# Name of Non-Unique Coordinate systems

Are there some examples (and a name) for non-unique coordinates (non-unique meaning may have multiple ways to represent the same point). Such as the one below. • Polar coordinates have infinitely many ways to express any point. – Tartaglia's Stutter Mar 5 at 20:18
• I knew about that one, but are there any others. And is there an actual name for the class of these objects. – Numoru Mar 5 at 20:25
• I don't know of such a name but it wouldn't surprise me to know that there was one. – Tartaglia's Stutter Mar 5 at 20:30

## 1 Answer

Well, coordinate systems are meant to be injective (one-to-one), and it is not usually good that this condition is not satisfied. For example, if we take a coordinate system defined by$$\Phi\colon\space[0,\infty]\times[0,2\pi) \longrightarrow \mathbb{R}^2$$ so that $$\Phi(r,\varphi)=(r \cos(\varphi), r \sin(\varphi));$$ we see (either by mere exploration or more rigorously by setting the jacobian determinant equal to 0) that it fails to be injective at $$(x,y)=(0,0)$$. And, indeed, many $$(r,\varphi)$$ can represent this point, we only need $$r=0$$ so any $$\Phi(0,\varphi)$$ equals $$(0, 0)$$.

Again, this is not something you usually seek, for wherever a coordinate system fails to be injective, many problems arise. To a physicist like me, the worst of them is that the nabla operator (the one with which one first defines gradient, divergence, curl and laplacians) might not be defined at these points. Indeed,

$$\vec{\nabla}\space\colon=\frac{\partial}{\partial x}\vec{u}_x+\frac{\partial}{\partial y}\vec{u}_y=\frac{\partial}{\partial r}\vec{u}_r+\frac{1}{r}\frac{\partial}{\partial \varphi}\vec{u}_\varphi.$$

This also happens with spherical coordinates at the origin, with cylindrical coordinates at every point in the $$z$$ axis, and with many more usual coordinate systems. Now, I don't know about any coordinate system made specifically to fail to be injective at specific points, nor do I know what that could be useful for, but I hope this helped you, OP, anyway.

EDIT: in all of those situations, the coordinate system fails to be injective at a 0 Lebesgue-measure subset of $$\mathbb{R}^n$$. If one happened to fail to be injective in a set with non-0 Lebesgue measure, it wouldn't even be possible to compute definite integrals within that region.