Generalized Curvatures

Question

I am sure that is: $$e_1'(t) = \chi_1(t)e_2(t)\Vert\gamma'(t)\Vert\qquad (1)$$
So it should be: $$\chi_1(t)=\frac{e_1'(t)}{e_2(t) \Vert\gamma'(t)\Vert}\qquad (2)$$ But I am also sure that is: $$\chi _{i}(t)={\frac {\langle \mathbf {e} _{i}'(t),\mathbf {e} _{i+1}(t)\rangle }{|\gamma'(t)|}}\qquad (3) [By\:\:wikipedia]$$ So how can you show this last result? (2) and (3) are compatible? What I did wrong? Thanks in advance.

Answer 1

What you did was to take relation (1) and "divide" by $e_2$, but this is not possible since $e_2$ is a vector. The correct approach is to consider the scalar product with $e_2 (t)$ in (1), and obtain

$$\langle e_1 ' (t), e_2(t) \rangle = \langle \chi_1(t) e_2(t) \| \gamma'(t)\|, e_2(t) \rangle = \chi_1(t) \| \gamma'(t)\| \langle e_2(t), e_2(t) \rangle = \chi_1(t) \| \gamma'(t)\|$$

because the Frenet frame is orthonormal, so $\| e_2 (t) \| = 1$. Hence, it is easy to deduce that $\chi_1(t) = \dfrac {\langle e_1 ' (t), e_2(t) \rangle} {\| \gamma'(t)\|}$, provided that $\| \gamma'(t)\| \ne 0$, i.e. provided that $\gamma$ be regular, which is almost always assumed.