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I'm currently taking an intro to Differential Geometry course, and am having trouble with the definition of a smooth curve. If you consider the curve $\lambda(t) = \left (\cos^3(t),\sin^3(t)\right )$ then clearly each of $x(t)$ and $y(t)$ are infinitely differentiable, and so by definition $\lambda(t)$ should be smooth. However plotting the curve shows that its graph has 4 cusps, and doesn't seem to be 'smooth' in any way that makes sense to me. Could somebody clarify for me the intuition behind what makes a curve smooth?

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    $\begingroup$ The cusps are a consequence of the fact that $\dot{\lambda}$ vanishes at some times. $\endgroup$ – user296602 Feb 21 at 18:48
  • $\begingroup$ Does the fact that $\dot \lambda$ vanishes sometimes not prevent $\lambda$ from being smooth? $\endgroup$ – Jim Feb 21 at 19:02
  • $\begingroup$ Write the equation of the curve as $x^{2\over 3}+y^{2\over 3}=1.$ From this implicit equation we can see easily that at points where $x=0$ or $y=0$ is the corresponding partial derivative not defined. $\endgroup$ – user376343 Feb 21 at 21:19
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    $\begingroup$ The function $y$ is smooth as a function of $t$ but not as a (local) function of $x$. $\endgroup$ – anomaly Feb 21 at 21:31
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The curve that you describe is indeed defined by smooth, differentiable functions in both coordinates.

Anyway, the tangent vector has a direction depending on the two derivatives at the point considered. And if these derivatives turn out to be zero simultaneously, the direction is undetermined and the curve is free to form an angular point or a cusp.

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As T. Bongers points out in the comments, the cusps you're getting reflect that the derivative is 0 at those points. I figured I'd expand on this and give a bit more intuition.

This is one of those times where it's important to distinguish between a manifold and its parametrization.

As you've already said, $\lambda(t)$ is a smooth map, $\Bbb{R}\to \Bbb{R}^2$, and we can call its image a curve (it is after all still a one dimensional manifold). However, its image is not a smooth curve.

What do I mean by this?

Well, we need to say precisely what we mean by a curve in $\Bbb{R}^2$. That is, we need to define the term submanifold. It turns out that there are some subtleties here, see the wikipedia page for the discussion. However, I suspect what would agree with your intuition here is the notion of an embedded (or regular) submanifold.

An embedded submanifold, $N$, of a smooth manifold $M$ is a subset $N\subseteq M$ which can be given a smooth atlas compatible with the subspace topology such that the inclusion $N\hookrightarrow M$ is a smooth immersion.

There is an equivalent local definition, which is that an embedded $k$-submanifold of an $n$-manifold $M$ is a subset $N$ of $M$ such that every point $p\in M$ has a coordinate neighborhood in $M$, $\phi: U\to \Bbb{R}^n$, such that $\phi(N\cap U)$ is the intersection of a $k$-plane with $\phi(U)$. These coordinate neighborhoods give the charts for the submanifold.

Now we can see that $\lambda$ fails to be a smooth immersion, since $d\lambda$ is zero at the cusps, which means that $d\lambda$ doesn't have full rank everywhere.

To see why the image can't be a smooth submanifold though, it's perhaps easier to look at the local definition. If we take $p$ to be a cusp, then there would need to be a neighborhood $U$ of $p$, and a smooth chart $\phi: U\to \Bbb{R}^2$ such that $\phi$ took the neighborhood of $p$ in the curve to a line in $\Bbb{R}^2$, which means that $\phi$ would have to "unfold" the cusp, which is impossible for a smooth map to do.

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  • $\begingroup$ @anonymous_downvoter I'm curious as to whether or not I've made an error in my answer. $\endgroup$ – jgon Feb 22 at 17:21
  • $\begingroup$ While I am not the "anonymous" downvoter, perhaps her disapproval is because this feels like nuking a mosquitoe. Surely, cannot a smooth curve be described without introducing the notion of a submanifold? $\endgroup$ – Jim May 15 at 9:30

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