Necessary and sufficient condition for a surface to be a sphere

1. Introduction

Let \(S\) be a \(C^{1}\)-smooth closed connected surface in \(\mathbb{R}^3\), the boundaryof the domain \(D\), \(N=N_s\) be the unit outer normal to \(S\) at the point \(s\). Throughout we assume that \(S\) satisfies these assumptions. Let \(P\) be the normal section of \(D\). A normal section is the intersection of \(D\) and the plane containing \(N\). Our result is the following:

Theorem 1.1. If all the normal sections for a fixed \(N\) are discs, then \(S\) is a sphere. Conversely, if \(S\) is a sphere then all its normal sections are discs.

There are several "characterizations" of the sphere in the literature. We will use the following.

Lemma 1.2. Let \(r=r(p,q)\) be a parametric representation of \(S\). If \([r(p,q), N_s]=0\) for all \(s=s(p,q)\) on \(S\), then \(S\) is a sphere. Here \([r,N]\) is the vector product of two vectors.

A proof of this result can be found in [ 1, 2]. For convenience of the reader a short proof of Lemma 1.2 is given in Section 2.

2. Proof

Theorem 1.1. Let \(s\in S\) be a fixed point and \(P\) be one of the normal sections of \(D\) corresponding to \(N_s\). By assumption, this section is a disc. Let \(O\) be its center and \(R\) be its radius. Rotate \(P\) about \(N_s\). Each of the resulting normal sections is a disc of radius \(R\) centered at \(O\). If \(r=r(p,q)\) is a parametric representation of \(S\) then \([r, N]=0\) for every point of \(S\) because each such point belongs to a boundary of a disc centered at \(O\) with radius \(R\). From Lemma 1.2 it follows that \(S\) is a sphere.

Lemma 1.2. One has \(N=[r_p(p,q), r_q(p,q)]/|[r_p(p,q),r_q(p,q)]|\), where \([a,b]\) is the vector product of \(a\) and \(b\), and \(|a|\) is the length of the vector. Therefore \([r,N]=0\) implies \([r,[r_p(p,q),r_q(p,q)]]=0\) or \(r_p(r,r_q)- r_q(r,r_p)=0\), where \((a,b)\) is the scalar product of two vectors. The vectors \(r_p\) and \(r_q\) are linearly independent since the surface \(S\) is smooth. Thus, \((r,r_q)=0\) and \((r,r_p)=0\). Consequently \((r,r)=const\), that is, \(S\) is a sphere.

Competing Interests

The author declares that he has no competing interests.