Note that $(\mathbb{Z}^2,d_W)$ where $d_W$ is word metric has asymptotic cone $$(\mathbb{R}^2,\| \ \|_1)=\lim_{t>0\rightarrow 0}\ t(\mathbb{Z}^2,d_W)$$

And Heisenberg group $\mathbb{H}^3$ has an asymptotic cone. It is a subRiemannian metric. But what is aymptotic cone of its discrete group ?

Thank you in advance.

  • 2
    $\begingroup$ You have to specify the generating subset, otherwise the asymptotic cone is not defined up to isometry; in particular for $\mathbf{Z}^2$ you implicitly mean the standard generating subset. For Heisenberg, see section 9 in Breuillard's paper "Geometry of groups of polynomial growth and shape of large balls", GGD, math.u-psud.fr/~breuilla/PolGrowth25.pdf. It will be a Finsler metric with respect to some norm on the 2-dimensional tangent plane at 1 of the contact structure, with some polyhedral 1-ball (depending on the choice of generating subset). $\endgroup$ – YCor Mar 24 '17 at 16:43
  • 1
    $\begingroup$ See also mduchin.math.tufts.edu/Current/heis-iumj.pdf $\endgroup$ – Ian Agol Mar 24 '17 at 19:19
  • $\begingroup$ @AntonPetrunin thanks, this was a typo $\endgroup$ – YCor Mar 24 '17 at 22:42
  • $\begingroup$ @YCor, Ian Agol : Thank you for introducing references. $\endgroup$ – Hee Kwon Lee Mar 26 '17 at 17:12

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.