How quickly does maximum sphericity of polyhedra with $n$ faces approach 1 as $n→∞$? I can show that sphericity $1 - \frac{5 \sqrt{3} π}{27n} - O(n^{-3/2})$ is possible. Is this, especially $O(n^{-3/2})$, optimal here? And has anyone got this sphericity before?
Update: I expanded the below discussion into a paper Asymptotic Optimal Sphericity.
Calculating optimal sphericity ($π^{1/3}(6V)^{2/3}/S$) for small $n$ is hard (even though it is an algebraic multiple of $π^{1/3}$), but asymptotics can give a sense of clarity, and here perhaps, polyhedra with a fractal-like distribution of nonhexagonal faces. $1 - \frac{5 \sqrt 3 π}{27n}$ is the ideal hexagonal limit (up to $O(n^{-2})$; see below for derivation), and the question is how closely can we approach it.
Maximum sphericity does not generally imply symmetry, but for some moderate $n$, when they exist, Goldberg polyhedra (these use hexagons and 12 pentagons with rotational icosahedral symmetry) likely give the best sphericity. See:
- Roundest Polyhedra by Alan Schoen (note the lack of symmetry).
- The Roundest Polyhedra with Symmetry Constraints (Lengyel, Gáspár, Tarnai 2017).
However, projecting a regular hexagonal grid on a sphere creates distortion, but for best sphericity, we want almost all faces to approximate regular hexagons of the same size.
- Approximately, local distortion is proportional to region width squared, with loss per unit area proportional to distortion squared.
- I conjecture that $n'$ nonhexagonal faces (and no holes) cannot get us within $o(\frac{1}{n'n})$ sphericity of the hexagon ideal (note: $Θ(\frac{1}{n'n})$ $(4≤n'<\sqrt{n})$ is possible by allowing nonadjacent pentagon-heptagon pairs below).
Instead, we can use an approximately regular hexagonal grid with $Θ(\sqrt n)$ topological defects (note that radius/typical_edge_length is $Θ(\sqrt n)$). A single adjacent pentagon-heptagon pair can delete a ray in a hexagonal grid; and intuitively, we need to delete $Ω(\sqrt n)$ rays to cover a large convex region of the sphere. At unit grid size, deleting a ray requires $Ω(1/r)$ relative distortion at distance $r$, which threatens to add a logarithmic factor to the $O(n^{-3/2})$ bound, which we can avoid by carefully grouping defects as follows.
The construction: Recursively partition the sphere into roughly $n^{1/3}$ pieces, each time cutting a piece along a geodesic (or otherwise with short length), and efficiently reducing maximum piece diameter. From top to bottom, given a piece, choose an approximately parallel unit vector field, and uniformly rotate it for low angle with the vector field of the parent piece. Fill the leaf pieces with an approximate regular hexagonal grid (with the right unit size) oriented parallel to the vector field. Then, recursively (bottom up), join sibling pieces, warping a narrow strip near the separating geodesic for alignment of the grids, with $O(\sqrt n d^3)$ topological defects per join, where $d$ is the diameter of the piece (for unit sphere radius). Note that the defects concentrate along the boundaries of the larger pieces, and the effective radius of distortion for each defect is limited by the distance between the defects.
Conjectures: I expect that as $n→∞$, optimal large scale behavior converges, giving us a picture of the topological defects in the optimal sphericity $n$-polyhedron. A question is how it looks like; I conjecture as follows. Let $T$ denote locations with high density of topological defects (in the limit).
* $T$ has 12 connected components, corresponding to equal area regions (60° holonomy). Each component is a tree. $T$ has zero area; no end of $T$ is in $T$. The number of components and distance minimization would suggest an approximate icosahedral symmetry, with 5 branches joined at each of the 12 roots. However, assuming 24° misalignment per unit length costs less than roughly $2 \cos \frac{π}{5}$ of 12° misalignment, it becomes favorable to join two or four branches into pairs (possibly with an approximate pyritohedral symmetry).
* Topologically (with a possible exception for the degree of the root), each connected component of $T$ looks like a binary tree that branches (i.e. every branch branches) at every rational interval, with large denominators corresponding to small branches. Each path is differentiable except at all branch points.
* We have conservation of flow — angle of misalignment between grids; corresponds to area (solid angle) of drainage. At each root, the flow adds up to 60° and cancels out.
* Branch angles depend on the parent flow $α$ and child flow $β$, locally minimizing cost of flow. Typically (in terms of diversion of flow) $\min(β,α-β)=Θ(α)$, in which case the angles are $Θ(\log^{-1/2} (1/α))$. For small $α$, we get approximate isotropy: Grid-induced relative cost anisotropy is $θ(\log^{-1}(1/α))$ (at flow $α$) per unit angle (deformations can use preferred alignments at scales below the inter-defect distance); also, $T$ has small typical angles, and a constant-speed detour does not change your average angle.
* The complement of $T$ is connected and contains a dodecahedral skeleton, but otherwise tree-like, with unique paths between points and the skeleton. Almost all points (all but zero area) lie on the ends. At almost all ends and on uncountably many points on each path, the compass direction of the path diverges with unbounded winding angles. I do not know whether path lengths are finite.
* For finite $n$, branching of defect lines ends at typical flow $Θ(n^{-1/3} \log^{5/6} n)$ (length $Θ(n^{-1/6} \log^{2/3} n)$ and defect-free region width $Θ(n^{-1/6} \log^{1/6} n)$). Due to these savings on branching, optimal sphericity is $1 - \frac{5 \sqrt 3 π}{27n} - C_1 n^{-3/2} + (C_2±o(1)) n^{-5/3} \log^{2/3} n$.
* Sphericity is but one of a class of problems for which most of the conjectures apply.
Computation of the hexagonal limit: For a polyhedron with an inscribed sphere (tangent to all faces) of radius 1, $V=S/3$ where $S$ is the surface area (and maximum sphericity $n$-polyhedron is known to have an inscribed sphere tangent at face centroids). Thus, given average distance from the sphere center to the surface $1+ε$ (with the averaging using the solid angle), we get sphericity $1 - ε + O(ε^2)$. In turn, $ε=d^2/2-O(d^4)$ at distance $d$ from the tangential point. Given a regular $m$-gon face tangent at its center and with $\frac{4π}{n}$ solid angle, side length $s=\sqrt{\frac{16π}{mn} \tan \frac{π}{m}}+O(n^{-1})$, and $ε=(\frac{1}{16}\cot^2 \frac{π}{m}+\frac{1}{48}) s^2+O(n^{-2}) = $$ (\cot \frac{π}{m} + {\frac 1 3}\tan \frac{π}{m}) \frac{π}{mn} + O(n^{-2})$. Thus, idealized (1-sphericity)*n approaches $\frac{5 \sqrt 3 π}{27}$ for $m=6$ (and $\frac{π}{3}$ for $m=4$, and $\frac{2\sqrt{3} π}{9}$ for $m=3$). For small $n$, some computations are in The Isoperimetric Problem for Polyhedra (Goldberg 1935).
Related problems:
* The construction also appears to work well for the Thomson problem (classical electrons on a sphere) if we use area preserving deformations for the regions and inter-region alignments; such deformations exist. This should also work well for some other Riesz $s$-potentials, and for minimal length equal-area partitions of the sphere.
* For unit circle packing on a sphere (Tammes problem), I do not see a way to smooth out fractional distances, so I only have $\frac{π\sqrt{3}}{6}-O(n^{-1/3})$ density, using regions of width $Θ(n^{-1/6}R)$, with each region approximately hexagonally packed, and with higher scale structure (above $Θ(n^{-1/6}R)$) less important for the packing density.
* If the maximum sphericity polyhedron must instead have a given number of vertices, we get sphericity within $O(n^{-3/2})$ of the idealized triangle lattice. Approximately square lattice appears best when $n_V+kn_F$ is fixed for $≈2.8<k<≈24.5$ (sphericity depends more on the number of faces $n_F$ than the number of vertices $n_V$); for that lattice, I did not verify that face planarity is consistent with the exponent $O(n^{-3/2})$.
