I wasn't sure how to make the title any more precise than that.

Let $(X,d)$ be a locally compact metric space. For any $\varepsilon>0$ let $\mathrm{Aut}_\varepsilon (X)$ be the subgroup of the automorphism group of $X$ generated by isometric automorphisms $f$ satisfying $d(x,f(x))<\varepsilon$ for every $x\in X$.

Suppose that for every $\varepsilon,\delta >0$ and any two $x,y\in X$ there exists an automorphism $f\in\mathrm{Aut}_\varepsilon (X)$ such that $d(f(x),y)<\delta$, does it follow that $X$ is bi-uniformly equivalent to $\mathbb{R}^n\times K$ for some finite $n$ and compact $K$? Where if $(Y,d_Y)$ and $(Z,d_Z)$ are metric spaces then $Y$ and $Z$ are bi-uniformly equivalent if there exists a uniformly continuous function $f:Y\rightarrow Z$ with uniformly continuous inverse.

  • 1
    $\begingroup$ Why do such automorphisms exist for $\mathbb R^n$? I mean, if $f(x)$ is near $x$ and $f(x)$ is near $y$, then $x$ and $y$ must be near too... $\endgroup$ – მამუკა ჯიბლაძე Feb 19 at 4:19
  • 2
    $\begingroup$ $f$ can in general be a finite composition of 'small' automorphisms. It doesn't need to be small itself. $\endgroup$ – James Hanson Feb 19 at 6:08
  • $\begingroup$ Oh I see thanks. So what is then this group for $\mathbb R^n$? All isometries? $\endgroup$ – მამუკა ჯიბლაძე Feb 19 at 6:18
  • 1
    $\begingroup$ On any metric space $X$, the subgroup generated by such isometries is included in the group $\mathrm{Aut}_{<\infty}(X)$ of isometries of bounded displacement, that is, those (surjective) self-isometries $f$ such that $\sup_xd(x,f(x))<\infty$. For $\mathbf{R}^n$ with any norm metric, this latter group is reduced to translations, and is equal to $\mathrm{Aut}_{\varepsilon}(X)$ for every $\varepsilon>0$. (The notation is unpractical, one should have a notation for the set of those isometries with displacement $\le\varepsilon$ and not only for the subgroup it generates.) $\endgroup$ – YCor Feb 19 at 7:33
  • $\begingroup$ @YCor Is there a standard notation for this? $\endgroup$ – James Hanson Feb 19 at 18:42

Your Answer

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

Browse other questions tagged or ask your own question.