Duality of a basis is not duality of individual vectors

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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 (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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