It is a theorem of Greene and Wu that a complete, simply-connected Kaehler manifold of everywhere nonpositive sectional curvature is a Stein manifold. I am curious about what kinds of additional assumptions one would need to make in order for something like the converse to hold. That is:

I'd like to know when, given a Stein manifold $X$, we are able to deduce the existence of a complete metric of nonpositive sectional curvature on $X$ (I don't need the metric to be Kaehler). In case it helps, I am especially interested in the case where $X$ is a subvariety of a hermitian symmetric domain of noncompact type.

I have some thoughts on this. For example, let's start with the observations that every Stein manifold can be embedded in $\mathbb{C}^N$ for some $N$, and that holomorphic sectional curvature decreases on complex subvarieties. Then the Euclidean metric on $\mathbb{C}^N$ induces a complete metric of nonpositive holomorphic sectional curvature on a Stein manifold $X\subset \mathbb{C}^N$. The wrinkle here is that nonpositive holomorphic sectional curvature is weaker than nonpositive sectional curvature.

  • 6
    $\begingroup$ I don't know anything about Stein manifold, but if many topologies can be realized, then the Cartan-Hadamard theorem will give you great constraints (any simply connected manifold admitting a metric with non-positive curvature must be diffeomorphic to $\mathbb{R}^n$). $\endgroup$ Sep 13, 2014 at 15:53
  • $\begingroup$ Ah, of course. Thank you for pointing that out. I'll admit my Riemannian geometry skills are a little bit rusty. $\endgroup$
    – Kevin
    Sep 13, 2014 at 16:14

1 Answer 1


Complete, simply connected manifolds of non-positive sectional curvature are diffeomorphic to $R^n$, by Cartan-Hadamard theorem. Conversely, if it's $R^n$, you can put the metric of negative curvature on it.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .