I don't know what a coordinate system *is*, rigorously. (But I suspect they're related to bases of vector spaces under some map, where the basis can be the standard basis and the map can be the identity map.)

On $\mathbb{R^2}$, I "know" that Cartesian coordinates $(x, y)$ and polar coordinates $(r, \theta)$ are related by the functions:

- $r : (x,y) \mapsto \sqrt{x^2 + y^2}$
- $\theta : (x,y) \mapsto \arctan(y/x)$.

Now, Wikipedia says that polar coordinates are related to Cartesian coordinates via:

- $r = \sqrt{x^2 + y^2}$
- $\displaystyle d\theta = -{y\over r^2} dx\ +\ {x \over r^2}dy$.

In particular, $d\theta$ looks suspiciously like a differential $1$-form, ie. a dual vector field, ie. a smooth section of the dual tangent bundle $T^* \mathbb{R^2}$ of $\mathbb{R^2}$, ie. a map that takes a point $(x,y)\in \mathbb{R^2}$ and outputs a dual tangent vector $\displaystyle -{y\over x^2 + y^2} dx\ +\ {x \over x^2 + y^2}dy$ expressed in terms of the dual standard basis $\{dx, dy\}$.

In terms of the de Rham complex of $\mathbb{R^2}$

$$\boldsymbol 0 \overset{d}{\longrightarrow} \Omega_0(\mathbb{R^2}) \overset{d}{\longrightarrow} \Omega_1(\mathbb{R^2}) \overset{d}{\longrightarrow} \Omega_2(\mathbb{R^2}) \overset{d}{\longrightarrow} \boldsymbol 0,$$

$d\theta$ seems to be an element of the vector space $\Omega_1(\mathbb{R^2})$. Also, $d\theta$ must be an *exact* form, being the image under $d$ of $\theta$ (I know this last statement is wrong, but why?).

How does this talk of vectors relates to coordinate systems? Why can the relationship between coordinate systems be differential? What does this differential relationship mean, and how does it relate to the previously stated one? If coordinate systems can be differentiated, does it mean they

aremaps?

(Please do correct the plethora of technical mistakes I must've made.)

## Best Answer

Careful: the point $(0,0)$ is a singularity of the transformation between rectangular and polar coordinates. Mostly, you should replace $\Bbb R^2$ by $\Bbb R^2-\{(0,0)\}$ to be on the safe side.

Schoolchildren love the formula $\theta=\tan^{-1}(y/x)$ but mathematicians realise that it is valid for $x>0$, but not $x<0$. The notation $d\theta$ is an abuse of notation. It really is a differential $1$-form on $\Bbb R^2-\{(0,0)\}$ but it is not the exterior derivative of any smooth function on that set. It is a closed $1$-form, but not exact, and is a generator of $H^1_{\text{de Rham}}(\Bbb R^2-\{(0,0)\})$ (one-dimensional).