Duality of a basis is not duality of individual vectors

Suppose we have a coordinate system $(x^\alpha)$ . This defines a coordinate basis $(\partial_\alpha)$ , where for each $\alpha$ the basis vector has components

$(\partial_\alpha)^\mu = \delta_\alpha^\mu$

in these coordinates. We also have the coordinate dual basis $(dx^\beta)$ where each dual vector or “1-form” has components

$(dx^\beta)_\mu = \delta^\beta_\mu.$

Now while these bases are dual in the sense of bases:

$dx^\beta(\partial_\alpha) = \delta^\beta_\alpha$

(by definition of dual basis), the individual vectors are not dual to the individual 1-forms in the sense of individual vectors. That is, for any given $\alpha$ , we have $\partial_\alpha$ and $dx^\alpha$ are not dual in general.

Instead, recall indices are raised and lowered using the metric components (in a coordinate basis, that is). Possibly the result could be seen by inspection, but for clarity let’s write $\mathbf e := \partial_\alpha$ for some chosen $\alpha$ . This vector has components $e^\nu = \delta^\nu_\alpha$ , hence the corresponding 1-form has components $e_\mu = g_{\mu\nu}e^\nu = g_{\mu\nu}\delta^\nu_\alpha = g_{\mu\alpha}$ . By the meaning of components this says the 1-form is $g_{\mu\alpha}dx^\mu$ . This is not $dx^\alpha$ , in general! In “musical isomorphism” notation, the result is:

$(\partial_\alpha)^\flat = g_{\mu\alpha}dx^\mu$

Similarly,

$(dx^\alpha)^\sharp = g^{\mu\alpha}\partial_\mu.$

To show the result another way, recall the metric defines the dual to our vector $\partial_\alpha$ to be $\mathbf g(\partial_\alpha,\cdot)$ . To examine this 1-form, feed it a vector (specifically, basis vectors $\partial_\mu$ ) and see how it acts on it:

$(\partial_\alpha)^\flat(\partial_\mu) := g(\partial_\alpha,\partial_\mu) = g_{\alpha\mu},$

which says $(\partial_\alpha)^\flat = g_{\alpha\mu}dx^\mu$ , as before.

In closing, another reason we cannot have $(\partial_\alpha)^\flat = dx^\alpha$ in general is that the coordinate basis vector $\partial_\alpha$ is not defined in terms of $x^\alpha$ alone, but also all the other coordinates chosen. More on that next.

(Schutz textbook (2009, §3.3, §3.5) makes a superb background to this discussion, and while the cited sections are for special relativity, in this case you can simply replace the Minkowski metric $\boldsymbol\eta$ with an arbitrary curved metric $\mathbf g$ .)

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