# What role does inertial frame play in general relativity?

The study of geometric aspect of Special Relativity is all about the geometry of Minkowski spacetime $$(M,\langle,\rangle)$$, a flat spacetime whose curvature vanishes at everywhere. The (Minkowski/Lorentz) inertial frame is defined based on Einstein's 2 postulates of Special Relativity, which in term of mathematic can be interpreted as a proper orthochronous transformation $$(t,x):\ M\longrightarrow\mathbb R^4$$, and its representation matrix $$L=(a_{ij})_{0\leq i,j\leq 3}$$ therefore is a proper orthochronous matrix, i.e $$L^\top\eta\,L\,=\,\eta\,=\,\text{diag}\,(-1,1,1,1),\ \ |L|=1,\ \ a_{00}\geq 1.$$ These transformation are called proper orthochronous Lorentz transformation.

So, it's clear that the inertial frames and Lorentz transformations take an important rule in the study of geometry of Minkowski spacetime. But, in the study of General Relativity where spacetimes need not to be flat and henceforth thet need not to be vector spaces or affine spaces, I wonder what is the role of inertial frames and Lorentz transformations ?

In general relativity, spacetime is a manifold which is locally flat. At any point $$P$$ in the manifold, it's possible to find a coordinate transformation where the components of the metric become that of Minkowski space, with vanishing partial derivatives. i.e. $$g_{\alpha\beta}\vert_P = \eta_{\alpha\beta}$$ and $$\partial_\mu g_{\alpha\beta}\vert_P = 0$$ (see Riemann normal coordinates). To put it physically, at any event we can find a local inertial (Lorentz) frame. On a 4-dimensional Riemannian manifold, this coordinate transformation can be determined with 6 degrees of freedom left over. These correspond to the 3 rotations and 3 boosts of the Lorentz group.