# When is a generalised Baumslag-Solitar group linear?

$$\DeclareMathOperator\BS{BS}$$The linearity of the Baumslag-Solitar groups $$\BS(m, n)=\langle a, t\mid t^{-1}a^mt=a^n\rangle$$ is completely understood, and it may be phrased as: $$\BS(m, n)$$ is linear if and only if it is residually finite*.

A generalised Baumslag-Solitar (GBS) group is a group which may be realised as the fundamental group of a graph of groups with all vertex and edge groups infinite cyclic. For example, Baumslag-Solitar groups themselves are GBS groups, with a single vertex and a single loop edge, while one-relator groups with center form another class of examples.

I am wondering if the above characterisation of linear Baumslag-Solitar groups extends to GBS groups:

Do there exist non-linear, residually-finite GBS groups?

*See either this old MO question, or combine the papers:

• [Classifying residual finiteness] S. Meskin Nonresidually finite one-relator groups Trans. Amer. Math. Soc. 164 (1972), 105–114
• [Classifying linearity, in Russian] R.T. Vol’vachev Linear representation of certain groups with one relation Vestsi Akad. Navuk BSSR Ser. Fi z.-Mat. Navuk 1985, no. 6, 3–11, 124
• For a GBS, you also need the vertex groups to be infinite cyclic, not just the edge groups. Commented Aug 10, 2022 at 10:05
• @Carl-FredrikNybergBrodda Whoops - fixed it! Thanks.
– ADL
Commented Aug 10, 2022 at 10:06
• For record, $\mathrm{BS}(m,n)$ is linear over $\mathbf{Q}$ iff it's linear in char. zero, iff it's linear over a field, iff it's linear over a product of fields, iff it's residually finite, iff [$|m|=|n|$ or $\min(|m|,|n|)=1$]. The equivalence linear = RF is not enough to be called a "classification".
– YCor
Commented Aug 10, 2022 at 10:52
• I think there are really two questions here: do we know which GBS groups are residually finite, and which are linear? I can’t imagine making progress without addressing at least the first of these. I’d start by looking at the work of Woodhouse and his coauthors on residual finiteness and linearity of the related class of tubular groups.
– HJRW
Commented Aug 11, 2022 at 16:00
• @ADL the RF question doesn't seem "extremely hard" at first sight, after a good understanding of the BS case. On should look at simple examples, e.g. graph of groups with a single vertex (1 selfloop is the BS case, so with more than 1 selfloop). And then 2 vertices, etc.
– YCor
Commented Aug 16, 2022 at 4:51

## 1 Answer

(1) I've been looking a little more. My belief is now:

Conjecture. Let $$G$$ be a generalized Baumslag-Solitar group (i.e. the Bass-Serre fundamental group of a nonempty finite graph of groups in which every vertex and edge group is infinite cyclic). Let $$\langle x\rangle$$ be one vertex group. Then we have one of the following:

(a) for some $$n\ge 1$$, $$\langle x^n\rangle$$ is a normal subgroup [this can be read on the graph, see below]. Then $$G$$ is virtually direct product of $$\mathbf{Z}$$ and a free group (and hence linear).

(b) $$G$$ is a solvable, Baumslag-Solitar group, and in particular is residually finite [assuming the graph is reduced, this means that there is a single vertex, at most a single self-loop with one inclusion being surjective]

(c) Otherwise, $$G$$ is not residually finite.

In (a) this can be read as follows: the partial isomorphism defined by any loop (including self-loops) is $$\pm$$ a partial identity. See also (4) below.

Reduced means that for every non-self loop, both inclusions are proper. If the (finite) graph of groups is not reduced, one can collapse non-reduced edges until one gets a reduced graph, without changing the Bass-Serre fundamental group.

(2) Now let me prove (sketch) the conjecture when the graph is a tree (with self-loops allowed).

Let me start with a graph with a single vertex (and finitely many self-loops). Such a group is thus defined by a family of pairs $$(n_i,m_i)_{i\in I}$$ of nonzero integers with $$I$$ finite.

Let us specify (a),(b),(c) to this case:

(a) for some $$n\ge 1$$, $$\langle x^n\rangle$$ is a normal subgroup: this means that $$|m_i|=|n_i|$$ for all $$i$$;

(b) $$G$$ is a solvable, Baumslag-Solitar group: this means that $$|I|=1$$ (say $$I=\{1\}$$) and $$\min(|m_1|,|n_1|)=1$$;

(c) other cases.

Since the case $$|I|\le 1$$ is already settled, we deal with $$|I|\ge 2$$. Assume $$G$$ is residually finite. By the case $$|I|=1$$, we know that for each $$i$$, either $$|n_i|=|m_i|$$ or $$\min(|n_i|,|m_i|)=1$$. We can assume that whenever $$\min(|n_i|,|m_i|)=1$$, we have $$1\in\{n_i,m_i\}$$.

Just to illustrate, a typical example would then be $$|I|=3$$, with the pairs $$(1,3),(-2,1),(4,-4)$$. The corresponding presentation is $$\langle t_1,t_2,t_3,x\mid t_1xt_1^{-1}=x^3,t_2x^{-2}t_2^{-1}=x,t_3x^4t_3^{-1}=x^{-4}\rangle.$$

Suppose that $$(1,m)$$ and $$(n,1)$$ both occur among the $$(n_i,m_i)$$, for some $$|n|,|m|\ge 2$$. So we have the subgroup with presentation $$\langle t,u,x\mid txt^{-1}=x^m,u^{-1}xu=x^n\rangle$$

In every finite quotient of this group, $$g\mapsto tgt^{-1}$$ defines an injective endomorphism of $$\langle x\rangle$$, hence bijective. Similarly for $$u$$. Hence $$\langle x\rangle$$ is normal in every finite quotient. Hence $$[t^{-1}xt,u^{-1}xu]$$ is trivial in every finite quotient. But it is not trivial in the group itself, by standard facts on HNN extensions.

A similar argument holds if $$(1,n_i)$$, $$(1,n_j)$$ both occur among the pairs with $$i\neq j$$ and $$|n_i|,|n_j|\ge 2$$: then $$[t_i^{-1}xt_i,t_j^{-1}xt_j]$$ is not trivial but vanishes in every finite quotient. (Same with $$(n_i,1)$$ and $$(n_j,1)$$.)

Next, if $$(1,n)$$ and $$(m,m)$$ both occur with $$|n|\ge 2$$: this corresponds to relators $$txt^{-1}=x^n$$, $$[u,x^m]=1$$: then $$[u,t^{-1}x^mt]$$ is nontrivial but vanishes in every proper quotient. Finally if $$(1,n)$$ and $$(m,-m)$$ both occur with $$|n|,|m|\ge 2$$, then $$[u,t^{-1}x^{2m}t]$$ is nontrivial but vanishes in every proper quotient (to check that it is nontrivial one has to check separately the case $$|n|=2$$).

If all these are excluded and $$|I|\ge 2$$, this means that we are in case (a).

(3) Here is the simplest case with a non-self loop: two vertices joined with two edges. This corresponds to a presentation $$\langle t,x,y\mid x^k=y^\ell,x^m=ty^nt^{-1}\rangle.$$

We get $$x^{kn}=y^{\ell n}=t^{-1}x^{\ell m}t$$. Indeed the discussion is:

If $$|kn|=|\ell m|$$, then $$\langle x^{kn}\rangle$$ is normal cyclic and we are in case (a). Otherwise, an argument similar to the Baumslag-Solitar case (and to the ones above) implies that the group is not residually finite.

(4) In general, let me clarify "this can be read on the graph" in (a): given a sequence of oriented edges $$e_1,\dots,e_p$$ forming a loop ($$p\ge 1$$). Let the left and right inclusion of $$e_j$$ be given by multiplication by $$n_j$$, $$m_j$$ respectively. Say that the loop is unimodular if $$\prod m_j=\prod n_j$$. Say the GBS graph of groups is unimodular if every loop is unimodular. Then (a) means that the graph is unimodular. (It is enough to check loops in a generating subset of the (ordinary, i.e; Poincaré) fundamental group of the graph.)

I haven't checked, but believe it's easy, [Edit: one easily checks] that if the graph is unimodular then $$G$$ is virtually direct product of $$\mathbf{Z}$$ and a free group. Indeed, after modding out by a normal cyclic subgroup contained in every edge group, the resulting group is Bass-Serre fundamental group of a finite graph of finite (cyclic) groups, and hence is virtually free.

And if the graph is not unimodular (and reduced with at least 2 edges) then one has to check the failure of residual finiteness. I did it above if the lack of unimodularity is witnessed by a self loop, and more quickly with it is witnessed by a loop of size 2. But I think there is no serious difficulty in general.

Eventually I believe the conjecture follows using routine arguments. Thus the linear examples are "obviously linear" (virtually $$\mathbf{Z}\times$$free or solvable BS groups) and the other ones are non-residually finite for a reason similar to the Baumslag-Solitar case, while the solvable examples appear as an exception.

• This is great! I wrote a previous comment with a doubt about one of your claims, but I since deleted it, since I convinced myself that what you wrote should be correct.
– HJRW
Commented Aug 16, 2022 at 12:39
• Thanks for this! This all makes sense, and the three cases in fact correspond to a classification of Whyte (arxiv.org/abs/math/0405272).
– ADL
Commented Aug 21, 2022 at 21:10