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Is there any example of a Kleinian group (acting on $\mathbb{H}^n$, $n \ge 3$) that contains a finite index isomorphic copy of itself? Here I don't consider Kleinian groups that only have parabolic elements.

My motivation is that in Euclidean spaces we have crystallographic groups that contain only translations. Such a group is isomorphic to a subgroup of finite index. Tilings of $\mathbb{R}^n$ coming from actions of those group will look the same on larger scale.

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    $\begingroup$ 1) Yes, every group admits itself as a subgroup of finite index. Possibly you mean a proper subgroup... 2) Yes with a proper subgroup, namely the free group on countably many generators. Possibly you restrict to finitely generated groups? $\endgroup$ – YCor May 5 '15 at 19:51
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    $\begingroup$ Yes I meant a proper subgroup of a finitely generated group. $\endgroup$ – Danny Nguyen May 5 '15 at 19:58
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    $\begingroup$ It you restrict to lattices in $PSL_2(\mathbf{C})$ (finite volume fundamental domain, e.g. compact), Mostow rigidity theorem implies that the lattice is not isomorphic to a proper subgroup of finite index. Indeed such an isomorphism would extend to an automorphism of $PSL_2(\mathbf{C})$, but automorphisms of the latter preserve the volume and cannot map a lattice to a proper subgroup of finite index. $\endgroup$ – YCor May 5 '15 at 23:19
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    $\begingroup$ In 3-dims. (discrete groups in $PSL_2(\mathbb{C})$), either a finitely generated group has finite covolume, in which case @YCor's comment applies, or it is infinite covolume, in which case if it is non-elementary, the Euler characteristic is $<0$ (by the Scott core theorem, finitely generated groups are finitely presented, so Euler characteristic makes sense), and therefore it cannot be isomorphic to a finite-index subgroup of itself. So Rivin's answer is overkill at least in 3D (co-Hopfian is much stronger than what you're asking for). $\endgroup$ – Ian Agol May 6 '15 at 4:14
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    $\begingroup$ @YCor, you also want Selberg's lemma (to show that they're virtually torsion-free) and the Sphere theorem (to show that the quotient is aspherical). Though perhaps the Sphere theorem is overkill... $\endgroup$ – HJRW May 6 '15 at 14:21
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Let us say that an abstract group $\Gamma$ is self-contained (I just made up this terminology) if $\Gamma$ is isomorphic to its proper subgroup of finite index.

Next, each discrete group of isometries of the euclidean space $E^n$ is also a Kleinian group (it is a discrete subgroup of $Isom(H^{n+1})$), since the isometry group $Isom(E^n)$ embeds into $Isom(H^{n+1})$. Thus, you already know examples of self-contained Kleinian groups (they come from discrete isometry groups of $E^n$). All these examples are virtually abelian (contain a free abelian subgroup of finite index), they are also known as elementary Kleinian groups, let us ignore them.

Now, there are self-contained nonelementary Kleinian subgroups $\Gamma$ of $Isom(H^2)$, namely each free group $F$ of infinite rank is self-contained, just take the kernel of an epimorphism of $F$ to $Z/n, n>1$. If $F$ is countable, it embeds as a discrete subgroup into $Isom(H^2)$. Using a similar construction, one can also construct self-contained infinitely generated Kleinian groups in $Isom(H^3)$ which are not free. If you insist on finite generation, then there are no finitely-generated self-contained Kleinian subgroups of $Isom(H^3)$, this is proven as follows. Consider, for simplicity, torsion-free Kleinian groups. Every such group is isomorphic to the fundamental group of a compact hyperbolizable 3-manifold (possibly with boundary). Now note that if $M$ is a compact hyperbolizable 3-dimensional manifold then either $M$ is homotopy-equivalent to a complete hyperbolic 3-manifold of finite volume or $\chi(M)<0$. In the latter case, no $d$-fold ($d>1$) finite cover $M'$ of $M$ is homotopy-equivalent to $M$ since $\chi(M')=d \chi(M)\ne \chi(M)$. If $\chi(M)=0$, then volume of $M$ is a topological invariant and you use the same argument as above using volume instead of the Euler characteristic.

Once you go to dimension $\ge 4$ and ask about any property of finitely generated Kleinian groups, not much is known (which is specific to finitely generated Kleinian groups as opposite to general Kleinian groups and general finitely generated matrix groups). In particular, it is unknown if there are self-contained finitely generated nonelementary Kleinian groups in $Isom(H^n)$, $n\ge 4$.

If you restrict further to geometrically finite nonelementary Kleinian groups $\Gamma< Isom(H^n), n\ge 4$, then, I think, they cannot be self-contained but I do not see a proof at the moment.

Edit. Is just noticed that all what I said about finitely generated Kleinian groups in dimension $\le 3$ was covered by Ian Agol's comments.

I see, however, how to deal with convex-cocompact groups in any dimension.

Theorem. Suppose that $\Gamma < Isom(H^n)$ is a convex-cocompact nonelementary group. Then $\Gamma$ is not self-contained.

I can write down a proof is you are interested (this is an old post after all). In the case of 1-ended groups, this result follows from a Zlil Sela's theorem.

Aside, there is an interesting related problem: For which (nonelementary) Kleinian groups $\Gamma< Isom(H^n)$ there exists $\alpha\in Isom(H^n)$ such that $$ \alpha \Gamma \alpha^{-1} < \Gamma $$ is a proper subgroup (one can add "of finite index"). The best result I know in this direction is due to Matsuzaki and Yabuki: The Patterson-Sullivan measure and proper conjugation for Kleinian groups of divergence type. Ergodic Theory Dynam. Systems 29 (2009), no. 2, 657–665. They proved that if $\Gamma$ is of divergence type then such proper conjugation cannot exist. On the other hand, for some groups of convergence type, proper conjugation is possible, but, in, the examples that I know, the index is infinite.

Question. Suppose that $\Gamma< Isom(H^n)$ is a nonelementary discrete group and $\alpha\in Isom(H^n)$ is such that $$ \alpha \Gamma \alpha^{-1} < \Gamma $$ is a subgroup of finite index. Is it true that this index equals $1$?

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This question is studied by Ohshika and Potyagailo (here is a link to the lucid, as usual, mathreview by McCullough.) A more definitive result is due to T. Delzant and L. Potyagailo (which has nice references, too).

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    $\begingroup$ This is a yes-or-no question and you provide a link without suggesting what you call "definitive result"... and a quick look at the link does not provide me what you have in mind, except that the link to "Delzant" is a paper by Delzant and Potyagailo (I hate Google for providing only the first name of authors in the list of math papers in Google requests). $\endgroup$ – YCor May 5 '15 at 19:54
  • $\begingroup$ The MathSciNet review has a typo: "A group is co-Hopfian if it is isomorphic to a subgroup of itself". $\endgroup$ – Danny Nguyen May 5 '15 at 23:07
  • $\begingroup$ @DannyNguyen I don't understand your comment and I would be grateful if you could help me clear up my misunderstandings. First, you have dropped the condition that the subgroup be proper, which seems essential to the definition of co-Hopfian. Second, I don't follow why "is" is in bold in your correction. $\endgroup$ – Neil Hoffman May 6 '15 at 4:05
  • $\begingroup$ @NeilHoffman: what Danny means is that the review says (1st sentence) "A group is co-Hopfian if it is isomorphic to a proper subgroup of itself" while it should be "it is not isomorphic"... $\endgroup$ – YCor May 6 '15 at 6:39
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    $\begingroup$ Another thing to mention is that there may be a problem with the proof of the Delzant--Potyagailo result. It relies on their Topology paper proving 'hierarchical accessibility', which contains an error. This has been fixed by Louder--Touikan (arxiv.org/abs/1302.5451), but one would need to check that the results needed for the Kleinian groups paper remain true. $\endgroup$ – HJRW May 6 '15 at 11:00

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