I'm looking for a proof of or counterexample to the following conjecture:
Every regular apeirohedron is genus 0 or ∞.
Here I mean an apeirohedron to mean a rank 3 abstract polytope with an infinite number of flags. And regular means that its automrophism group acts transitively on its flags.
Of course if this holds for abstract polytopes is should hold for concrete ones as well.
What I've tried
I have proven that the genus cannot be any finite value greater than 1. However I still feel like it is impossible for the genus to be 1 and I can't rule that out. I'll first describe my partial proof, and then several dead end attempts.
We can show that every abstract polytope corresponds to a smooth-connected Riemann surface with an isomorphic automorphism group. That is if there is a regular polytope with automorphism group $G$, then there is a smooth connected Riemann surface with automorphism group $G$. By the Hurwitz automorphism theorem a smooth connected Riemann surface of genus $g \geq 2$ must have a finite automorphism group. Thus a regular apeirohedron must have genus 0, 1 or ∞.
If a regular apeirohedron admits a non-trivial loop, then, with equivalence being the set of flags the loop passes through, there must be an infinite number of non-trivial loops. However it is possible that all these non-trivial loops are homotopic.