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It seems like the differentiability of a curve should morally speaking be part of the curve itself. More specifically, whether or not a curve is differentiable should not depend on the coordinates.

So the problem is that $x^{3}$ is differentiable everywhere and $x^{1/3}$ is not, even though the latter is just the former in different coordinates. Why should a function lose differentiability just by changing coordinates?

Fomalhaut
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  • I'm not sure where the moral argument here is, but the definition of differentiability of a function is dependent on the point. A function is differentiable if it is so for every point. – Rushabh Mehta Mar 17 '19 at 20:00
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    @DonThousand should a curve stop being differentiable if you change the coordinates? – Fomalhaut Mar 17 '19 at 20:02
  • Yes it should, because a horizontal slope in a function doesn't translate well in the inverse. A vertical slope corresponds to an undefined derivative. – Rushabh Mehta Mar 17 '19 at 20:03
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    $x\mapsto x^{1/3}$ is not a diffeomorphism – wjmolina Mar 17 '19 at 20:05
  • @Cleric that is probably the best explanation of all the answers here. – Rushabh Mehta Mar 17 '19 at 20:06
  • Remember that the tangent line to $y=x^3$ is horizontal at the origin so when you flip the $x$ and the $y$ to get $y=x^{\frac{1}{3}}$ that horizontal tangent becomes a vertical tangent which does not correspond to a derivative. – John Douma Mar 17 '19 at 20:10
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    That is a very good observation and is related to very deep questions. There are two or three things that need to be distinguished. (1) There is a definition of differentiability of a function. You know, the limit of the slopes thing The function $y(x)=x^{1/3}$ is not differentiable in that definition. (2) There is a notion of differentiability for the locus of points that satisfy $y=x^{1/3}$. But it is not only the set of point, but the set of points together with extra structure. What is called a differentiable manifold. In this context the locus $y=x^{1/3}$ can be given one of those. – user647486 Mar 17 '19 at 20:11
  • @Cleric you can take $y = |x|$ and perform any coordinate transform but it never becomes differentiable at the sharp edge. But you can transform $y = x^{1/3}$ and it does become differentiable at the flat part. So they're not non-differentiable in the same sense. – Fomalhaut Mar 17 '19 at 20:11
  • A re-parametrization or a change of coordinate of the curve appens to change some of the regularity-related characteristics of the curve (as your example shows): it is to avoid so that we use the natural parameter (see https://en.wikipedia.org/wiki/Differential_geometry_of_curves): parametrized in this way, your curve becomes regular: you still won't be able to treat it as a function thought: $y'(x)$ does not give you information about regularity, only on $\frac{y'(t)}{x'(t)}$ which is of course non indipendent of the coordinates –  Mar 17 '19 at 20:12
  • Now, in general, given a topological space, you might give it many non-equivalent differentiable structures. However, in your case you are in dimension $1$. You could say, then that the possibility of giving it tangents everywhere is a property of the topological space that is the locus $y=x^{1/3}$ with the induced topology from the plane. – user647486 Mar 17 '19 at 20:13

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There are two different points in your question that I think need to be answered:

  1. Is $y'(x)=\frac{y'(t)}{x'(t)}$ invariant for a change of variable?

No, it isn't: the slope of the tangent vector is dipendent on your frame of references. Even simply rotating it can change your result, leading to the non existence of the ratio (that is, to a vertical vector, which is not even that patological). Changing the coordinates by something that is not a diffeomorphism changes the results too (for more information, look at the notes)

You adress as regularity this ratio, but it is not; A curve can be regular even if its tangent vector is vertical. As an example, think of the line $y=3$: it is indeed quite regular, even thought it cannot be written as $y=mx+q$

Note 1

You might now be curious if the effective regularity can be "damaged" by a change of coordinates: I will try to answer to this question

  1. Does a simple change of coordinates modify the regularity of the curve itself?

The question is not so easy to answer: if you allow any kind of parametrization then yes, it does, as you example shows (changing $t'=t^{\frac{1}{3}}$, or making the inversion $y'=x,\ x'=y$). On the other hand, curves are usually studied with a special parameter, called arc-length or natural parameter (see here for more informations) that avoids this problem: equipped with such a parameter, the tangent vector has always length $1$ (if the curve satisfies certain requirements, such as being $C^1$, $\gamma'\neq 0$, not self intersecting and so on), and the regularity of the curve is not changed by a diffeomorfism

As for your example, the regularity is indeed different: if $y=|x|$ gets parametrized by arc lenght, we get: $\gamma(t)=\frac{1}{\sqrt{2}}(t,|t|)$, and this curve does not have a tangent vector defined in $t=0$, so it is indeed non regular. Computing the vector tangent to $y=x^{\frac{1}{3}}$ with the natural parameter, we get $\dot{\gamma}(0)=(0,1)$

Note 2 (on the natural parameter): As user647486 pointed out, in more than $1$ dimension a set can be equipped with a lot of different differentiable structures (see here for an example by John Milnor). In one dimension, however, the differentiable (and metric) structure is unique: this is (for the metric structure) due to the fact that we are able to make a regular curve isomorphic to a interval, thing that we are un-able to do in more than one dimension, as the Theorema Egregium shows

  • you can take $y=|x|$ and perform any coordinate transform but it never becomes differentiable at the sharp edge. But you can transform $y=x^{1/3}$ and it does become differentiable at the flat part. So they're not non-differentiable in the same sense. – Fomalhaut Mar 17 '19 at 21:01
  • @TomislavOstojich actually, if you take $x'=x^{\frac{1}{3}},\ y'=y$, the function becomes $y'=|x'|(x')^2$, which is indeed differentiable at 0 –  Mar 17 '19 at 21:05
  • @TomislavOstojich : you cannot change the regularity of $|x|$ by a diffeomorphism, that's true, while with $x^{\frac{1}{3}}$ you can indeed make it differentiable: that is connected with the regularity of the natural parameter, as I explained in my answer. I hope it is clear –  Mar 17 '19 at 21:09
  • Is th notion of differentiable dependent on the underlying geometry? Your transformation is not an isometry. – Fomalhaut Mar 21 '19 at 09:18