Let $γ : I → S^2$ be a regular curve in the 2-sphere. Let $\kappa_g$ denote the geodesic curvature. Regarding the curve $γ$ as a space curve $S^2 ⊂ \mathbb R^3$ and assuming it to be Frenet, calculate its curvature $κ$ and torsion $τ$ in terms of $\kappa_g$.

So we have $$\kappa_g = \langle T', \gamma \times T\rangle = \langle\gamma'', \gamma \times \gamma'\rangle$$ where the brackets denote the inner product, where $$T = \frac {d\gamma}{dt}, $$

and where the Frenet frame is the 3-dimensional orthonormal basis $(\gamma', \gamma \times \gamma', \gamma)$.

The curvature of a space curve is given by $$\kappa = \frac{\lVert \gamma \times \gamma'\rVert}{\lVert\gamma'\rVert^3}$$

The Torsion is given by $$\tau = \frac{\langle\gamma' \times \gamma'',\gamma'''\rangle}{\lVert\gamma' \times \gamma''\rVert^3}$$

So basically the objective is to write $\kappa$ and $\tau$ in terms of $\kappa_g$? I'm not really sure how it's possible to get something like $\lVert\gamma'\rVert^3$ out of the definition of $\kappa_g$.

On the other hand it sort of looks like we could obtain the numerator for $\tau$ by just differentiating $\kappa_g$ and noticing that both are of the form of a scalar triple product with the inner product between the two lower order derivatives and the highest order derivative with the numerator of $\kappa_g$ is one degree less than that of the numerator of $\tau$.

Can anyone let me know if I'm on the right track or perhaps suggest how I might acquire the entire expression(s) for $\kappa$ and $\tau$ from $\kappa_g$?

  • 1
    $\begingroup$ First of all, assume $\gamma$ is arclength-parametrized (I assume that's what the problem means anyhow). Use Frenet formulas and remember that $\kappa_n=\pm 1$ for any curve lying on the unit sphere. $\endgroup$ Mar 17, 2017 at 3:42

1 Answer 1


Here are a couple of explicit hints. For a theoretical problem, you (most) always want to assume arclength parametrization and avoid those cross-product expressions. I will use $T,N,B$ for the Frenet frame. Differentiation is with respect to arclength.

(1) If $\gamma$ is a curve on the unit sphere, show that $\gamma\cdot N = -1/\kappa$.

(2) By definition (since the unit normal to the sphere is the position vector), $\kappa N = \kappa_g (\gamma\times T) + \kappa_n\gamma$, and $\kappa_n = -1$. First, this tells you that $\kappa = \sqrt{\kappa_g^2+1}$.

(3) Next, differentiate this and precisely one term will involve $\tau B$. We can isolate $\tau$ by dotting with $B$. You should then solve for $\tau$ in terms of $\kappa_g'$ and $\kappa$.

  • $\begingroup$ Thanks for the sustained guidance @Ted Shifrin! Just a few questions; what exactly is $\kappa_n$? Is that the curvature in the direction of the normal vector? That doesn't sound right... anyways, I'm working on your first point and have found the following: $$\gamma N = \gamma*(\frac{\gamma^{''}}{||\gamma^{''}||})$$ but am unsure as to where to go from here. I was looking at Frenet Formulae and saw that $\gamma^{''} = \frac{dT}{ds} = \kappa N$ so I tried substituting but to no real avail. I tried starting from $-\frac{1}{\kappa}$ by substitution but cross product was messing me up $\endgroup$ Mar 20, 2017 at 2:48
  • $\begingroup$ @KyleO'Connell: Yes, $\kappa_n=\kappa N\cdot n$ always. Remember next that a curve on a sphere centered at the origin is characterized by $\gamma\cdot T=0$ (why?). At some point, you may find my differential geometry text (free .pdf linked in my profile) helpful. $\endgroup$ Mar 20, 2017 at 3:29
  • $\begingroup$ Ah, I see. I know now from Lemma 2.1 of your book that $\gamma \centerdot \gamma^{'} = 0$ iff $||\gamma||$ is constant. Which makes sense and the way you put it in the remark is very illuminating. "Any component of the tangent vector in the direction of the position vector would push the curve off the surface" So I get that I am unsure how I'm applying it given the formula I'm working with is in terms of the original function and second derivative. Could we just say $\gamma \centerdot \gamma^{''} = 0$ and, if so, I'm not sure how that helps me, just trying to think of possibilities. $\endgroup$ Mar 20, 2017 at 21:23
  • $\begingroup$ Kyle, the fundamental principle in differential geometry is that whenever you see some quantity, you should differentiate :) Especially when you have a quantity that is constant. So if $\gamma\cdot T = 0$, what does differentiation tell you? And, as I said ages ago, really keep the Frenet frame in there explicitly, rather than various derivatives of $\gamma$. $\endgroup$ Mar 20, 2017 at 21:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .