[proofplan] Fix a Kähler class $[\omega]$ and split every real $(1,1)$-class into its component along $[\omega]$ and its primitive component. The Kähler direction is positive because the top-degree form $\omega\wedge\omega$ evaluates positively on the complex orientation fundamental class. The Hodge--Riemann bilinear relations give the opposite sign on every nonzero primitive real $(1,1)$-class. Since the decomposition is orthogonal for the intersection form, these two sign computations determine the signature. [/proofplan]

[step:Define the intersection form and the primitive hyperplane] Let \begin{align*} V:=H^{1,1}(X;\mathbb R) \end{align*} and define the symmetric bilinear map \begin{align*} Q:V\times V\to\mathbb R \end{align*} by \begin{align*} Q(\alpha,\beta):=\langle \alpha\smile\beta,[X]\rangle. \end{align*} Here $\alpha$ and $\beta$ denote real cohomology classes of degree $2$, and $[X]\in H_4(X;\mathbb R)$ denotes the fundamental class determined by the complex orientation. Fix a Kähler form $\omega\in A^{1,1}(X;\mathbb R)$, where $A^{1,1}(X;\mathbb R)$ denotes the space of smooth real closed $(1,1)$-forms on $X$, and let $[\omega]\in V$ be its cohomology class. Define the primitive real $(1,1)$-subspace relative to $[\omega]$ by \begin{align*} P:=\{\alpha\in V:\alpha\smile[\omega]=0\in H^4(X;\mathbb R)\}. \end{align*} Since $X$ is connected, compact, and has complex dimension $2$, the group $H^4(X;\mathbb R)$ is one-dimensional, generated by the orientation class dual to $[X]$. Therefore the condition $\alpha\smile[\omega]=0$ is equivalent to \begin{align*} Q(\alpha,[\omega])=\langle \alpha\smile[\omega],[X]\rangle=0. \end{align*} Thus $P$ is exactly the $Q$-orthogonal hyperplane to $[\omega]$ inside $V$, once $Q([\omega],[\omega])\neq 0$ is known. [/step]

[step:Show that the Kähler class gives a positive direction]Because $\omega$ is a Kähler form on a complex surface, $\omega\wedge\omega$ is a positive smooth top-degree form. Therefore its evaluation on the complex orientation fundamental class is strictly positive: \begin{align*} Q([\omega],[\omega])=\langle [\omega]\smile[\omega],[X]\rangle>0. \end{align*} In particular $[\omega]\neq 0$, and the line $\mathbb R[\omega]\subset V$ is positive for $Q$.[/step]

[guided]We first isolate the one positive direction. The class $[\omega]$ lies in $H^{1,1}(X;\mathbb R)$ because $\omega$ is a real closed $(1,1)$-form. The intersection form evaluates it against itself as \begin{align*} Q([\omega],[\omega])=\langle [\omega]\smile[\omega],[X]\rangle. \end{align*} Since $X$ has complex dimension $2$, the form $\omega\wedge\omega$ is a top-degree form. Positivity of the Kähler form means that, in every holomorphic coordinate chart, this top-degree form is a positive multiple of the complex orientation volume form. Because $X$ is compact and $\omega\wedge\omega$ is not identically zero, its evaluation on the complex orientation fundamental class is strictly positive: \begin{align*} \langle [\omega]\smile[\omega],[X]\rangle>0. \end{align*} Thus $Q([\omega],[\omega])>0$. This proves that every nonzero vector in the line $\mathbb R[\omega]$ has positive square, since for $c\in\mathbb R$ with $c\neq 0$, \begin{align*} Q(c[\omega],c[\omega])=c^2Q([\omega],[\omega])>0. \end{align*}[/guided]

[step:Split every real $(1,1)$-class into a Kähler component and a primitive component] Let $\beta\in V$ be arbitrary. Define the real scalar \begin{align*} c:=\frac{Q(\beta,[\omega])}{Q([\omega],[\omega])}. \end{align*} Define \begin{align*} \alpha:=\beta-c[\omega]\in V. \end{align*} Then \begin{align*} Q(\alpha,[\omega])=Q(\beta,[\omega])-cQ([\omega],[\omega])=0. \end{align*} Hence $\alpha\in P$, and \begin{align*} \beta=c[\omega]+\alpha. \end{align*} The decomposition is unique: if $c[\omega]+\alpha=0$ with $c\in\mathbb R$ and $\alpha\in P$, then pairing with $[\omega]$ gives \begin{align*} 0=cQ([\omega],[\omega])+Q(\alpha,[\omega])=cQ([\omega],[\omega]). \end{align*} Since $Q([\omega],[\omega])>0$, we get $c=0$, and then $\alpha=0$. Therefore \begin{align*} V=\mathbb R[\omega]\oplus P. \end{align*} Moreover this direct sum is $Q$-orthogonal because $Q(\alpha,[\omega])=0$ for every $\alpha\in P$. [/step]

[step:Apply the Hodge--Riemann bilinear relations on primitive real $(1,1)$-classes] Let $\alpha\in P$ be nonzero. The hypotheses of the [Hodge--Riemann bilinear relations for primitive Hodge classes](/theorems/8065) are satisfied: $X$ is compact Kähler of complex dimension $2$, the class $[\omega]$ is Kähler, and $\alpha$ is a primitive class of type $(1,1)$. Applying [citetheorem:8065] with $p=q=1$, $k=p+q=2$, and $n=2$, the sign factor in the primitive form is \begin{align*} i^{p-q}(-1)^{k(k-1)/2}=i^0(-1)^1=-1. \end{align*} The theorem therefore gives \begin{align*} -\langle \alpha\smile\overline{\alpha},[X]\rangle>0 \end{align*} for every nonzero primitive class $\alpha\in H^{1,1}(X)$. Because $\alpha$ is real, $\overline{\alpha}=\alpha$, so \begin{align*} -\langle \alpha\smile\alpha,[X]\rangle>0. \end{align*} Since $Q(\alpha,\alpha)=\langle \alpha\smile\alpha,[X]\rangle$, this is exactly \begin{align*} Q(\alpha,\alpha)<0. \end{align*} Thus $Q$ is negative definite on $P$. [/step]

[step:Read off the signature from the orthogonal decomposition] The orthogonal direct sum \begin{align*} V=\mathbb R[\omega]\oplus P \end{align*} has one positive summand, namely $\mathbb R[\omega]$, and the restriction of $Q$ to $P$ is negative definite. Complex conjugation preserves $H^{1,1}(X)$, and the real subspace $H^{1,1}(X;\mathbb R)$ complexifies to $H^{1,1}(X)$. Hence \begin{align*} \dim_{\mathbb R}V=h^{1,1}(X). \end{align*} Since $\dim_{\mathbb R}\mathbb R[\omega]=1$, we have \begin{align*} \dim_{\mathbb R}P=h^{1,1}(X)-1. \end{align*} Therefore the signature of $Q$ on $H^{1,1}(X;\mathbb R)$ is \begin{align*} (1,h^{1,1}(X)-1). \end{align*} This proves both the stated signature and the refined assertion that $\mathbb R[\omega]$ is the positive direction while the primitive real $(1,1)$-classes form a negative definite subspace. [/step]