4
$\begingroup$

Let $(X,d)$ be a metric space. There are always other metrics on $X$ which generates the same topology, and have the same isometry groups, for instance $\tilde d =\sqrt d$. (The same will be true for any injective continuous function $f:\mathbb{R}^{\ge 0}\to \mathbb{R}^{\ge 0}$, taking zero to itself and which satisfying some constraints to ensure triangle inequality).

Question: Is there any example when $\text{Iso}(X,d)$ determines $d$ up to "functional-dependence"? (i.e, any other metric $\tilde d$, which generates the same topology as $d$, and with the same isometry group is of the form of $f\circ d$ for some suitable $f$)

$\endgroup$
3
$\begingroup$

It turns out the answer is positive. Take for instance $\mathbb{R}^n$ with the Euclidean metric $d$. Let $\tilde d$ be any other metric on $\mathbb{R}^n$ satisfying $\text{Iso}(\mathbb{R}^n,d)=\text{Iso}(\mathbb{R}^n,\tilde d)$. We will show $\tilde d(x,y)$ depends only on $d(x,y)$.

Let $c \in \mathbb{R}^{> 0}$, and consider any two pairs of points of distance $c$ of each other, i.e: $a,b,\tilde a,\tilde b \in \mathbb{R}^n$ such that $d(a,b)=d(\tilde a,\tilde b)=c$. Then there exists an isometry $\phi$ of $(\mathbb{R}^n,d)$ such that $\phi(a)=\tilde a,\phi(b)=\tilde b$. (First translate by $\tilde a -a$ and then compose with a suitable orthogonal linear map).

By our assumption, $\phi$ is also an isometry of $(\mathbb{R}^n,\tilde d)$, so: $\tilde d(a,b)=\tilde d(\tilde a,\tilde b):=f(c)$

Thus, $\tilde d(x,y)$ indeed depends only on $d(x,y)$, as required.

This proofs works for any metric space whose isometry group is 2-transitive, i.e it acts transitively on pairs of points with identical distances. In particular, it works for the round sphere, and for any finite dimensional normed space whose isometry group acts transitively on the unit sphere. (It is known that any such normed space is in fact an inner product space*, so this is essentially just the case of Euclidean $\mathbb{R}^n$ again)

I wonder if 2-transitivity of $\text{Iso}(X,d)$ is necessary for determining the metric in such a way. I suspect it's not.


*This follows essentially because Isometry group of a norm is always contained in some Isometry group of an inner product. For details , see here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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