# Thread: Parametrization of a smooth surface

1. ## Parametrization of a smooth surface

The reason that I am posting this is because I'm having a hard time figuring out how to pose the question in a manner that makes any sense. But I thought I'd post it here and see if anyone can help guide me.

Suppose we have a smooth surface defined in 3-space that is given by $X^3 = X^3 ( X^1 , ~X^2 )$. (The superscripts indicate a contravariant index, not a power.)

Suppose we have a parametrization of the surface, $X^1 = v^1$, $X^2 = v^2$, and thus $X^3 = X^3(v^1,~ v^2 )$. My question is: "Is there a transformation of coordinates of this surface such that the transformed coordinates are not allowable, but still have the same domain as the original parametrization?"

I haven't been able to do much in the way of a proof but geometrically speaking I don't think such a transformation can be done. The surface is, of course, independent of the parametrization so it has the same number of critical points, the same curvature at a given point, etc. But I can't see how to make a proof out of it.

I'm probably making a mountain out of a molehill but it bothers me that I can't come up with a way to write this thing down in clear terms.

Thanks!

-Dan

2. ## Re: Parametrization of a smooth surface

can you expand on what you mean by "not allowable" ?

3. ## Re: Parametrization of a smooth surface

Originally Posted by romsek
can you expand on what you mean by "not allowable" ?
Sorry. I don't know what terms are standard. This is from "Introduction to Vector and Tensor Analysis," Wrede, pg. 223.

An allowable transformation is of the form $X^{ \mu } = X^{ \mu } \left ( \overline{X} ^{ \nu } \right )$, where the $\overline{X} ^{ \nu }$ are the new coordinates obeying the following:
1) Each of the functions $X^{ \mu } \left ( \overline{X} ^{ \nu } \right )$ has continuous partial derivatives at least of first order

2) The determinant $| X^{ \mu } / \overline{X} ^{ \nu } | \neq 0$ for every coordinate triple of the domain determined by the function $X^{ \mu }$

-Dan