I am having a lot of doubts when computing covariant derivatives and lie brackets without coordinates. In particular I have the following situation:
I have (in coordinates), $ds^2 = Xd\phi ^2 + g_{ab}dx^a dx^b$ where $X$ and $g_{ab}$ don't depend on $\phi$. I define $\xi = \partial_\phi$ and $e_\phi = X^{-1/2}\xi$. I consider an orthonormal frame $e_\phi, e_1, e_2, e_3$ in which $[\xi,e_b] = 0$. I want to show $g(D_a e_b, e_\phi) = 0$ where $D_a e_b$ is the covariant derivative w.r.t. $e_a$ of $e_b$. If I was using coordinates this would be simple with the Christoffel symbols. Since I am not, I use the following formula:
$$g(Z, D_Y X) = \frac{1}{2}\{X(g(Y,Z)) + Y(g(Z,X)) - Z(g(X,Y)) - g([X,Z],Y) - g([Y,Z],X) - g([X,Y],Z)\}$$
Now, for the parts that are of the form $e_b(g(e_\phi, e_a))$, $e_\phi(g(e_a,e_b))$ or $g([e_a,e_b],e_\phi)$ I can easily prove they are $0$. My problem is with $-g([e_b,e_\phi],e_a) - g([e_a,e_\phi], e_b)$
Is it as simple as say that since the coefficients for $e_a,e_b,e_\phi$ are all constants the derivatives must be $0$? How does one compute covariant derivatives and Lie Brackets in general when defining a base of the tangent space instead of working with coordinates?
Thank you.
You will need to compute these in coordinates since you defined them using coordinates. The only way to be able to compute them without coordinates is to have some information about this basis. Now we know that it is orthonormal so all we know is that $g(e_i,e_j)=\delta_i^j$ which you can use to prove that the first quantities you mentioned vanish. To compute the resst note that $e_a=e_a^i\partial_{x_i}$ where the functions $e_a^i$ don't depend on $\phi$. So to compute $[e_b,e_\phi]$ just use the "algebraic" definition $$ [e_b,e_\phi](f) = e_b(e_\phi(f))-e_\phi(e_b(f)) $$ and then using the independence of many terms and considering canselations you get $$ [e_b,e_\phi] = e_b(X)\partial_\phi $$ which concludes your proof. Hope this helps