# Showing that a curve $\Gamma^*$ is spherical

I'm having troubles with this differential geometry problem:

Let $\Gamma : \overrightarrow{x} = \overrightarrow{x}(s)$ be a curve of $E^3$ with natural parameter $s$ which does not pass through the origin $O$. Now consider the curve

$$\Gamma^* : \overrightarrow{x}^*(s) = \frac{\overrightarrow{x}(s)}{f(s)},$$

where $f(s) = \left|\overrightarrow{x}(s)\right|$.

I need to show that $\Gamma^*$ is spherical and that $s$ is a natural parameter of $\Gamma^*$ if and only if $f^2 + (f')^2 = 1$.

Usually when we want to show that a curve is spherical it is sufficient to show that

$$\frac{\rho(s)}{\tau(s)} + (\rho'(s)\tau(s))' = 0$$

where $\rho(s) = \frac 1{\kappa(s)}$ is the radius of curvature and $\tau(s) = \frac 1{\sigma(s)}$ is the radius of torsion. However here I have no way of calculating these scalar quantities. I'm also lost with the second part of the question.

• Do we agree that you do not have to prove that the curve is spherical ($\dfrac{x}{\|x\|}$ has, by definition, a unit norm). The issue, as I understand it,is to impose that arc length is preserved in the "normaization" operation. – Jean Marie Aug 27 '16 at 15:23
• @JeanMarie Yes, I agree. Since $\frac{\overrightarrow{x}(s)}{\left|\overrightarrow{x}(s)\right|}$ has unit norm, then $\Gamma^*$ must be spherical. I think this is what my professor want us to notice. Thanks. – user347616 Aug 27 '16 at 17:13
• @TedShifrin Ok. Also, I don't understand how do we differentiate $\vec x^*$ is this particular case. Would you mind explain it to me? – user347616 Aug 28 '16 at 22:19
• Yes. Use that to simplify the expression you get when you differentiate $|\vec x^{*\prime}(t)|^2$, and you should be in good shape. – Ted Shifrin Aug 28 '16 at 23:32
• Oddly, there was a question some weeks ago about the general ($C^1$) solution to differential equation $f^2 + (f')^2 = 1$ see my answer. I would be interested if somebody finds a connection with the present question. – Jean Marie Aug 30 '16 at 4:55

In general, a curve is spherical if and only if it lives on some sphere - in our case, the sphere of radius $1$ and center $0$. Since

$$\| \vec x ^* (s) \| = \left\| \frac {\vec x(s)} {f(s)} \right\| = \left\| \frac {\vec x(s)} {\| \vec x (s) \|} \right\| = \frac {\| \vec x(s) \|} {\| \vec x (s) \|} = 1 ,$$

it is clear that $\vec x ^* (s)$ sits on the unit sphere for every $s$. (Notice that the condition that $\Gamma$ should not pass through $0$ insures that $f(s) \ne 0$, so that we may divide by $f(s)$.)

Moving on to the second part, notice that if $\vec x ^* = \dfrac {\vec x} f$, then $\vec x = f \vec x^*$, so deriving with respect to $s$ gives

$$\vec x ' = f' \vec x ^* + f {\vec x ^*} ' ,$$

whence

$$\tag {*}\| \vec x ' \| ^2 = \langle \vec x ' , \vec x '\rangle = \langle f' \vec x ^* + f {\vec x ^*} ' , f' \vec x ^* + f {\vec x ^*} '\rangle = (f')^2 \langle \vec x ^*, \vec x ^* \rangle + 2 f' f \langle \vec x ^*, {\vec x ^*} '\rangle + f^2 \langle {\vec x ^*} ', {\vec x ^*} ' \rangle .$$

Since $\Gamma ^*$ is spherical, we have $\langle \vec x ^*, \vec x ^* \rangle = 1$. But we can squeeze even more out of this fact! Deriving it with respect to $s$ gives

$$0 = \langle {\vec x ^*} ' , {\vec x ^*} \rangle + \langle {\vec x ^*} , {\vec x ^*} ' \rangle = 2 \langle {\vec x ^*} ' , {\vec x ^*} \rangle ,$$

whence $\langle {\vec x ^*} ' , {\vec x ^*} \rangle = 0$, a result that we shall also plug in $(*)$.

Finally, remember that $\Gamma$ carries the natural parametrization, which implies that $\| \vec x ' \| = 1$, yet another fact that we shall plug in $(*)$.

After all these remarks, then, formula $(*)$ can be rewritten as

$$\tag {**} 1 = (f')^2 + f^2 \langle {\vec x ^*} ', {\vec x ^*} ' \rangle .$$

Things are easy now. If $\Gamma ^*$ carries the natural parametrization, then $\| {\vec x ^*} ' \| = 1$, so $(**)$ becomes $1 = (f')^2 + f^2$. Conversely, if $1 = (f')^2 + f^2$, then $(**)$ becomes $f^2 = f^2 \langle {\vec x ^*} ', {\vec x ^*} ' \rangle$ and, since $\Gamma$ does not pass through $0$, we have $f(s) \ne 0 \ \forall s$ so we may divide by $f^2$ and obtain $1 = \langle {\vec x ^*} ', {\vec x ^*} ' \rangle = \| {\vec x ^*} ' \| ^2$, whence $\| {\vec x ^*} ' \| = 1$, showing that $\Gamma ^*$ carries the natural parametrization.

It is interesting to notice that if you had started your proof by deriving $\vec x ^* = \dfrac {\vec x} f$, you would have obtained

$$\| {\vec x ^*} ' \| ^2 = \frac {(f') ^2} {f^4} \| x \| ^2 - 2 \frac {f'} {f^3} \langle \vec x, \vec x ' \rangle + \frac 1 {f^2} \| \vec x ' \| ^2$$

and the only thing that you could have done here would have been replacing $\| \vec x ' \|$ by $1$ (because of $\Gamma$ being naturally parametrized), and then you would have been stuck.

The thing that you learn from here is that the most straightforward and intuitive approach to a problem might quickly get you stuck, while a more convoluted approach might lead to the solution. (Unfortunately, this happens almost always in mathematics, which is why work in this field is so difficult.)

• This is a wonderful solution, thank you. I was able to follow all your demonstration with ease. The footnote is also extremely interesting, especially because it explains why my approach could not have led anywhere and therefore was wrong. There is still one big question which worries me: how can I get all of these intuitions to solve these kinds of exercices? If I had thought about the problem a bit more maybe I could have answered the first part, but I could never come up with the correct solution for the second part. – user347616 Aug 30 '16 at 13:08
• @ManonVd: At first, I also attempted to solve the problem in the naive way, quickly realizing that this wasn't leading anywhere. Then I thought about the fact that this naive approach doesn't make use of the fact that $\Gamma ^*$ is spherical and canonically parametrized, so something was fishy. Fortunately, there weren't many options left: I just backtracked to the beginning and took the alternative route. – Alex M. Aug 30 '16 at 14:14