Spherical Curves

When does a curve in space lie on a sphere? That is, is there a way to tell if some given curve actually lies on a sphere, without being given that information in advance?

My students asked this nice question, whose answer shows off the basic techniques of classical differential geometry very clearly. So, I figure it is helpful to share the answer, and also the way we found it in class.



I suppose that it is worth mentioning that I knew in advance that the question had an answer, though I couldn't recall the exact formulation. This gave me some confidence to change ahead. I had completely forgotten both the answer and the path to it, but there is a short list of powerful tools that one can use in classical differential geometry. Often, you can make headway by just trying a bunch of different permutations of the tricks and seeing what you can do with the results.

Tricks to keep in mind:

1. When in doubt, take a derivative.
2. The Frenet-Serret apparatus $\{T,N,B\}$ forms an orthonormal basis of ${\mathbb{R}}^3$ for every value of the parameter.
3. When working with theoretical curves, making the assumption of arclength parametrization is a good idea.

Let's sharpen our question and the desired outcome to make it fit the expectations that classical differential geometers would recognize.

Let $\gamma$ be a unit-speed curve in ${\mathbb{R}}^3$. Find a criterion which is computable from $\gamma$ and ensures that the image of $\gamma$ lies on some sphere, which is unknown in advance.

We seek the answer in the form of a differential equation which such a curve must satisfy. In what follows, we shall use a prime to denote differentiation with respect to the arclength parameter $s$.

We begin by searching for a necessary criterion. Assume that $\gamma$ lies on a sphere with center $p$ and radius $R$. Then, for each value of the parameter $s$, we see that

$$R^2=⟨\gamma (s)-p,\gamma (s)-p⟩.$$

Taking a derivative, using the product rule and symmetry and clearing a constant, we find that 

$$\begin{array}{}\text{(1)}& 0=⟨\gamma (s)-p,T⟩\end{array}$$

(Recall that $T=T(s)$ is really a function. But things are easier to write if you suppress some of the functional notation.) This still leaves $p$ in the equation, so we can't be done. Let's try another derivative and see if we get lucky. This leads to 

$$0=⟨\gamma (s)-p,T^{\prime }⟩+⟨T,T⟩=\kappa ⟨\gamma (s)-p,N⟩+1,$$

which is conveniently rearranged to read

$$\begin{array}{}\text{(2)}& -{\displaystyle \frac{1}{\kappa }}=⟨\gamma (s)-p,N⟩.\end{array}$$

At this point, it is easy to get discouraged. One can clearly see that the product rule works against us, and we will never get rid of the factor $\gamma (s)-p$, no matter how many times we take a derivative. Rather than submit to despair, let us take stock. Equations (1) and (2) give us expressions for the inner product of $\gamma (s)-p$ with the vectors $T$ and $B$. But that is two out of three elements of an orthonormal basis! 

An important property of an orthonormal basis is that for any vector, one can write write that vector as a linear combination of the basis vectors by computing inner products. In our situation, the exact fact is this:

$$\gamma (s)-p=⟨\gamma (s)-p,T⟩T+⟨\gamma (s)-p,N⟩+⟨\gamma (s)-p,B⟩.$$

So, we are really two thirds of the way to computing this nice representation. Maybe it can help us. Let's finish it. We need to find the component of $\gamma (s)-p$ in the binormal direction.

We start from Equation (2) and take a derivative to find

$${\displaystyle \frac{{\kappa }^{\prime }}{{\kappa }^2}}=⟨\gamma (s)-p,N^{\prime }⟩+⟨T,N⟩.$$

Using Equation (1) and the Frenet-Serret Equations, we see that

$${\displaystyle \frac{{\kappa }^{\prime }}{{\kappa }^2}}=⟨\gamma (s)-p,-\kappa T+\tau B⟩+0=\tau ⟨\gamma (s)-p,B⟩.$$

We rearrange this to find our third component:

$$\begin{array}{}\text{(3)}& ⟨\gamma (s)-p,B⟩={\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}.\end{array}$$

So we see that $\gamma (s)-p=-{\displaystyle \frac{1}{\kappa }}N+{\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}B$ for each value of $s$. It is now rather straightforward to compute that 

$$R^2=⟨\gamma (s)-p,\gamma (s)-p⟩={\displaystyle \frac{1}{{\kappa }^2}}+{\left({\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}\right)}^2$$

This is almost what we want! The quantity on the right-hand side is a quantity which can be computed from the original curve using only differentiation, and the dreaded center $p$ of the sphere no longer makes an appearance. To get the radius $R$ out of the equation, we need only take a single derivative. 

We deduce a spherical curve satisfies the following differential equation: 

$$\begin{array}{}\text{(4)}& {\{{\displaystyle \frac{1}{{\kappa }^2}}+{\left({\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}\right)}^2\}}^{\prime }=0\end{array}$$

Because it will be more convenient in a moment, we note that with some gentle reorganizing this equation can be rewritten as

$$\begin{array}{}\text{(5)}& {\left({\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}\right)}^{\prime }={\displaystyle \frac{\tau }{\kappa }}.\end{array}$$

Strictly speaking, we have only derived this as a necessary condition. What is left to do?

1. Check what might go wrong. Both $\kappa$ and $\tau$ appear in a denominator, so they must not vanish.
2. Check that the condition is also sufficient.

If $\kappa$ vanishes on any interval, then that portion of the curve is a straight line, so it is clearly not spherical. If $\tau$ vanishes on any interval, then that portion of the curve is planar. In this case, we become unable to form equation (3), and instead the line above it reads as ${\kappa }^{\prime }/{\kappa }^2=0$. This implies that ${\kappa }^{\prime }=0$, so that portion of the curve must be part of a circle. This all conforms to our expectations.

How do we see that the condition we derived is sufficient? Things are always easier when you know what the answer is supposed to be in advance. Suppose that $\gamma$ is a unit-speed curve satisfying equation (4). Take any small portion of the curve where neither the curvature nor the torsion vanish, and consider the associated curve

$$p(s)=\gamma (s)+{\displaystyle \frac{1}{\kappa }}N-{\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}B.$$

This curve is not necessarily parametrized by arclength! One then checks in a direct manner that

$${\displaystyle \frac{dp}{ds}}=T+{\left({\displaystyle \frac{1}{\kappa }}\right)}^{\prime }N+{\displaystyle \frac{1}{\kappa }}{N}^{\prime }-{\left({\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}\right)}^{\prime }B-{\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}{B}^{\prime }.$$ 

By using the Frenet-Serret equations and gathering terms, one eventually finds that the coefficients of $T$ and $N$ vanish, and 

$${\displaystyle \frac{dp}{ds}}=\{{\displaystyle \frac{\tau }{\kappa }}-{\left({\displaystyle \frac{{\kappa }^{\prime }}{\tau {\kappa }^2}}\right)}^{\prime }\}B.$$

So by equation (5), $dp/ds$ vanishes. This means that $p$ is a constant vector. Of course, it is the center of our sphere. It is then not hard to similarly compute that $||\gamma (s)-p||$ is a constant, which means that $\gamma$ lies on a sphere centered at $p$.