ON THE DIMENSION OF ARCHIMEDEAN SOLIDS

NEW

Problem: Reconstructed conjecture: in the Erdős–Harary–Tutte unit-distance sense, the two unresolved Archimedean-solid graphs in Madaras–Široczki—especially the rhombicosidodecahedral graph—have graph dimension $2$. A planar dimension-$2$ representation would in particular give an injective map of vertices to $\mathbb R^2$ with every graph edge drawn as length $1$.

Result: The conjecture is false. The rhombicosidodecahedral graph has no injective planar unit-distance representation.

Let $D$ be the dodecahedron. Model the rhombicosidodecahedral graph $R$ by vertices $(v,F)$, where $v$ is a vertex of $D$ incident with a face $F$. Two such vertices are adjacent if either they share a face and the corresponding vertices of $D$ are adjacent, or they share a vertex and the corresponding faces are incident there.

Assume $p:V(R)\to\mathbb R^2$ is an injective unit-distance representation. If $uv$ is an edge of $D$ with incident faces $F,G$, then

$$(u,F),(v,F),(v,G),(u,G)$$

is a 4-cycle of unit edges. Four distinct points forming an equilateral quadrilateral form a parallelogram, hence

$$p(u,G)-p(u,F)=p(v,G)-p(v,F). \tag{1}$$

Now pass to the dual $D^*$, the icosahedral graph. For an oriented dual edge $F\to G$, define

$$a(F,G)=p(v,G)-p(v,F),$$

where $v$ is either endpoint of the primal edge shared by $F,G$. Equation (1) makes this well-defined, and $\|a(F,G)\|=1$.

At each vertex $v$ of $D$, the three incident faces $F,G,H$ form a triangular face of $D^*$, and

$$a(F,G)+a(G,H)+a(H,F)=0.$$

Since the facial triangles generate the cycle space of the planar triangulation $D^*$, the cochain $a$ integrates to points $q(F)\in\mathbb R^2$ satisfying

$$q(G)-q(F)=a(F,G).$$

Thus $q$ is a planar unit-distance representation of the icosahedral graph.

But this is impossible. In the icosahedral graph, the five neighbors of any vertex induce a 5-cycle. If the central vertex is placed at $O$, its five neighbors lie on the unit circle. Consecutive neighbors in the 5-cycle are also distance $1$, so their angular differences must each be $\pm\pi/3$. Around a closed 5-cycle, the sum of five signs $\pm1$ would have to be divisible by $6$, impossible.

Therefore $R$ has no planar unit-distance representation, so its dimension is not $2$. This disproves the conjecture.

Citation: Conjecture from T. Madaras and P. Široczki, “On the dimension of Archimedean solids,” Opuscula Mathematica 34(1) (2014), 123–138, Section 6. The disproof above is self-contained.