Tiling by the ideal right-angled octahedron

Prove that every ideal, right-angled polyhedron assembled from copies of the Scharlau–Walhorn polyhedron SW#1{2} is tiled by the ideal, right-angled hyperbolic octahedron.

Background

The paper classifies ideal, right-angled polyhedra assembled from the 49 Scharlau–Walhorn arithmetic Coxeter polyhedra. For SW#1{2}, Proposition \ref{Bipyramid} proves that every such polyhedron is tiled by smaller pieces derived from the octahedron: the triangular bipyramid \frac14\cO, the intermediate polyhedron \frac12\cO, and \cO itself. The unresolved strengthening is that the decomposition should always be obtainable directly from copies of the full ideal, right-angled octahedron \cO.

Establishing this would sharpen the classification by showing that the octahedron, square antiprism, and rhombicuboctahedron are the unique minimal ideal, right-angled hyperbolic polyhedra with arithmetic reflection groups.

References

We conclude with a formal statement of the conjecture mentioned after Thm. {Thm:MainConverse}: Let $\cP$ be the polyhedron SW#1{2}, and let $\cO$ be the ideal, right-angled hyperbolic octahedron. Then any ideal, right-angled $\cP$-polyhedron is tiled by~$\cO$.

Arithmetic Polyhedra  (2609.05349 - Allcock et al., 4 Sep 2026) in Conjecture 5.?, Section 5, subsection “The case SW#1{2}”; reiterated in Section 6, Further Directions, item 1