16
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ If the connectedness of the manifold were not required then we could have examples being unions of 5 spheres of any fixed dimension. $\endgroup$ Commented Oct 6, 2014 at 22:54
  • 1
    $\begingroup$ Do we a priori require that isometry between $B_r(x)$ and $B_r(y)$ maps $x$ into $y$? $\endgroup$ Commented Oct 6, 2014 at 23:33
  • 1
    $\begingroup$ @FedorPetrov no -- note that in Włodzimierz Holsztyński's answer there is no such isometry $\endgroup$ Commented Oct 7, 2014 at 0:14
  • 1
    $\begingroup$ 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). $\endgroup$ Commented 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? $\endgroup$ Commented Aug 10, 2016 at 17:09

1 Answer 1

12
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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