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Classical Theory of Particles and Fields, by Boris Kosyakov, has the following in appendix A:

This completely mystifies me, since the gradient of a scalar is what I would think of as the prototypical example of a covector, not a vector. The partial derivative operator [itex]\partial_i=\partial/\partial x^i[/itex] has a lower index; it's a covector.

My best guess would be that this is occurring because of the notational collision between the interpretation of upper/lower-index quantities as vectors/covectors and the use of lower/upper-index quantities as basis vectors for the space of vectors/covectors. That is, we often use [itex]\partial_i[/itex] as a basis for the space of vectors, and we then have expressions like [itex]a^i\partial_i[/itex], which require context in order to see that they are vectors expressed using the Einstein summation convention, rather than scalars. Similarly, [itex]dx^i[/itex] can be taken as the covector to [itex]\partial_i[/itex]. However, none of this resolves the craziness (AFAICT) of referring to a gradient as a vector rather than a covector.

In this total derivative, I would give [itex]dx^i[/itex] the context-dependent interpretation of being an infinitesimally small [itex]\Delta x^i[/itex], which is the prototypical example of a vector (not covector). For a total derivative involving a finite change, we would have [itex]\Delta F\approx \Delta x^i \partial F/\partial x^i[/itex]. I don't see what Kosyakov could do here, since [itex]\Delta x^i[/itex] is clearly a vector, not a covector.

Is Kosyakov's point of view unusual? Is there something I'm missing?

(I don't own a copy of the book. This came to my attention through this physics.SE question, and it turned out that this part of the book is accessible through Amazon's peephole.)Elie Cartan proposed to use differential coordinates [itex]dx^i[/itex] as a convenient basis of 1-forms. The differentials [itex]dx^i[/itex] transform like covectors [...] Furthermore, when used in the directional derivative [itex]dx^i \partial F/\partial x^i[/itex], [itex]dx^i[/itex] may be viewed as a linear functional which takes real values on vectors [itex]\partial F/\partial x^i[/itex]. The line elements [itex]dx^i[/itex] are called [...] 1-forms.

This completely mystifies me, since the gradient of a scalar is what I would think of as the prototypical example of a covector, not a vector. The partial derivative operator [itex]\partial_i=\partial/\partial x^i[/itex] has a lower index; it's a covector.

My best guess would be that this is occurring because of the notational collision between the interpretation of upper/lower-index quantities as vectors/covectors and the use of lower/upper-index quantities as basis vectors for the space of vectors/covectors. That is, we often use [itex]\partial_i[/itex] as a basis for the space of vectors, and we then have expressions like [itex]a^i\partial_i[/itex], which require context in order to see that they are vectors expressed using the Einstein summation convention, rather than scalars. Similarly, [itex]dx^i[/itex] can be taken as the covector to [itex]\partial_i[/itex]. However, none of this resolves the craziness (AFAICT) of referring to a gradient as a vector rather than a covector.

In this total derivative, I would give [itex]dx^i[/itex] the context-dependent interpretation of being an infinitesimally small [itex]\Delta x^i[/itex], which is the prototypical example of a vector (not covector). For a total derivative involving a finite change, we would have [itex]\Delta F\approx \Delta x^i \partial F/\partial x^i[/itex]. I don't see what Kosyakov could do here, since [itex]\Delta x^i[/itex] is clearly a vector, not a covector.

Is Kosyakov's point of view unusual? Is there something I'm missing?

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