Geometry of Curves and Surfaces — Problem 3.1

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

A convex polyhedron $P$ here is the boundary of the convex hull of finitely many points of $\mathbb R^3$ which do not all lie in a plane. Its edge graph consists of its vertices and edges. A spanning edge tree $T$ contains all vertices and is connected and has no cycles. Cutting along $T$ produces a compact polyhedral disk $P_T$. Its planar development is a map $u_T:P_T\to\mathbb R^2$ that is locally an isometry and preserves lengths on each face. An unfolding is simple if $u_T$ is one-to-one; in that case the developed surface has no overlaps. A polyhedron is unfoldable if some spanning edge tree gives a simple unfolding. The source explains that allowing any collection of edge cuts with connected planar development yields the same question. No affine deformation of $P$ is part of this definition.

Statement recorded in the supplied source.

Is every convex polyhedron in $\mathbb R^3$ unfoldable? That is, does every such $P$ admit a spanning tree of its actual edges whose cut surface has an injective planar development? [S2]

Short English statement

Can every convex polyhedron be cut along its edges into one connected piece that unfolds flat without overlap?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 7000014 (AMR-069-0014). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Mohammad Ghomi, Affine unfoldings of convex polyhedra, arXiv:1305.3231v2 (2014), Introduction, pp. 2--4, including Note 1.5; Section 2.1, p. 5. https://arxiv.org/pdf/1305.3231v2.