How to choose a set of non-orthonormal basis vectors for the absolute space of a stationary and axisymmetric space-time in General Relativity?

Question

In General Relativity, the space-time is described by the metric tensor $g_{\mu\nu}$, where $\mu,\nu=0,1,2,3$ and the interval is written as $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$. A 3+1 split allows to write the above interval as $$dl^2=\gamma_{ij}dx^i dx^j,$$ where $\gamma_{ij}=\left(-g_{ij}+\dfrac{g_{0i}g_{0j}}{g_{00}}\right)$ is the $\textit{absolute}$ space metric and $i,j=1,2,3$ (The Classical Theory of Fields (Landau & Lifshitz), Chapter 10).

I am dealing with a problem in a stationary and axisymmetric space-time (Kerr metric). In some references, I found that for the 3-dimensional metric tensor $\gamma_{ij}$, it is preferable to choose a set of non-orthonormal basis vectors (instead of an orthonormal set) as follows: $$\mathbf{e_1}=\partial/\partial r,\quad \mathbf{e_2}=\partial/\partial\theta,\quad \mathbf{e_3}=\partial/\partial\phi-(g_{t\phi}/g_{tt})(\partial/\partial t)$$.

However, I couldn't understand the reason for choosing this particular set of basis vectors and this is not discussed in details in any of the references.

My question is: What is the logic behind choosing the above set of non-orthonormal basis vectors for the metric $\gamma_{ij}$? And is this a unique set of basis vectors that corresponds to $\gamma_{ij}$?

Answer 1

My answer is only on the pure geometric side of the question.\ A time slice give you a 3-mFranifold. Locally, a 3 manifold is an euclidean 3 ball. By taking $r$ as the radius and by taking a level set of that radius function, you get a 2 sphere (locally). $\theta$ and $\phi$ are the standard coordinate of the euclidean 2-sphere. The term in front of the time vector coordinate is a gauge, determining how far the time slicing is from that given by a product manifold. That explains the logic.\ As for the uniqueness of the choice, the answer is in the manner of doing the choice: there are an infinite number.