Limit Group decomposition

I would need a clarification about a statement in the article Limit groups and groups acting freely on $\mathbb{R}^n$-trees by Vincent Guirardel.

First recall that a limit group is a finitely generated group that is a limit of free groups in the space of marked groups (regardless of a choice of marking). This is (non-trivially) equivalent to: a finitely generated group that is fully residually free. Such groups are automatically finitely presented. (And they are obviously torsion-free, so that in the next few lines "cyclic" means "either trivial or infinite cyclic".)

Here are now the relevant excerpts. First, Corollary 3.4 (p1429), with reference to Remeslennikov: Any limit group has a free action on an $\mathbf{R}^n$-tree [for some $n$].

Then, Theorem 7.1 (p1430): "dévissage theorem": if a finitely generated, freely indecomposable group has a free action on an $\mathbf{R}^n$-tree ($n\ge 2$), then it decomposes as the Bass-Serre fundamental group of a finite graph of groups, with cyclic edge groups and each of whose vertex groups admits a free action on an $\mathbf{R}^{n-1}$-tree.

Just after the statement of this theorem, the author recalls Rips' result that freely indecomposable finitely generated groups with a free action on an $\mathbf{R}$-tree are either free abelian or fundamental groups of closed surfaces (of negative curvature).

Then he deduces Hence, a limit group can be obtained from abelian and surface groups by a finite sequence of free products and amalgamations over $\mathbf{Z}$.

Question: Are HNN-extensions (with cyclic edge groups) really unnecessary in this decomposition?

Here is the full reference to the cited article:
Vincent Guirardel, Limit groups and groups acting freely on $\mathbb{R}^n$-trees, Geometry & Topology 8 (2004), pages 1427-1470
DOI: 10.2140/gt.2004.8.1427, arXiv: math/0306306 [math.GR]

• At the very least, please provide enough context so we know what you mean and what you know already. What's your background? – David Roberts Apr 23 '15 at 2:16
• I am also having trouble understanding what you are asking. Here is a discussion that might help you refine your question: berstein.wordpress.com/2011/04/27/…. It has a number of references that might be helpful. – Neil Hoffman Apr 23 '15 at 4:14
• The question is very vaguely stated ("can be obtained" is unprecise). And regardless of the precise meaning, the question sounds like "Here is a theorem (vaguely stated). Why isn't it false?". I'd answer: "because it's true"! – YCor Apr 23 '15 at 19:00
• (To clarify, my last comment refers to the current version of the question, which is fine. The original question was incomprehensible.) – HJRW Apr 24 '15 at 8:42
• @HJRW The question is still incomprehensible without looking at the linked paper. I'll edit so that it makes sense. – YCor Apr 25 '15 at 10:13

It certainly isn't the case that every group which acts freely on an $\mathbb{R}^{n}$-tree can be written as an amalgamated free product; rather, I think the author meant the term `amalgamation' as an informal way of referring to graphs of groups, and hence to include HNN extensions.
On the other hand, because of special properties of limit groups, it is true that you don't need to consider the case of HNN extensions if you don't mind passing to subgroups. Recall that for any group $G$ and element $g$, the corresponding extension of centralizers is the amalgamated free product $G*_{C(g)}(C(g)\times \mathbb{Z})$ (where $C(g)$ of course denotes the centralizer of $g$). The class of iterated extensions of centralizers is the class of groups obtained by starting with finitely generated free groups and inductively extending centralizers. Kharlampovich--Myasnikov proved: