Complete Spherical Convex Bodies

Similarly to the classic notion in Euclidean space, we call a set on the sphere $S^d$ complete, if adding any extra point increases the diameter. Complete sets are convex bodies of $S^d$. We prove that on $S^d$ complete bodies and bodies of constant width coincide.


On spherical geometry
Let S d be the unit sphere in the (d + 1)-dimensional Euclidean space E d+1 , where d ≥ 2. By a great circle of S d we mean the intersection of S d with any two-dimensional subspace of E d+1 . The common part of the sphere S d with any hyper-subspace of E d+1 is called a (d − 1)-dimensional great sphere of S d . By a pair of antipodes of S d we mean any pair of points of intersection of S d with a straight line through the origin of E d+1 .
Clearly, if two different points a, b ∈ S d are not antipodes, there is exactly one great circle containing them. By the arc ab connecting a and b we mean the "smaller" part of the great circle containing a and b. By the spherical distance |ab|, or shortly distance, of these points we understand the length of the arc connecting them. The diameter diam(A) of a set A ⊂ S d is the number sup a,b∈A |ab|. By a spherical ball B ρ (r) of radius ρ ∈ (0, π 2 ], or shorter a ball, we mean the set of points of S d having distance at most ρ from a fixed point, called the center of this ball. Spherical balls of radius π 2 are called hemispheres. Two hemispheres whose centers are antipodes are called opposite hemispheres.
We say that a subset of S d is convex if it does not contain any pair of antipodes and if together with every two points a, b it contains the arc ab. By a convex body, or shortly body, on S d we mean any closed convex set with non-empty interior.
Recall a few notions from [6]. If for a hemisphere H containing a convex body C ⊂ S d we have bd(H) ∩C = ∅, then we say that H supports C. If hemispheres G and H of S d are different and not opposite, then L = G ∩ H is called a lune of S d . The (d − 1)-dimensional hemispheres bounding the lune L and contained in G and H, respectively, are denoted by G/H and H/G. We define the thickness of a lune L = G ∩ H as the spherical distance of the centers of G/H and H/G. For a hemisphere H supporting a convex body C ⊂ S d we define the width width H (C) of C determined by H as the minimum thickness of a lune of the form H ∩ H ′ , where H ′ is a hemisphere, containing C. If for all hemispheres H supporting C we have width H (C) = w, we say that C is of constant width w.

Spherical complete bodies
Similarly to the traditional notion of a complete set in the Euclidean space E d (for instance, see [1], [2], [3] and [10]) we say that a set Theorem 1. Arbitrary set of a diameter δ ∈ (0, π) on the sphere S d is a subset of a complete set of diameter δ on S d .
We omit the proof since it is similar to the proof by Lebesque [9] in E d (it is recalled in Part 64 of [1]). Let us add that earlier Pál [12] proved this for E 2 by a different method.
The following fact permits to use the term a complete convex body for a complete set.
Then K coincides with the intersection of all balls of radius δ centered at points of K. Moreover, K is a convex body.
Proof. Denote by I the intersection of all balls of radius δ with centers in K.
Since diam(K) = δ, then K is contained in every ball of radius δ whose center is at a point of K. Consequently, K ⊂ I.
Let us show that I ⊂ K, so let us show that x ∈ K implies x ∈ I. Really, from x ∈ K we get |xy| > δ for a point y ∈ K, which means that x is not in the ball of radius δ and center y, and thus x ∈ I.
As an intersection of balls, K is a convex body.
Proof. Suppose the contrary, i.e., that |pq| < δ for a point p ∈ bd(K) and for every point q ∈ K. Since K is compact, there is an ε > 0 such that |pq| ≤ δ − ε for every q ∈ K. Hence, in a positive distance below ε from p there is a point s ∈ K such that |sq| ≤ δ for every q ∈ K. Thus diam(K ∪ {s}) = δ, which contradicts the assumption that K is complete. Consequently, the thesis of our lemma holds true.
For different points a, b ∈ S d at a distance δ < π from a point c ∈ S d define the piece of circle P c (a, b) as the set of points v ∈ S d such that cv has length δ and intersects ab.
We show the next lemma for S d despite we apply it later only for S 2 .
Lemma 3. Let K ⊂ S d be a complete convex body of diameter δ. Take P c (a, b) with |ac| and |bc| equal to δ such that a, b ∈ K and c ∈ S d . Then P c (a, b) ⊂ K.
Proof. First let us show the thesis for a ball B of radius δ in place of K. There is unique S 2 ⊂ S d with a, b, c ∈ S 2 . Consider the disk D = B ∩ S 2 . Take the great circle containing P c (a, b) and points a * , b * of its intersection with the circle bounding D. There is unique c * ∈ S 2 such that P c (a, b) ⊂ P c * (a * , b * ). Clearly, P c * (a * , b * ) ⊂ D ⊂ B. Hence P c (a, b) ⊂ B. By the preceding paragraph and Lemma 1 we obtain the thesis of the present lemma.

Complete and constant width bodies on S d coincide
Here is our main result presenting the spherical version of the classic theorem in E d proved by Meissner [11] for d = 2, 3 and by Jessen [5] for arbitrary d. Proof. (⇒) Prove that if K ⊂ S d of diameter δ is complete, then K is of constant width δ.
Suppose the opposite, i.e., that width I (K) = δ for a hemisphere I supporting K. By Theorem 3 and Proposition 1 of [6] width I (K) ≤ δ. So ∆(K) < δ. By lines 1-2 of p. 562 of [6] the thickness of K is equal to the minimum thickness of a lune containing K. Take such a lune L = G ∩ H, where G, H are different and non-opposite hemispheres. Denote by g, h the centers of G/H and H/G, respectively. Of course, |gh| < δ. By Claim 2 of [6] we have g, h ∈ K. By Lemma 2 there exists a point g ′ ∈ K in the distance δ from g. Since the triangle ghg ′ is non-degenerate, there is a unique two-dimensional sphere S 2 ⊂ S d containing g, h, g ′ . Clearly, ghg ′ is a subset of M = K ∩ S 2 . Hence M is a convex body on S 2 . Denote by F this hemisphere of S 2 such that hg ′ ⊂ bd(F ) and g ∈ F . There is a unique c ∈ F such that |ch| = δ = |cg ′ |. By Lemma 3 for d = 2 we have P c (h, g ′ ) ⊂ M.
We intend to show that c is not on the great circle E of S 2 through g and h. In order to see this, for a while suppose the opposite, i.e. that c ∈ E. Then from |g ′ g| = δ, |g ′ c| = δ and |hc| = δ we conclude that ∠gg ′ c = ∠hcg ′ . So the spherical triangle g ′ gc is isosceles, which together with |gg ′ | = δ gives |cg| = δ. Since |gh| = ∆(L) = ∆(K) > 0 and g is a point of ch different from c, we get a contradiction. Hence, really, c ∈ E.
By the preceding paragraph P c (h, g ′ ) intersects bd(M) at a point h ′ different from h and g ′ . So the non-empty set P c (h, g ′ ) \ {h, h ′ } is out of M. This contradicts the result of the paragraph before the last. Consequently, K is a body of constant width δ.
(⇐) Let us prove that if K is a spherical body of constant width δ, then K is a complete body of diameter δ. In order to prove this, it is sufficient to take any point r ∈ K and show that diam(K ∪ {r}) > δ.
Take the largest ball B ρ (r) disjoint with the interior of K. Since K is convex, B ρ (r) has in common with K exactly one point p. By Theorem 3 of [8] there exists a lune L ⊃ K of thickness δ with p as the center of one of the two (d − 1)-dimensional hemispheres bounding this lune. Denote by q the center of the other (d − 1)-dimensional hemisphere. By Claim 2 of [6] also q ∈ K. Since p and q are the centers of the two (d − 1)-dimensional hemispheres bounding L, we have |pq| = δ. From the fact that rp and pq are orthogonal to bd(H) at p, we see that p ∈ rq. Moreover, p is not an endpoint of rq and |pq| = δ, Hence |rq| > δ. Thus diam(K ∪ {r}) > δ. Since r ∈ K is arbitrary, K is complete.
We say that a convex body D ⊂ S d is of constant diameter δ provided diam(D) = δ and for every p ∈ bd(D) there is a point p ′ ∈ bd(D) with |pp ′ | = δ (see [8]).
The following fact is analogous to the result in E d given by Reidemeister [13]. Proof. Take a complete body D ⊂ S d of diameter δ. Let g ∈ bd(D) and G be a hemisphere supporting D at g. By Theorem 2 our D is of constant width δ. So width G (D) = δ and a hemisphere H exists that the lune G ∩ H ⊃ D has thickness δ. By Claim 2 of [6] centers h of H/G and g of G/H belong to D. So |gh| = δ. Thus D is of constant diameter δ. Consider a body D ⊂ S d of constant diameter δ. Let r ∈ D. Take the largest B ρ (r) whose interior is disjoint with D. Denote by p the common point of B ρ (r) and D. A unique hemisphere J supports B ρ (r) at p. Observe that D ⊂ J (if not, there is point v ∈ D out of J; clearly vp passes through intB ρ (r), a contradiction). Since D is of constant diameter δ, there is p ′ ∈ D with |pp ′ | = δ. Observe that ∠rpp ′ ≥ π 2 . If it is π 2 , then |rp ′ | > δ. If it is over π 2 , the triangle rpp ′ is obtuse and then by the law of cosines |rp ′ | > |pp ′ | and hence |rp ′ | > δ. By |rp ′ | > δ in both cases we see that D is complete.
By Theorem 2, in Theorem 3 we may exchange "complete" to "constant width". This form is proved earlier as follows. Any body of constant width δ on S d is of constant diameter δ and the inverse is shown for δ ≥ π 2 , and for δ < π 2 if d = 2 (see [8]). By [4] the inverse holds for any δ. Our short proof of Theorem 3 is different from these in [8] and [4].