0
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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 $\frac{\partial f}{\partial \theta}= \frac{\partial f}{\partial x} \frac{\partial x}{ \partial \theta}+ \frac{\partial f}{\partial y} \frac{\partial y}{ \partial \theta}$.

$\endgroup$
1
  • $\begingroup$ Please do not vandalize your answers. $\endgroup$
    – Chris
    Commented Feb 25, 2021 at 10:25
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Yes, but my chart was written $(r, \theta) \mapsto (x,y)$ which would imply that we take the derivative w.r.t. $x$ and $y.$ If I understand @Expert Nonexpert then the chart should actually be viewed as $p \in M \mapsto (r, \theta)$. $\endgroup$
    – Jbag1212
    Commented Feb 26, 2021 at 20:02
  • $\begingroup$ "my chart was written $(r, \theta) \mapsto (x, y)$". That's not a chart. You could think of it as a transition map, but definitely not a chart. It looks like maybe your confusion is because you take $O = \mathbb{R}^2$, and any chart is a map $\psi: O \to U$, with $U = \mathbb{R}^2$ also. But those are two different $\mathbb{R}^2$s. It's best to think of $M$ and $O$ as nebulous spaces. Coordinates like $x$, $y$, $r$, $\theta$ are never actually in those spaces; they merely correspond to points in those spaces. $\endgroup$
    – Mike
    Commented Feb 26, 2021 at 20:27
  • $\begingroup$ @Jbag1212 The chart is the labeling. It's a tricky notion because we perform this labeling for the surface of the sphere by slapping labels on it. We don't stop and think about how to express that mathematically. When we do that we are effectively mapping the points that are solutions to the equation r^2 = x^2 + y^2 +z^2 to the labels. That isn't usually stressed. That's why it can be confusing. $\endgroup$
    – user288901
    Commented Feb 28, 2021 at 5:33
  • $\begingroup$ @Jbag1212 then is f is defined on those points (x,y,z) =(x(theta, phi),y(theta,phi),z(theta,phi)). Wald refers to a $\psi$ and that's tricky too because $\psi$ is the mapping to the (theta, phi) chart. but we hardly ever really work with that mathematical function in practice. we are much more familiar $\psi^{-1}$ with x=rcos(theta) and y=rsin(theta) $\endgroup$
    – user288901
    Commented Feb 28, 2021 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.