On Leighton’s graph covering theorem

We give Leighton’s proof, mildly modified to clarify its structure and to make explicit the fact that it handles colored graphs. To ease comparison with his version, we have copied some of his notation.

We define a concept of a “symmetry-restricted graph.” The underlying data consist of a graph of colors $C$ together with a collection, indexed by the vertex-colors $i\in I$, of finite permutation groups $\Delta_{i}$ with an indexing of the orbits of $\Delta_{i}$ by the edge colors $k$ with $\partial_{0}k=i$. A symmetry-restricted graph for these data is a $C$-colored graph $G$ and for each $i$–vertex $v\in\operatorname{V}(G)$ a representation of $\Delta_{i}$ as a color-preserving permutation group on the set $\operatorname{star}(v)$ of edges departing $v$. A morphism of symmetry-restricted graphs $\phi\colon G\to G^{\prime}$ is a colored graph homomorphism $\phi$ which restricts to a weakly equivariant isomorphism from $\operatorname{star}(v)$ to $\operatorname{star}(\phi(v))$ for each $v$. (A map $\phi\colon X\to Y$ of $\Delta$–sets is weakly equivariant if it is equivariant up to conjugation, i.e., there is a $\gamma\in\Delta$ such that $\phi(\delta z)=\gamma\delta\gamma^{-1}\phi(x)$ for each $x\in X$ and $\delta\in\Delta$.) Note that a morphism is a covering map; if it is bijective it is an automorphism.

An example of a symmetry-restricted graph in this sense is a $k$-regular oriented fat graph; we have just one vertex color and the group $\Delta$ is a cyclic group of order $k$ acting transitively on each $\operatorname{star}(v)$. Another example is the following:

For Example 2.1 one has a simple geometric proof similar to the fat-graph proof for $k$–regular graphs. Create a $3$–dimensional “fat graph” from $G$ by truncating the corners of the dodecahedra and cubes to form small triangles and thickening each edge of $G$ to be a rod with triangular cross-section joining these triangles. The rods should have a fixed length and thickness, and be attached rigidly to the truncated polyhedra which play the rôle of vertices. Then the universal covering is a 3-dimensional fat tree whose isometry group acts properly discretely, so the result follows as before.

But if we have a graph $G$ made, say, of icosahedra connected to cubes by edges, then it is less obvious how to create a rigid fat-graph version, since the vertex degrees of icosahedron and cube are $5$ and $3$, which do not match.

To prove the above theorem we will need the graph of groups approach of Bass and Kulkarni.

We give a simplified version of the Bass-Kulkarni proof of Leighton’s theorem in its colored version, Theorem 1.1.

We retain the notation of Section 1. In particular, $C=\tilde{G}/\operatorname{Aut}(\tilde{G})$ is the (refined) graph of colors, with vertex set $I$ and edge set $K$. For the moment we assume that $\operatorname{Aut}(\tilde{G})$ acts without inversions, so $C$ has no edge with $k=\bar{k}$.

We use this graph as the underlying graph for a graph of groups, associating a group $\text{A}_{i}$ of size $a_{i}$ to each vertex $i$ and a group $\text{B}_{k}=\text{B}_{\bar{k}}$ of size $b_{k}$ to each edge $k$, along with an injection $\phi_{k}\colon\text{B}_{k}\to\text{A}_{\partial_{0}k}$. Of course we have to choose our groups so that $\text{B}_{k}$ embeds in $\text{A}_{\partial_{0}k}$ for each $k$; one such choice is the one of Remark 1.2.

Let Γ be the fundamental group of this graph of groups and $T$ the Bass-Serre tree on which Γ acts; this action has quotient $T/\text{G}=C$, vertex stabilizers $\text{A}_{i}$, and edge stabilizers $\text{B}_{k}$. Then $T$ is precisely the tree $\tilde{G}$. Now Γ acts properly discretely on $T$. So, if we can express $G$ and $G^{\prime}$ as quotients $T/\text{L}$ and $T/\text{L}^{\prime}$ with Λ and $\text{L}^{\prime}$ in Γ, then Λ and $\text{L}^{\prime}$ are finite index in Γ so $T/(\text{L}\cap\text{L}^{\prime})$ is the desired common covering.

To complete the proof we must show that such Λ and $\text{L}^{\prime}$ exist in Γ. This is the content of Bass’s Conjugacy Theorem ([3], see also [6]). We replace it for now by a “fat graph” argument (but see Theorem 4.2).

For each finite group $\Delta$ choose a finite complex $B\Delta$ with fundamental group $\Delta$ and denote its universal covering by $E\Delta$. We also assume that any inclusion $\phi\colon\text{F}\to\Delta$ of finite groups that we consider can be realized as the induced map on fundamental groups of some map $B\phi\colon B\text{F}\to B\Delta$ (this is always possible, for example, if $B\text{F}$ is a presentation complex for a finite presentation of Φ). We now create a “fat graph” version of our graph of groups by replacing vertex $i$ by $B\text{A}_{i}$, edges $k$ and $\bar{k}$ by $B\text{B}_{k}\times[0,1]$ (with the parametrization of the interval $[0,1]$ reversed when associating this to $\bar{k}$), and gluing each $B\text{B}_{k}\times[0,1]$ to $B\text{A}_{\partial_{0}k}$ by the map $B\phi_{k}\colon B\text{B}_{k}\times\{0\}\to B\text{A}_{\partial_{0}k}$ which realizes the inclusion $\phi_{k}\colon\text{B}_{k}\to\text{A}_{\partial_{0}k}$.

This is a standard construction which replaces the graph of groups by a finite complex $K$ whose fundamental group is Γ. The universal covering of $K$ is a fat-graph version $\mathcal{T}$ of the tree $T$, obtained by replacing $i$–vertices by copies of $E\text{A}_{i}$ and $k$–edges by copies of $E\text{B}_{k}\times[0,1]$. The “fat edges” $E\text{B}_{k}\times[0,1]$ are glued to the “fat vertices” $E\text{A}_{i}$ by the lifts of the maps $B\phi_{k}$. An automorphism of $\mathcal{T}$ will be a homeomorphism which is an isomorphism on each piece $E\text{A}_{i}$ and $E\text{B}_{k}\times[0,1]$ (where the only isomorphisms allowed on an $E\Delta$ are covering transformations for the covering $E\Delta\to K\Delta$).

We can similarly construct fat versions of the graphs $G$ and $G^{\prime}$, replacing each $i$–vertex by a copy of $E\text{A}_{i}$ and each $k$–edge by a copy of $E\text{B}_{k}\times[0,1]$. There is choice in this construction: if $\partial_{0}k=i$ then at the $E\text{A}_{i}$ corresponding to an $i$–vertex $v$ there are $r_{k}$ edge-pieces $E\text{B}_{k}\times[0,1]$ to glue to $E\text{A}_{i}$ and $r_{k}$ “places” on $E\text{A}_{i}$ to do the gluing, and we can choose any bijection between these edge-pieces and places; moreover, each gluing is then only determined up to the action of $\text{B}_{k}$. Nevertheless, however we make these choices, we have:

Since the fattened versions of $G$ and $G^{\prime}$ each have universal covering $T$, they are each given by an action of a subgroup of $\text{G}=\operatorname{Aut}(T)$, as desired, thus completing the proof of the colored Leighton’s theorem for the case that Γ has no inversions.

If Γ does invert some edge, so $C$ has an edge $k=\bar{k}$, then the edge stabilizer is an extension $\bar{\text{B}}_{k}$ of the cyclic group $C_{2}$ by $\text{B}_{k}$. We can assume that the inclusion $\text{B}_{k}\subset\bar{\text{B}}_{k}$ is represented by a double covering $K\text{B}_{k}\to\bar{\text{B}}_{k}$. The complex $(K\text{B}_{k}\times[0,1])/C_{2}$ (diagonal action of $C_{2}$) is then the object that the “half-edge” $k$ of $C$ should be replaced by in fattening $C$. The proof then goes through as before.∎

Our earlier fat graph proofs for the $k$–regular case and for Example 2.1 are special cases of the proof we have just given if we generalize the proof to allow orbifolds.

It is not hard to extend the above proof to prove the following theorem, which we leave to the reader; we do not know if the general symmetry-restricted version of Leighton’s theorem is true.

One of the two ingredients of the original Bass-Kulkarni proof of Leighton’s theorem is Bass’s Conjugacy Theorem. This theorem holds for symmetry-restricted graphs (see below), but this appears not to help extend the above results. The other ingredient in the Bass-Kulkarni proof is to find a subgroup of $\operatorname{Aut}(T)$ that acts properly discretely on $T$ with quotient $C$. Such a group would necessarily be given by a graph of finite groups with underlying graph $C$, and we are back in the situation of the proof we have already given, which appears to need strong conditions on $C$.

The Conjugacy Theorem says, in our language, that if $T$ is a colored tree whose colored automorphism group acts without inversions (i.e., the graph of colors $C=T/\operatorname{Aut}(T)$ has no loops $k=\bar{k}$), and $H\subset\operatorname{Aut}(T)$ is a subgroup with $T/H=C$, then any Γ that acts freely on $T$ can be conjugated into $H$ by an element of $\operatorname{Aut}(T)$.