Given a smooth, compact complex surface with ample canonical bundle satisfying $c_1^2=3c_2$, is it true that every Kahler class is a multiple of $c_1$? This seems to be the case for fake projective planes, because $b_2=1$.
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The answer is no.
A counterexample is provided by the so-called Cartwright-Steger surface $X$, namely a surface of general type with $p_g(X)=q(X)=1$ and $K_X^2=9$.
Such a surface has Picard rank $3$, and moreover $\mathrm{NS}(X)=H^{1, 1}(X)$, in other words all Hodge classes in $X$ are algebraic. In particular, the vector space $$H^{1, 1}_{\mathbb{R}}(X):=H^{1,1}(X) \cap H^1(X, \, \mathbb{R})$$ has real dimension $3$, because its complexification is $H^{1, 1}(X).$
On the other hand, by general facts the Kähler cone $\mathcal{K}_X$ of $X$ is an open, convex cone in $H^{1,1}_{\mathbb{R}}(X)$, hence it cannot be of dimension $1$.
answered Jul 19, 2016 at 13:44
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6$\begingroup$ In fact, for a surface with $c_1^2=3c_2$, an easy computation gives $h^{1,1}=p_g+q+1$. So the answer is no for any such surface which is not a fake projective plane. $\endgroup$– abxCommented Jul 19, 2016 at 13:51