1
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • 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
    Commented Mar 24, 2017 at 16:43
  • 1
    $\begingroup$ See also mduchin.math.tufts.edu/Current/heis-iumj.pdf $\endgroup$
    – Ian Agol
    Commented Mar 24, 2017 at 19:19
  • $\begingroup$ @AntonPetrunin thanks, this was a typo $\endgroup$
    – YCor
    Commented Mar 24, 2017 at 22:42
  • $\begingroup$ @YCor, Ian Agol : Thank you for introducing references. $\endgroup$
    – Hee Kwon Lee
    Commented Mar 26, 2017 at 17:12

0

Reset to default

You must log in to answer this question.