I am looking for an example of a compact complex manifold with negative sectional (not holomorphic) curvature which is not Kählerian. Can such an example exist?
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1$\begingroup$ Aside from Riemann surfaces of genus at least two, are you aware of any examples of compact complex manifolds (Kähler or non-Kähler) which admit a metric of negative sectional curvature? $\endgroup$– Michael AlbaneseCommented Sep 27, 2020 at 3:09
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1$\begingroup$ Any Kahler manifold with constant negative holomorphic sectional curvature (and complex dimension at least 2) has negatively quarter-pinched sectional curvature, so you can obtain compact Kahler examples by taking compact quotients of complex hyperbolic space. $\endgroup$– Gabe KCommented Sep 27, 2020 at 4:28
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$\begingroup$ @GabeK: Thanks. For anyone else who is wondering where the quarter-pinching comes from, it follows from the formula in this answer. $\endgroup$– Michael AlbaneseCommented Sep 28, 2020 at 17:55
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$\begingroup$ In particular, if $X$ and $Y$ are unit length and orthogonal, then $K(X, Y) = -\frac{1}{4}[1 + 3g(X, JY)^2]$. The inequality then follows as $0 \leq g(X, JY)^2 \leq 1$. $\endgroup$– Michael AlbaneseCommented Apr 22, 2023 at 13:42
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1 Answer
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If the compact complex manifold $X$ has a Hermitian metric $\omega = g(J\cdot, \cdot)$ of negative sectional curvature for the Chern connection, then the first Chern Ricci curvature $$\text{Ric}^{(1)}(\omega) =_{\text{loc}} \sqrt{-1} g^{k \bar{\ell}} R_{i \bar{j} k \bar{\ell}} dz^i \wedge d\bar{z}^j = - \sqrt{-1} \partial \bar{\partial} \log(\omega^n)$$ is negative, and hence, $\eta := - \text{Ric}^{(1)}(\omega)$ defines a closed real $(1,1)$--form that is positive-definite, i.e., a Kähler form. Hence, the manifold must be Kähler. In fact, since the first Chern Ricci curvature is the curvature form of a Hermitian metric on the anti-canonical bundle, negative (Chern) sectional curvature of a Hermitian metric forces the canonical bundle $K_X$ to be a positive line bundle. Kodaira's embedding theorem then implies that $X$ must be projective.
If the compact complex manifold admits a Riemannian (not compatible with the complex structure) metric of negative sectional curvature, then it is not at all clear that such a manifold must be Kähler. It is known that if a compact Kähler manifold is homotopic to a compact Riemannian manifold with negative sectional curvature then it has ample canonical bundle (and is therefore, projective), see this paper.
Some further results: Carlson--Toledo showed that if $N$ is a compact manifold of constant negative curvature and real dimension $4$, then $N$ has no complex structure.
