The reparametrization theorem says the following:
If $α:I\to\mathbb{R}^n$ is a regular curve in $\mathbb{R}^n$, then there exists a reparametrization $\beta$ of $\alpha$ such that $β$ has unit speed.
My question is this:
If the curve is not regular, then is there no arc length parameterization?.
What I tried was to get the following example $t\mapsto (|t|t,t^2)$ for $t\in[-1,1]$ whose graph would approximate this:
It is understood that when reaching point $(0,0)$ the particle that follows this route stops instantly and then he continues its journey, but if there were a parameterization $\beta$ by arc length, it means that when reaching that point it would continue with $||\beta'||=1$, is there such a possibility? How would it be explained if it existed.
Thanks.
Best Answer
In order to have an arclength parametrization, a curve must have a tangent line at each point (since, as you pointed out, an arclength parametrization gives you a length-one velocity vector at each point, and that vector spans the tangent line).
However, if you have a non-regular parametrization whose image is still a $1$-dimensional manifold, then that manifold will have a tangent line at each point, even if the parametrization fails to give it. The example I gave in the comments was $\alpha(t) = (t^3,t^3)$, which fails to be regular at $t=0$. Nevertheless, the image of $\alpha$ can obviously be parametrized regularly by $\beta(u)=(u,u)$. The example you gave has no tangent line at the origin and so there can be no arclength parametrization.
It is perhaps not obvious that any connected $1$-dimensional manifold sitting in $\Bbb R^n$ can be parametrized (globally) by arclength. By the implicit function theorem, it can locally be parametrized as a graph. You obtain open intervals of the curve which are smoothly parametrized. That's good enough: Write down arclength functions on those open intervals and, since they have to agree (up to a constant, by which we adjust) on the overlaps of those intervals, the arclength function can be defined smoothly on the union of the intervals.