[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 $Q$.
[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]
[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]
