# Complete spherical convex bodies

## Abstract

Similarly to the classic notion in Euclidean space, we call a set on the sphere $$S^d$$ complete, provided adding any extra point increases its diameter. Complete sets are convex bodies on $$S^d$$. Our main theorem says that on $$S^d$$ complete bodies of diameter $$\delta$$ coincide with bodies of constant width $$\delta$$.

## On spherical geometry

Let $$S^d$$ be the unit sphere in the $$(d+1)$$-dimensional Euclidean space $$E^{d+1}$$, where $$d\ge 2$$. 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 $$a, b \in S^d$$ are not antipodes, there is exactly one great circle containing them. By the arc ab connecting a with b 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 mean the length of the arc connecting them. The diameter $$\mathrm{diam}(A)$$ of a set $$A \subset S^d$$ is the number $$\mathrm{sup}_{a, b \in A} |ab|$$. By a spherical ball $$B_{\rho }(r)$$of radius $$\rho \in (0, {\pi \over 2}]$$, or shorter a ball, we mean the set of points of $$S^d$$ having distance at most $$\rho$$ from a fixed point, called the center of this ball. Spherical balls of radius $$\pi \over 2$$ 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 ab 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 $$C \subset S^d$$ we have $$\mathrm{bd}(H) \cap C \not = \emptyset$$, then we say that H supports C. If hemispheres G and H of $$S^d$$ are different and not opposite, then $$L = G \cap H$$ 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 $$L = G \cap H$$ as the spherical distance of the centers of G/H and H/G. For a hemisphere H supporting a convex body $$C \subset S^d$$ we define the width $$\mathrm{width}_H(C)$$of C determined by H as the minimum thickness of a lune of the form $$H \cap H'$$, where $$H'$$ is a hemisphere, containing C. If for all hemispheres H supporting C we have $$\mathrm{width}_H(C) = w$$, 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 $$K \subset S^d$$ of diameter $$\delta \in (0, \pi )$$ is complete provided $$\mathrm{diam}(K \cup \{x\}) > \delta$$ for every $$x \not \in K.$$

### Theorem 1

An arbitrary set of diameter $$\delta \in (0, \pi )$$ on the sphere $$S^d$$ is a subset of a complete set of diameter $$\delta$$ on $$S^d$$.

We omit the proof since it is similar to the proof by Lebesgue [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.

### Lemma 1

Let $$K \subset S^d$$ be a complete set of diameter $$\delta$$. Then K coincides with the intersection of all balls of radius $$\delta$$ centered at points of K. Moreover, K is a convex body.

### Proof

Denote by I the intersection of all balls of radius $$\delta$$ with centers in K.

Since $$\mathrm{diam}(K) = \delta$$, then K is contained in every ball of radius $$\delta$$ whose center is a point of K. Consequently, $$K \subset I$$.

Let us show that $$I \subset K$$; so let us show that $$x \not \in K$$ implies $$x \not \in I$$. Really, from $$x \not \in K$$ we get $$|xy| > \delta$$ for a point $$y \in K$$, which means that x is not in the ball of radius $$\delta$$ with center y, and thus $$x \not \in I$$.

The first thesis implies that K is a convex body. $$\square$$

### Lemma 2

If $$K \subset S^d$$ is a complete body of diameter $$\delta$$, then for every $$p \in \mathrm{bd}(K)$$ there exists $$p' \in K$$ such that $$|pp'|=\delta$$.

### Proof

Suppose the contrary, i.e., that $$|pq| < \delta$$ for a point $$p \in \mathrm{bd}(K)$$ and for every point $$q \in K$$. Since K is compact, there is an $$\varepsilon > 0$$ such that $$|pq| \le \delta - \varepsilon$$ for every $$q \in K$$. Hence there is a point $$s \not \in K$$ in a positive distance from p which is smaller than $$\varepsilon$$ such that $$|sq| \le \delta$$ for every $$q \in K$$. Thus $$\mathrm{diam}(K \cup \{s\}) = \delta$$, which contradicts the assumption that K is complete. Consequently, the thesis of our lemma holds true. $$\square$$

For different points $$a, b \in S^d$$ at a distance $$\delta < \pi$$ from a point $$c \in S^d$$ define the piece of the circle $$P_c(a,b)$$ as the set of points $$v \in S^d$$ such that cv has length $$\delta$$ and intersects ab.

We show the next lemma for $$S^d$$ despite we apply it later only for $$S^2$$.

### Lemma 3

Let $$K \subset S^d$$ be a complete convex body of diameter $$\delta$$. Take $$P_c(a,b)$$ with |ac| and |bc| equal to $$\delta$$ such that $$a, b \in K$$ and $$c \in S^d$$. Then $$P_c(a,b) \subset K$$.

### Proof

First let us show the thesis for a ball B of radius $$\delta$$ in place of K. There is unique $$S^2 \subset S^d$$ with $$a, b, c \in S^2$$. Consider the disk $$D= B \cap S^2$$. Take the great circle containing $$P_c(a,b)$$ and points $$a^*, b^*$$ of its intersection with the circle bounding D. There is a unique $$c^* \in S^2$$ such that $$P_c(a,b) \subset P_{c^*}(a^*,b^*)$$. Clearly, $$P_{c^*}(a^*,b^*) \subset D \subset B$$. Hence $$P_c(a,b) \subset B.$$

By the preceding paragraph and Lemma 1 we obtain the thesis of the present lemma. $$\square$$

## 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.

### Theorem 2

A body of diameter $$\delta$$ on $$S^d$$ is complete if and only if it is of constant width $$\delta$$.

### Proof

($$\Rightarrow$$) Let us prove that if a body $$K \subset S^d$$ of diameter $$\delta$$ is complete, then K is of constant width $$\delta$$.

Suppose the opposite, i.e., that $$\mathrm{width}_I(K) \not = \delta$$ for a hemisphere I supporting K. By Theorem 3 and Proposition 1 of [6] $$\mathrm{width}_I(K) \le \delta$$. So $$\varDelta (K) < \delta$$. 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 $$L = G \cap H$$, where GH are different and non-opposite hemispheres. Denote by gh the centers of G/H and H/G, respectively. Of course, $$|gh| < \delta$$. By Claim 2 of [6] we have $$g, h \in K$$. By Lemma 2 there exists a point $$g' \in K$$ in the distance $$\delta$$ from g. Since the triangle $$ghg'$$ is non-degenerate, there is a unique two-dimensional sphere $$S^2 \subset S^d$$ containing $$g, h, g'$$. Clearly, $$ghg'$$ is a subset of $$M= K \cap S^2$$. Hence M is a convex body on $$S^2$$. Denote by F this hemisphere of $$S^2$$ such that $$hg' \subset \mathrm{bd}(F)$$ and $$g \in F$$. There is a unique $$c \in F$$ such that $$|ch| =\delta = |cg'|$$. By Lemma 3 for $$d=2$$ we have $$P_c(h,g') \subset M$$.

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 $$c \in E$$. Then from $$|g'g| = \delta$$, $$|g'c| =\delta$$ and $$|hc| = \delta$$ we conclude that $$\angle gg'c = \angle hcg'$$. So the spherical triangle $$g'gc$$ is isosceles, which together with $$|gg'|= \delta$$ gives $$|cg| = \delta$$. Since $$|gh| = \varDelta (L) = \varDelta (K) >0$$ and g is a point of ch different from c, we get a contradiction. Hence, really, $$c \not \in E$$.

By the preceding paragraph $$P_c(h,g')$$ intersects $$\mathrm{bd}(M)$$ at a point $$h'$$ different from h and $$g'$$. So the non-empty set $$P_c(h,g') \setminus \{h,h'\}$$ is out of M. This contradicts the result of the paragraph before the last. Consequently, K is a body of constant width $$\delta$$.

($$\Leftarrow$$) Let us prove that if K is a spherical body of constant width $$\delta$$, then K is a complete body of diameter $$\delta$$. In order to prove this, it is sufficient to take any point $$r \not \in K$$ and to show that $$\mathrm{diam}(K \cup \{r\}) > \delta$$.

Take the largest ball $$B_{\rho }(r)$$ disjoint with the interior of K. Since K is convex, $$B_{\rho }(r)$$ has exactly one point p in common with K. By Theorem 3 of [8] there exists a lune $$L \supset K$$ of thickness $$\delta$$ 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 $$q \in K$$. Since p and q are the centers of the two $$(d-1)$$-dimensional hemispheres bounding L, we have $$|pq|= \delta$$. From the fact that rp and pq are orthogonal to $$\mathrm{bd}(H)$$ at p, we see that $$p \in rq$$. Moreover, p is not an endpoint of rq and $$|pq| = \delta$$, Hence $$|rq| > \delta$$. Thus $$\mathrm{diam}(K \cup \{r\}) > \delta$$. Since $$r \not \in K$$ is arbitrary, K is complete. $$\square$$

We say that a convex body $$D \subset S^d$$ is of constant diameter $$\delta$$ provided $$\mathrm{diam}(D) = \delta$$ and for every $$p \in \mathrm{bd}(D)$$ there is a point $$p' \in \mathrm{bd}(D)$$ with $$|pp'| = \delta$$ (see [8]). The following fact is analogous to the result in $$E^d$$ given by Reidemeister [13].

### Theorem 3

Bodies of constant diameter on $$S^d$$ coincide with complete bodies.

### Proof

Take a complete body $$D \subset S^d$$ of diameter $$\delta$$. Let $$g \in \mathrm{bd}(D)$$ and G be a hemisphere supporting D at g. By Theorem 2 the body D is of constant width $$\delta$$. So $$\mathrm{width}_G(D) = \delta$$ and there exists a hemisphere H such that the lune $$G \cap H \supset D$$ has thickness $$\delta$$. By Claim 2 of [6] the centers h of H/G and g of G/H belong to D. So $$|gh| =\delta$$. Thus D is of constant diameter $$\delta$$.

Consider a body $$D \subset S^d$$ of constant diameter $$\delta$$. Let $$r \not \in D$$. Take the largest $$B_{\rho }(r)$$ whose interior is disjoint with D. Denote by p the common point of $$B_{\rho }(r)$$ and D. A unique hemisphere J supports $$B_{\rho }(r)$$ at p. Observe that $$D \subset J$$ (if not, there is a point $$v \in D$$ out of J; clearly vp passes through $$\mathrm{int}B_{\rho }(r)$$, a contradiction). Since D is of constant diameter $$\delta$$, there is $$p' \in D$$ with $$|pp'| =\delta$$. Observe that $$\angle rpp' \ge \frac{\pi }{2}$$. If it is $$\frac{\pi }{2}$$, then $$|rp'| > \delta$$. If it is larger than $$\frac{\pi }{2}$$, the triangle $$rpp'$$ is obtuse and then by the law of cosines $$|rp'| > |pp'|$$ and hence $$|rp'| > \delta$$. By $$|rp'| > \delta$$ in both cases we see that D is complete. $$\square$$

Theorem 2 permits to change “complete” to “constant width” in Theorem 3. This form is proved earlier as follows. In [8] it is shown that any body of constant width $$\delta$$ on $$S^d$$ is of constant diameter $$\delta$$ and the inverse is argued for $$\delta \ge \frac{\pi }{2}$$, and for $$\delta < \frac{\pi }{2}$$ if $$d=2$$. By [4] the inverse holds for any $$\delta$$. Our short proof of Theorem 3 is quite different from the considerations in [4, 8] and [7].

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Correspondence to Marek Lassak.

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Lassak, M. Complete spherical convex bodies. J. Geom. 111, 35 (2020). https://doi.org/10.1007/s00022-020-00547-2

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### Keywords

• Sphere
• lune
• convex body
• complete body
• constant width
• constant diameter

• 52A55