I asked this question on StackExchange, but until now there is not any answer or hint. I hope I can get some help here.

When I am reading ''A course in differential geometry'' of Klingenberg, I cannot be sure the Frenet frame defined in this book is independent of the choice of parameter of a curve. As a result, the definition of curvature of curve in this book confuses me.

Let $I$ be a interval in $\mathbb{R}$, and $c: I \rightarrow \mathbb{R}^n$ be a smooth map. We regard $c$ as a parameter of a smooth curve $C$, and we always assume that the derivatives $c'(t), c''(t), \dots c^{(n-1)} (t)$ are linearly independent.

According to Proposition 1.2.2 of this book, there is a unique Frenet frame $e_1, \dots, e_n$ along $C$ such that (1) $c'(t), c''(t), \dots, c^{(k)}(t)$ and $e_1(t), \dots, e_k(t)$ have the same orientation for $1 \le k \le n-1$, and that (2) $e_1(t), \dots, e_n(t)$ has the positive orientation.

Using this special Frenet frame, we can calculate the coefficient functions $\omega_{ij}: I \rightarrow \mathbb{R}$ such that $e_i'(t) =\sum \omega_{ij} (t) e_j(t)$ for any $t \in I$. And in Definition 1.3.3 of this book, the $i$th curvature $\kappa_i$ of $C$ is defined to be the quotient $$\kappa_i(t) = \frac{\omega_{i,i+1}(t)}{|c'(t)|}.$$

Now, if we have another interval $\tilde{I}$ and another smooth map $\tilde{c}: \tilde{I} \rightarrow \mathbb{R}^n$, as well as a diffeomorphism $\varphi: I \rightarrow \tilde{I}$ with $\varphi'(t) >0$, then we can regard $\tilde{c}$ as another parameter of the same curve $C$.

By the mathematical induction, the linear independence of $c'(t), c''(t), \dots, c^{(n-1)}(t)$ implies the linear independence of $\tilde{c}' (\tilde{t}), \tilde{c}'' (\tilde{t}), \dots, \tilde{c}^{(n-1)} (\tilde{t})$. Hence, Proposition 1.2.2 can give another Frenet frame, so we can compute the new coefficient functions $\tilde{\omega}_{ij}: \tilde{I} \rightarrow \mathbb{R}$, and define the new curvatures $\tilde{\kappa}_i (\tilde{t})$.

I am not quite familiar with differential geometry, but I think that the curvatures of a fixed curve should be independent of the choice of its parameter. That is, we should have $\kappa_i =\tilde{\kappa}_i \circ \varphi$, namely $\kappa_i (t) =\tilde{\kappa}_i (\varphi(t))$ for any $t \in I$. (Review that $\varphi: I \rightarrow \tilde{I}$ is defined to be the change of parameters.)

But how to prove this? The ''proof'' in Klingenberg's book really confuses me.

The part (ii) of Proposition 1.3.2 tells us that if the change of variables $\varphi$ is given as above, and if we let $\tilde{e}_i =e_i \circ \varphi$, then the new $\tilde{\kappa}_i$ calculated by the $\tilde{e}_i$'s is equal to the old $\kappa_i$ calculated by the $e_i$'s. (The proof is quite simple.)

However, these $\tilde{e}_i$'s are not calculated in the way that the Frenet frame is defined in Proposition 1.2.2, that is, they are not calculated by the Schmidt orthogonalization of the derivatives $\tilde{c}', \tilde{c}'', \dots, \tilde{c}^{(n-1)}$. They are just given by $\tilde{e}_i =e_i \circ \varphi$, and even Klingenberg himself does not use the name ''Frenet frame'' in Proposition 1.3.2.

My question is whether these $\tilde{e}_i =e_i \circ \varphi$ are just equal to the Frenet frame associated to the new parameter $\tilde{c}: \tilde{I} \rightarrow \mathbb{R}$, i.e., the result of Schmidt orthogonalization of $\tilde{c}' (\varphi(t)), \tilde{c}'' (\varphi(t)), \dots, \tilde{c}^{(n-1)} (\varphi(t))$ is just the $e_1 \circ \varphi (t), \dots, e_n \circ \varphi(t)$?

If the answer is YES, how can we prove this? For $e_1$ and $\tilde{e}_1$, this is easy by the chain rule. But for general, this seems quite complex because we need to use the Faà di Bruno's formula to compute the higher derivatives, and Faà di Bruno's formula is quite complex. (I tried very very hard but failed.)

If the answer is NO, then in order to show that the definition of curvatures $\kappa_i$ are the independent of choices of parameters, we need to compute the real Frenet frame under the new parameter. But the computations seem quite complex for the same reason.

Thank you for your attention.