# Something weird! (diff geom)

1. Oct 23, 2006

### quasar987

Here's something really weird. As can be read in Pressley's "Elementary Differential Geometry":

Proposition 1.3: Any reparametrization of a regular curve is regular.

And 4 pages later:

Exemple 1.8: For the parametrization $\gamma(t)=(t,t^2)$ of the parabola y=x², $\dot{\gamma}$ is never 0 so $\gamma$ is regular. But $\tilde{\gamma}(t)=(t^3,t^6)$ is also a parametrization of the same parabila. This time, $\dot{\tilde{\gamma}}=(3t^2,6t^5)$ and this is zero when t=0, so $\tilde{\gamma}$ is not regular.

Just to make sure that $\tilde{\gamma}$ is a reparametrization of $\gamma$, consider the reparametrization map $\phi:(-\infty,+\infty)\rightarrow (-\infty,+\infty)$ define by $\phi(t)=t^3$. Then $\phi$ is a smooth bijection with a smooth inverse such that $\gamma \circ \phi = (\phi(t),\phi(t)^2)=(t^3,t^6)= \tilde{\gamma}$, so $\tilde{\gamma}$ is really a reparametrization of $\gamma$ but it is not regular, contradicting proposition 1.3.

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

### AKG

It's inverse is not smooth. It's inverse isn't even once differentiable (look at what would happen to the derivative of the inverse at 0).

3. Oct 23, 2006

### quasar987

mmh, yes.

What is a little strange though is that in another book on differential geometry, the author makes the definition that a curve is a reparametrization of another curve is there exist a reparametrization map btw them that is continuous, bijective (from the domain of one curve to the domain of the other) and monotonous increasing. In other words, he does not require of the reparametrization map to be smooth.

So these two ways of defining when two curves differ by a change of param really aren't equivalent, because in the second case, proposition 1.3 above is not true. :-O

Last edited: Oct 23, 2006
4. Oct 24, 2006

### ObsessiveMathsFreak

No, the reparameterisation has to be differentiable as well. I.e., the whole thing has to be a "diffeomorphism", not just a straight isomorphism. Not that a striaght isomorphism would be completely terrible. It just wouldn't be a very good curve anymore.

5. Oct 29, 2006

### quasar987

That the reparametrization map and its inverse be one time differentiable is enough for prop.1.3 to be true. Is "smoothness" a luxury, or is it important for some other reason that it really be indefinitely differentiable?

6. Oct 29, 2006

### matt grime

Smoothness is a luxury - almost all of differential geometry can be done with C^2 functions.

7. Oct 29, 2006

### quasar987

Good to know.