Isn't a differential curve with corners contradictory?

Question

While reading "Riemannian Geometry" by M. Do Carmo, I've ran into a confusing remark following the definition of a curve. The author presents the following definition (2.8, chapter 1) for a curve:

A differential mapping $c:I \rightarrow M$ of an interval $I \subset \mathbb{R}$ into a differentiable manifold M is called a (parametrized) curve

Following the definition, Do Carmo remarks:

Observe that a parametrized curve can admit self-intersections as well as corners.

My Question is - Isn't the geometric interpretation of a "differentiable" is "smooth"? How come a differentiable curve admits a corner?

I'd be glad to see an example for a differentiable curve with a corner.

Many Thanks!

Answer 1

I already gave an example in the comment section showing that a curve can be smooth although it have a cusp at some point: the curve $t\in \Bbb R \mapsto (t^2,t^3)$ is such a curve.

However, one could argue that the cusp in the latter example isn't a corner, since there is no angle between the two parts of the curve meeting there. Indeed, one can still define a tangent to the curve at the cusp geometrically, or by using sufficiently high order derivatives. Hence, I would like to give an example with a proper angle, where no notion of tangent can exist.

You surely already know that the following function $$ \begin{array}{r|ccc} f\colon & \Bbb R &\longrightarrow &\Bbb R \\ & x &\longmapsto &\begin{cases} e^{-\frac{1}{x^2}} & \text{ if } x \neq 0, \\ 0 & \text{ if } x = 0, \end{cases} \end{array} $$ is smooth, with $f^{(k)}(0)=0$ for all $k\geqslant 0$. If not, take a look at this.

Consider the parametrized curve $$ \begin{array}{r|ccc} \gamma\colon & \Bbb R &\longrightarrow& \Bbb R^2 \\ & t & \longmapsto & \begin{cases} (f(t),f(t)) & \text{ if } t>0, \\ (0,0) & \text{ if } t=0,\\ (-f(t),f(t)) &\text{ if } t<0. \end{cases} \end{array} $$ It can be easily shown that $\gamma$ is a smooth curve, even at the origin. Its support is the same as that of the curve $s\in (-1,1) \mapsto (s,|s|)$, and thus has a $90°$ corner at the origin. Still, it is a smooth curve.

What is important here is that no smooth parametrization of this curve can have a non-zero derivative at the origin. In other word, this curve has no regular parametrization. It is the main reason why textbooks usually state their results beginning with "Let $\gamma$ be a regular curve".

Answer 2

Pick a smooth increasing function $\phi : [0, 1] \to [0, 1]$ such that $\phi = 0$ on $[0, 1/4]$ and $\phi = 1$ on $[3/4, 1]$. Take any two smooth curves $\gamma_1$ from $p$ to $q$ and $\gamma_2$ from $q$ to $r$, both defined on $[0, 1]$. Then the concatenation $\gamma$ of $\gamma_1 \circ \phi$ and $\gamma_2 \circ \phi$ is a smooth curve. Yet visually, while the curve is stuck at $q$, it can turn around and make a corner. After all, $\gamma_1$ and $\gamma_2$ can be arbitrary smooth curves.