0
$\begingroup$

This question comes from section 1.3 of Topological Graph Theory by Tucker and Gross.

Let $B$ be a group acting freely on a graph $G$ meaning that the only automorphism defined from the group action with a fixed point (on both vertices and edges) is the identity automorphism. The quotient $G/B$ is the graph with vertex orbits as vertices and edge orbits as edges. The endpoints of an edge orbit $[e]$ are the vertex orbits containing vertices incident to any edges in $[e]$. We define the quotient map $q_B \colon G \rightarrow G/B$ to be the map that maps vertices and edges to their orbits.

A covering projection is defined to be a map between graphs $p \colon \tilde{G} \rightarrow G$ such that there exists an isomorphism $i \colon \tilde{G}/B \colon \rightarrow G$ with $p = i \circ q_B$.

The star of a vertex $v$, $\text{star}(v)$, is the subgraph consisting of $v$, the neighbors of $v$, and each edge connecting $v$ to a neighbor. A map between graphs $f$ is a local isomorphism if for each vertex $v$ the image $f(\text{star}(v))$ is isomorphic to $\text{star}(v)$.

The exercise asks to show that if $p \colon \tilde{G} \rightarrow G$ is a covering projection and $G$ is a simple graph then $p$ is a local isomorphism.

Here's my approach so far. It should be sufficient to show that $q_B$ is a local isomorphism. This means that no two vertices or edges in $\text{star}(v)$ should lie in the same orbit under the action of $B$. I have been trying to show that two vertices from $\text{star}(v)$ lie in the same orbit, then the action defined by $B$ must have a fixed point coming from an automorphism other than the identity. If anyone could help me fill in the details for this argument, or has a completely different approach, it would be much appreciated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.