If we do not have the metric $g_{\mu\nu}$ for a given spacetime, are vectors $x^\mu$ more fundamental than covectors $x_\mu$ or vice versa?


(if the metric were given we could just raise/lower the indices and convert a vector into a covector and vice versa, hence my specification)

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    $\begingroup$ What do you mean by "more fundamental"? $\endgroup$ – Physicist137 Nov 27 '14 at 19:19
  • $\begingroup$ Which one would you be able to write down first $\endgroup$ – SuperCiocia Nov 27 '14 at 19:21
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    $\begingroup$ Vectors and covectors both exist on arbitrary manifolds. None depends on the existence of a metric. $\endgroup$ – ACuriousMind Nov 27 '14 at 19:24
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    $\begingroup$ Both are duals of one another: think of the Riesz representation theorem. On finite dimensional differentiable manifolds, the classes of vector and covector fields can each be (and equally well be) thought of as the class of linear functionals of the other. So they're very alike and equally "fundamental" in this way. Of course, only one (vector field) is a section of the tangent space. $\endgroup$ – WetSavannaAnimal Nov 28 '14 at 0:28

Given $\bar{x}^i(x^j)$ transformation law of coordinates. The vector components $X^i$ and the covector $X_i$ are defined to transform like: $$ \bar X^i = \frac{\partial\bar x_i}{\partial x_j} X^j,\quad\quad\quad \bar X_i = \frac{\partial x_i}{\partial\bar x_j} X_j $$

Both are simultaneously defined independent of the metric. And their definitions is independent of each other. Therefore, both are "equally fundamental".

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    $\begingroup$ Though physicists will oddly insist that this is a definition, it is not. Vectors and covectors are members of the (co)tangent spaces, which are at every point the space spanned by the $\partial_{x_i}$ (tangent space) and its dual. The transformation law follows from the definition, but it is not a definition by itself. $\endgroup$ – ACuriousMind Nov 27 '14 at 19:29
  • $\begingroup$ To tell the truth, the book I've read is not that formal. It defines vector and co-vector this way. Feel free to edit (or to delete) the answer (or to tell me to delete). Or to post a new answer. I would appreciate if you give me references to the mathematical formalism of this. $\endgroup$ – Physicist137 Nov 27 '14 at 19:36
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    $\begingroup$ Any account of differential geometry will do. I also show the definition of (co)tangent spaces in my answer here. $\endgroup$ – ACuriousMind Nov 27 '14 at 19:40
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    $\begingroup$ @ACuriousMind you may be interested in the final section of chapter 3 of Lee's book Introduction to Smooth Manifolds, where it is noted that the definitions of the vectors and covectors based on their transformation laws are, in fact, the oldest way of conceptualizing (co)tangent spaces, and is equivalent to the other, more modern definitions. See also my question on HSM.SE here. $\endgroup$ – Danu Nov 27 '14 at 23:01

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