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In the textbook "Gravitation" by Misner,Thorne and Wheeler (page 717), when the isotropy of the universe is considered, it is stated that in the comoving frame the metric tensor can be written as:

$g_{\alpha \beta} =\frac{\partial}{\partial x^\alpha}\frac{\partial}{\partial x^\beta}$

How does one come at this conclusion?

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  • $\begingroup$ I would have sais that by definition, changing frame from a free falling to another would have raised a metric tensor $g_{\alpha \beta}={\partial x^\mu \over \partial x^\alpha}{\partial x^\nu \over \partial x^\beta}\eta_{\mu \nu}$, wouldn't it? $\endgroup$
    – Matt
    Jun 28 '19 at 12:29
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As Matt says in a comment, this is just a notation stating the definition of the metric. MTW notate it with a dot, to make it clear that the right-hand side is the inner product of two vectors. $\partial/\partial x^\alpha$ is a notation for the vector corresponding to a unit change in the coordinate $x^\alpha$. In this notation, the partial derivative is just being used as a basis vector for the space of vectors. By taking the dot product of unit two vectors, you get a component of the metric. This isn't anything special about cosmology. It's generic.

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  • $\begingroup$ I thought $\frac{\partial}{\partial x^\alpha}$ is tangent vector to the manifold, whereas a covector is $dx^\alpha$. As basis vectors of the space of covector I would expect $dx^{1}, dx^{3}, \ldots dx^{n}$ with $n$ the dimension of the manifold with a coordinate chart $(x^1, x^2, \ldots, x^n)$. $\endgroup$
    – Frederic Thomas
    Jun 28 '19 at 13:23
  • $\begingroup$ @FredericThomas: You're right, thanks for the correction. $\endgroup$
    – user4552
    Jun 28 '19 at 15:17

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