# Examples of quasi-negative but not negative holomorphic sectional curvature

Let $(X,\omega)$ be a compact Kähler manifold and call $\operatorname{HSC}_{\omega}(x,[v])$ the holomorphic sectional curvature of the Chern connection of $\omega$ at the point $x\in X$ in the direction of $v\in T_{X,x}\setminus\{0\}$. Since $\omega$ is Kähler, it coincides with the usual (Riemannian) sectional curvature of the complex $2$-plane spanned by $(v,Jv)$.

Now, $\operatorname{HSC}_{\omega}$ is said to be quasi-negative if it is non positive at every point and every direction, and there exists at least one point $x_0\in X$ such that $\operatorname{HSC}_{\omega}(x_0,[v])<0$ for every $v\in T_{X,x_0}\setminus\{0\}$.

Since $X$ is compact, if $\operatorname{HSC}_{\omega}$ is strictly negative, then it is bounded away from zero and it is classically known that $X$ is Kobayashi-hyperbolic.

Also, it has been recently shown that quasi-negative holomorphic sectional curvature (and therefore also strict negativity, but this was shown a little bit earlier by Wu and Yau in the projective case and extended to the Kähler setting by Tosatti and Yang) implies the ampleness of the canonical bundle.

Question. Does anybody know an example of a compact Kähler manifold which admits a Kähler metric with quasi-negative holomorphic sectional curvature but that is not Kobayashi hyperbolic?

Of course, such an example would have dimension at least two and would also give an example of a compact Kähler manifold admitting a Kähler metric with quasi-negative holomorphic sectional curvature but with no (even) Hermitian metrics with negative holomorphic sectional curvature.

Here is where I naturally stared to look at, but I wasn't able (or didn't have enough perseverance...) to push the computations till the end.

(Possible beginning of an) Example. Take a three dimensional abelian variety $Y$ containing a (say smooth) elliptic curve $E$. Now, cut $Y$ with a high degree projective hypersurface containing $E$, and call this section $X$. If the hypersurface is general enough and its degree is high enough, some refined version of Bertini's theorem ensures that $X$ is in fact smooth. Since $X$ contains an elliptic curve, it cannot be hyperbolic. Now, endow $X$ with the Kähler metric $\omega$ induced by the flat metric on $Y$. Thanks to the decreasing property of the holomorphic (bi)sectional curvature, we surely have $\operatorname{HSC}_{\omega}\le 0$. Moreover, by adjunction, $K_X$ is ample. Can we find a point of $X$ where $\operatorname{HSC}_{\omega}$ is strictly negative in all directions?

Please observe that there might be some subtleties to overcome: for instance, if one takes the embedding of a compact Riemann surface of higher genus $C$ into its Jacobian, then the induced metric in this case is not necessarily strictly negative! If the Riemann surface is hyperelliptic, then this induced metric will have zero curvature exactly on the Weierstrass points of $C$...

Of course any other (hopefully even more trivial...) example (or counterexample...) is very welcome!

• Sorry if I said anything wrong, try to understand the example you proposed. the 3-dim abelian variety $Y$ has to be isogenous to the produce of $E$ and an abelian surface. Could you give some details on how to choose a high degree hypersurface to ensure its insection with $Y$ is a surface $X$ of $K_X>0$, and can we always realize such a $X$ as some Theta divisor of $Y$? – Bo_Y Jan 8 '17 at 14:19
• @Bo_Y Hi. I am thinking t the abelian variety as embedded in some $\mathbb P^n$. Then I cut it with a projective hypersurface, which is an ample divisor. The intersection is thus an ample divisor in the abelian variety. If smooth, its canonical bundle is just the restriction of the line bundle defined by this divisor onto itself, by adjunction. Therefore it is ample. As for your second question, I actually don't know, I should think about it. Why do you ask? – diverietti Jan 9 '17 at 13:40
• First, I thought there are no abelian varieties whose theta divisor can contain elliptic curves, I might be wrong here. I would like to know how to pick the degree of your hypersurface to ensure it contains elliptic curves. Second, in your example, assume other property you mentioned holds, we can see the bisectional curvature of the induced metric has lots of zeros in the following sense: given any point, and any direction $u$, there exists a direction $v$ such that $R_{u \bar{u} v \bar{v}}=0$, this can be derived since second segre form $c_1^2-c_2$ is 0. – Bo_Y Jan 10 '17 at 8:06
• The existence of a projective hypersurface of high degree cutting the abelian variety along a smooth surface and containing the fixed elliptic curve is guaranteed by the generalized Bertini's theorem contained in "Bertini theorems for hypersurface sections containing a subscheme" by Kleiman and B. Altman. – diverietti Jan 12 '17 at 15:59
• Thanks for the remark on the bisectional curvature. But I am not convinced that it suggests the holomorphic sectional curvature tends to be zero in a lot of points: think at the bidisk with the product Poincaré metric... – diverietti Jan 12 '17 at 16:01