Sorry that my question is a bit vague. I have a space $X$ and two metrics that are both complete and induce the same topology on $X$. Certainly the metrics do not need to be Lipschitz equivalent, but is there any other reasonable definition to say that the metrics are "basically" equivalent?
Are they actually "basically" equivalent? -- in the sense "is there maybe a meaningful property of one metric that does hold for the other one"?
Let me illustrate where my problem stems from: As I read, one of the standard ways to define the space of Sobolev maps between two manifolds $M$ and $N$ is by embedding the target manifold $N$ by $\iota$ into an $\mathbb{R}^n$ for large $n$. We then say that a map $u\colon M \to N$ is Sobolev if $\iota\circ u\colon M \to \mathbb{R}^n$ is Sobolev in the classical sense. It is also clear that the Sobolev norm on $W^{k,p}(M;\mathbb{R}^n)$ induces a nice distance function on the space of Sobolev maps between $M$ and $N$.
I was trying to show that this definition does not depend on the choice of the embedding $\iota\colon M \to N$ used. I do manage to prove that neither the space itself nor the topology depend on this choice, but unfortunately the resulting distance functions are different and in fact not even bi-Lipschitz.
I guess my proper question is: What notion of equivalence intermediate between "induce the same topology" and "bi-Lipschitz" should I be looking for -- I do believe that the metrics should be similar -- for example all metrics are clearly complete...
Thank you.
