When we map a vector to its corresponding covector with the metric:


is there a second (implied) metric being used to convert the basis vectors too? Written explicitly:

$$g_{\mu\nu}(x^\mu \hat e_\mu)=g_{\mu\nu}(x^\mu)g^{\mu\nu} (\hat e_\mu)=x_\nu \hat e^\nu$$

Or does the metric also map the basis to the corresponding dual basis at the same time?


Physicists play a bit fast and loose when we say $x^\mu$ corresponds to a vector; $x^\mu$ should really be interpreted as the coordinates of some vector $x = x^\mu \hat{e}_\mu$.

In the same way, $g_{\alpha\beta}x^\beta \equiv x_\alpha$ is a matrix multiplication corresponding to the tensor contraction $$(g_{ab} \hat{e}^a \otimes \hat{e}^b) \cdot (x^c \hat{e}_c) = g_{ab}x^c \hat{e}^a \delta^b_c = (g_{ab}x^b )\hat{e}^a$$ where we are working in an orthogonal basis. The "fast and loose" notation of dropping the basis vectors is highly computationally convenient, because the uncontracted indices automatically tell us how to put the basis vectors back in.

Because of this, the question asked is not exactly well-posed, in that if one wants to reintroduce the basis vectors for $x^\mu$ into the problem, one also needs to do the same for $g_{\mu\nu}$. In other words, the equation $g_{\mu\nu} \hat{e}^\nu \stackrel{?}{=} \hat{e}_\mu$ doesn't make sense. The object on the LHS is a linear combination of covectors, while the RHS is a vector.

However, via contraction we have $(g_{ab} \hat{e}^a \otimes \hat{e}^b) \cdot (\hat{e}_c) = g_{ac}\hat{e}^a$, which maps the covector $\hat{e}_c \rightarrow g_{ac}\hat{e}^a$, so one can "schematically" think that the metric "lowers" the indices of the basis vectors. As explained, one needs to be careful with this line of thought. This subtlety is similar to $|\psi \rangle^* \neq \langle \psi |$, but complex conjugation does indeed induce a map $|\psi \rangle \stackrel{*}{\rightarrow} \langle \psi |$.

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  • $\begingroup$ In your first equation, is there a difference between $\hat e^a$ acting on $\hat e_c$ or $\hat e^b$ acting on $\hat e_c$? $\endgroup$ – Charlie Feb 26 at 19:39
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    $\begingroup$ Yes, there is a difference in general; it determines which tensor index you are contracting. Though in this case, $g_ab$ is symmetric so it turns out to not matter. $\endgroup$ – Aaron Feb 27 at 15:53
  • $\begingroup$ That makes sense, thank you for taking the time to help :) $\endgroup$ – Charlie Feb 27 at 15:59

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