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Good evening,

i try to understand following proof from Carmo: Differential Geometry of Curves and Surfaces

enter image description here

I have two Questions:

  1. Why is $ e_1 \times e_2=N$ and $N \times e_1 =e_2$ Is it because $e_1, e_2$ are the unit vectors tangent to the coordinate curve and and so with $N$ a moving trihedron? And is it important for that, that $F=0$

  2. Why is this equation true?

$$ \left \langle \left ( \frac{x_u}{\sqrt{E}}\right )_u,\frac{x_v}{\sqrt{G}} \right \rangle = - \frac{1}{2}\frac{E_v}{\sqrt{EG}}$$

how can I generate it from

$$\langle x_{uu},x_{v} \rangle=-\frac{1}{2}E_v $$

Thanks

EDIT: Solved the first question with Geodesic curvature for orthogonal parametrization

Walter Frosch
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  • P.S. Now that I have your attention, you never responded to our queries about your post. Where did that (erroneous?) formula come from? – Ted Shifrin Dec 04 '22 at 22:23
  • https://math.stackexchange.com/questions/748974/gaussian-curvature-k-of-of-orthogonal-parametrization-x is a Question in Carmo: Differential Geometry of Curves and Surfaces. And you get it from the Gauß-Formular and Theorema egregium – Walter Frosch Dec 05 '22 at 09:08
  • This formula holds only for $F=0$, as I already discussed in the previous comments. – Ted Shifrin Dec 05 '22 at 16:15

1 Answers1

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Yes, $F=0$ is essential in all this. $e_1,e_2,N$ form an (oriented) orthonormal basis.

For your second question, just use the product rule and the fact that $\langle x_u,x_v\rangle = 0$. It falls out immediately.

Ted Shifrin
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