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!

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    $\begingroup$ The map $t \mapsto (t^2,t^3)$ is a smooth curve with a corner at the origin. See this $\endgroup$
    – Didier
    Oct 16, 2022 at 8:13
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    $\begingroup$ What is important to notice is that the derivative of any smooth parametrization of this curve has to vanish there, since it has to do an instant U-turn $\endgroup$
    – Didier
    Oct 16, 2022 at 8:56
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    $\begingroup$ Your conjecture that there can't be corners would be correct if you parameterized the curve by its arc length, for example, or any other parameterization in which the speed never goes to zero. $\endgroup$ Oct 16, 2022 at 22:22
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    $\begingroup$ "The derivative of any smooth parametrization of this curve has to vanish" (at a cusp point) is the main thing here. A "curve" is not a one-dimensional submanifold; it is a function with a given parametrization. It's perfectly possible that a smooth curve travels to a single point, slows down, comes to a stop, and then starts again in a different direction. (The tangent vector is well-defined at all points – including the cusp point, where it is zero.) And as the answer below shows, it is possible for such a curve to be $C^\infty$ (but not analytic!) $\endgroup$
    – printf
    Oct 16, 2022 at 23:02
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    $\begingroup$ We could summarize all this by the following analogy. An object's movement is continuous and differentiable on $t$. An object can follow a non-differentiable path. But at a point where the path is non-differentiable, e.g. a 90° turn, the speed must be $0$. $\endgroup$ Oct 16, 2022 at 23:21

2 Answers 2


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".

  • $\begingroup$ Thanks @Didier, the construction of the 90-degree-cornered-curve was very educating for me. $\endgroup$
    – NG_
    Oct 16, 2022 at 10:36
  • $\begingroup$ One can consider the function $f(x)=0$ if $x\leqslant0$, $f(x)=e^{-1/x}$ if $x>0$; then construct $\gamma:\mathbb{R}\to\mathbb{R}^2$ as $\gamma(x) = (f(x), f(-x))$. This is a curve that enters the origin from the positive $y$ axis, halts at $(0,0)$, then exits via the positive $x$ axis. See p.45 of Bishop & Goldberg, Tensor Analysis on Manifolds, Dover Publications, New York (1980) $\endgroup$
    – printf
    Oct 16, 2022 at 22:50
  • $\begingroup$ @printf Yes, this also works $\endgroup$
    – Didier
    Oct 17, 2022 at 6:38

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.


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