Is it possible to parametrize a chart by arc-length?

Question

Let $M$ be smooth connected 1-dimentional manifold.

I think that there is an atlas of $M$ such that every chart of it is parametrized by arc-length

I know that there is an atlas of $M$ such that every chart of it is diffeomorphism from an open interval $I$ in $\mathbb{R}$ to an open subset of $M$.

So, I want to show that we can regard the chart $\phi: I \to M$ as a parametrized chart by arc-length. But I cannot prove that $\phi'(t) \neq 0$ for all $t \in I$. This is needed to define the arc-length function.

Is this correct? And how to show this?

Answer 1

I think you need to be more precise about your definition of an atlas and of your definition of arc length. For example, does your atlas is supposed maximal (or saturated) ? How do you define arc length ?

I suppose you are working on an embedded 1-dimensional manifold with the induced euclidean metric (ie: the norm of a tangent vector is equal to it's usual norm on the euclidean space).

Now, I will try to guide you for your questions. With your notations :

• You application $\phi : I \to M$ is a diffeomorphism as you said. What can you say about it's differential ? Does it imply $\phi '(t) \neq 0$ for all $t \in I$ ? I will spoil you : the answer is yes! But why?
• To obtain an arc-length parametrization of $\phi$, try to compose $\phi$ with $F^{-1}$ where $F : I \to \mathbb{R}$ is defined by $$F(u) = \int_{u_0}^u |\phi '(t)| dt$$ for all $u \in I$ and for a fixed $u_0 \in I$. Why does $\phi \circ F^{-1}$ is still a diffeomorphism? Why does $|(\phi \circ F^{-1})'(t)| = 1$ ? It works because $$|(\phi \circ F^{-1})'(t)| = |\phi'(F^{-1}(t))| |(F^{-1})'(t)| = \frac{|\phi'(F^{-1}(t))|}{|F'(F^{-1}(t))|} = \;? = 1. $$

I hope it will help you.