# If you are not given a metric, which one is more fundamental: a vector or a covector? [closed]

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?

Why?

(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)

• What do you mean by "more fundamental"? – Physicist137 Nov 27 '14 at 19:19
• Which one would you be able to write down first – SuperCiocia Nov 27 '14 at 19:21
• Vectors and covectors both exist on arbitrary manifolds. None depends on the existence of a metric. – ACuriousMind Nov 27 '14 at 19:24
• 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. – 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: $$\bar X^i = \frac{\partial\bar x_i}{\partial x_j} X^j,\quad\quad\quad \bar X_i = \frac{\partial x_i}{\partial\bar x_j} X_j$$
• 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. – ACuriousMind Nov 27 '14 at 19:29