# Is there a criterion for the regularness of a curve?

1. Sep 14, 2006

### quasar987

I've read the chapters on curves from two different books on diff. geometry and both say that if a curve is regular, then there exists a unit-velocity reparametrization. But regularness depends on the choice of parametrization. For instance, both the curves (t,t²) and (t³,t^6) are parametrization of the parabola y=x², but only the first is regular.

So I ask, is there a criterion to determine whether or not a parametrized curve admits a regular reparametrization?

2. Sep 15, 2006

### jbusc

iirc, a parameterization is regular if the derivative of the parameterization never vanishes.

and, any piecewise smooth curve admits a regular parameterization. consider each smooth segment of a that piecewise smooth curve, then, obviously it must have some finite arc length at any point along the curve to the beginning or end. since it is smooth, this means it can be parameterized by arc length which is obviously regular.

the important distinction is that curves are not regular, it is their parameterizations which are (or are not) regular.

i don't know more since i never covered non-regular parameterizations in detail. as far as i can tell, they're fairly useless.

anyone, please correct me if i'm wrong.

3. Sep 18, 2006

### quasar987

what does iirc mean?

4. Sep 19, 2006

### TD

If I Recall/Remember Correctly

5. Sep 19, 2006

### HallsofIvy

Note that your definition of "regular" was "there exist a parametrization". It is not the parametrization that is regular, it is the curve. The one curve given by both (t,t2) and (t3, t6) is regular because of the (t, t2) parametrization.

6. Oct 22, 2006

### quasar987

I did not include the definition of regularness in the OP. The definition used by both books was

A curve of parametrization $\gamma$ is regular if $\dot{\gamma}$ is nowhere vanishing.

The "then there exists a unit-velocity param." is a thm following from the definition. It is not a iff statement so the existence of a unit-velocity param. is not a caracterization of regularness either.

So I think my question remains unanswered. The reason I revived this thread is because I ran into a similar problem but in the topics of surfaces, which I made the subject of a disctinct thread.

Last edited: Oct 22, 2006
7. Oct 23, 2006

### cliowa

Well, if there is a parametrization $\gamma$, where $\dot{\gamma}$ nowhere vanishes, you can construct a new parametrization $\beta:=\frac{\gamma}{\dot{\gamma}}$. As you can easily see, this parametrization has unit speed. In fact, you could make a parametrization with just any speed $a\in\mathbb{R}, a\neq 0$, by setting $\alpha:=a\cdot \frac{\gamma}{\dot{\gamma}}$. Unit speed is just a choice of convenience, simplicity.

Now to your main question: Very often curves and surfaces are given as level sets of some function $f$. Let $C$ be your curve, then this would mean that $C=f^{-1}(c), c\in\mathbb{R}$. If now the derivative $df(x)$ is regular for all $x\in C$, then $C$ is said to be regular (Note that this is indeed the same thing as you stated).
The only thing left to determine now is whether $df(x)$ is regular or not. Do you know how to do that?

If you know the implicit function theorem you can see that $df(x)$ regular implies that there is a local parametrization of $C$ around some $x\in C$, for every $x\in C$. Now you could try to patch things together.
Best regards...Cliowa

P.S.: If you don't know the implicit function theorem, you absolutely need to have a look at it, else you won't get far.

Last edited: Oct 23, 2006
8. Oct 23, 2006

### quasar987

Does this cover curves in 3D also? I.e. given any parametrized curve, can we find a function f(x,y,z) such that the level set f(x,y,z)=c is the curve?

9. Oct 24, 2006

### cliowa

This often actually is the more natural way of attacking a parametrization. Usually you won't have any parametrization, but simply a surface (surface in 2 dim=curve) expressed as a level set. Take for example things like an circle, ellipse, hyperbola, etc. There are quite nice parametrizations, but one very nice defintion of an ellipse is a description as a level set (stating a relation which for all points must hold).
As you can see, this covers surfaces in n dimensions (finite!). But that's precisely the implicit function theorem.