Show that for a graph of groups all of whose edge homomorphisms are injective maps \(\mathbb{Z} \to \mathbb{Z}\), we can choose \(K\) to have universal cover a product \(T \times \mathbb{R}\) with \(T\) a tree. Work out in detail the case that the graph of groups is the infinite sequence \(\mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{3} \mathbb{Z} \xrightarrow{4} \mathbb{Z} \to \cdots\) where the map \(\mathbb{Z} \xrightarrow{n} \mathbb{Z}\) is multiplication by \(n\). Show that \(\pi_1(K\Gamma)\) is isomorphic to \(\mathbb{Q}\) in this case. How would one modify this example to get \(\pi_1(K\Gamma)\) isomorphic to the subgroup of \(\mathbb{Q}\) consisting of rational numbers with denominator a power of 2?
Solution
Consider first an arbitrary graph of groups in which every vertex and edge group is infinite cyclic and every edge homomorphism is injective. After choosing generators, every edge homomorphism has the form \[ \mathbb Z\longrightarrow\mathbb Z, \qquad k\longmapsto m k \] for some nonzero integer \(m\).
1. Its universal graph-of-spaces cover is a tree of copies of \(\mathbb R\) joined by strips \(\mathbb R\times I\).
The classifying space of each cyclic vertex or edge group can be chosen to be \(S^1\). Its universal cover is \(\mathbb R\). Hence each lifted vertex space is a line and each lifted edge mapping cylinder is a strip. The incidence graph of the lifted pieces is the Bass–Serre tree \(T\).
2. Every lifted boundary map of an edge strip is an affine homeomorphism of \(\mathbb R\).
A map \(S^1\to S^1\) inducing multiplication by a nonzero integer \(m\) on fundamental groups lifts to \[ t\longmapsto mt+c \] on universal covers. This is a homeomorphism of \(\mathbb R\).
3. The universal cover is homeomorphic to\[ \boxed{T\times\mathbb R}. \]
Choose a coordinate on one vertex line and propagate coordinates across adjacent edge strips. By 2, each boundary attachment can be straightened to the identity by an affine reparametrization of the new vertex line. Since the incidence graph \(T\) is a tree, each new vertex is reached by a unique path from the initial vertex, so no conflicting coordinate choices arise. The resulting tree of lines and strips is exactly the product cell structure on \(T\times\mathbb R\).
Now specialize to the ray \[ G_1\xrightarrow{\times2}G_2\xrightarrow{\times3}G_3\xrightarrow{\times4}\cdots, \qquad G_k\cong\mathbb Z. \]
4. Since the underlying graph is a tree, the graph product is the direct limit\[ \pi_1(K\Gamma) \cong \varinjlim \bigl( \mathbb Z\xrightarrow{2}\mathbb Z\xrightarrow{3}\mathbb Z\xrightarrow{4}\cdots \bigr). \]
There are no stable letters because a maximal tree is the whole underlying ray. The fundamental group presentation therefore takes the free product of the vertex groups and imposes exactly the edge identifications. This is the colimit of the displayed directed system.
5. This direct limit is isomorphic to \(\mathbb Q\).
Embed the \(k\)th copy of \(\mathbb Z\) into \(\mathbb Q\) by \[ \phi_k(r)=\frac{r}{k!}. \] These maps are compatible because \[ \phi_{k+1}((k+1)r) =\frac{(k+1)r}{(k+1)!} =\frac r{k!} =\phi_k(r). \] Thus the direct limit identifies with \[ \bigcup_{k\ge1}\frac1{k!}\mathbb Z. \] Every rational number \(a/b\) lies in this union because \(b\) divides \(k!\) for all sufficiently large \(k\). Hence \[ \varinjlim G_k\cong\mathbb Q. \]
6. To obtain the subgroup of rationals whose denominator is a power of \(2\), replace the sequence by\[ \mathbb Z\xrightarrow{2}\mathbb Z\xrightarrow{2}\mathbb Z\xrightarrow{2}\cdots. \]
The same direct-limit calculation embeds the \(k\)th copy as \[ 2^{-(k-1)}\mathbb Z. \] Therefore the limit is \[ \bigcup_{k\ge0}2^{-k}\mathbb Z =\mathbb Z[1/2] =\left\{\frac a{2^k}:a\in\mathbb Z,\ k\ge0\right\}. \]