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This question is partly connected with the following Connection between Stalling's end theorem and Seifert-van Kampen Theorem.

By Stalling's Theorem a group with more than one end splits over a finite subgroup, i.e. can be written as an HNN-Extension or a free product with amalgamation (over a finite subgroup).

However when I was working through the proofs of Dunwoody, Dunwoody & Krön and Krön, which all use Bass-Serre Theory, I was wondering if there is any way to detect if we are dealing with an HNN-Extension or a free product with amalgamation. Primarily I was thinking of some property of the Cayley graph or the action on it by the group in question. For example one could consider the induced action on the end space of the Cayley graph.

In my opinion this is a natural question and I would be grateful for any comment or references regarding this.

EDIT: I am looking for properties of the group, its Cayley graph, the action of the group on its Cayley graph and/or its end space etc., which help to distinguish if the group, supposed to split by Stalling's Theorem, is indeed an HNN-Extension or a free product with amalgamation.

For some statements one probably has to make additional assumptions such as the group is finitely generated or finitely presented etc.

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    $\begingroup$ What do you mean by "detect" do you mean by an algorithm? By the way, a group splits as an HNN extension if and only if it admits a surjection onto $\mathbb{Z}$. $\endgroup$
    – NWMT
    Commented Aug 18, 2015 at 23:36
  • $\begingroup$ This is Bieri-Strebel right? However as far as I know this holds only for finitely presented groups. $\endgroup$
    – M.U.
    Commented Aug 19, 2015 at 11:07
  • $\begingroup$ By "detect" I meant properties of e.g. the group, the Cayley graph or the action of the group onto the Cayley graph and iths end space etc. that help to distinguish if when the group splits by Stalling's theorem we have an HNN-Extension or a free product with amalgamation. Maybe I should be more precise in the original post. $\endgroup$
    – M.U.
    Commented Aug 19, 2015 at 11:12
  • $\begingroup$ First of all, the surjection onto Z thing is actually straight up Bass-Serre theory and requires only finite generation. One direction is obvious, for the other direction simply note that the Cayley graph of Z is a somewhat silly tree.) Bieri-Strebel answers more subtle questions. This is unrelated to your question though. Your new edit is much clearer. $\endgroup$
    – NWMT
    Commented Aug 19, 2015 at 13:29

2 Answers 2

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There is nothing intrinsic about whether a group splits as a free product with amalgamation or an HNN extension. For example free groups split both ways $$ \langle a,b \rangle = \langle a \rangle * \langle b \rangle = \langle a,b ; t \mid t^{-1} a t = b \rangle $$ as do many other groups. If you're cutting up a Cayley graph like in the papers you cite, one cut may give you an HNN extension, while the other can give you a free product with amalgamation.

Actually free products with amalgamation and HNN extensions aren't that different. It's just that because of the celebrated Seifert Van-Kampen theorem, HNN extensions don't get no respect! Bass-Serre theory unifies and generalizes these notions.

If your group acts on a simplicial tree with one edge orbit and two vertex orbits, then it splits as an amalgamated free product. If it acts with one edge orbit and one vertex orbit then it splits as an HNN extension. Any characterization must follow somehow from this definition. For example an HNN extension is dsitinguished from a FPA if the group cannot be generated by elements that act elliptically (i.e. fix a vertex) on the the Bass-Serre tree.

With respect to ends, the most canonical cut set may give you neither a free product with amalgamation nor an HNN extension, but a more general graph of groups decomposition, namely a Dunwoody decomposition. Such cut sets would consist of disjoint orbits of finite cut sets.

Edit: By the way it is "obvious" that the ends of the Bass-Serre tree dual to the Dunwoody decomposition are in natural bijective corresponsdence with the set of ends of your group; at least in the f.p. (or accessible) case.

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While there's no single answer to this question, there are a lot of partial results.

To start with, there is the still unresolved Kropholler-Roller conjecture, although various special cases are known, see for example this math overflow question.

Also, there are various quasi-isometric rigidity theorems which say that if such-and-such a finitely generated group $G$ splits as an HNN-extension of a free product with amalgamations then any group quasi-isometric to $G$ also splits. Stallings theorem itself has this format, because the number of ends of a finitely generated group is a quasi-isometry invariant. For further examples of such theorems see Papasoglu's paper "Group splittings and asymptotic topology" J. Reine Angew. Math. 602 (2007), 1–16, and my paper with Whyte and Sageev, "Quasi-actions on trees II: Finite depth Bass-Serre trees", Mem. Amer. Math. Soc. 214 (2011), no. 1008

Of particular importance is the vast literature on JSJ decompositions, which can be thought of as the theory of splittings of one-ended groups over two-ended groups. Bowditch's paper "Cut points and canonical splittings of hyperbolic groups", Acta Math. 180 (1998), no. 2, 145–186, characterizes which hyperbolic groups $G$ have a nontrivial splitting over a two-ended group, in terms of the topology of the Gromov boundary of $G$, which gives an example of a quasi-isometric rigidity theorem in the context of JSJ decompositions.

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  • $\begingroup$ Thanks for your quick answer. Could you please add the titles of the papers you mention since I cannot open the documents from the links you posted. $\endgroup$
    – M.U.
    Commented Aug 18, 2015 at 18:42
  • $\begingroup$ @M.U.: Oops, sorry about that. Fixed. $\endgroup$
    – Lee Mosher
    Commented Aug 19, 2015 at 11:56

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