I'm currently studying group theory to delve into the Bianchi classification of 3-dimensional spaces, to then learn more about homogeneous spaces.
Instead of going for the quite orthodox way that Bianchi employed in his work, I'm trying to follow the algebraic construction from Landau by constructing the problem in a local inertial frame of reference. When doing so, we erect a set of basis vectors $\{e_i^{(a)}\}$ with $i = 1,2,3$ being the spatial coordinates and $(a) = 1,2,3$ a label to identify the vector. This triad uniquely defines the tensorial quantities in said frame of reference when we project those into them, i.e.
$$T^{(a)} = e^i_{(a)} T_i$$
also valid for arbitrary ranks when we use a dual triad $\{e^i_{(a)}\}$ such that
$$e_i^{(a)} e^j_{(a)} = \delta_i^j$$ $$e_i^{(a)} e^i_{(b)} = \delta_b^a$$
When we project a vector from the tangent space into the local frame, that is
$$ X_a := e_{(a)}^i \frac{\partial}{\partial x^i}$$
we can show that those satisfy a Lie algebra
$$[X_a, X_b] = C^d_{ab} X_d \quad \text{,} \quad C^d_{ab} = -C^d_{ba}$$
and those can be reduced to a set of 9 unique algebras, which can be found in the table of the same chapter in Landau.
My question then is: How does one proceed to find the basis vectors $\{e_{(a)}^i\}$ given a Lie algebra?
For instance, if we have a Bianchi-II algebra
$$[X_1, X_2] = 0 = [X_1, X_3]$$ $$[X_2, X_3] = X_1$$
which leads to, for a test function $f(x^k)$
$$\bigg(e_{(1)}^i (\partial_i e_{(2)}^j) - e_{(2)}^i (\partial_i e_{(1)}^j) \bigg)\partial_j f = 0$$ $$\bigg(e_{(1)}^i (\partial_i e_{(3)}^j) - e_{(3)}^i (\partial_i e_{(1)}^j) \bigg)\partial_j f = 0$$ $$\bigg(e_{(2)}^i (\partial_i e_{(3)}^j) - e_{(3)}^i (\partial_i e_{(2)}^j) \bigg)\partial_j f = e_{(1)}^j \partial_j f$$
It boils down to solving the PDEs above. This is solvable. Taub 1950 computed all the unique Lie brackets and put them on appendix, so I'm trying to reproduce them to understand how to approach similar problems.