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I'm reading a section [Core Principles of Special and General Relativity by Luscombe] on how the derivative of a basis vector in a manifold is related to connection coefficient. Quoting (the notation $A^{\alpha}_{\beta'}$ means $\partial x^{\alpha}/\partial x^{\beta'}$ - instead of using primed and unprimed variables, we use primed/unprimed indices to distinguish different bases):

Consider the total differential of a vector field $\mathbf{t}=t^{\alpha}(x^1,\ldots,x^n)\mathbf{e}_{\alpha}$ (without specifying what "$\text{d}$" means): $$\text{d}\mathbf{t}=\text{d}t^{\alpha}\mathbf{e}_{\alpha}+t^{\alpha}\text{d}\mathbf{e}_{\alpha}=\bigg(\frac{\partial t^{\alpha}}{\partial x^{\beta}}\text{d}x^{\beta}\bigg)\mathbf{e}_{\alpha}+t^{\alpha}\bigg(\frac{\partial\mathbf{e}_{\alpha}}{\partial x^{\beta}}\text{d}x^{\beta}\bigg)$$ The second term involves derivatives of vectors - the very quantiuty we're trying to formulate with the covariant derivative. We expect a change $\mathbf{e}_{\beta}\to\mathbf{e}_{\beta}+\text{d}\mathbf{e}_{\beta}$ under $x^{\alpha}\to x^{\alpha}+\text{d}x^{\alpha}$ because coordinate basis vectors are tangent to coordinate curves. We know that the manifold $M$ is locally flat. Thus in a sufficiently small neighborhood $p\in M$ there is a local inertial frame, the basis vectors of which are constants; call them $\{\mathbf{e}^0_{\beta}\}$. The coordinate basis $\{\mathbf{e}_{\alpha}\}$ in a neighborhood of $p\in M$ can be expressed in the basis $\{\mathbf{e}^0_{\beta}\}$, $\mathbf{e}_{\alpha}(x)=A^{\beta'}_{\alpha}(x)\mathbf{e}^0_{\beta'}$. Differentiating this formula (the $\mathbf{e}^0_{\beta'}$ are constants), $$\partial_{\mu}\mathbf{e}_{\alpha}=(\partial_{\mu}A^{\beta'}_{\alpha})\mathbf{e}^0_{\beta'}=(\partial_{\mu}A^{\beta'}_{\alpha})A^{\rho}_{\beta'}\mathbf{e}_{\rho}$$ where we've inverted the basis transformation, $\mathbf{e}^0_{\beta'}=A^{\rho}_{\beta'}\mathbf{e}_{\rho}$

Now this "we know that the manifold $M$ is locally flat" doesn't seem like a good enough explanation for why basis vectors at two different points were related by the transformation law. The book covers the explicit, mathematical derivation for why basis vectors in the same tangent space but under different charts $(U,x)$ and $(U',x')$ can be related by $\mathbf{e}_{\beta'}=A^{\rho}_{\beta'}\mathbf{e}_{\rho}$.

But in the above quote, basis vectors at different points, and hence belonging to different tangent spaces have been related by the exact same formula. Can anyone help me with a proper mathematical proof on how the same transformation law comes about in this context? Without that, "$M$ is locally flat" is a handwavy argument.

Edit: For more context, I quote the text after the above quoted paragraph:

Thus there is a three-index symbol, call it $\gamma^{\rho}_{\mu\alpha}\equiv(\partial{\mu}A^{\beta'}_{\alpha})A^{\rho}_{\beta'}$, effecting the derivative of $\mathbf{e}_{\alpha}$ along the coordinate curve $x^{\mu}$:$$\partial_{\mu}\mathbf{e}_{\alpha}=\gamma^{\rho}_{\mu\alpha}\mathbf{e}_{\rho}$$ which we note involves all other vectors of the basis. We'll show that $\gamma^{\nu}_{\alpha\mu}=\Gamma^{\nu}_{\alpha\mu}$, the Christoffel symbols. [footnote: Many books define the connection coefficient with the above equation] While we invoked a physical argument to get to this point (spacetime manifold is locally flat), it's not necessary to do so. We'll reach the same conclusion once we define geodesic curves on a manifold, which in turn require parallel transport and the covariant derivative. It's all connected!

Shirish Kulhari
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  • I can't explain it because it doesn't make sense, at least not without a good deal more context. For instance, on a cylinder in 3-space, which is certainly locally flat, there is no "locally constant" set of basis vectors for the tangent space in any neighborhood. There may be a set of $n$ vector fields that we'd like to declare "constant" for some imagined differentiation operator, whose other properties we can then derive, but... well, one possible answer here is "Consider reading a different book." – John Hughes Aug 15 '20 at 16:28
  • @JohnHughes: I've quoted the relevant section from the start. Maybe I should also mention that the author is talking about spacetime manifold in particular being "locally flat". But then whatever he claims after that would only be valid for that specific type of manifold. Also, in the section he's trying to explain the fact that connection coefficients are somehow related to change in basis vectors as we move from one point to another in a neighborhood. He does mention in some sources, the connection coefficients are defined as derivatives of basis vectors. Maybe I should check that out – Shirish Kulhari Aug 15 '20 at 16:39
  • @JohnHughes: I have added the paragraph just after the part I'd already quote in the form of an edit to the question. – Shirish Kulhari Aug 15 '20 at 16:50
  • I don't understand what A is, exactly, here. But this type of thing is usually called "parallel transport", and I think the author is appealing to the so-called "geodesic normal coordinates" that always exist in a small enough neighborhood on a manifold. – pseudocydonia Aug 18 '20 at 06:06
  • @pseudocydonia: $A^{\alpha}_{\beta'}$ means $\partial x^{\alpha}/\partial x^{\beta'}$ (just for notational convenience). By this point the author hasn't talked about parallel transport and geodesics at all. So then I guess he's using the consequence of concepts/results he hasn't introduced yet. (I got the same impression from the Edit addendum paragraph in my question) – Shirish Kulhari Aug 18 '20 at 06:26
  • Honestly -- as someone who started out in physics, it shocks me that physicists (like, I assume, the author of the book) seem to prefer to understand differential geometry without precisely defining things. It's all well and good to have non-rigorous infinitesimals; but differential geometry is a giant mess with dozens of objects floating around. I second the recommendation to read a different book. – pseudocydonia Aug 18 '20 at 07:23
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    It may not directly help you, but Lee's Intro to Smooth Manifolds is extremely careful and clear. Lee even frequents stackexchange. – pseudocydonia Aug 18 '20 at 07:24
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    Anyway, some people seem to be able to figure out different styles of differential geometry more easily. I find this heavy computational style opaque (but also the excessively abstract nonsense version doesn't quite work for me). I do think what he is saying follows from the basic facts of geodesic normal coordinates. That is: in a small enough neighborhood around $p \in M$, we can work in "polar coordinates" where a point is uniquely specified by a tangent vector $v$ at $p$ and a positive $r>0$, because one follows the geodesic starting from $p$, in the direction $v$, for distance $r$. – pseudocydonia Aug 18 '20 at 07:30
  • In these coordinates one does indeed have an isomorphism induced by the coordinate chart between the small neighborhood of $p$ and the patch of Euclidean space in the domain of the chart, hence one can use the fact that in Euclidean space, the tangent spaces at different points are related by translation. – pseudocydonia Aug 18 '20 at 07:32
  • But this is not obvious! There are a bunch of things that need to be checked carefully. I think Jost's Riemannian Geometry discusses some of this early on. – pseudocydonia Aug 18 '20 at 07:34
  • @pseudocydonia: Thanks so much for the recommendation - I'll check out Lee's book. As for your last comment, I think I'll understand it when I learn about geodesics and all, so I'll definitely get back to the comment after that. Thanks again! – Shirish Kulhari Aug 18 '20 at 07:34
  • @pseudocydonia: One thing - I'd say write your comments down as an answer? Will be glad to mark it as accepted – Shirish Kulhari Aug 18 '20 at 07:37
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    Sure thing. Oh, also: apparently there is a terminological conflict between GR and math-differential geometry, on the meaning of "locally flat". So be careful! This has tripped me up before. https://math.stackexchange.com/a/1971496/381572 – pseudocydonia Aug 20 '20 at 04:30
  • @pseudocydonia: Aha that's extremely useful! I did suspect that's what the author meant by "locally flat" - a region covered by a single chart. What I'm kind of struggling with is a proof that the transformation law for basis vectors (as referred to in the question) holds for vectors at different points in the same open set $U$. Vectors in different coordinate systems at the same point - no problem. Vectors in different coordinate systems at different point in the same open set $U$ that admits both coordinate systems - that part is confusing – Shirish Kulhari Aug 20 '20 at 04:36
  • Hmm. Unfortunately I forget how to do this type of calculation. Best of luck – pseudocydonia Aug 20 '20 at 05:18

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