# Relating basis vectors at different points in a neighborhood

• B
I'm reading a section on the derivative of a vector in a manifold. 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.

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strangerep

[I'll never understand why questioners think it's a good idea not to post their sources (sigh).]

[I'll never understand why questioners think it's a good idea not to post their sources (sigh).]
Core Principles of Special and General Relativity, Luscombe

strangerep
Core Principles of Special and General Relativity, Luscombe
OK, so it's Luscombe sect 14.2 on p250 for those playing along at home.

I, too, don't like the handwavy reasoning. But it really all just falls out from a simple change of coordinate variables, and the chain rule. E.g., for a change of variables $$x^\mu ~\to~ x'^\mu ~=~ x'^\mu(x) ~,$$ we can use the chain rule to figure out the corresponding transformation of derivative operators,. Let ##f(x)## be an arbitrary scalar field. Then $$\frac{\partial f}{\partial x'^\mu} ~=~ \frac{\partial x^\alpha}{\partial x'^\mu} \; \frac{\partial f}{\partial x^\alpha} ~.$$ The vector basis elements ##{\mathbf e}_\mu## are just the coordinate derivative operators here. You don't even need to think about tangent spaces It's just an elementary change of variable.

HTH.

OK, so it's Luscombe sect 14.2 on p250 for those playing along at home.

I, too, don't like the handwavy reasoning. But it really all just falls out from a simple change of coordinate variables, and the chain rule. E.g., for a change of variables $$x^\mu ~\to~ x'^\mu ~=~ x'^\mu(x) ~,$$ we can use the chain rule to figure out the corresponding transformation of derivative operators,. Let ##f(x)## be an arbitrary scalar field. Then $$\frac{\partial f}{\partial x'^\mu} ~=~ \frac{\partial x^\alpha}{\partial x'^\mu} \; \frac{\partial f}{\partial x^\alpha} ~.$$ The vector basis elements ##{\mathbf e}_\mu## are just the coordinate derivative operators here. You don't even need to think about tangent spaces It's just an elementary change of variable.

HTH.
It helps for sure! As far as I've read (so by my limited understanding), we interpret $$\bigg(\frac{\partial f}{\partial x^i}\bigg)_p=\partial_i(f\circ x^{-1})(x(p))$$ And here's the derivation of chain rule I wrote that is as explicit as I could make it \begin{align} \bigg(\frac{\partial}{\partial x^i}\bigg)_pf&=\partial_i(f\circ x^{-1})(x(p)) =\partial_i\big((f\circ x'^{-1})\circ(x'\circ x^{-1})\big)(x(p)) \\&=\big(\partial_j(f\circ x'^{-1})\big)(x'(p))\cdot \big(\partial_i(x'\circ x^{-1})\big)^j(x(p)) \\&=\big(\partial_j(f\circ x'^{-1})\big)(x'(p))\cdot \big(\partial_i(x'^j\circ x^{-1})\big)(x(p)) \\&=\bigg[\bigg(\frac{\partial}{\partial x'^j}\bigg)_pf\bigg]\cdot\bigg[\bigg(\frac{\partial}{\partial x^i}\bigg)_px'^j\bigg] \\&=\bigg(\frac{\partial x'^j}{\partial x^i}\bigg)_p\bigg(\frac{\partial}{\partial x'^j}\bigg)_pf \end{align} The point is that while employing the chain rule, we're calculating the derivatives at a single point only. It doesn't work if we try calculating derivative w.r.t. ##x## chart map at one point and w.r.t. the ##x'## chart map at another point. So that was the primary source of confusion.

Having said that, I know that the chain rule that you wrote in your answer definitely works in flat spaces where there's a global chart. So just to be clear, your version of chain rule follows because of the assumption that the neighborhood of a point ##p## is locally flat, right? And the precise mathematical description for "locally flat" would be that that neighborhood can be covered by a single chart, correct?

strangerep