I'm reading the book Gravitation (Misner, Thorne and Wheeler), and going over some of the exercises. The exercise that I'm particularly interested in at the moment is Exercise 9.7, which poses the following question:
Compute the commutator $[e_{\hat{\theta}},e_{\hat{\phi}}]$ of the vector fields
$$e_{\hat{\theta}}=\frac{1}{r}\frac{\partial}{\partial\theta},\space\space\space e_{\hat{\phi}}=\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}.$$
Express your result as a linear combination of $e_{\hat{\theta}}$ and $e_{\hat{\phi}}$.
The book has been interchangeably using the notation that partial derivative operators are equal/equivalent to basis vectors. As a result I'm treating $\frac{\partial}{\partial\theta}$ as $e_{\theta}$, and $\frac{\partial}{\partial\phi}$ as $e_{\phi}$ (I'm not sure if I'm abusing notation here accidently). Also, since only $\theta$ and $\phi$ "vector fields" have been provided should I make the assumption that the coordinate index corresponding to $r$ is not being summed over for repeated dummy indices $\alpha$ and $\beta$. I figure that even if it was to be summed over, it wouldn't contribute any non-vanishing terms as $r$ is orthogonal to $\theta$ and $\phi$ (?).
I've attempted to find the commutator by substituting $e_{\hat{\theta}}$ and $e_{\hat{\phi}}$ into a formula provided in a previous section (eq. 9.20) where the commutator of vectors $u$ and $v$ was derived:
$$[u,v]=(u^\alpha v^\beta_{,\alpha}-v^\alpha u^\beta_{,\alpha})\frac{\partial}{\partial x^\beta}$$
Below is what I've tried so far:
$$[e_{\hat{\theta}},e_{\hat{\phi}}]=\left[\left(\frac{1}{r}e_\theta\right)^{\alpha}\left(\frac{1}{r\sin\theta}e_\phi\right)^{\beta}_{,\alpha}-\left(\frac{1}{r\sin\theta}e_\phi\right)^{\alpha}\left(\frac{1}{r}e_\theta\right)^{\beta}_{,\alpha}\right]\frac{\partial}{\partial x^\beta}$$
At this next line I suspect I might be slightly abusing notation by replacing $\alpha$ in the first and second term with $\theta$ and $\phi$ respectively to indicate the component:
$$=\left[\left(\frac{1}{r}e_\theta\right)^{\theta}\left(\frac{1}{r\sin\theta}e_\phi\right)^{\beta}_{,\theta}-\left(\frac{1}{r\sin\theta}e_\phi\right)^{\phi}\left(\frac{1}{r}e_\theta\right)^{\beta}_{,\phi}\right]\frac{\partial}{\partial x^\beta}$$
$$=\left[\left(\frac{1}{r}\right)\left(\frac{1}{r\sin\theta}e_\phi\right)^{\beta}_{,\theta}-\left(\frac{1}{r\sin\theta}\right)\left(\frac{1}{r}e_\theta\right)^{\beta}_{,\phi}\right]\frac{\partial}{\partial x^\beta}$$
Expressing the two indices of $\beta$ explicitly:
$$=\left[\left(\frac{1}{r}\right)\left(\frac{1}{r\sin\theta}e_\phi\right)^{\phi}_{,\theta}-\left(\frac{1}{r\sin\theta}\right)\left(\frac{1}{r}e_\theta\right)^{\phi}_{,\phi}\right]\frac{\partial}{\partial \phi}+\left[\left(\frac{1}{r}\right)\left(\frac{1}{r\sin\theta}e_{\phi}\right)^{\theta}_{,\theta}-\left(\frac{1}{r\sin\theta}\right)\left(\frac{1}{r}e_\theta\right)^{\theta}_{,\phi}\right]\frac{\partial}{\partial \theta}$$
$$=\left[\left(\frac{1}{r}\right)\left(\frac{1}{r\sin\theta}\right)_{,\theta}-\left(\frac{1}{r\sin\theta}\right)(0)_{,\phi}\right]\frac{\partial}{\partial \phi}+\left[\left(\frac{1}{r}\right)(0)_{,\theta}-\left(\frac{1}{r\sin\theta}\right)\left(\frac{1}{r}\right)_{,\phi}\right]\frac{\partial}{\partial \theta}$$
$$=\left[\left(\frac{1}{r}\right)\left(\frac{1}{r\sin\theta}\right)_{,\theta}\right]\frac{\partial}{\partial \phi}+\left[-\left(\frac{1}{r\sin\theta}\right)\left(\frac{1}{r}\right)_{,\phi}\right]\frac{\partial}{\partial \theta}$$
Assuming $\frac{1}{r}$ is constant with respect to $\phi$, causing the second term to vanish:
$$=\left[\left(\frac{1}{r}\right)\left(\frac{1}{r\sin\theta}\right)_{,\theta}\right]\frac{\partial}{\partial \phi}=\left[\left(\frac{1}{r}\right)\left(-\frac{r\cos\theta}{r^2\sin^2\theta}\right)\right]\frac{\partial}{\partial \phi}=-\frac{\cos\theta}{r^2\sin^2\theta}\frac{\partial}{\partial \phi}$$
$$\therefore [e_{\hat{\theta}},e_{\hat{\phi}}]=-\frac{\cos\theta}{r\sin\theta}\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}=-\frac{1}{r}\cot\theta \space e_{\hat{\phi}}$$
I have no one to check this with and was hoping someone could take a look on here. Thanks in advance!
Best Answer
You are making hard work of this: $[e_\theta,e_\phi]=e_\theta e_\phi-e_\phi e_\theta$. Let's see what this differential operator does to a function $f$. $$e_\theta e_\phi f=\frac1r\frac{\partial}{\partial\theta}\frac1{r\sin\theta} \frac{\partial f}{\partial\phi}=\frac1{r^2\sin\theta} \frac{\partial^2 f}{\partial\theta\,\partial\phi}-\frac{\cos\theta}{r^2\sin^2\theta}\frac{\partial f}{\partial\phi}$$ while $$e_\phi e_\theta f=\frac1{r\sin\theta}\frac{\partial}{\partial\phi}\frac1r \frac{\partial f}{\partial\theta}=\frac1{r^2\sin\theta} \frac{\partial^2 f}{\partial\phi\,\partial\theta}$$ so that $$[e_\theta,e_\phi]f=-\frac{\cos\theta}{r^2\sin^2\theta}\frac{\partial f}{\partial\phi}=-\frac{\cot\theta}re_\phi f.$$