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I am trying to find a Riemannian metric $G:\mathbb{R}^3\rightarrow\mathbb{R}^{3\times{3}}$ on the manifold $\mathbb{R}^3$ such that $G$ is not uniformly positive definite, and there is no isometry $\phi:\mathbb{R}^3\rightarrow\mathbb{R}^3$ satisfying $G=\phi^*I_3$, where $I_3$ is the identity matrix. Moreover, I want $(\mathbb{R}^n,G)$ be geodesically complete. It seems easy to find $G$, but I could not succeed to find any.

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The formulation seems to be wrong --- If you need a metric which is not isometric to Euclidean take one with non-zero curvature, one can make it "not uniformly positive definite" by perturbing parametrization... –  Anton Petrunin Oct 26 '10 at 2:23

1 Answer 1

Try $ds^2 = |d{\bf x}|^2/(1+|{\bf x}|^2)$. The resulting metric space is complete since the "horizon" is infinitely far away. But maybe I have misunderstood the notion "not uniformly positive definite".

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