Questions in Geometric Group Theory — Q 13.3

Definitions and mathematical statement

A finite graph is planar if it admits an embedding in the two-sphere in which edges meet only at common endpoints. For connected finite graphs, a covering map $p:H\to G$ is a graph map such that for each vertex $v\in V(H)$ the induced map on incident edges is a bijection onto the edges incident to $p(v)$. Its degree is the common finite cardinality of the vertex fibers. A double cover has degree two; no regularity assumption is imposed on a general finite cover.

Is the following implication true for every connected finite graph $G$? \[ \text{$G$ has a finite planar cover} \quad\Longrightarrow\quad \text{$G$ is planar or has a planar double cover.} \] Equivalently, the least degree of a planar finite cover, if any exists, is one or two. This is the source's $1$--$2$--$\infty$ formulation of the planar-cover problem.

The covering graph is not an arbitrary graph mapping onto $G$: the local bijection condition must hold at every vertex. In particular, merely drawing a subdivision or a branched cover does not establish a planar cover. One can formulate the question for simple graphs; loops and parallel edges require the same incidence-based covering convention. For a disconnected base graph, the issue is tested component by component, so the connected formulation contains the substantive problem.

Short English statement

If a finite graph has any finite planar cover, must it already have one of degree at most two?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 1101303 (AMR-010-1303), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Bestvina - Questions in Geometric Group Theory (pdf) (2004), Question 13.3 (PDF page 24). Original source indexed by AMR-010-1303. https://www.math.utah.edu/~bestvina/eprints/questions-updated.pdf.