formula for a finite direct sum of finite-dimensional complex vector spaces: the dimension of the direct sum is the sum of the dimensions. Applying that formula to the displayed decomposition yields \begin{align*} \dim_{\mathbb C}H^k(X;\mathbb C)=\sum_{\substack{0 \le p,q \le 2 \;:\; p+q=k}}\dim_{\mathbb C}H^{p,q}(X). \end{align*} By definition, the left-hand side is $b_k(X)$. By the Dolbeault identification from the previous step, the summand $\dim_{\mathbb C}H^{p,q}(X)$ is exactly $h^{p,q}(X)$. Hence \begin{align*} b_k(X)=\sum_{\substack{0 \le p,q \le 2 \;:\; p+q=k}}h^{p,q}(X). \end{align*} This proves the Betti number formula.[/guided]

[step:Apply conjugation symmetry to the Hodge summands] Since $X$ is compact Kähler, [Symmetry of Hodge Numbers]([citetheorem:8081]) applies to $X$. Hence, for all integers $p,q$ with $0 \le p,q \le 2$, \begin{align*} h^{p,q}(X)=h^{q,p}(X). \end{align*} [/step]

[step:Use Serre duality to reflect across the middle of the diamond] Let $K_X$ denote the canonical sheaf of $X$. Since $X$ has complex dimension $2$, one has \begin{align*} K_X=\Omega_X^2. \end{align*} Fix integers $p,q$ with $0 \le p,q \le 2$. Let $(\Omega_X^p)^\vee:=\mathcal{H}om_{\mathcal O_X}(\Omega_X^p,\mathcal O_X)$ denote the holomorphic dual sheaf of $\Omega_X^p$. Serre duality for compact complex manifolds of dimension $2$ applied to the locally free sheaf $\Omega_X^p$ gives a perfect pairing and therefore a complex-linear duality \begin{align*} H^q(X,\Omega_X^p)^\vee\cong H^{2-q}\left(X,(\Omega_X^p)^\vee\otimes K_X\right). \end{align*} The wedge product of holomorphic forms gives a nondegenerate sheaf pairing \begin{align*} \Omega_X^p\otimes \Omega_X^{2-p}\to \Omega_X^2. \end{align*} Equivalently, it induces an isomorphism of locally free sheaves \begin{align*} \Omega_X^{2-p}\cong(\Omega_X^p)^\vee\otimes \Omega_X^2. \end{align*} Using $K_X=\Omega_X^2$, Serre duality becomes \begin{align*} H^q(X,\Omega_X^p)^\vee\cong H^{2-q}(X,\Omega_X^{2-p}). \end{align*} Taking complex dimensions and using $\dim_{\mathbb C}V^\vee=\dim_{\mathbb C}V$ for finite-dimensional complex vector spaces gives \begin{align*} h^{p,q}(X)=h^{2-p,2-q}(X). \end{align*} [/step]

[step:Combine the three identities] The first symmetry $h^{p,q}(X)=h^{q,p}(X)$ was proved by conjugation symmetry, the second symmetry $h^{p,q}(X)=h^{2-p,2-q}(X)$ was proved by Serre duality, and the Betti number formula was obtained by taking dimensions in the Kähler Hodge decomposition. These are exactly the three asserted conclusions for all $0 \le p,q \le 2$ and all $0 \le k \le 4$. [/step]