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Warning: this may be totally trivial, or totally wrong.

I've been working through Sean Carroll's lecture notes, and I've got to http://preposterousuniverse.com/grnotes/grnotes-two.pdf [Broken]. I follow the derivation for showing that the tangent space bases are the partial derivatives (Carroll's equation 2.9 on the 13th page of the PDF, numbered 43). However, he only sketches the proof for the cotangent basis, and I'm not sure I've filled in the gaps correctly.

Carroll says that the gradient is the canonical 1-form and quotes its action on the vector ##d/d\lambda## as (Carroll's 2.14, 14th page, numbered 44):

[tex]

\mathrm{d}f\left(\frac{d}{d\lambda}\right) =\frac{df}{d\lambda}

[/tex]

You can expand this using Carroll's 2.9 as

[tex]

\mathrm{d}f\left(\frac{d}{d\lambda}\right) =\frac{df}{d\lambda} =\frac{dx^\mu}{d\lambda}\partial_\mu f

[/tex]

from which I deduce that ##d/d\lambda## should be read as ##dx^\mu/d\lambda## in this context, and that ##\mathrm{d}f## is what I would have written in a non-relativistic context as ##\nabla f##. I presume that the difference in notation is because ##\nabla## is reserved for covariant differentiation.

Carroll then says that the gradient of the coordinate functions is the basis, and proves it by acting it on the partials to get a Kronecker delta (Carroll's 2.15, 15th page numbered 45):

[tex]

\mathrm{d}x^\mu\left(\partial_\nu\right) = \frac{\partial x^\nu}{\partial x^\mu}\partial_\nu x^\mu

[/tex]

Obviously this expression is a delta function.

Is this chain of reasoning correct? I'm none too sue that I'm handling operators and index notation correctly.

I've been working through Sean Carroll's lecture notes, and I've got to http://preposterousuniverse.com/grnotes/grnotes-two.pdf [Broken]. I follow the derivation for showing that the tangent space bases are the partial derivatives (Carroll's equation 2.9 on the 13th page of the PDF, numbered 43). However, he only sketches the proof for the cotangent basis, and I'm not sure I've filled in the gaps correctly.

Carroll says that the gradient is the canonical 1-form and quotes its action on the vector ##d/d\lambda## as (Carroll's 2.14, 14th page, numbered 44):

[tex]

\mathrm{d}f\left(\frac{d}{d\lambda}\right) =\frac{df}{d\lambda}

[/tex]

You can expand this using Carroll's 2.9 as

[tex]

\mathrm{d}f\left(\frac{d}{d\lambda}\right) =\frac{df}{d\lambda} =\frac{dx^\mu}{d\lambda}\partial_\mu f

[/tex]

from which I deduce that ##d/d\lambda## should be read as ##dx^\mu/d\lambda## in this context, and that ##\mathrm{d}f## is what I would have written in a non-relativistic context as ##\nabla f##. I presume that the difference in notation is because ##\nabla## is reserved for covariant differentiation.

Carroll then says that the gradient of the coordinate functions is the basis, and proves it by acting it on the partials to get a Kronecker delta (Carroll's 2.15, 15th page numbered 45):

[tex]

\mathrm{d}x^\mu\left(\partial_\nu\right) = \frac{\partial x^\nu}{\partial x^\mu}\partial_\nu x^\mu

[/tex]

Obviously this expression is a delta function.

Is this chain of reasoning correct? I'm none too sue that I'm handling operators and index notation correctly.

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