Calculating the coordinate basis

Question

In Robert M. Wald's General Relativity the definition of the "coordinate basis" (of the tangent space) of a manifold is given by:

Let $\psi: O \to U \subset \mathbb{R}^n$ be a chart with $p \in O.$ If $f \in \mathcal{F},$ then by definition $f \circ \psi^{-1}: U \to \mathbb{R}$ is $C^{\infty}.$ For $\mu = 1, ... , n$ define $X_\mu: \mathcal{F} \to \mathbb{R}$ by

$$X_{\mu} (f) = \frac{\partial}{\partial x^ \mu} (f \circ \psi^{-1})\Big|_{\psi(p)}.$$

---

The basis $\{ X_\mu \}$ of $V_p$ is called a coordinate basis. Had we chosen a different chart, $\psi^{'},$ we would have obtained a different coordinate basis $\{ X'_\nu \}.$ We can, of course, express $X_\mu$ in terms of the new basis $\{ X'_\nu \}.$ Using the chain rule of advanced calculus, we have

$$X_{\mu} = \sum_{\nu =1}^{n} \frac{\partial x'^\nu}{\partial x^ \mu} \Big|_{\psi(p)} {X'}_\nu.$$

However, this definition does not seem to be "right".

Let $O$ be $\mathbb{R}^2$ and let $\Psi$ be the identity chart. In this case, we get that $$X_1 (f) = \frac{\partial f}{\partial x}$$ $$X_2 (f) = \frac{\partial f}{\partial y}$$ which is fine. If we now choose a different chart though, $\psi^{'}$, which is the "polar coordinate system" given by $\psi(r, \theta) = (r \cos \theta, r \sin \theta);$ $\psi^{-1}(x,y) = (\sqrt{x^2 + y^2}, \arctan {y/x}).$

From the chain rule it seems like what we want is $$X_1 = \frac{\partial f}{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x} = \frac{\partial r}{\partial x} X'_r + \frac{\partial \theta}{\partial x}X'_\theta $$

Or, in other words, $X'_1 = X'_r = \frac{\partial f}{\partial r}, X'_2 = X'_\theta = \frac{\partial f}{\partial \theta}.$ But if we calculate $X'_1$ and $X'_2$ from the definition we get:

$$X'_1 = \frac{\partial}{\partial x} (\tilde{f} \circ \psi^{'-1})\Big|_{\psi^{'}(p)}.$$ (where $f(x,y) = \tilde{f} (r\cos \theta, r \sin \theta) = \tilde{f}(r, \theta).$) $$X'_1 = \frac{\partial \tilde{f}}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial \tilde{f}}{\partial \theta}\frac{\partial \theta}{\partial x}.$$

How exactly are we supposed to apply the definition given for the polar coordinate chart?

Answer 1

The diagram on page 15 is helpful. The axes of your chart are not x and y. Your axes are labeled $\theta$ and $r$. It is very important to understand that the chart is literally just a labeling in $R^n$.

$f$ is defined on the manifold. Not the chart. But you have a mapping from the manifold to the chart.

Therefore,

$f|_p$ = $f(r,\theta)_\ |_{\psi^{-1} (r,\theta)}$.

${X}_\theta (f)= \frac{\partial}{\partial x^\theta} (f \circ \psi^{-1} (r,\theta))$

or

$\frac{\partial f}{\partial \theta}= \frac{\partial}{\partial \theta} (f \circ \psi^{-1} (r,\theta))$

If $g:(r,\theta) \rightarrow (x(r,\theta),y(r,\theta))$ then

$\frac{\partial f}{\partial \theta}= \frac{\partial}{\partial \theta} (f \circ g \circ \psi^{-1} (r,\theta))$

Then apply the chain rule to get $\frac{\partial f}{\partial \theta}= \frac{\partial f}{\partial x} \frac{\partial x}{ \partial \theta}+ \frac{\partial f}{\partial y} \frac{\partial y}{ \partial \theta}$.

Answer 2

While @ExpertNonexpert is right that $\tilde{f}$ is an unnecessary misdirection, the mistake that actually matters is in this line:

$$X'_1 = \frac{\partial}{\partial x} (\tilde{f} \circ \psi^{'-1})\Big|_{\psi^{'}(p)}.$$

Specifically, the derivative is taken with respect to the wrong parameter. It should be

$$X'_1 = \frac{\partial}{\partial {x'}^1} (f \circ \psi^{'-1})\Big|_{\psi^{'}(p)}.$$

And since ${x'}^1$ is just $\theta$, everything works out as you'd expect.