Width of spherical convex bodies

For every hemisphere K supporting a convex body C on the sphere Sd we define the width of C determined by K. We show that it is a continuous function of the position of K. We prove that the diameter of every convex body $${C \subset S^d}$$C⊂Sd equals the maximum of the widths of C provided the diameter of C is at most $${\frac{\pi}{2}}$$π2. In a natural way, we define spherical bodies of constant width. We also consider the thickness Δ(C) of C, i.e., the minimum width of C. A convex body $${R \subset S^d}$$R⊂Sd is said to be reduced if Δ(Z) < Δ(R) for every convex body Z properly contained in R. For instance, bodies of constant width on Sd and regular spherical odd-gons of thickness at most $${\frac{\pi}{2}}$$π2 on S2 are reduced. We prove that every reduced smooth spherical convex body is of constant width.


Introduction
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 . In particular, for S 2 the (d − 1)-dimensional great spheres are great circles. 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 . Observe that if two different points are not antipodes, there is exactly one great circle containing them.
If two different points a, b ∈ S d are not antipodes, by the arc ab connecting them we mean the shorter 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. Moreover, we put π, if the points are antipodes and 0 if the points coincide. By a spherical ball 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. An open ball is the set of points of S d having distance smaller than ρ from a point. Balls on S 2 are called disks. Spherical balls of radius π 2 are called hemispheres. In other words, by a hemisphere of S d we mean the common part of S d with any closed half-space of E d+1 . We denote by H(m) the hemisphere whose center is m. Two hemispheres whose centers are antipodes are called opposite hemispheres. By an open hemisphere we mean the set of points having distance less than π 2 from a fixed point. By a spherical (d − 1)-dimensional ball of radius ρ ∈ (0, π 2 ] we mean the set of points of a (d − 1)-dimensional great sphere of S d which are at distance at most ρ, from a fixed point, called the center of this ball. The (d−1)-dimensional balls of radius π 2 are called (d − 1)-dimensional hemispheres. If d = 2, we call them semicircles.
We say that a set C ⊂ S d is convex if it does not contain any pair of antipodes and if together with every two points it contains the whole arc connecting them. By a convex body on S d we mean a closed convex set with non-empty interior. Observe that a set C ⊂ S d is a convex body if and only if it is contained in an open hemisphere and is an intersection of hemispheres. For a short survey of definitions of convexity on S d we refer to Sect. 9.1 of [1]. The literature concerning this subject is very large. For instance see [2,3] and [4].
Clearly, the intersection of every family of convex sets is also convex. Thus for every set Q ⊂ S d contained in an open hemisphere of S d there exists the unique smallest convex set containing Q. It is called the convex hull of Q and it is denoted by conv(Q).

Lemma 1. If Q ⊂ S d is a closed subset of an open hemisphere, then conv(Q)
is also closed. This lemma follows by applying an analogous theorem for compact sets in E d+1 .
If a (d − 1)-dimensional great sphere G of S d has a common point t with a convex body C ⊂ S d and if its intersection with the interior of C is empty, we say that G is a supporting (d − 1)-dimensional great sphere of C passing through t. We also say that G supports C at t. If H is the hemisphere bounded by G and containing C, we say that H supports C at t. If at every boundary point of a convex body C ⊂ S d exactly one hemisphere supports C, we say that the body is smooth.
By the well known fact that a set C ⊂ S d is convex if and only if the cone generated by it in E d+1 is convex and from the classic separation theorem in Euclidean space we obtain the following analogous fact for S d . Let P ⊂ S d be a convex body. Let Q ⊂ S d be a convex body or a hemisphere. We say that P touches Q from outside if P ∩ Q = ∅ and int(P ) ∩ int(Q) = ∅. We say that P touches Q from inside if P ⊂ Q and bd(P ) ∩ bd(Q) = ∅. In both cases, points of bd(P ) ∩ bd(Q) are called points of touching.
The convex hull V of k ≥ 3 points on S 2 such that none of them belongs to the convex hull of the remaining points is called a spherical convex k-gon. The mentioned points are called the vertices of V . We write . . , v k are successive vertices of V when we go around V on the boundary of V . In particular, when we take k ≥ 3 successive points in a spherical circle of radius less than π 2 on S 2 with equal distances of every two successive points, we obtain a regular spherical k-gon.

Lunes
If hemispheres G and H of S d are different and not opposite, then L = G ∩ H is called a lune of S d . This notion is considered in many books and papers, for lunes on S 2 see e.g. [5], p. 18. The (d − 1)-dimensional hemispheres bounding the lune L and contained in G and H, respectively, are denoted by G/H and H/G. Proof. If |ab| ≤ π 2 , then on the great circle containing the arc ab we find points p and q such that a ∈ pb, b ∈ aq, |pb| = |qa| = π 2 . If |ab| > π 2 , then on the great circle containing the arc ab we find points p and q such that q ∈ pb, p ∈ aq, |pb| = |qa| = π 2 . The lune L = H(p) ∩ H(q) is the one that we are looking for.
Since every lune L determines exactly one pair of centers of the (d − 1)dimensional hemispheres bounding L, from Claim 1 we see that there is a one-to-one correspondence between lunes and pairs of points (different and not antipodes) of S d . We omit the simple proof of the following lemma.
For a convex body C ⊂ S d they matter lunes containing it, and in particular such lunes for which both (d − 1)-dimensional hemispheres bounding the lune have non-empty intersection with C. We say that a lune passes through a boundary point p of a convex body C ⊂ S d if the lune contains C and if the boundary of the lune contains p. If the centers of both (d − 1)-dimensional hemispheres bounding a lune belong to C, then we call such a lune an orthogonally supporting lune of C.
By applying the classical Blaschke selection theorem (e.g., see [6], p. 64) in E d+1 we easily obtain its spherical analogue and also the following lemma.

Width and thickness of a spherical convex body
For every hemisphere K supporting a convex body C ⊂ S d we are looking for hemispheres K * supporting C such that the lunes K ∩ K * are of the minimum thickness, i.e., which are the "narrowest" lunes of the form K ∩ K over all hemispheres K supporting C. By compactness arguments we immediately see that at least one such hemisphere K * exists, and thus at least one corresponding lune K ∩ K * exists. Denote by width K (C) its thickness and we call it the width of C determined by K. This notion of width of C ⊂ S d is an analogue of the notion of width of a convex body of E d . How to find the width of C determined by a given hemisphere K? Theorem 1 presented below and its proof present a procedure for establishing width K (C).
First let us prove a lemma needed for the proof of Theorem 1.

Lemma 5.
Let G and H be different and not opposite hemispheres, and let g denote the center of G. If g ∈ bd(H), then by B denote the ball with center g which touches H (from inside or outside) and by t the point of touching. If g ∈ bd(H), we put t = g. We claim that t is always at the center of the Width of spherical convex bodies 559 Proof. If g ∈ bd(H) consider any two corners r 1 and r 2 of the lune G ∩ H. Look at the triangles gtr 1 and gtr 2 . Below we explain that they have three equal elements. They have the length of the common side gt. Since g is the center of G, we have |gr 1 | = π 2 = |gr 2 |. By the orthogonality of gt and H it follows that ∠gtr 1 = π 2 = ∠gtr 2 . Since these three elements are equal, we have |tr 1 | = |tr 2 | (by the way, they are both equal to π 2 since |r 1 r 2 | = π). When g ∈ bd(H), we have |gr 1 | = |gr 2 |; still r 1 , r 2 belong to the boundary of H. Thus t is always the center of the hemisphere H/G.
Denote by k the center of K.
I. If k ∈ C, then there exists a unique hemisphere K * supporting C such that the lune L = K ∩ K * contains C and has thickness width K (C). This hemisphere supports C at the point t at which the largest ball B with center k touches C from outside. We have where B denotes the largest ball with center k contained in C, and for every such t this hemisphere K * , denoted K * t , is unique. For every t we have Δ(K ∩ K * t ) = π 2 + ρ B , where ρ B denotes the radius of B. Proof. Figures 1 and 2 illustrate this theorem and its proof. They show the orthogonal look to the hemisphere K from outside. Part I. Since C is a convex body and B is a ball, we see that B touches C from outside and the point of touching is unique. Denote it by t (see Fig. 1). By Lemma 2, the bodies C and B are in some two opposite hemispheres. What is more, since B is a ball touching C from outside, this pair of hemispheres is unique. Denote by K * t the one which contains C. We intend to show that K * t is nothing else but the promised K * . Denote by k * the center of K * t . Since k is also the center of B and since B and K * t touch from outside at t, we have t ∈ kk * . From Lemma 5 we see that t is the center of the (d − 1)-dimensional hemisphere K * t /K. Analogously, from this lemma we conclude that the common point u of kk * and the boundary of K is the center of K/K * t . Since t and u are centers of the (d − 1)-dimensional hemispheres bounding the lune K ∩ K * t , we have |tu| = Δ(K ∩ K * t ). This and |kt| Part II. Clearly, there is at least one hemisphere K * supporting C at k. Of course, Δ(K ∩ K * ) = π 2 . By Lemma 5 we see that k is the center of K * /K. Part III. Take the largest ball B ⊂ C with center k. Clearly, there is at least one boundary point t of C which is also a boundary point of B (see Fig. 2). We find a hemisphere K * t which supports C at t. Of course, it also supports B and thus, for given t, it is unique.
For every t there is a unique point u ∈ K/K * t such that k ∈ tu. This, |ku| = π 2 and |kt| = ρ B imply |tu| = π 2 + ρ B . Hence the facts, resulting from Lemma 5, that t is the center of K * t /K and that u is the center of K/K * t give Δ(K ∩ K * t ) = π 2 + ρ B . If we assume that there exists a hemisphere M ⊃ C such that the lune K ∩ M is narrower than π 2 + ρ B , then this lune does not contain B, and hence it does not contain C either. A contradiction. Thus the narrowest lunes of the form K ∩ N containing C are of the form K ∩ K * t . Let us point out that in Part I, so if the center k of K does not belong to C, the lune K ∩ K * is unique. In Part II this narrowest lune K ∩ K * containing C is sometimes unique and sometimes not. This depends on the point k = t of C which belongs to the boundary of B. In Part III for any given point t of touching C by B from inside (we may have one, or finitely many, or infinitely many such points t), the lune K ∩ K * t is unique. For instance, if C ⊂ S 2 is a regular spherical triangle of sides π 2 and the circle bounding a hemisphere K contains a side of this triangle, then K ∩ K * is not unique. Namely, as K * we may take any hemisphere containing C, whose boundary contains this vertex of C which does not belong to K. The thickness of every such lune K ∩ K * equals π 2 . If C is a regular spherical triangle of sides over π 2 and the boundary of K contains a side of this triangle, then K ∩ K * is not unique either. This time the boundary of K * contains a side of C different from the side which is in K. So we have exactly two positions of K * .
Here are two corollaries from Theorem 1 (for the second we also apply Lemma 5).

Corollary 2. The point t of support in Theorem 1 is the center of the
We define the thickness Δ(C) of a convex body C ⊂ S d as follows: Δ(C) = inf{width K (C); K is a supporting hemisphere of C}.
Compactness arguments show that the infimum is realized. As a consequence, Δ(C) = min{width K (C); K is a supporting hemisphere of C}. By the definitions of width and thickness we conclude that the thickness of every convex body C ⊂ S d is equal to the minimum thickness of a lune containing C.
At this moment observe that our definition of width K (C) has an advantage, when applied to find the thickness of a convex body C ⊂ S d . Namely, it is sufficient to find the minimum of the values of width K (C) over all hemispheres K supporting C. Theorem 1 helps to establish every width K (C).
Example 1. Applying Theorem 1 we easily find the thickness of any regular triangle T α of angles α. Formulas of spherical trigonometry imply that Δ(T α ) = arccos cos α sin α/2 for α < π 2 . If α ≥ π 2 (but, of course, α < 2 3 π), then Δ(T α ) = α. In both cases Δ(T α ) is realized for width K (T α ), where K is a hemisphere whose bounding semicircle contains a side of T α . In the first case T α is symmetric with respect to the arc A connecting the centers of K/K * and K * /K, while in the second T α is symmetric with respect to the arc passing through the middle of A and having endpoints at the corners of the lune K ∩ K * . For α = π 2 there are infinitely many positions in K * . Theorem 2. As the position of the (d − 1)-dimensional supporting hemisphere of a convex body C ⊂ S d changes, the width of C determined by this hemisphere changes continuously.
Proof. We keep the notation of Theorem 1. Of course, the positions of k and thus of B depend continuously on K. Hence π 2 − ρ B and π 2 + ρ B change continously. This and Corollary 1 imply the thesis of our theorem. It does not matter here that for a fixed K sometimes the lunes K ∩ K * are not unique; still they are all of equal. This claim results immediately from Corollary 2 applied twice: for each of the two (d − 1)-dimensional hemispheres bounding L.
If for every hemisphere supporting a convex body C ⊂ S d the width of C determined by K is the same, we say that C is a body of constant width. In particular, spherical balls of radius smaller than π 2 are bodies of constant width.
Also every spherical Reuleaux odd-gon is a convex body of constant width. Recall this notion. Take a regular spherical k-gon v 1 . . , k are equal (the indices are taken modulo k). Denote them by δ. Assume that δ ≤ π 2 . Let B i , where i = 1, . . . , k, be the disk with center v i and radius δ. The set B 1 ∩ · · · ∩ B k is just a spherical Reuleaux k-gon.
By the definition of width and by Claim 2, if C ⊂ S d is a body of constant width, then every supporting hemisphere G of C determines a supporting Vol. 89 (2015) Width of spherical convex bodies 563 hemisphere H of C for which G ∩ H is a lune such that the centers of G/H and H/G belong to the boundary of C. Is the opposite true? More precisely, is a convex body C ⊂ S d of constant width provided every supporting hemisphere G of C determines at least one hemisphere H supporting C such that G ∩ H is a lune with the centers of G/H and H/G in the boundary of C?

Diameter
By the diameter diam(C) of a set C ⊂ S d we mean the supremum of the spherical distances between pairs of points of C. Clearly, if C is closed, the diameter of C is realized for at least one pair of points of C.

Claim 3.
Let diam(C) ≤ π 2 for a convex body C ⊂ S d and assume that diam(C) = |ab| for some points a, b ∈ C. Denote by L the lune such that a and b are the centers of the (d − 1)-dimensional hemispheres bounding L. We have C ⊂ L.
Remark 1. In general, Claim 3 does not hold true without the assumption that diam(C) ≤ π 2 . A simple counterexample is the triangle T = abc with |ab| = 2 3 π ≈ 2.0944, |bc| = π 6 ≈ 0.5236 and ∠abc = 95 • . From the Al Battani formulas, also called law of cosines for sides, (see, e.g., [5], p. 45), we get |ac| ≈ 2.0609. Consequently, |ab| = 2 3 π is the diameter of T . Since ∠abc = 95 • , the lune with centers a and b of the semicircles bounding it does not contain c. Still its thickness is 2 3 π. Thus this lune does not contain T . Compactness arguments lead to the conclusion that for every convex body C ⊂ S d the supremum of width H (C) over all hemispheres H supporting C is realized for a supporting hemisphere of C, that is, we may take here the maximum instead of supremum.
The following theorem is an analog of the classic theorem for Euclidean space.

Theorem 3.
Let diam(C) ≤ π 2 for a convex body C ⊂ S d . We have max{width K (C); K is a supporting hemisphere of C} = diam(C).
Proof. Let K be an arbitrary hemisphere supporting C and let s ∈ C be a point of support by K (see Fig. 1). Take k, t , u and K * like in Parts I and II of Theorem 1. By Lemma 3 we have |st| ≥ |ut|. Hence diam(C) ≥ |st| ≥ |ut| = width K (C). This and the assumption that K is an arbitrary hemisphere 564 M. Lassak AEM supporting C imply that diam(C) is at least the maximum of width K (C) over all supporting hemispheres K of C. Let a, b ∈ C be such that |ab| = diam(C). Take the lune L from Claim 3, i.e., H(p) ∩ H(q) like in its proof. Thus diam(C) equals the thickness of L, i.e., width H(p) (C). Hence diam(C) is at most the maximum of width K (C) over all supporting hemispheres K of C.
The following example shows that Theorem 3 requires the assumption diam(C) ≤ π 2 . Example 2. Let T be an isosceles triangle with base of length λ close to 0 and the height perpendicular to it of length μ ∈ ( π 2 , π). Denote by w the center of the base and by v the opposite vertex of T . Lemma 3 implies that wv is the diametrical segment of T . Take the hemisphere K supporting T at w. Denote by k the center of K. Clearly, k ∈ wv, so k is in the interior of T . Let ρ be the radius of the largest disk B with center k contained in T . The radius ρ of B is arbitrarily close to 0, as λ is sufficiently small. Applying Part III of Theorem 1 we conclude that the width of T determined by K is π 2 + ρ. Hence it may be arbitrarily close to π 2 , as λ is sufficiently small. On the other hand, the diameter |wv| of T may be arbitrarily close to π, as μ is sufficiently close to π.
Proof. Let K be an arbitrary hemisphere supporting C and let s ∈ C be a point of support by K (see Fig. 2). Take k, t and K * like in Parts I-III of Theorem 1.
If k ∈ int(C), so if we apply Parts I and II of Theorem 1, we repeat the consideration of the first paragraph of the proof of Theorem 3 which gives width K (C) ≤ diam(C).
Assume that k ∈ int(C), so that we apply Part III of Theorem 1. Clearly, |sk| = π 2 . Take the largest ball B with center k contained in C. Denote by ρ its radius. By Part III we have width K (C) = π 2 + ρ. Provide the great circle through s and k. It intersects the boundary of B at two points. Denote by z this from these two points for which k ∈ sz. We have |sz| = |sk| + |kz| = π 2 + ρ, which, by Part III, equals width K (C). This and |sz| ≤ diam(C) lead to the conclusion that width K (C) ≤ diam(C).
Since K is an arbitrary hemisphere supporting C, we get the thesis.

Reduced bodies
In analogy to the definition of reduced bodies in Euclidean space E d introduced in [7] (see also [8][9][10] and [11]), we define reduced convex bodies on S d . We Vol. 89 (2015) Width of spherical convex bodies 565 say that a convex body R ⊂ S d is reduced if Δ(Z) < Δ(R) for every convex body Z ⊂ R different from R. By our definition of bodies of constant width on S d we see that they are reduced bodies. In particular, every Reuleaux polygon on S 2 is a reduced body.
It is easy to show that all regular odd-gons on S 2 of thickness at most π 2 are reduced bodies. The assumption that the thickness is at most π 2 matters here. For instance take the regular triangle T α of angles α > π 2 (see Example 1). Take the hemisphere K whose boundary contains a side of T α and apply Part III of Theorem 1. The corresponding ball B ⊂ T α touches T α from inside at exactly two points t 1 , t 2 . Cutting off a part of T α by the shorter arc of the boundary of B between t 1 and t 2 we obtain a convex body Z ⊂ T α . We have Δ(Z) = Δ(T α ), which implies that T α is not reduced.
Dissect a disk on S 2 by two orthogonal great circles through its center. The four obtained parts are called quarters of disks. In particular, the triangle of sides and angles π 2 is a quarter of a disk. It is easy to see that every quarter of a disk is a reduced body and that the thickness of it is equal to the radius of the original disk. More generally, each of the 2 d parts of a spherical ball on S d dissected by d pairwise orthogonal great (d − 1)-dimensional spheres through the center of this ball is a reduced body of S d . We call it 1 2 d -th part of a ball. Clearly, its thickness is equal to the radius of the above ball.
We say that e is an extreme point of a convex body C ⊂ S d provided the set C \ {e} is convex. From the analogue of the Krein-Milman theorem for convex cones (e.g., see [12]) its analogue for spherical convex bodies follows: every convex body C ⊂ S d is the convex hull of its extreme points. This and the fact that the common part of any closed convex body C ⊂ S d with any of its supporting (d − 1)-dimensional great sphere is a closed convex set imply the following lemma.  Proof. Let B i be the open ball of radius Δ(R)/i centered at e and let R i = conv(R\B i ) for i = 2, 3, . . .. By Lemma 1 every R i is a convex body. Moreover, since e is an extreme point of R, R i is a proper subset of R. So, since R is reduced, Δ(R i ) < Δ(R). By the definition of thickness of a convex body, R i is contained in a lune L i of thickness Δ(R i ).
Let m i , m i be the centers of the (d − 1)-dimensional hemispheres H i , H i bounding L i . We claim that at least one of these two centers, say m i , belongs to the closure of R \ R i . The reason is that in the opposite case, there would be a neighborhood N i of m i such that N i ∩ R i = N i ∩ R, which would imply that H i supports R at m i . Moreover, H i supports R at m i . Hence Δ(R) = Δ(L i ) = Δ(R i ), in contradiction with Δ(R i ) < Δ(R).
Since m i ∈ R \ R i for i = 2, 3, . . . , we see that the sequence of points m 2 , m 3 , . . . tends to e. Consequently, e is the center of a (d − 1)-dimensional hemisphere bounding L.
Remark 2. Besides the lune from Theorem 4, sometimes we have additional lunes L ⊃ R of thickness Δ(R) through e for which e is not in the middle of a (d − 1)-dimensional hemisphere bounding L . This happens, for instance, when R is a spherical regular triangle T α with α ≤ π 2 . Theorem 4 leads to the following questions. Is it true that through every boundary point p of a reduced body R ⊂ S 2 a lune L ⊃ R of thickness Δ(R) passes? A consequence would be that every reduced body R ⊂ S 2 is an intersection of lunes of thickness Δ(R). Is a stronger version of the preceding question true, namely, that there is always such a lune L with p at the center of one of the two (d − 1)-dimensional hemispheres bounding L?
There are more questions on spherical reduced bodies. For instance, which properties of reduced bodies in E d , and especially in E 2 (see [9] and [10]), may be reformulated and proved for reduced bodies on S d ? Are there reduced spherical polytopes on S d , where d ≥ 3, different from 1 2 d -th part of a ball? Or at least a spherical simplex different from 1 2 d of S d (comp. [11]). By Theorem 4 we obtain the following spherical analog of a theorem from [8], see also Corollary 1 in [9] and [10].

Theorem 5. Every smooth reduced body on S d is of constant width.
Proof. Let R ⊂ S d be a smooth reduced body. Take any supporting hemisphere K of R. By Lemma 6 the boundary of K contains an extreme point e of R. Since R is smooth, K is the unique supporting hemisphere of R at e. Moreover, from Theorem 4 we see that through e a lune L ⊃ R of thickness Δ(R) passes. Thus L = K ∩ K * and hence width K (R) = Δ(R). This and the arbitrariness of K imply the thesis of our theorem.
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