My question is related to several notions of hyperbolicity, applied to Kähler manifolds (projective, in general). Kähler hyperbolicity was introduced in [this paper of Gromov's](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-33/issue-1/K%C3%A4hler-hyperbolicity-and-L_2-Hodge-theory/10.4310/jdg/1214446039.full). He calls a Kähler manifold *Kähler hyperbolic* if the lift to the universal cover of the symplectic form(=imaginary part of hermitian metric) is the differential of a bounded $1$-form. A $1$-form is bounded if its norm is pointwise absolutely bounded (the norm is induced by the pull-back of the metric to the tangent space).


As he notes in the introduction, this notion implies Kobayashi hyperbolicity. My question is regarding the converse, namely to find an example of a Kobayashi hyperbolic manifold that isn't Kähler hyperbolic.




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My guess is that an example of an algebraic variety with the properties described below exists (so it would be nice if an algebraic geometer could say a few things here).


For a Kähler hyperbolic manifold, Gromov proves that its Euler characteristic is $(-1)^n$, where $n$ is its *complex* dimension. He also shows a Kähler hyperbolic manifold has quasi-ample canonical bundle (namely, it has Kodaira dimension $n$). Another observation is that a Kähler hyperbolic manifold cannot have amenable fundamental group. So any algebraic variety which fails one of the above tests but still is Kobayashi hyperbolic would fit the bill.


For somewhat different reasons, I would be happier if this would be a projective variety, which also doesn't have any $(2,0)$ cohomology. As a side question, just to check my understanding, is it true that for a variety with no $(2,0)$ classes in cohomology, any homology class that comes from a map of a surface (pushing forward the fundamental class) can in fact be realized by some algebraic curve in the variety?



Hyperbolicity for algebraic varieties and relation to curves on them
Take a very general hypersurface $j \colon X \hookrightarrow \mathbb{P}^n$ of sufficiently high degree, $3 \leq n \leq 4$. Then $X$ is Kobayashi hyperbolic (Kobayashi actually conjectured that this is true for all $n$ and Siu recently outlined a strategy for proving this, see Diverietti's comment below).


On the other hand, Lefschetz hyperplane theorem says that the natural map


$\pi\_1(X) \stackrel{j\_\*}{\longrightarrow} \pi\_1(\mathbb{P}^n)$


is an isomorphism. Then $X$ is simply connected, in particular its fundamental group is amenable and so $X$ cannot be Kahler hyperbolic.


If $n=4$, the projective variety $X$ also satisfies your second request. If fact, again by Lefschetz theorem, see for instance [Dimca, Singularities and topology of hypersurfaces, Theorem 2.6 p. 151], for all $n \geq 4$ we also have an isomorphism


$H^2(\mathbb{P}^n) \stackrel{j^\*}{\longrightarrow} H^2(X)$.


Since $H^2(\mathbb{P}^n)=\mathbb{C}$, Hodge decomposition yields


$H^{2,0}(X)=H^{0,2}(X)=0, \quad H^{1,1}(X)=\mathbb{C}$.


If Kobayashi's conjecture were true, then the hypersurfaces $X$ would provide examples in all dimensions.


