Systematic way to find the 1-form of the spin connection?

Question

I'm currently reading Nakahara's book "Geometry, Topology and Physics, 2nd Edition". Section 7.8 discusses Cartan's structure equations, which can be expressed, for a torsion free connection, as: $$de^a + \omega^a_{\ \ b}\wedge e^b=0$$ where $\omega^a_{\ \ b}$ is the spin connection 1-form and $e^a$ are the 1-form vierbeins for the metric.

Then, he gives a simple example (Example 7.14.) in which he calculates the connection for the case of a sphere $S^2$: $$ds^2 = d\theta^2 + \sin^2\theta d\phi^2$$ in this case we have: $$e^1 = d\theta\qquad e^2=\sin\theta d\phi$$ so the Cartan's first equation reads as: $$\begin{align} \sin\theta\omega^1_{\ \ 2}\wedge d\phi &=0\\ \cos\theta d\theta\wedge d\phi + \omega^2_{\ \ 1}\wedge d\theta&=0 \end{align}$$ Of course from the second one, we can immediately infer $\omega^2_{\ \ 1}=\cos\theta d\phi$ and then, by antisymmetry, $\omega^1_{\ \ 2}=-\cos\theta d\phi$.

Okay, in this case it's very simple and we can find it with the naked eye, but what if the form of the metric is more complicated? In that case we'll have a larger number of terms, possibly with crossed coordinates, and we'll need another method to solve it.

For example, if we have something like: $$ds^2 = -e^{2f(\rho)}dt^2+d\rho^2+\alpha_1^2(\rho)\left(d\theta^2+\sin^2\theta d\phi^2 \right)+\alpha_2^2(\rho)\left(d\bar{\theta}^2+\sin^2\bar{\theta} d\bar{\phi}^2 \right)+\alpha_3^2(\rho)\left(d\tilde{\theta}^2+\sin^2\tilde{\theta} d\tilde{\phi}^2 \right)$$ the first Cartan equation involves a sum of about 8 terms (see the imagen below), we can't act like the previous example. Would there be any systematic way to calculate it?

where I've introduced the Maurer-Cartan 1-forms $\sigma_{1}=\sin \theta d \phi, \quad \sigma_{2}=d \theta, \quad \sigma_{3}=\cos \theta d \phi$

Answer 1

I'll show you a simpler example (with only 4 variables, not 8). Writing out all those equations is not informative. You have to think about what the basis $2$-forms are. Remember that the connection forms will satisfy $\omega^i_j = -\omega^j_i$ and $\omega^0_j = \omega^j_0$ for $i,j\ge 1$. (In particular, $\omega^i_i = 0$ for all $i\ge 1$.) Since \begin{align*} e^0 &= e^{f(\rho)}dt \\ e^1 &= d\rho \\ e^2 &= \alpha(\rho)d\theta \\ e^3 &= \alpha(\rho)\sin\theta\,d\phi, \end{align*} we begin by differentiating. \begin{align*} de^0 &= e^{f(\rho)}f'(\rho)\,d\rho\wedge dt \\ de^1 &= 0 \\ de^2 &= \alpha'(\rho)\,d\rho\wedge d\theta \\ de^3 &= \alpha'(\rho)\sin\theta\,d\rho\wedge d\phi+\alpha(\rho)\cos\theta\,d\theta\wedge d\phi. \end{align*}

Let's just start with the obvious. For $de^0$, we need the $e^1$ term: $$e^{f(\rho)}f'(\rho)\,d\rho\wedge dt = -e^{f(\rho)}f'(\rho)\,dt\wedge d\rho = -e^{f(\rho)}f'(\rho)\,dt\wedge e^1,$$ and so $\color{red}{\omega^0_1 = e^{f(\rho)}f'(\rho)\,dt}$ (and $\omega^0_j$ for $j>1$ will all turn out to be $0$ by symmetry).

From $de^1=0$, we see that $\omega^1_0 = \omega^0_1$ is indeed a multiple of $e^0$. $\omega^1_2 = -\omega^2_1$ and $\omega^1_3=-\omega^3_1$ will be decided momentarily.

From $de^2=\alpha'(\rho)d\rho\wedge d\theta$, we infer that $\color{red}{\omega^2_1=\alpha'(\rho)d\theta}$. Moreover, $\omega^2_0=0$ since no $dt$ appears.

Now it gets a bit more interesting. We have $de^3 = \alpha'(\rho)\sin\theta\,d\rho\wedge d\phi + \alpha(\rho)\cos\theta\,d\theta\wedge d\phi$, and so we see $e^1$ and $e^2$ appearing. Indeed, from $$de^3 = -\omega^3_1\wedge e^1 - \omega^3_2\wedge e^2$$ we see that $\color{red}{\omega^3_1 = \alpha'(\rho)\sin\theta\,d\phi}$ and $\color{red}{\omega^3_2 = \cos\theta\,d\phi}$. Note that $\omega^3_1$ and $\omega^3_2$ are both multiples of $e^3$, so that $\omega^1_3\wedge e^3 = \omega^2_3\wedge e^3 = 0$, as needed, and, indeed, no more terms show up in $de^1$ and $de^2$. Note that, once again, no $dt$ appears, and so $\omega^3_0 = 0$.

All the connection forms are now determined.

Answer 2

I will try to answer it as clear and concise as posible, using what I think is a more intuitive notation. \begin{equation} 0 \rightarrow t,\ 1 \rightarrow \rho,\ 2 \rightarrow 2,\ 3\rightarrow 1,\ 4\rightarrow\bar{2},\ 5\rightarrow\bar{1},\ 6\rightarrow\tilde{2},\ 7\rightarrow\tilde{1} \end{equation} From the equation where $a=t$, one can deduce that $\omega_\rho^t=f'e^f dt$, which also cancels out one term in $a=\rho$, and that $\omega^t_a$ should be $0$ for every numerical a.

One can also see easily that the $\omega^r_a$ terms must compensate the term that shows up when differenciating the $\alpha(\rho)$ in $de^a$,taking the form of $\omega^\rho_a=\alpha'(\rho)\cdot\sigma^b$ where $b=1$ when $a=2$ and so on. This can be applied to every 2-sphere indices. It is also compatible with the $a=\rho$ equation.

In order to counter the $d\sigma^a$ term in the $a=$"a number", you can set $\omega_2^1=\sigma^3$.

The other components of the 1-form are $0$. I will not discuss this part of the problem in detail because it is trivial when you assume that they can be built by a linear combination of $dt$, $d\rho$ and the Maurer-Cartan 1-forms (or other 1-forms involved in your definition of the space), which is something aproximating a general way to find the 1-form of spin, at least in relatively simple cases from my experience.