Spherical bodies of constant width

The intersection L of two different non-opposite hemispheres G and H of the d-dimensional unit sphere Sd\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$S^d$$\end{document} is called a lune. By the thickness of L we mean the distance of the centers of the (d-1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(d-1)$$\end{document}-dimensional hemispheres bounding L. For a hemisphere G supporting a convex body C⊂Sd\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C \subset S^d$$\end{document} we define widthG(C)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{width}_G(C)$$\end{document} as the thickness of the narrowest lune or lunes of the form G∩H\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$G \cap H$$\end{document} containing C. If widthG(C)=w\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm{width}_G(C) =w$$\end{document} for every hemisphere G supporting C, we say that C is a body of constant width w. We present properties of these bodies. In particular, we prove that the diameter of any spherical body C of constant width w on Sd\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$S^d$$\end{document} is w, and that if w<π2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$w < \frac{\pi }{2}$$\end{document}, then C is strictly convex. Moreover, we check when spherical bodies of constant width and constant diameter coincide.


Introduction
Consider the unit sphere S d in the (d + 1)-dimensional Euclidean space E d+1 for d ≥ 2. The intersection of S d with any two-dimensional subspace of E d+1 is called a great circle of S d . By a (d − 1)-dimensional great sphere of S d we mean the common part of S d with any hyper-subspace of E d+1 . The 1-dimensional great spheres of S 2 are called great circles. By a pair of antipodes of S d we understand a 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. As the arc ab connecting a and b we define the shorter part of the great circle containing these points. The length of this arc is called the spherical distance |ab| of a and b, or shortly distance. Moreover, we agree that the distance of coinciding points is 0, and that of any pair of antipodes is π. that the lune K ∩ K * is of the minimum thickness. In other words, there is a "narrowest" lune of the form K ∩ K over all hemispheres K supporting C. The thickness of the lune K ∩ K * is called the width of C determined by K. We denote it by width K (C).
We define the thickness Δ(C) of a spherical convex body C as the smallest width of C. This definition is analogous to the classical definition of thickness (also called minimal width) of a convex body in Euclidean space.
By the relative interior of a convex set C ⊂ S d we mean the interior of C with respect to the smallest sphere S k ⊂ S d that contains C. Now we intend to show that the intersection of all hemispheres containing A is contained in conv(A). Assume the opposite, i.e., that there is a point x / ∈ conv(A) which belongs to every hemisphere containing A. Since A is closed, by Lemma 1 of [6] the set conv(A) is also closed. Hence there is an ε > 0 such that B ε (x) ∩ conv(A) = ∅. Since these two sets are convex, we may apply the following more general version of Lemma 2 of [6]: any two convex disjoint sets on S d are subsets of two opposite hemispheres (which is true again by the separation theorem for convex cones in E d+1 ). So B ε (x) and conv(A) are in some two opposite hemispheres. Hence x does not belong to the one which contains conv(A). Clearly, that one also contains A. This contradicts our assumption on the choice of x, and thus the proof is finished.

A few lemmas on spherical convex bodies
We omit the simple proof of the next lemma, which is analogous to the situation in E d and needed a few times later. Here our hemisphere plays the role of a closed half-space there. If p = b then it is obvious. Otherwise there is a unique point q ∈ K/M such that p ∈ bq. Moreover, there exists x ∈ (K/M ) ∩ (M/K) such that q ∈ ax. The reader can easily show that points p, q belong to the triangle abx and thus observe that there exists y ∈ ab such that p ∈ xy, which confirms the statement from the first paragraph of the proof.

Lemma 2. Let
We have |by| ≤ |ba| < π 2 . The inequality |by| < π 2 means that y is in the interior of H(b). Of course, |bx| = π 2 , which means that x ∈ bd(H(b)). From the two preceding sentences we conclude that xy is a subset of H(b) with x being its only point on bd (H(b)). Thus, if |pb| = π 2 , we conclude that p ∈ bd(H(b)), and consequently p = x, which implies that p is a corner of K ∩ M . The last sentence means that the statement of our lemma holds true.
For every x ∈ S d at distance π 2 from o denote by x the point of the arc ox at distance μ from x. Consider two points Proof. Let o, m be points of S d and ρ be a positive number less than π 2 . Let us show that First assume that B ρ (o) ⊂ H(m). Let b be a boundary point of B ρ (o) such that o ∈ mb. We have: |om| = |bm| − |ob| = |bm| − ρ ≤ π 2 − ρ, which confirms the "only if" part of (1). Assume now that |om| ≤ π 2 − ρ. Let b be any point of B ρ (o). We have: |bm| ≤ |bo| + |om| ≤ ρ + π 2 − ρ = π 2 . Therefore every point of B ρ (o) is at a distance at most π 2 from m. Hence B ⊂ H(m), which confirms the "if" part of (1). So (1) is shown. Lemma 1 of [6] guarantees that Y = conv(B μ (x 1 ) ∪ B μ (x 2 )) is a closed set as a convex hull of a closed set. Consequently, from Lemma 1 we see that Y is the intersection of all hemispheres containing Y . Moreover, observe that an arbitrary hemisphere contains a set if and only if it contains the convex hull of it. Hence Y is the intersection of all hemispheres containing As a result of the preceding paragraph, in order to prove the statement of our lemma it is sufficient to show that every hemisphere H(m) containing . Thus, having (1) in mind we see that in order to verify this it is sufficient to show that for any m ∈ S d Let us assume the first two of these inequalities and show the third one. Observe that x, x 1 and x 2 belong to the spherical triangle x 1 x 2 o. Therefore the arcs xo and x 1 x 2 intersect. Denote the point of intersection by g.
In this paragraph we consider the intersection of S d with the threedimensional subspace of E d+1 containing x 1 , x 2 , m. Observe that this intersection is a two-dimensional sphere concentric with S d . Denote this sphere by S 2 . Denote by o the other unique point on S 2 such that the triangles x 1 x 2 o and x 1 x 2 o are congruent. By the first two inequalities of (2) we obtain into two parts so that x 1 belongs to one of them and x 2 belongs to the other. Therefore at least one of the arcs x 1 m and x 2 m, say x 1 m intersects go or go, say go. Denote this point of intersection by s. Taking the first assumption of (2) into account and using two times the triangle inequality we obtain Applying the just obtained inequality and looking now again on the whole S d we get |x m| ≤ |x g| + |gm| ≤ |x g| + |og| = |x o| = π 2 − μ which is the required inequality in (2). Thus by (2) also our lemma holds true.

Lemma 5. Let C ⊂ S d be a convex body. Every point of C belongs to the convex hull of at most d + 1 extreme points of C.
Proof. We apply induction with respect to d. For d = 1 the statement is trivial since every convex body on S 1 is a spherical arc. Let d ≥ 2 be a fixed integer. Assume that for each k = 1, 2, . . . , d − 1 every boundary point of a spherical convex body C ⊂ S k belongs to the convex hull of at most k + 1 extreme points of C.
Let x be a point of C. Take an extreme point e of C. If x is not a boundary point of C, take the boundary point f of C such that x ∈ ef . In the opposite case put f = x.
If f is an extreme point of C, the statement follows immediately. In the opposite case take a hemisphere K supporting C at f . Put C = bd(K) ∩ C. Observe that every extreme point of C is also an extreme point of C. Let Q be the intersection of the smallest linear subspace of E d+1 containing C with S d . Clearly, Q is isomorphic to S k for k < d. Moreover, C has non-empty relative interior with respect to Q, because otherwise there would exist a smaller linear subspace of E d+1 containing C . Thus, by the inductive hypothesis we obtain that f is in the convex hull of a set F of at most d extreme points of C. Therefore x ∈ conv({e} ∪ F ) which means that x belongs to the convex hull of d + 1 extreme points of C. This finishes the inductive proof.
The proof of the following d-dimensional lemma is analogous to that of the two-dimensional Lemma 4.1 from [8] shown there for a wider class of reduced spherical convex bodies. AEM Lemma 6. Let C ⊂ S d be a spherical convex body with Δ(C) > π 2 and let L ⊃ C be a lune such that Δ(L) = Δ(C). Each of the centers of the (d − 1)dimensional hemispheres bounding L belongs to the boundary of C and both are smooth points of the boundary of C.
Having the next lemma in mind, we note the obvious fact that the diameter of a convex body C ⊂ S d is realized only for some pairs of points of bd(C).

Lemma 7.
Assume that the diameter of a convex body C ⊂ S d is realized for points p and q. The hemisphere K orthogonal to pq at p and containing q ∈ K supports C.
Proof. Denote the diameter of C by δ.
Assume first that δ > π 2 . The set of points at distance at least δ from q is the ball B π−δ (q ), where q is the antipode of q. Clearly, K has only p in common with B π−δ (q ).
Since the diameter δ of C is realized for pq, every point of C is at distance at most δ from q. Thus C has empty intersection with the interior of B π−δ (q ).
Assume that K does not contain C. Then C contains a point b ∈ K. Observe that the arc bp has nonempty intersection with the interior of B π−δ (q ) [the reason: K is the only hemisphere touching B π−δ (q ) from outside at p]. On the other hand, by the convexity of C we have bp ⊂ C. This contradicts the fact from the preceding paragraph that C has empty intersection with the interior of B π−δ (q ). Consequently, K contains C. Now consider the case when δ ≤ π 2 . For every y ∈ K we have |pq| < |yq| which by |pq| = δ implies y ∈ C. Thus if y ∈ C, then y ∈ K.
Let us apply our Lemma 7 for a convex body C of diameter larger than π 2 . Having in mind that the center k of K is in pq and thus in C, by Part III of Theorem 1 in [6] we obtain that Δ(K ∩ K * ) > π 2 . This gives the following corollary which implies the other one. The symbol diam(C) denotes the diameter of C. Corollary 1. Let C ⊂ S d be a convex body of diameter larger than π 2 and let diam(C) be realized for points p, q ∈ C. Take the hemisphere K orthogonal to pq at p which supports C. Then width K (C) > π 2 . Corollary 2. Let C ⊂ S d be a convex body of diameter larger than π 2 and let K denote the family of all hemispheres supporting C. Then max K∈K width K (C) > π 2 .

Spherical bodies of constant width
If for every hemisphere supporting a convex body W ⊂ S d the width of W determined by K is the same, we say that W is a body of constant width (see [6] and for an application also [5]). In particular, spherical balls of radius smaller than π 2 are bodies of constant width. Also every spherical Reuleaux odd-gon (for the definition see [6], p. 557) is a convex body of constant width. Each of the 2 d+1 parts of S d dissected by d + 1 pairwise orthogonal (d − 1)dimensional spheres of S d is a spherical body of constant width π 2 , which easily follows from the definition of a body of constant width. The class of spherical bodies of constant width is a subclass of the class of spherical reduced bodies considered in [6] and [8], and mentioned by [3] in a larger context, (recall that a convex body R ⊂ S d is called reduced if Δ(Z) < Δ(R) for every body Z ⊂ R different from R, see also [7] for this notion in E d ).
By the definition of width and by Claim 2 of [6], if W ⊂ S d is a body of constant width, then every supporting hemisphere G of W determines a supporting hemisphere H of W for which G ∩ H is a lune of thickness Δ(W ) such that the centers of G/H and H/G belong to the boundary of W . Hence every spherical body W of constant width is an intersection of lunes of thickness Δ(W ) such that the centers of the (d − 1)-dimensional hemispheres bounding these lunes belong to bd(W ). Recall the related question from p. 563 of [6] if a convex body W ⊂ S d is of constant width provided every supporting hemisphere G of W determines at least one hemisphere H supporting W such that G ∩ H is a lune with the centers of G/H and H/G in bd(W ).
Here is an example of a spherical body of constant width on S 3 .
Example. Take a circle X ⊂ S 3 (i.e., a set congruent to a circle in E 2 ) of a positive diameter κ < π 2 , and a point y ∈ S 3 at distance κ from every point x ∈ X. Prolong every spherical arc yx by a distance σ ≤ π 2 − κ up to points a and b so that a, y, x, b are on one great circle in this order. All these points a form a circle A, and all points b form a circle B. On the sphere on S 3 of radius σ whose center is y take the "smaller" part A + bounded by the circle A. On the sphere on S 3 of radius κ + σ with center y take the "smaller" part B + bounded by B. For every x ∈ X denote by x the point on X such that |xx | = κ. Prolong every xx up to points d, d so that d, x, x , d are in this order and |dx| = σ = |x d |. For every x provide the shorter piece C x of the circle with center x and radius σ connecting b and d determined by x and also the shorter piece D x of the circle with center x of radius κ + σ connecting a and d determined by x. Denote by W the convex hull of the union of A + , B + and all the pieces C x and D x . It is a body of constant width κ + 2σ (hint: for every hemisphere H supporting W and every H * the centers of H/H * and H * /H belong to bd(W ) and the arc connecting them passes through one of our points x, or through the point y). Proof. Observe that if a ball touches W at p from inside, then there exists a unique hemisphere supporting W at p such that our ball touches this hemisphere at p. So for any ρ ∈ (0, π 2 ) there is at most one ball of radius ρ touching W from inside at p. Our aim is to show that we can always find one.
In the first part of the proof consider the case when p is an extreme point of W . By Theorem 4 of [6] there is a lune L = K ∩M of thickness w containing W such that p is the center of K/M . Denote by m the center of M and by k the center of K. Clearly, m ∈ pk and |pm| = w − π 2 . Since width M (W ) = w, by the third part of Theorem 1 of [6] the ball B w−π/2 (m) touches W from inside. Moreover, it touches W from inside at the center of M * /M . Since K is one of these hemispheres M * , our ball touches W at p.
In the second part consider the case when p is not an extreme point of W . From Lemma 5 we see that p belongs to the convex hull of a finite set E of extreme points of W . We do not lose the generality assuming that E is a minimum set of extreme points of W with this property. Hence p belongs to the relative interior of conv(E).
Take a hemisphere K supporting W at p and denote by o the center of K. Since p belongs to the relative interior of conv(E), by Lemma 2 we obtain conv(E) ⊂ bd(K). Moreover, conv(E) is a subset of the boundary of W .
We intend to show that for every x ∈ conv(E) the inclusion holds true, where x denotes the point on ox at distance w − π 2 from x. By Lemma 4 for w = μ, if (3) holds true for x 1 , x 2 ∈ conv(E), then (3) is also true for every point of the arc x 1 x 2 . Applying this lemma a finite number of times and considering the first part of this proof, we conclude that (3) is true for every point of conv(E), so in particular for p. Clearly, the ball B w− π 2 (p ) supports W at p from inside.
Both parts of the proof confirm the statement of our theorem.
By Lemma 6 we obtain the following proposition generalizing Proposition 4.2 from [8] for arbitrary dimension d. We omit an analogous proof.

Proposition 1. Every spherical body of constant width larger than π
2 (and more generally, every reduced body of thickness larger than π 2 ) of S d is smooth. From Corollary 2 we obtain the following corollary which implies the two other ones.  Proof. Take a body W of constant width w < π 2 and assume it is not strictly convex. Then there is a supporting hemisphere K of W that supports W at more than one point. By Claim 2 of [6] the centers a of K/K * and b of K * /K belong to bd(W ). Since K supports W at more than one point, K/K * contains also a boundary point x = a of W . By the first statement of Lemma 3 of [6] we have |xb| > |ab| = w. Hence diam(W ) > w.
By Corollary 4 we have diam(W ) ≤ π 2 . By Theorem 3 of [6] we see that w = diam(W ). This contradicts the inequality diam(W ) > w from the preceding paragraph. The contradiction means that our assumption that W is not strictly convex must be false. Consequently, W is strictly convex.
On p. 566 of [6] the question is put if for every reduced spherical body R ⊂ S d and for every p ∈ bd(R) there exists a lune L ⊃ R fulfilling Δ(L) = Δ(R) with p as the center of one of the two (d − 1)-dimensional hemispheres bounding this lune. The following theorem gives a positive answer in the case of spherical bodies of constant width. It is a generalization of the version for S 2 given as Theorem 5.3 in [8]. The idea of the proof of our theorem below for S d substantially differs from the one given for S 2 . Proof. Part I for w < π 2 . By Theorem 2 the body W is strictly convex, which means that all its boundary points are extreme. Thus the statement follows from Theorem 4 of [6].
Part II for w = π 2 . If p is an extreme point of W we again apply Theorem 4 of [6]. Consider the case when p is not an extreme point. Take a hemisphere G supporting W at p. Applying Corollary 5 we see that W ⊂ H(p). Clearly, the lune H(p) ∩ G contains W . The point p is at distance π 2 from every corner of this lune and also from every point of the opposite (d − 1)-dimensional hemisphere bounding the lune. Hence this is a lune that we are looking for.
Part III, for w > π 2 . By Lemma 5 the point p belongs to the convex hull conv(E) of a finite set E of extreme points of W . We do not lose the generality by assuming that E is a minimum set of extreme points of W with this property. Hence p belongs to the relative interior of conv(E). By Proposition 1 we know that there is a unique hemisphere K supporting W at p. Since p belongs to the relative interior of conv(E), by Lemma 2 we have conv(E) ⊂ bd(K). Moreover, conv(E) is a subset of the boundary of W .
By Theorem 4 of [6] for every e ∈ E there exists a hemisphere K * e (it plays the part of K * in Theorem 1 of [6]) supporting W such that the lune K ∩ K * e is of thickness Δ(W ) with e as the center of K/K * e . By Proposition 1, for every e the hemisphere K * e is unique. For every e ∈ E denote by t e the center of K * e /K and by k e the boundary point of K such that t e ∈ ok e , where o is the center of K. So e, k e are antipodes. Denote the set of all these points k e by Q.
Clearly, the ball B = B Δ(W )− π 2 (o) (as in Part III of Theorem 1 in [6]) touches W from inside at every point t e . Moreover, from the proof of Theorem 1 of [6] and from the earlier established fact that every e ∈ E is the center of K/K * e and every t e is the center of K * e /K we obtain that o belongs to all the arcs of the form et e .
Put U = conv(Q ∪ {o}). Denote by U B the intersection of U with the boundary of B, and by U W the intersection of U with the boundary of W . Having this construction in mind we see the following one-to-one correspondence between some pairs of points in U B and U W . Namely, between the pairs of points of U B and U W such that each pair is on the arc connecting o with a point of conv(Q). Now, we will show that U W = U B . Assume the opposite. By the preceding paragraph, our opposite assumption means that there is a point x which belongs to U W but not to U B . Hence |xo| > Δ(W ) − π 2 . Moreover, there is a boundary point y of the (d − 1)-dimensional great sphere bounding K such that o ∈ xy and a point y ∈ oy at distance Δ(W ) − π 2 from y. We have |xy | = |xo| + |oy| − |yy | > Δ(W ) − π 2 + π 2 − Δ(W ) − π 2 = π 2 . By Lemma 5 the point x belongs the convex hull of a finite set of extreme points of W . Assume for a while that all these extreme points are at distance at most π 2 from y . Therefore all of them are contained in H(y ). Thus their convex hull is contained in H(y ) and so x ∈ H(y ). This contradicts the fact established in the preceding paragraph that |xy | > π 2 . The contradiction shows that at least one of these extreme points is at distance larger than π 2 from y . Take such a point z for which |zy | > π 2 . Since z is an extreme point of W , by Theorem 4 of [6] it is the center of one of the (d − 1)-dimensional hemispheres bounding a lune L of thickness Δ(W ) which contains W . Hence by the third part of Lemma 3 of [6] every point of L different from the center of the other (d − 1)-dimensional hemisphere bounding L is at distance smaller than Δ(W ) from z. Taking into account that the distance of these centers is Δ(W ) we see that the distance of every point of L, and in particular of W , from z is at most Δ(W ).
By Theorem 1 the ball B Δ(W )− π 2 (y ) touches W from inside at y.