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Let $(X,d)$ be a metric space and $x,y \in X$. Assume that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric.

Is it true that there exists an isometry of $X$ sending $x$ to $y$?

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No. Let $x$ and $y$ be connected by an edge and let's use the graph distance as our metric. At $x$, connect paths of length $n$ for each $n\in\mathbb N$. At $y$, do the same, but also connect an infinite path.

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    $\begingroup$ Thank you. I guess i need to figure out the appropiate conditions i might have in my situation. $\endgroup$ – wspin Oct 6 '14 at 17:20
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    $\begingroup$ I upvoted your answer but you could do a bit more to help to interpret it; I mean to replace interpreting your answer with a clear and complete description. $\endgroup$ – Włodzimierz Holsztyński Oct 6 '14 at 18:01
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    $\begingroup$ For my part, I found this answer to be elegant and easy to understand. While thinking about it, I got a "model-theoretic flash": it reminds me of issues of "local isomorphism" which arise in that subject. I wonder if one can make a real connection there... $\endgroup$ – Pete L. Clark Oct 6 '14 at 20:59
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    $\begingroup$ I can't turn this example into a (Riemannian, say) manifold example. Is it me, or is this worth a new question? $\endgroup$ – Marco Golla Oct 6 '14 at 21:03
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    $\begingroup$ The subtle way in which infinity enters @Bjørn's example is very nice (I should have said it earlier, from the start--better late than not at all :-); and this example is only amplified by the later 5-point finite example. $\endgroup$ – Włodzimierz Holsztyński Oct 8 '14 at 18:22
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There is a 5-point example $\ X := \{x\ y\ a\ b\ c\},\ $ with (symmetric) metrics $\ d\ $ as follows:

$$d(x\ y) = d(a\ b) = 1$$

$$d(x\ a) = d(y\ b) = 2$$

$$d(x\ b) = d(y\ a) = 3$$

$$d(x\ c) = d(y\ c) = 6$$ $$d(a\ c) = 5\qquad\qquad d(b\ c) = 4$$

Oviously, $\ B_r(x)\ $ and $\ B_r(y)\ $ are isometric for every $\ r>0,\ $ while there is no isometry $\ f:X\rightarrow X\ $ for which $\ f(x)=y$.

REMARK 1  This example is ironic because while the respective balls are isometric, the isometry of the balls doesn't respect the centers. In this sense every bounded (and especially--finite) required example would be ironic.

REMARK 2  Number $\ 5\ $ is minimal.

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    $\begingroup$ Excellent. Sometimes the number 5 is more complicated than the number $\infty$. $\endgroup$ – Bjørn Kjos-Hanssen Oct 6 '14 at 22:31
  • $\begingroup$ @Bjørn, thank you. You have produced a nice quotation (proverb, wisdom, ...). $\endgroup$ – Włodzimierz Holsztyński Oct 6 '14 at 22:35

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