Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the Riemannian manifold $\mathbb{R}^n$ and a smooth Riemannian metric $G:\mathbb{R}^n\rightarrow\mathbb{R}^{n\times{n}}$. We know that if $w_1I_n\leq{G}(x)\leq{w_2}I_n$ for some $w_1,w_2\in\mathbb{R}^+$, any $x\in\mathbb{R}^n$, and identity matrix $I_n$, then we have $\sqrt{w_1}\Vert{x_1}-x_2\Vert_2\leq\mathbf{d}_G(x_1,x_2)\leq\sqrt{w_2}\Vert{x_1}-x_2\Vert_2$, where $\mathbf{d}_G$ is the Riemannian distance function with respect to the metric $G$. Now, I am interested in the following equivalency between $\mathbf{d}_G$ and the usual Euclidean distance: $\alpha_1(\Vert{x_1}-x_2\Vert_2)\leq\mathbf{d}_G(x_1,x_2)\leq\alpha_2(\Vert{x_1}-x_2\Vert_2)$, where $\alpha_i:\mathbb{R}^+_0\rightarrow\mathbb{R}^+_0$ are continuous, strictly increasing functions, $\alpha_i(0)=0$ and $\alpha_i(r)\rightarrow\infty$ as $r\rightarrow\infty$, for $i=1,2$. I tried to find a characterization on the metric $G$ to reach to this version of equivalency, but I could not find anything. I am just wondering if there is any nice result for this kind of equivalency.

share|improve this question

1 Answer 1

This equivalence between distances (even on metric space more in general) is exactly the uniform equivalence: the distance $d$ and $d'$ (in your case the Euclidean distance) induce the same uniform structure. In other words, the identity map on $\mathbb{R}^n$ is uniformly continuous from $d$ to $d'$ and from $d'$ to $d$, with modulus of continuity respectively $\alpha_1^{-1}$ and $\alpha_2.$ The first equivalence you mentioned is indeed a special case, with linear moduli of continuity: it's the "Lipschitz equivalence", meaning that the identity map is bi-lipschitz; you may define analogously an Hoelder equivalence etc).

Also, it is not difficult to see that in your case $d_G$ may be uniformly, but not Lipschitz equivalent to the Euclidean distance (that is, the latter notion is really weaker than the former). It is possible to give simple sufficient conditions in terms of point-wise inequalities on $G(x)$, that ensure the uniform equivalence to the Euclidean distance. However, I don't see an easy characterization of this, the reason being that the distance $d_G$ is defined in terms of minimization on paths, and there may be uniform equivalence even if $G(x)$ has very high bumps at some points (say, a sequence with $G(x_k)\geq kI_n$).

share|improve this answer
I just want to add that I also assume $\mathbb{R}^n$ with respect to the metric $G$ is geodesically complete. Hence, for computing $\mathbf{d}_G$, I am just taking integral along the geodesic. –  Majid Dec 2 '10 at 1:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.