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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.

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    $\begingroup$ 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. $\endgroup$ Jan 12 at 16:12
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    $\begingroup$ 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. $\endgroup$ Jan 12 at 16:44

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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.

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