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Can a topological vector space have two metrics both of which give rise to same topology as the space has, but one is translation invariant but the other is not?

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up vote 6 down vote accepted

Yes. $\mathbb R$ with $d(x,y) = |x^3-y^3|$ has the same topology as with the standard metric.

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Yes. Take $\mathbb{R}$ with $d_1(x,y) = |x-y|$ and $d_2(x,y) = |\arctan x - \arctan y|$.

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A further condition you might want is that the identity map be uniformly continuous in both directions. (For example, this will mean that you have the same Cauchy sequences in either case.) For that, you could choose a uniformly continuous bijective map $f:\mathbb R\to\mathbb R$ with uniformly continuous inverse, like $f(x)=\frac{x^3}{x^2+1}$ or $f(x)=xe^{1/(x^2+1)}$, and then define a new metric on $\mathbb R$ by $d(x,y)=|f(x)-f(y)|$.

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Ah..this seems a nice general way! – Nirakar Neo Nov 23 '12 at 4:37

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