here I come with a question about inertial frames as defined in General Relativity, and how to prove that the general definition is consistent with the particular case of Special Relativity.
So to contextualize, I have found that one can define inertial frames in General Relativity, as follows: given a 4-dimensional lorentzian manifold $M$ with metric tensor $g$ and signature (-1,1,1,1), a frame field is defined to be a set of four vector fields $\{e_{0},e_{1},e_{2},e_{3}\}$ such that
$g(e_{i},e_{j}) = \eta_{ij}$ ($\eta$ being the Minkowski matrix with the signature given above)
Further, one says that the frame field in question is inertial and nonrotating if the following condition is satisfied:
$\nabla_{e_{0}}e_{i} = 0$ for $i=0,1,2,3$
where $\nabla$ is the Levi-Civita connection on the manifold (let me call this "Definition A"). All this straight out of Wikipedia, and quite elegant and nice.
Now, my question is how does this correspond to the definition of inertial frames in Special Relativity as usually found in textbooks. So now assume that g is the Minkowski metric and that we are in Minkowski spacetime, where there is a given global coordinate chart $(t,x,y,z)$ (I am setting $c=1$ for simplicity) such that:
$ g = -dt\otimes dt + dx\otimes dx + dy\otimes dy + dz\otimes dz$
Then it is straightforward to see that the vector fields $\partial/\partial t, \partial/\partial x, \partial/\partial y, \partial/\partial z$ are indeed an inertial nonrotating frame field in the abstract sense of Definition A. But in Minkowski spacetime one has Definition B (the one usually found in textbooks), where we are given an initial "precursor" inertial frame (the one used by the scientist writting the textbook, or whatever), which in this context is a coordinate chart with certain physical properties, namely that there are no inertial forces (i.e. the Christoffel symbols vanish identically), and can of course be identified with the coordinate chart $(t,x,y,z)$ I mentioned earlier. And then -crucially- we are told that "inertial frames" are all those coordinate charts related to this precursor coordinate chart by a Lorentz transformation, which is a linear coordinate transformation $\Lambda$ such that $\eta_{ij} = \Lambda^{k}_{i}\Lambda^{\ell}_{j}\eta_{k\ell}$ (this guarantees that the spacetime interval is preserved).
So, here comes my question: in Minkowski spacetime, take an arbitrary nonrotating inertial frame in the sense of Definition A; that is, a set of four vector fields $e_{0},e_{1},e_{2},e_{3}$ satistying the requisites of Definition A. Is it the case, then, that there exists a coordinate chart $x^{0},x^{1},x^{2},x^{3}$ and a suitable Lorentz transformation $\Lambda$ from $(t,x,y,z)$ (the "precursor coordinate chart" inherent to the construction of Minkowski spacetime) to $(x^{0},x^{1},x^{2},x^{3})$ such that:
$e_{i} = \partial/\partial x^{i}$ for $i=0,1,2,3$ ?
Consider this smooth vector field: $$ e_0 = \partial_t, \ \ \ e_1 = \cos z \partial_x - \sin z \partial_y, \ \ e_2=\sin z \partial_x + \cos z \partial_y, \ \ e_3 = \partial_z. $$ This obeys $\nabla_{e_0} e_0 = \nabla_{e_0} e_i = 0$.
However, there exists no coordinate system $(x^0, x^1, x^2, x^3)$ such that $e_i= \partial_{x^i}$ for $i = 0,1,2,3$.
Indeed, suppose for contradiction that such a coordinate system exists. Then the commutator $[e_1,e_3] = \partial_{x^1}\partial_{x^3} - \partial_{x^3}\partial_{x^1}$ is identically zero, since partial derivatives commute. However, by explicit calculation, $$ [e_1, e_3] = \sin z \partial_x + \cos z \partial_y \neq 0. $$