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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

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

By Hodge decomposition, this implies

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

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

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.

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

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

By Hodge decomposition, this implies

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

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

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

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

By Hodge decomposition, this implies

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

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

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 (Siu actually conjectured that this is true for all $n$, 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

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

By Hodge decomposition, this implies

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

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

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

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

By Hodge decomposition, this implies

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

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