- #1
Shirish
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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):
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.
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}##.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}##
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.