4
$\begingroup$

What are examples of finitely generated groups $\Gamma$ and $\Lambda$ such that the metric space $\Lambda$ embeds into $\Gamma$ quasi-isometrically but such that $\Lambda$ is very much not a subgroup of $\Gamma$?

Attempt 1 at ``$\Lambda$ is very much not a subgroup of $\Gamma$'': $\Gamma$ has no subgroup commensurable to $\Lambda$.

Edit: the question is answered in the comments by YCor and Ian Agol. More examples are welcome. Specifically, I would be interested in $\Gamma$ that have many such $\Lambda$ (in particular, further $\Lambda$ that are neither QI to $\Gamma$ nor $\mathbb{Z}$).

Context: in comparing group theory to geometry one often looks at whether homomorphic embeddings are also quasi-isometric embeddings. This question is about the other way around. There are probably tons of examples, the problem is to verify that $\Lambda$ is very much not a subgroup of $\Gamma$. For example, any two points of the Cantor space $\partial F_2$ define a quasi-isometric embedding of $\mathbb{Z}$ into $F_2$ and only very few (countably many) of these are anywhere close to being homomorphic images, yet $\Gamma$ does have a subgroup isomorphic to $\mathbb{Z}$.

$\endgroup$
  • 7
    $\begingroup$ (1) Two cocompact lattices in the same simple Lie group of rank $\ge 2$, or rank 1 and superrigid (2) Burger Mozes, or irreducible lattice in $SL_2(Q_p)^2$, vs product of two free groups; similarly, product of two surface groups vs irreducible lattice in $SL_2(R)^2$ (3) lamplighter on $C_p$ and $C_q$ for $p,q$ distinct primes (not QI, but each QI-embeds into the other one), etc. (4) two lattices in the same nilpotent Lie group (a real form having two rational forms: this exists in dim$\ge 6$, for instance this can be found inside the product of two Heisenberg groups); two lattices in SOL. $\endgroup$ – YCor Oct 22 '18 at 13:39
  • 7
    $\begingroup$ (5) every infinite f.g. group $\Gamma$ has a QI-embedded $\mathbf{Z}$, but if $\Gamma$ is torsion (e.g., Golod-Shafarevich, Grigorchuk, Burnside, etc) then it has no subgroup commensurable to $\mathbf{Z}$. Also $\Gamma\wr\Gamma$ then has a QI-embedded $\mathbf{Z}^d$ for every $d\ge 0$, but no subgroup commensurable to $\mathbf{Z}^d$ for any $d\ge 1$. (6) every non-amenable fg group, or virtually solvable group of exponential growth, has a QI-embedded copy of $F_2$, but there's not always a free subgroup (initial examples by Olshanskii; more recent examples by Monod acting on the interval). $\endgroup$ – YCor Oct 22 '18 at 13:55
  • 4
    $\begingroup$ In addition to @YCor's examples, there are hyperbolic planes embedded in sol, which gives quasi-isometric embeddings of surface groups. mathoverflow.net/a/308560/1345 More generally, Fisher-Whyte have explored q.i. embeddings between symmetric spaces. arxiv.org/abs/1407.0445 $\endgroup$ – Ian Agol Oct 22 '18 at 22:29
  • $\begingroup$ As regards your edit, several of the examples already given have $\Lambda$ neither QI to $\mathbf{Z}$ nor to $\Gamma$. $\endgroup$ – YCor Oct 23 '18 at 9:56
  • 2
    $\begingroup$ I have further examples, but it's a bit endless... $\endgroup$ – YCor Oct 23 '18 at 11:40

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.