I'm wondering if there is any literature on graphs of groups where the maps $G\to H$ from an edge group $G$ to its endpoint group $H$ are not necessarily $\pi_1$-injective. Or is this just too general to actually say anything meaningful? Are there any results on this subject?
$\begingroup$
$\endgroup$
9
-
$\begingroup$ I guess you mean "injective", not "$\pi_1$-injective" $\endgroup$– YCorCommented Feb 17, 2018 at 8:13
-
$\begingroup$ Too general does not mean it should not be defined. When the graph is a tree, this is a tree of groups and the "Bass-Serre fundamental group" is just the colimit, a very general categorical construction, which makes sense without injectivity condition. $\endgroup$– YCorCommented Feb 17, 2018 at 8:14
-
2$\begingroup$ The simplest example that is not a tree is the HNN case: A group $H$ endowed with two homomorphisms $f,g:H\to G$. Then the resulting Bass-Serre fundamental group is the quotient of the free product of $G$ with a cyclic group $<t>$ by the relations $tf(x)t^{-1}=g(x)$, $x\in H$. It be obtained as follows: consider the smallest quotient $G/N$ of $G$ on which $f,g:H\to G/N$ have the same kernel. Then this is the HNN extension of $G/N$ with respect to the induced partial isomorphism. The quotient $G/N$ can be described by a somewhat complicated iterated procedure of killing kernels... $\endgroup$– YCorCommented Feb 17, 2018 at 8:18
-
$\begingroup$ I should add another simple example in the tree case (with infinite graphs) this encodes arbitrary inductive limits of sequences of groups, while in the injective case we can only encode inductive limit of sequences of groups with injective homomorphisms. $\endgroup$– YCorCommented Feb 17, 2018 at 10:16
-
$\begingroup$ Thank you! Can the same process you described for HNN extensions be applied to amalgamated free products? Where can I learn more about this? $\endgroup$– Harry ReedCommented Feb 17, 2018 at 19:44
|
Show 4 more comments
3 Answers
$\begingroup$
$\endgroup$
2
Let us call an "HNN extension" with non-injective homomorphism between associated subgroups ni-HNN extension as opposite to the ordinary HNN extensions. I know three examples where ni-HNN extensions were used.
Ilya Kapovich result that an ascending ni-HNN extension of a free group $F$ with non-injective homomorphism $\phi : F\to F$ is actually isomorphic to an ascending HNN extension of some free group.
A result of Igor Lysenok and Rostislav Grigorchuk that a finitely presented amenable group containing the Grigorchuk group is a ni-HNN extension of some finitely presented group ("A set of defining relations for the Grigorchuk group", Mat. Zametki 38 (1985), no. 4, 503–516, "An example of a finitely presented amenable group that does not belong to the class EG". Mat. Sb. 189 (1998), no. 1, 79--100).
Our with A. Yu. Olshanskii finitely presented non-amenable torsion-by-cyclic group is a ni-HNN extension of some finitely presented group containing the free Burnside group ("Non-amenable finitely presented torsion-by-cyclic groups." Publ. Math. Inst. Hautes Études Sci. No. 96 (2002), 43–169).
-
$\begingroup$ Having a look at I. Kapovich's paper (faculty.math.illinois.edu/~kapovich/PAPERS/en3.pdf), it seems to me he exclusively considers injective endomorphisms. Is there a difference with the published version? $\endgroup$– YCorCommented Feb 18, 2018 at 0:39
-
$\begingroup$ Sorry! Wrong reference. I have fixed that. $\endgroup$– user6976Commented Feb 18, 2018 at 1:13
$\begingroup$
$\endgroup$
2
As Yves has already indicated in comments, any "non-injective" graph of groups $\mathcal{G}$ canonically describes a graph of groups $\overline{\mathcal{G}}$ with the usual injectivity hypothesis. I'll briefly explain the construction.
For $\mathcal{G}$ a "non-injective" graph of groups, the fundamental group $\pi_1\mathcal{G}$ still makes sense -- you can either go through Serre's definition in Trees and check that injectivity wasn't used, or a quick and dirty justification is to notice that you can build a corresponding graph of spaces $\mathcal{X}$ in the usual way, and define $\pi_1\mathcal{G}=\pi_1\mathcal{X}$.
At this point, for every edge or vertex $x$ of $\mathcal{G}$, there is a homomorphism $G_x\to\pi_1\mathcal{G}$; let $\overline{G}_x$ denote its image. The data $\overline{G}_x$ attached to the underlying graph of $\mathcal{G}$ now define an "injective" graph of groups of the usual kind.
For this reason, there isn't much literature on "non-injective" graphs of groups, although I can think of a few places where the above construction is used: in Bestvina and Feighn's Inventiones paper on the Rips machine, for instance.
Of course, passing from $\mathcal{G}$ to $\overline{\mathcal{G}}$ is a very destructive process, which often yields something trivial (in the sense that the inclusion of some vertex map is surjective). So for this construction to be useful, you have to have some reason why $\overline{\mathcal{G}}$ is non-trivial in your particular case. One useful remark is that HNN extensions, or more generally graphs of groups with underlying graphs that aren't trees, are always non-trivial.
-
$\begingroup$ Ah, I see. Unfortunately I have an example where the image $\bar G_x$ is equal to $\pi_1(\cal G)$ for every vertex. $\endgroup$ Commented Feb 17, 2018 at 19:50
-
$\begingroup$ @HarryReed, then I think graphs of groups won't help you with that example, unfortunately. $\endgroup$– HJRWCommented Feb 19, 2018 at 11:27
$\begingroup$
$\endgroup$
1
You can interpret $\pi_1$ of the graph of groups as $\pi_1$ of the homotopy colimit of the corresponding diagram of classifying spaces. When the maps in the graph are not injective, this homotopy colimit is not necessarily a $K(\pi_1 \mathcal G, 1)$. In other words, even if $\pi_1 \mathcal G$ is boring, a graph of groups with non injective maps can have a nontrivial $\pi_2$!
answered Feb 17, 2018 at 20:29
-
1$\begingroup$ For instance, the 2-sphere cut along its equator. $\endgroup$– HJRWCommented Feb 18, 2018 at 7:51