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I've been wondering if we can attribute any physical meaning to the inverse metric. I mean when we talk about the metric itself, there are lots of insights we can have towards its role in spacetime, yet I cannot see any physical meaning for the inverse metric. For now, I just see it as tensor with the special property of giving the identity when joined with the metric. Rigorously speaking, I would say it is not even an "inverse" actually, as it doesn't map like one. But still, is there any physical way of interpreting this tensor?

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Another way of looking at user40330's answer is to think of the inverse metric as the map from the space of one-forms (or differentials, if you prefer) and mapping them to the space of vectors (or directional derivatives, if you prefer that language), and then thinking of the metric as the inverse of this map.

Namely

$$g^{-1}({ d}v)=g^{ab}v_{b} = v^{a} = {\vec v}$$

and

$$g({\vec v}) = g_{ab}v^{b} = v_{a} = dv$$

This map is obviously invertible, thanks to the properties of matrices, and you obviously have $g^{-1}(dv,dv) = g({\vec v},{\vec v})$.

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The tensor algebra is symmetric between one-forms and vectors. One could start with defining any of them first and then obtain the rest of the things.

The inverse metric tensor is a linear map that takes two one forms on a manifold and maps into $\mathbb{R}.$

$g^{\mu \nu}: A_\mu,B_\nu \rightarrow \mathbb{R}$

It of course tranforms like a vector with respect to both the indices.

So to answer your question, the inverse metric tensor is as physical as the metric tensor. Just as metric tensor provides a way to measure length of vector fields, inverse metric provides a way to measure length of one-form fields.

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