Let $X$ be an $n$-dimensional complex manifold, let $\omega$ be a Kahler metric on $X$ and let $R$ be the $(4,0)$ curvature tensor of $\omega$. We can simplify the tensor $R$ in different ways, two of which are:

  • The holomorphic bisectional curvature is $b(\xi,\eta) = R(\xi,\eta,\xi,\eta)/|\xi|^2|\eta|^2$.

  • The holomorphic sectional curvature is $h(\xi) = R(\xi,\xi,\xi,\xi)/|\xi|^4$.

Now, what exactly is the holomorphic sectional curvature?

To make sense of the question, consider the Ricci curvature of $\omega$. In the Kahler case, this can be defined as the curvature form of the Hermitian metric that $\omega$ defines on the canonical bundle $K_X$. That's a quite nice geometric object that can be interpreted in algebro-geometric ways.

We can make similar sense of the holomorphic bisectional curvature. Consider the projectivized bundle $\pi:\mathbb P(T_X) \to X$. It admits the tautological bundle $\mathcal O(-1) \hookrightarrow \pi^* T_X$ and $\omega$ defines a Hermitian metric on the tautological bundle, whose curvature form at a given point is basically (modulo abuse of notation, sign errors) $$ i\Theta = -|\cdot|^2 \pi^* b + \omega_{\mathrm{FS},\mathbb P(T_X)}, $$ so the negativity of the holomorphic bisectional curvature controls the positivity of $T_X^*$.

By contrast, I know no similar way of thinking about the holomorphic sectional curvature, other than perhaps "the thingy that makes the Schwarz lemma work", which is a much more analytic approach to the situation. It of course has the same average as the Ricci curvature (the scalar curvature), both are controlled by the holomorphic bisectional curvature and neither controls the other, but is that the end of the story? Is the holomorphic sectional curvature a purely analytic object that cannot be attached to any bundle or sheaf?

  • $\begingroup$ Over a Riemann surface, the holomorphic sectional curvature coincides with the Gaussian curvature. $\endgroup$
    – user94803
    Jul 18, 2016 at 8:45

2 Answers 2


Maybe you already were aware of that, or maybe it really doesn't answer to your question, but I'll try anyhow...

Take a look at this, Subsection 7.5 on page 39. The construction you talk about in your question was not so far from the answer!

What you have to do is to check the negativity of the curvature of the induced metric on $\mathcal O(-1)$, but just along the "contact" subbundle (what Demailly calls $V_1$).

I don't know if this looks "algebraic" to your taste, but to me it is more than "the thingy that makes the Schwarz lemma work".

If you want more details, don't hesitate to ask!


  • $\begingroup$ Small remark: The holomorphic sectional curvature makes the Schwarz lemma work if the metric is Kähler. In the Hermitian setting, at present, you need a slight strengthening of the holomorphic sectional curvature -- what is called the "real bisectional curvature" -- which reduces to the holomorphic sectional curvature in the Kähler case. Of course, this depends on what is meant by "Schwarz lemma". $\endgroup$
    – user105074
    Dec 5, 2020 at 21:50

This is perhaps not algebraic either, but it may lead to an algebraic formulation: By an old result of Wu, the holomorphic sectional curvature of $(M,g)$ (at a point $p$ in the direction of some tangent vector $v$) is the Gauss curvature of the induced metric $\gamma^{\ast} g$ on the immersed curve $\gamma(\mathbb{D}) \subset M$. Here, $\gamma : \mathbb{D} \to M$ is the analytic disk which supremizes(?) the Gauss curvature: $$K_{\gamma^{\ast}g} \ = \ -\frac{1}{\gamma^{\ast} g} \frac{\partial^2 \log(\gamma^{\ast}g)}{\partial t \partial \overline{t}},$$ where $t$ is the coordinate on $\mathbb{D}$ and is subject to the condition $\gamma(0) =p$ and $\gamma'(0)=v$.

This is analogous to: ``plurisubharmonic = subharmonic on each complex line".


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