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 are maps?

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

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    $\begingroup$ Knowing all this, are you not familiar with definition of smooth manifold? Specifically, given any point $p$ of manifold $M$, we can find smooth chart $(U,\varphi)$ containing $p$. By definition, $\varphi\colon U\to \mathbb R^n$ is homeomorphism onto open subset of $\mathbb R^n$. Now, we often write $\varphi = (x^1,\ldots, x^n)$ and call this local coordinates, where $x^i = \pi_i\circ \varphi$ and $\pi_i\colon \mathbb R^n\to \mathbb R$ is standard projection map. (cont'd) $\endgroup$ – Ennar Sep 2 '17 at 9:35
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    $\begingroup$ With specified local coordinates, $\{\left.\frac{\partial}{\partial x^1}\right|_p,\ldots,\left.\frac{\partial}{\partial x^n}\right|_p\}$ is base of tangent space $T_pM$, while $\{dx^1|_p,\ldots,dx^n|_p\}$ is the corresponding dual base for $T_p^*M$. $\endgroup$ – Ennar Sep 2 '17 at 9:35
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    $\begingroup$ $d\theta$ is not defined when $r=0$, so in truth it is not an element of $\Omega_1(\mathbb{R}^2)$. $\endgroup$ – Alessio Di Lorenzo Sep 2 '17 at 9:36
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    $\begingroup$ étale-cohomology, note that the emphasis is not on one chart, but on whole atlas. What you describe is called local parametrization and it is exactly how you think of parametrization (as in parametrized curve, or surface). But to return to the point, it's not about one chart, it's about collection of charts that cover whole space. If they are smoothly compatible, we call that smooth atlas. Maximal atlas is called smooth structure and you can think of maximal atlas as containing all charts that are smoothly compatible. $\endgroup$ – Ennar Sep 4 '17 at 14:50
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    $\begingroup$ About orthonormality. You see, on general smooth manifolds, there is no notion of scalar product on tangent spaces. If you have scalar product on each tangent space that varies smoothly, it's called Riemannian metric. In general case, you can't guarantee orthonormality, but in case of $\mathbb R^n$ if you choose standard coordinates, orthonormality condition gives rise to Euclidean metric. $\endgroup$ – Ennar Sep 4 '17 at 14:51

To answer your questions in order, given a finite-dimensional vector space $V$, we can use the basis vectors to determine a coordinate representation of $V$. There is a natural isomorphism between $V$ and its coordinate representation say $[V]_{\mathcal{B}}$. From here we construct a coordinate system which is just a bijection from $[V]_{\mathcal{B}}$ to $\mathbb{E}^{\textbf{dim}(V)}$.

If we switch our attention to the real objects of study i.e manifolds then $[V]_{\mathcal{B}}$ is just the tangent space at a point $p$. When defining what it means for a map to be smooth between manifolds, we noticed that it would be nice if smoothness was independent of charts, but how do we do it? Well,

$$ f \circ \phi^{-1} = f \circ \psi^{-1} \circ (\psi \circ \phi^{-1})$$

Hence we need to require that $ \psi \circ \phi^{-1}$ is a diffeomorphism for any $\phi, \psi$ with overlapping domains. It now follows that $\psi \circ \phi^{-1}$ is a reparametrization. Now given $(U, \phi=x^1,...,x^n)$ and $(V, \psi=y^1,...,y^m)$ to be local charts with overlapping domains, we know that $\frac{\partial}{\partial x^j}$ and $\frac{\partial}{\partial y^k}$ for basis for the tangent spaces. You also know that given $\psi \circ \phi^{-1}: \phi(U) \to \psi(V)$ is a diffeomorphism then $D(\psi \circ \phi^{-1}): T_{\phi(p)} \phi(U) \to T_{\psi(p)} \psi(V)$ is an isomorphism.

In the first paragraph we made an identification of the vector space with its coordinate representation. Notice that if we let $V = \mathbb{R}^n$ then the only distinction between $V, \mathbb{E}^n$ is that we've equipped the coordinate representation of $V$ with the topology induced by the standard euclidean metric. In this case, the map from $[\mathbb{R}^n]_{\mathcal{B}}$ to $\mathbb{E}^n$ can just be thought of as the identity map. This is the reason why we say $\mathbb{R}^n$ can be seen as an $n$-manifold just by using $\phi=(x^1,...,x^n)$ where $x^j(p_1,...,p_n) = p_j$ i.e $\phi(\textbf{p}) = \textbf{p}$ is just the identity.

Now lets get to your specific example. Producing new coordinate systems will just be a matter of thinking about how to parametrize sub-manifolds i.e each sub-maniold in the plane will give rise to a coordinate system. Here we take circles centered at the origin. From this we see,

$$p = (x,y) \Rightarrow p = (r, \theta) : r = \sqrt{x^2+y^2}, \theta = \textbf{arctan} (y,x)$$

(using this definition arctan via wiki) i.e the maps $\textbf{id}:= \phi=(x,y), \psi:=(r, \theta)$ have overlapping domains and it follows that,

$$(\phi \circ \psi^{-1})(r,\theta) = \phi(x,y) = (x,y)$$

$$( \psi \circ \phi^{-1})(x,y) = \psi(x,y) = (r, \theta)$$

i.e the transition map is smooth.

Lastly, $\theta = \theta(x,y)$ and so its total differential $D\theta (p_0) = (\theta_x(p_0), \theta_y(p_0))^T$ i.e given $q$ sufficiently close to $p_0:=(p_1,p_2)$ say $q=(p_1+\Delta x/\Delta t,p_2+\Delta y/\Delta t)$ then $q - p_0 = (\Delta x/\Delta t, \Delta y/ \Delta t)$ and so to get the vector at $p_0$ i.e the velocity of some curve passing through $p_0$, we let $\Delta t \to 0$ i.e we get the vector $(dx,dy)$ and hence,

$$ D \theta (p_0) \begin{pmatrix} dx \\ dy \end{pmatrix}_{p_0} = \theta_x(p_0) dx(p_0) + \theta_y(p_0) dy(p_0)$$

Again, the reason for dividing through by $\Delta t$ is that the matrix representation for the differential can also be achieved by just defining $D f(p) \cdot \gamma'(0) = (f \circ \gamma)'(0) \cdot \gamma'(0)$ where $\gamma$ is a curve passing through $p$. The differential relationship is just showing you how to approximate the directional change in $\theta$ given a change of $(x,y)$ in some direction.

  • $\begingroup$ Careful: Your formula for $\theta$ only works on $\{x>0\}$. $\endgroup$ – Ted Shifrin Sep 5 '17 at 16:46
  • $\begingroup$ @TedShifrin: I inserted the definition I'm using. Should be fixed now. $\endgroup$ – Faraad Armwood Sep 8 '17 at 16:23

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).


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