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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 ?

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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.

Inertial frames are also crucial in the equivalence principle, which allowed Einstein to generalize special relativity. The equivalence principle states a freely falling observer in general relativity is an inertial observer, and therefore they "carry" with them a local Lorentz frame in their fall.

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  • $\begingroup$ Do you mean in "at any event we can find a local inertial (Lorentz) frame" that at any event, the tangent space is a Minkowski space ? $\endgroup$
    – PermQi
    Commented Feb 11 at 22:09
  • $\begingroup$ @PermQi No. Minkowski space is a manifold, not a tangent space. However in special relativity, we can often get away with identifying the tangent space with the manifold itself- but we can’t do this in GR. I mean that a small neighborhood about a point in curved spacetime resembles the manifold of special relativity. For example, if you zoom in close enough to the surface of a sphere, it looks like flat space. $\endgroup$
    – Aiden
    Commented Feb 11 at 23:36
  • $\begingroup$ so, that means at any event, there exists a neighborhood which is isometric to a Minkowski spacetime ? $\endgroup$
    – PermQi
    Commented Feb 12 at 1:00
  • $\begingroup$ but, does it mean that neighborhood is a vector space ? Because Minkowski spacetime should be a vector space $\endgroup$
    – PermQi
    Commented Feb 12 at 1:21
  • $\begingroup$ @Aiden I would suggest rewording the first sentence to remove “…which is locally flat”. Because “locally” has a standard meaning in topology and manifold theory, which conflicts with the Physics usage (which is what you intend), and it seems OP is (understandably so) confused by this usage. $\endgroup$
    – peek-a-boo
    Commented Feb 12 at 2:52

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