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I have come across the following exercise (the context is curves and surfaces in $\mathbb{R}^3$ and the Gauss map):

If $C=\alpha(I)$ is a line of curvature, and $k$ is its curvature at $p$, then $$ k = \mid k_n k_N \mid $$ where $k_n$ is the normal curvature at $p$ along the tagent line of $C$, and $k_N$ is the curvature of the spherical image $N(C) \subset S^2$ at $N(p)$.

I am not sure I understand the question, though... because it seems to me that if $C$ is a line of curvature, the normal curvature should be identical to the curvature of $C$ itself, hence $k = k_n$. Am I mistaken?

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up vote 3 down vote accepted

$C$ being a line of curvature means that the tangent vector of $C$ at every point is a principal direction, not that its normal curvature is identical to its curvature.

To solve your problem, you will need a formula for curvature that doesn't use parameterization by arc-length since the gauss map doesn't always give a curve parameterized by arc length. Let's use

$k(t) = \frac{|\alpha' \times \alpha''|}{|\alpha'|^3}$ (exercise 12 in section 1-5 of Do Carmo, which you are probably using?)

$C$ being a line of curvature means that the Gauss map $N$ is such that

$N'(t) = \lambda(t) \alpha'(t)$ where $-\lambda(t)$ is the curvature in the direction of $\alpha'(t)$

If you compute the curvature $k_N$ using these two formulas you should get the result after rearranging. Comment if you have more questions :)

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Thanks, that helped me to solve the exercise! Though I am still having a hard time visualizing this curve... – koletenbert Jul 25 '11 at 23:03
I tried to find a way to "see" the result too but was unsuccessful so far... – Vhailor Jul 26 '11 at 3:06

Suspect that is when line of curvature not always being a geodesic. When surface normal and normal (from Frenet-Serret frame) do not coincide, it is due to a tangential $ k_g $ component in general.. $ k^2 = k_g^2 + k_n^2 $.

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