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Let $(X,d)$ be a metric space, whose metric $d$ is not known.

Let $G=(f,\circ)$ be its group of isometries (that is, distance preserving functions $f:X\rightarrow X$, with the usual function composition as group operation).

Is $d$ uniquely specified by $G$?

If the answer is yes: how can we explicitely know the form of $d$ from $G$?

If the answer is no: under what simplifying assumptions will $d$ be specified by $G$?

user420196
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    Imagine $n$ points in a sufficiently high-dimensional Euclidean space, all equidistant from one another, and then put $k$ more points in "random places" so that no two of them have the same distances to the fixed $n$ points. Then for any $k$, the group of isometries $G$ will be $S_n$. – hunter Feb 26 '17 at 19:30

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No. If $d$ is a metric, then so is $d/(1+d)$. There are probably many functions other than $x\mapsto x/(1+x)$ that could be composed with a metric to produce another metric.

So the best you might hope for, is that two metrics with the same isometries are related by composition with some function. Whether that is true or not, I don't know.

Harald Hanche-Olsen
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    Maybe I'm missing something, but does this new metric space have the same isometries as the old space? – pjs36 Feb 26 '17 at 15:51
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    Sure, because not only is the new metric a function of the original, but the original is a function of the new one! – Harald Hanche-Olsen Feb 26 '17 at 15:55
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    Oh, and by the way, I expect that for many metric spaces, the isometry group is trivial. That makes the isometry group a rather poor determinant of the metric. – Harald Hanche-Olsen Feb 26 '17 at 15:57
  • I see, thank you for that! – pjs36 Feb 26 '17 at 15:59
  • Thanks for the answer, I wanted to think of it overnight before commenting back. Then, this implies, e.g. that the euclidean metric is not the only one that has (and only has) the euclidean group as a group of isometries, which is a bit striking to me!

    Btw, as pjs36 asks, the key for $d$ and $d/(1+d)$ to have the same isometries is that the map $d/(1+d)$ is invertible (in $[0,\mathbb{R}) \leftrightarrow [0,1)$ ), isn't it?

     – user420196 Feb 27 '17 at 11:51
  • You're welcome, and yes. – Harald Hanche-Olsen Feb 27 '17 at 16:44
  • Sorry to come back to this. I just read that – user420196 Mar 08 '17 at 17:04