Does "$s'(t_0) = ||c'(t_0)||$" actually mean "$s'(t_0) \cong ||c'(t_0)||$" or "$\dot s(t_0) = ||c'(t_0)||$"?

Question

Follow-up question: Chain rule: Does "$\gamma'(s) = c'(t(s))t'(s)$" actually mean "$\gamma'(s) = c'(t(s))\dot t(s)$" (or "$\gamma'(s) \cong c'(t(s)) t'(s)$")?

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My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

Here are Section 2.1 and Section 2.2.

Question: Does the "$s'(t) = ||c'(t)||$" before Proposition 2.3 actually mean "$s'(t) \cong ||c'(t)||$" or "$\dot s(t) = ||c'(t)||$"? See Volume 1 Section 8.6 for the notation $\dot s(t)$.

This is what I understand:

1. $||c'||$ is a map $||c'||:[a,b] \to [0,\infty)$ that satisfies the assumptions for the fundamental theorem of calculus (I follow the one from wikipedia: Continuous map $f: [a,b] \to A$ with $A \subseteq \mathbb R$). (I ask about this here.)

   • 1.1. If $c$ is regular/an immersion, then $||c'||:[a,b] \to (0,\infty)$ ($||c'||$ will never be $0$) is smooth, by this. Thus, $||c'||$ would be continuous and so satisfy the assumptions for the fundamental theorem of calculus. I'm not sure $||c'||$ satisfies the assumptions even when $c$ is not regular/an immersion (but maybe $||c'||$ does since Paulo Mourão can prove the smoothness part without immersion). (I also ask about this here.)

2. Let $\dot s$ be calculus derivative, with notation form Volume 1 Section 8.6.

3. Use $t$ to denote the standard coordinate (Volume 1 Section 8.6) on $[a,b]$, and use $t_0$ to denote a point in $[a,b]$. Let $x$ be the standard coordinate on $[0,l]$.

4. $\dot s = ||c'||$, by fundamental theorem of calculus, (1) and (2).

5. For each $t_0 \in [a,b]$, $\dot s(t_0) = ||c'(t_0)||$, by (3) and (4).

6. $s'(t_0)=\dot s(t_0) \frac{d}{dx}|_{s(t_0)}$, by Volume 1 Exercise 8.14, (2) and (3).

7. $s'(t_0)$ "$\cong$" $\dot s(t_0)$, where "$\cong$" is in the sense of and Volume 1 Proposition 8.15, by (5) and (6) .

8. Therefore, $s'(t_0)$ "$\cong$" $||c'(t_0)||$, by (7).

9. However, technically the isomorphic relation in (8) is between $s'(t_0)$, a tangent vector, and $||c'(t_0)||$, a real number. So, we don't exactly have equality, kind of like here.

Answer 1

You are right, by the definitions of the book it makes no sense, it should be either $$ \dot s(t)=\|c'(t)\|~~\text{ or }~~ s'(t)=\|c'(t)\|\frac{\partial}{\partial t} $$ as the dot denotes the scalar value of the derivative of a scalar function and the prime the vector in the tangent space. It is only that the identification of the tangent space of a vector space with that vector space itself is so natural that the distinction between both gets usually neglected.

Note that there can be no dotted variant of the derivative of $c$, as the manifold in general is no vector space or affine space (with a canonical flat tangent bundle $M\times V$). In connection with a chart one can again use the dotted variant for the ephemere distinction between an element of the tangent space of $\Bbb R^n$ and the collection of its coordinates.