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What is the minimum number of points on the sphere $S^d \subset \mathbb{R}^{d+1}$ which cannot be covered by $d+1$ double caps? A double cap is defined to be a set $\{x \in S^d: |\langle x,a \rangle| > 1/\sqrt{2}\}$ where $a \in S^d$.

I am particularly interested in the case $d=2$.

The following theorem by Alexandru Damian can be used to construct a lower bound. See here for a sketch of a proof and here for the original related question by Joseph O'Rourke.

Theorem. Let $P \subset S^d$ be a finite subset with cardinality $|P|$ and $A \subset S^d$ be a subset with spherical measure $\sigma^d(A)$ (normalised such that $\sigma^d(S^d)=1$). Then the following holds:

$$\nexists R \in SO(3):R(P) \subseteq A \Rightarrow |P|>\frac{1}{1-\sigma^d(A)}.$$

So (for $d=2$) choosing $A$ to be three disjoint double caps centred on three mutually orthogonal points we have $\sigma^2(A)=3-\frac{3}{\sqrt2}$, so any point set $P$ which cannot be covered by three double caps must satisfy:

$$|P|>\frac{1}{1-(3-3/\sqrt2)}=8.2426\ldots$$

i.e. any eight points on $S^2$ can be covered by three double caps.

Given this argument makes no use of the geometry of the set $A$ it seems to me to be a fairly weak lower bound. I am interested in finding a higher lower bound or indeed an exact solution.

Tommy Williams
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  • Unless I am missing something, your "double caps" are invariant under the antipodal map, so your question is really about $\mathbb R P^n$ rather than $S^n$ and furthermore it has a simpler formulation there, in terms of spherical caps (i.e., distance balls) rather than double caps. – Mikhail Katz Aug 13 '23 at 14:05
  • Yes, I believe that is an equivalent way of formulating the question. It is not immediately clear to me how this simpler formulation lends itself to an improved lower bound or solution though. – Tommy Williams Aug 13 '23 at 14:40
  • The regular dodecahedron has 20 vertices. Choosing one out of each antipodal pair, we get 10 points. Did you try to figure out if these 10 points can be covered by your "double caps"? – Mikhail Katz Aug 13 '23 at 16:15
  • Yes, the 10 points defined by a regular dodecahedron can be covered by three double caps. Consider w.l.g. the points on $S^2$ defined by the directions $(\pm1,\pm1,1),(\pm\phi,\pm\frac{1}{\phi},0),(0,\pm\phi,\frac{1}{\phi}),(\pm\frac{1}{\phi},0,\phi,)$. The double caps centred on the directions $(1,0,0),(0,\sqrt3,1),(0,-1,\sqrt3)$, for example, cover these 10 points. – Tommy Williams Aug 13 '23 at 17:50
  • Interesting. What's the smallest configuration that you have that cannot be covered by three double caps? What about 10 of the vertices of the dodecahedron plus six of the vertices of the dual icosahedron? – Mikhail Katz Aug 14 '23 at 11:10
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    The smallest set I have found is the following 13 directions: $$(1,0,0),(0,1,0),(0,0,1),(0,\pm1,1),(\pm1,0,1),(\pm1,1,0),(\pm1,\pm1,1).$$ This cannot be covered by three double caps by virtue of being not 3-colourable. Since no two orthogonal vertices can be covered by the same double cap and not 3-colourable implies the vertices cannot be partitioned into 3 subsets that each contain no two orthogonal vertices. – Tommy Williams Aug 14 '23 at 14:40

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