# Can a complete manifold have both positive and negative curvature?

Let $$M$$ be a smooth manifold. Can there exist two Riemannian metrics $$g$$ and $$h$$ on $$M$$ such that:

1. $$(M,g)$$ and $$(M,h)$$ are complete.
2. The curvature of $$g$$ is everywhere positive.
3. The curvature of $$h$$ is everywhere negative.

There are many different notions of curvature (sectional, Ricci, scalar) and I didn't say which one I wanted to look at, but I'd be happy with an answer in dimension 2 where they are all the same. If the manifold is compact this isn't possible by Gauss-Bonnet and friends, so let's assume $$M$$ is not compact.

This came up in the context of a problem about the existence of certain holomorphic functions on the unit disk. I thought the problem lead to having a complete positively curved metric on the disk, which is obviously not possible because the disk also has the negatively curved Poincaré metric, and then I couldn't figure out why that was obvious.

• This is impossible for compact manifolds in all dimensions (and sectional curvature). On the other hand, every compact manifold of dimension at least 3 admits a metric of negative Ricci curvature. Commented Jan 12 at 16:12
• Following on from Moishe Kohan's comment, if a compact manifold of dimension at least 3 admits a metric of positive scalar curvature, then it also admits a metric of negative scalar curvature (in fact every function arises as the scalar curvature of some metric in this case). This is due to Kazdan and Warner, see here. Commented Jan 12 at 16:44

Consider $$\mathbb{R}^2$$ embedded in $$\mathbb{R}^3$$ via the map $$\varphi : \mathbb{R}^2 \to \mathbb{R}^3$$, $$(x_1, x_2) \mapsto (x_1, x_2, \sqrt{1 + x_1^2 + x_2^2})$$. The image of $$\varphi$$ is $$\{(y_1, y_2, y_3) \in \mathbb{R}^3 \mid y_1^2 + y_2^2 - y_3^2 = -1, y_3 > 0\}$$; i.e. one connected component of a hyperboloid of two sheets.
If $$\hat{g}$$ denotes the Euclidean metric on $$\mathbb{R}^3$$, then $$g := \varphi^*\hat{g}$$ has non-constant positive scalar curvature. On the other hand, if $$\hat{h}$$ denotes the Minksowski metric on $$\mathbb{R}^3$$ (with signature $$++-$$), then $$h:=\varphi^*\hat{h}$$ is a Riemannian metric of constant scalar curvature $$-1$$, i.e. a hyperbolic metric.