This question is a natural follow-up of this other question, asked earlier today by wspin.

Let's say that a metric space $(X,d)$ has two poles if:

  • there are two distinct points $x$, $y$ such that for every $r\ge 0$ the open balls $B_r(x)$ and $B_r(y)$ are isometric;
  • there is no isometry $f$ of $(X,d)$ such that $f(x)=y$.

Bjørn Kjos-Hanssen gave a very pretty example of a space with two poles (it's the metric space associated to an infinite graph). I've tried to turn this example into a surface by "inflating" it (imagine it in the 3-space, and take the boundary of a regular neighourhood), but failed. Hence my question:

Are there connected Riemannian manifolds with two poles?

If one drops the connectedness assumption, then Włodzimierz Holsztyński's answer to the question linked above is a 0-dimensional example. One can also ask for more: compactness, trivial $\pi_1$, curvature restrictions, etc., but I'll stick with the broader question for now.

The rough meta-question in the back of my mind is: "How nice can a bipolar space be?", so answers/comments in this direction are very welcome.

  • $\begingroup$ If the connectedness of the manifold were not required then we could have examples being unions of 5 spheres of any fixed dimension. $\endgroup$ Oct 6, 2014 at 22:54
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    $\begingroup$ Do we a priori require that isometry between $B_r(x)$ and $B_r(y)$ maps $x$ into $y$? $\endgroup$ Oct 6, 2014 at 23:33
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    $\begingroup$ @FedorPetrov no -- note that in Włodzimierz Holsztyński's answer there is no such isometry $\endgroup$ Oct 7, 2014 at 0:14
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    $\begingroup$ If we do a priori require that the isometry $\phi_r$ between $B_r(x)$ and $B_r(y)$ maps $x$ into $y$, and if balls are relatively compact, the answer is no. Indeed, by the Ascoli-Arzelà theorem, a subsequence of the isometries $\phi_{r_j}$ converges uniformly on compact sets to an isometry $\phi:X\to X$ mapping $x$ to $y$ (surjectivity follows by compactness again). $\endgroup$ Oct 7, 2014 at 8:59
  • $\begingroup$ @Marco Let $ M_{g}$ be a compact surface with constant negative curvature. the constant curvature implyies local isometry but global isometries are rare since the isometry group has at most 84(g-1) elements. So this gives an example. Am I mistaken? $\endgroup$ Aug 10, 2016 at 17:09

1 Answer 1


The answer is no if $M$ is assumed to be complete:

Let $x, y \in M$ such that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric via $$f_r : B_r(x) \to B_r(y).$$ Set $R := \inf \{ r \mid B_r(x) = M\}$ (here $R = \infty$ if $M$ is unbounded). First I claim that

For all $0 < r < R$ we have $f_r(x) = y$:

Assume on the contrary that $f_r(x) \neq y$. Let $z \in M \setminus \overline B_r(y)$ and $\gamma_1 : [0,l] \to M$ be a minimal arc length geodesic from $y$ to $z$. Then $\gamma_1(r) \in \partial B_r(y)$. Also let $\gamma_2$ be a minimal geodesic from $f_r(x)$ to $\gamma_1(r)$. Since $f_r^{-1}(\gamma_1(r)) \in \partial B_r(x)$ (here we extend $f$ to $\overline B_r(x)$ by metric completion), $f_r^{-1}(f_r(x)) = x$, $\gamma_2$ is minimal and $f_r$ is an isometry it follows that $\gamma_2$ has length $r$. But then also the curve obtained by concatenating $\gamma_2$ and ${\gamma_1}_{\vert [r,l]}$ is minimal between $f_r(x)$ and $z$. This is a contradiction, since geodesics do not branch.

Now, consider a sequence $f_{r_n}$ of such isometries with $f_{r_n}(x) = y$ and ${r_n} \to R$ for $n \to \infty$. After passing to a subsequence we may assume that $f_{r_n}$ converges to an isometry $f : M \to M$ with $f(x) = y$ (the isometries are determined by their differentials at $x$, which converge after some subsequence by Ascoli-Arzelà (or simply by linearity)).

Two remarks:

  • This argument holds more generally for complete length spaces with curvature bounded below.

  • I am not sure whether the completeness assumption is necessary.


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