Cartan's vielbein

Question

This is from Anthony Zee's Einstein Gravity in a Nutshell on page 600, where he explains how to derive the connection form $\omega$ in Cartan's vielbein formalism.

The example in the lower part is $\Bbb S^2$, given by $\mathrm{d}s^2=\mathrm{d}\theta^2+\sin^2 \theta\,\mathrm{d}\phi^2$, and $e^1 = \mathrm{d}\theta$ and $e^2 = \sin\theta\, \mathrm{d}\phi$.

I don't understand how $\omega^{12}$ can be $cos\theta\, \mathrm{d}\phi$. Since $e^1=\mathrm{d}\theta$, we must have:

$$ \mathrm{d} e^1 = 0 =-{\omega^1}_\alpha e^\alpha = -{\omega^1}_1 e^1 - {\omega^1}_2 e^2 $$

This can only be true if ${\omega^1}_1 = {\omega^1}_2 = 0$.

But Zee's result (from $\omega$'s antisymmetry) is ${\omega^1}_2 = -\cos\theta\,\mathrm{d}\phi$.

Answer 1

Not quite, $\omega _{ij}$ is an antisymmetric matrix of 1-forms. S0 $\omega_{ii}=0$ always, but $\omega_{12}=f e^2$ with $f$ some function to be determined solves your equation.

Also the correct equation has a wedge product, \begin{equation} de^i+ \omega^i_{\ k} \wedge e^k=0 \end{equation}

Answer 2

That's the pitfall of omitting the $\wedge$ product. The $1$-forms $e^1$ and $e^2$ need not be "linearly independent" in the sense that $$\alpha_1\wedge e^1+\alpha_2\wedge e^2=0\implies \alpha_1=\alpha_2=0, $$ since the $e^1$-component of $\alpha_1$ will get wedged with the $e^2$-component of $\alpha_2$, and so on. These partial results need not be zero or cancel each other. The correct consequence of the expression in display above is known as Cartan's Lemma.$\newcommand\d{\mathrm{d}}$

The expression $$\d e^2 = \cos\theta\, \d\theta\wedge \d\phi = -(\cos \theta\,\d\phi)\wedge e^1$$tells us that $\omega^2_{\;1} = \cos \theta\,\d\phi$, but it is not always easy to recognize $e^1$ and $e^2$ here. They might be messy and not pop up immediatelly.

The general procedure goes as follows: by anti-symmetry we have $\omega^1_{\;1} = 0$. Write $\omega^1_{\;2}=A_1\,\d\theta+A_2\,\d\phi$, where $A_1$ and $A_2$ are smooth functions (the $\omega^\alpha_{\;\beta\mu}$, if you want). Since $\d e^1=0$, we write $$0=(A_1\,\d\theta+A_2\,\d\phi)\wedge (\sin \theta\,\d\phi) = A_1\sin \theta \,\d\theta\wedge \d\phi, $$and now $A_1=0$. Using the expression for $\d e^2$, we see that $$\cos \theta \,\d\theta\wedge \d\phi =-\omega^2_{\;1}\wedge e^1 =-(-A_2 \d\phi)\wedge \d\theta $$gives $A_2=-\cos \theta$, hence $\omega^1_{\;2}=-\cos \theta\,\d\phi$ as wanted.