[proofplan] We use the [Hodge decomposition](/theorems/2745) of a compact Kähler manifold to compute $b_1(X)$ and $b_2(X)$ by summing Hodge numbers along total degrees $1$ and $2$. Hodge symmetry identifies the conjugate summands and gives the two stated dimension formulas. For $b_2^+(X)$, we decompose real degree-two cohomology into the Kähler line, the real part of $H^{2,0}(X)\oplus H^{0,2}(X)$, and the primitive real $(1,1)$ part; the Hodge-Riemann bilinear relations give positive definiteness on the first two pieces and negative definiteness on the primitive $(1,1)$ piece. The parity conclusions then follow immediately from the formulas. [/proofplan]

[step:Sum the Hodge decomposition in degrees one and two]Let \begin{align*} H^{p,q}(X)\subset H^{p+q}(X;\mathbb C) \end{align*} denote the Hodge summand of type $(p,q)$ furnished by [[Hodge Decomposition for Compact Kähler Manifolds](/theorems/8066)][citetheorem:8066]. Since $X$ is compact Kähler, the same theorem gives, for every $k\ge 0$, \begin{align*} H^k(X;\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X). \end{align*} The canonical extension of scalars identifies $H^k(X;\mathbb C)$ with $H^k(X;\mathbb R)\otimes_{\mathbb R}\mathbb C$, so \begin{align*} b_k(X)=\dim_{\mathbb C}H^k(X;\mathbb C). \end{align*} Therefore, in degree $1$, \begin{align*} b_1(X)=h^{1,0}(X)+h^{0,1}(X). \end{align*} In degree $2$, \begin{align*} b_2(X)=h^{2,0}(X)+h^{1,1}(X)+h^{0,2}(X). \end{align*}[/step]

[guided]The point of the [Hodge decomposition](/theorems/3941) is that it splits complex cohomology into pieces indexed by type. Since $X$ is a compact Kähler manifold, [Hodge Decomposition for Compact Kähler Manifolds][citetheorem:8066] applies directly and gives complex subspaces \begin{align*} H^{p,q}(X)\subset H^{p+q}(X;\mathbb C) \end{align*} such that \begin{align*} H^k(X;\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X) \end{align*} for every $k\ge 0$. We also need to connect complex cohomology dimensions with the real Betti numbers. By definition, \begin{align*} b_k(X)=\dim_{\mathbb R}H^k(X;\mathbb R). \end{align*} The canonical scalar-extension map identifies \begin{align*} H^k(X;\mathbb R)\otimes_{\mathbb R}\mathbb C \end{align*} with $H^k(X;\mathbb C)$. Hence complexification preserves dimension in the sense that \begin{align*} \dim_{\mathbb C}H^k(X;\mathbb C)=\dim_{\mathbb R}H^k(X;\mathbb R)=b_k(X). \end{align*} For $k=1$, the pairs $(p,q)$ with $p+q=1$ are $(1,0)$ and $(0,1)$. Thus \begin{align*} H^1(X;\mathbb C)=H^{1,0}(X)\oplus H^{0,1}(X), \end{align*} and taking complex dimensions gives \begin{align*} b_1(X)=h^{1,0}(X)+h^{0,1}(X). \end{align*} For $k=2$, the pairs $(p,q)$ with $p+q=2$ are $(2,0)$, $(1,1)$, and $(0,2)$. Thus \begin{align*} H^2(X;\mathbb C)=H^{2,0}(X)\oplus H^{1,1}(X)\oplus H^{0,2}(X), \end{align*} and taking complex dimensions gives \begin{align*} b_2(X)=h^{2,0}(X)+h^{1,1}(X)+h^{0,2}(X). \end{align*}[/guided]

[step:Use Hodge symmetry to simplify the first two formulas] By [[Hodge Symmetry for Compact Kähler Manifolds](/theorems/8081)][citetheorem:8081], complex conjugation identifies $H^{p,q}(X)$ anti-linearly with $H^{q,p}(X)$. Hence \begin{align*} h^{p,q}(X)=h^{q,p}(X) \end{align*} for all $p,q$. Applying this with $(p,q)=(1,0)$ gives \begin{align*} b_1(X)=h^{1,0}(X)+h^{0,1}(X)=2h^{1,0}(X). \end{align*} Applying it with $(p,q)=(2,0)$ gives \begin{align*} b_2(X)=h^{2,0}(X)+h^{1,1}(X)+h^{0,2}(X)=h^{1,1}(X)+2h^{2,0}(X). \end{align*} [/step]

[step:Decompose real degree-two cohomology into positive and negative Hodge-Riemann pieces]Let $\omega$ be a Kähler form on $X$, and let \begin{align*} \kappa:=[\omega]\in H^2(X;\mathbb R)\cap H^{1,1}(X) \end{align*} be its Kähler class. Define the real intersection form \begin{align*} Q_X:H^2(X;\mathbb R)\times H^2(X;\mathbb R)&\to\mathbb R, & Q_X(\alpha,\beta)&=\langle \alpha\smile\beta,[X]\rangle. \end{align*} Let $P_{\mathbb R}^{1,1}\subset H^2(X;\mathbb R)\cap H^{1,1}(X)$ denote the primitive real $(1,1)$ subspace \begin{align*} P_{\mathbb R}^{1,1}:=\{\alpha\in H^2(X;\mathbb R)\cap H^{1,1}(X):Q_X(\alpha,\kappa)=0\}. \end{align*} Since $X$ is connected and $\omega$ is Kähler, the class $\kappa$ is nonzero and satisfies $Q_X(\kappa,\kappa)>0$. Therefore \begin{align*} H^2(X;\mathbb R)\cap H^{1,1}(X)=\mathbb R\kappa\oplus P_{\mathbb R}^{1,1}. \end{align*} Let \begin{align*} V_{\mathbb R}:=H^2(X;\mathbb R)\cap\left(H^{2,0}(X)\oplus H^{0,2}(X)\right) \end{align*} be the real subspace underlying the conjugate pair $H^{2,0}(X)\oplus H^{0,2}(X)$. Complex conjugation identifies $H^{2,0}(X)$ with $H^{0,2}(X)$, so \begin{align*} \dim_{\mathbb R}V_{\mathbb R}=2h^{2,0}(X). \end{align*} The real Hodge decomposition in degree $2$ is \begin{align*} H^2(X;\mathbb R)=\mathbb R\kappa\oplus P_{\mathbb R}^{1,1}\oplus V_{\mathbb R}. \end{align*} By the [Hodge-Riemann bilinear relations on primitive cohomology](/theorems/8068), in the surface case the form $Q_X$ is positive definite on $\mathbb R\kappa\oplus V_{\mathbb R}$ and negative definite on $P_{\mathbb R}^{1,1}$. Equivalently, this is the degree-two surface case of [Hodge-Riemann Bilinear Relations on Primitive Cohomology][citetheorem:8068], with the complex orientation used in the definition of $Q_X$. It follows that a maximal positive-definite subspace of $H^2(X;\mathbb R)$ has dimension \begin{align*} \dim_{\mathbb R}(\mathbb R\kappa\oplus V_{\mathbb R})=1+2h^{2,0}(X). \end{align*} Thus \begin{align*} b_2^+(X)=1+2h^{2,0}(X). \end{align*}[/step]

[guided]The intersection form in degree two is the real symmetric bilinear map \begin{align*} Q_X:H^2(X;\mathbb R)\times H^2(X;\mathbb R)&\to\mathbb R, & Q_X(\alpha,\beta)&=\langle \alpha\smile\beta,[X]\rangle, \end{align*} where $[X]\in H_4(X;\mathbb R)$ is the fundamental class determined by the complex orientation. The number $b_2^+(X)$ is, by definition, the largest possible dimension of a real subspace on which this form is positive definite. Let $\omega$ be the Kähler form of the Kähler metric on $X$, and define its cohomology class by \begin{align*} \kappa:=[\omega]\in H^2(X;\mathbb R)\cap H^{1,1}(X). \end{align*} Because $\omega$ is positive and $X$ is connected and compact, $\kappa$ gives one distinguished real $(1,1)$ direction. More precisely, the Hodge-Riemann bilinear relations imply \begin{align*} Q_X(\kappa,\kappa)>0. \end{align*} In particular $\kappa\ne 0$, so $\mathbb R\kappa$ is a one-dimensional positive subspace. Next define the primitive real $(1,1)$ subspace by \begin{align*} P_{\mathbb R}^{1,1}:=\{\alpha\in H^2(X;\mathbb R)\cap H^{1,1}(X):Q_X(\alpha,\kappa)=0\}. \end{align*} This is exactly the $Q_X$-orthogonal complement of the Kähler line inside the real $(1,1)$ part. Since $Q_X(\kappa,\kappa)>0$, every element $\alpha\in H^2(X;\mathbb R)\cap H^{1,1}(X)$ has a unique decomposition into its component along $\kappa$ and its primitive component: \begin{align*} \alpha=\frac{Q_X(\alpha,\kappa)}{Q_X(\kappa,\kappa)}\kappa+\left(\alpha-\frac{Q_X(\alpha,\kappa)}{Q_X(\kappa,\kappa)}\kappa\right). \end{align*} The second term lies in $P_{\mathbb R}^{1,1}$ because its pairing with $\kappa$ is \begin{align*} Q_X\left(\alpha-\frac{Q_X(\alpha,\kappa)}{Q_X(\kappa,\kappa)}\kappa,\kappa\right)=Q_X(\alpha,\kappa)-\frac{Q_X(\alpha,\kappa)}{Q_X(\kappa,\kappa)}Q_X(\kappa,\kappa)=0. \end{align*} Thus \begin{align*} H^2(X;\mathbb R)\cap H^{1,1}(X)=\mathbb R\kappa\oplus P_{\mathbb R}^{1,1}. \end{align*} Now define the real subspace coming from the conjugate Hodge types $(2,0)$ and $(0,2)$: \begin{align*} V_{\mathbb R}:=H^2(X;\mathbb R)\cap\left(H^{2,0}(X)\oplus H^{0,2}(X)\right). \end{align*} Complex conjugation carries $H^{2,0}(X)$ anti-linearly onto $H^{0,2}(X)$ by [Hodge Symmetry for Compact Kähler Manifolds][citetheorem:8081]. Therefore the real subspace fixed by conjugation inside $H^{2,0}(X)\oplus H^{0,2}(X)$ has real dimension twice the complex dimension of $H^{2,0}(X)$: \begin{align*} \dim_{\mathbb R}V_{\mathbb R}=2h^{2,0}(X). \end{align*} Combining this with the real form of the Hodge decomposition gives \begin{align*} H^2(X;\mathbb R)=\mathbb R\kappa\oplus P_{\mathbb R}^{1,1}\oplus V_{\mathbb R}. \end{align*} It remains to determine the sign of $Q_X$ on these three pieces. The surface case of [Hodge-Riemann Bilinear Relations on Primitive Cohomology][citetheorem:8068], with the complex orientation used to define $[X]$, says that $Q_X$ is positive definite on the Kähler line $\mathbb R\kappa$, positive definite on the real subspace $V_{\mathbb R}$ coming from $H^{2,0}(X)\oplus H^{0,2}(X)$, and negative definite on the primitive real $(1,1)$ part $P_{\mathbb R}^{1,1}$. Hence the positive directions are precisely the directions in \begin{align*} \mathbb R\kappa\oplus V_{\mathbb R}. \end{align*} No vector in $P_{\mathbb R}^{1,1}$ can be added to a positive-definite subspace as an additional positive direction, because $Q_X$ is negative definite there. Therefore a maximal positive-definite subspace has dimension \begin{align*} \dim_{\mathbb R}(\mathbb R\kappa\oplus V_{\mathbb R})=1+2h^{2,0}(X). \end{align*} This proves \begin{align*} b_2^+(X)=1+2h^{2,0}(X). \end{align*}[/guided]

[step:Read off the parity consequences] The formula \begin{align*} b_1(X)=2h^{1,0}(X) \end{align*} shows that $b_1(X)$ is even. The formula \begin{align*} b_2^+(X)=1+2h^{2,0}(X) \end{align*} shows that $b_2^+(X)$ is odd. This proves all asserted formulas and parity statements. [/step]