Let $(\mathbb G,\Gamma)$ be a graph of groups. A $G$-path from $u_0$ to $u$ is $$g_0e_1g_1\cdots e_{n}g_n,$$ where $e_1\cdots e_{n}$ is a walk in $\Gamma$ from $u_0$ to $u$ and each $g_i\in G_{s(e_i)}$. We denote the set of all $\mathbb G$-paths from $u_0$ to $u$ by $\pi[u_0,u]$. Let $\mathbb F(\mathbb G,\Gamma)$ be the group generated by the vertex groups and the elements $e$ of edge $\Gamma$, subject to the relations $$\{s_e(g)e=et_e(g)\mid g\in G_e,e\in E(\Gamma)\},\{e^{-1}=\bar{e}\mid e\in E(\Gamma)\}$$

Recall that the fundamental group of $(\mathbb G,\Gamma)$ on the base point $u_0$ is the set of all elements $g_0e_1g_1\cdots e_{n}g_n$ in $\mathbb F(\mathbb G,\Gamma)$, where $e_1\cdots e_{n}$ is a closed walk from $u_0$ to $u_0$. In other words, the fundamental group of $(\mathbb G,\Gamma)$ is $\pi[u_0,u_0]$.

My question is regarding the universal cover (Bass-Serre tree) of $(\mathbb G,\Gamma)$.

In the book "Trees'' by Serre page 51, the universal cover is defined in the following way:

The vertices are the left cosets of the vertex groups in $\pi[u_0,u_0]$, i.e.

$$\bigcup_{u\in V(\Gamma)} \pi[u_0,u_0]G_u$$

However in a paper by Bass "Covering theory for graphs of groups ", he defined in the following way:

The vertices are $$\bigcup_{u\in V(\Gamma)} \pi[u_0,u]G_u$$

My question is that

Are they $\pi[u_0,u_0]$-isomorphic?or they are completely different?

If yes, I really wonder to know the $\pi[u_0,u_0]$-isomorphism map.


To address a small point: the sets $\pi[u_0,u]$ should really be considered as their images in $\mathbb F(\mathbb G,\Gamma)$. This has the effect of doing some algebraic version of passing to homotopy classes of paths rel endpoints. For example, if $e$ is an edge of $\Gamma$, then $e\bar e e$ and $e$ should be considered as equivalent. This is already what happens when constructing the universal cover of a graph, say.

Modulo agreement on that point, my claim is that the definitions of the vertex sets of the universal cover of $(\mathbb G,\Gamma)$ are equivalent. Here is how to see it. Fix a choice of spanning tree $T$ in $\Gamma$. For every vertex $u$ of $\Gamma$, there is a unique path $\sigma_u$ from $u_0$ to $u$ that stays entirely within $T$. We can think of $\sigma_u$ as an element of $\pi[u_0,u]$ by sending $\sigma_u$ to (the equivalence class of) the ordered sequence of edges it traverses.

We get maps $\pi[u_0,u_0] \to \pi[u_0,u]$ and $\pi[u_0,u]\to \pi[u_0,u_0]$. The former sends a class $\tau \in \pi[u_0,u_0]$ to the class of the concatenation $\tau\sigma_u$. The latter sends $\rho \in \pi[u_0,u]$ to the class of the concatenation $\rho\bar\sigma_u$. Since the classes $\tau$ and $\tau\sigma_u\bar\sigma_u$ are equal, it’s easy to see that these maps are inverse bijections.

  • 1
    $\begingroup$ Don't you mean that for every vertex $u$ there is a unique path that says entirely within $T$ (not $\Gamma$)? $\endgroup$ – Max Horn Dec 30 '19 at 8:44
  • $\begingroup$ I absolutely do! good catch $\endgroup$ – Rylee Lyman Dec 30 '19 at 16:34

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