Suppose $g_{ij}$ is a tensor of type $(0,2)$, then what type of object is $\partial_{i}g_{jk}$? Is it even a tensor, and if so, of what type? Is the $\partial_{i}$ still a differential with respect to the $i$-th variable in this context?
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1$\begingroup$ Only in flat spacetime. It is a 3-rd order tensor, called the gradient of g. $\endgroup$– DanielCCommented Feb 26, 2021 at 9:54
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2$\begingroup$ In general, the covariant derivative of a tensor is a tensor. However, in the special case of the metric tensor, it vanishes identically because of the properties of the Levi-Civita connection (metric compatibility and null torsion). $\endgroup$– freshCommented Feb 26, 2021 at 9:56
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$\begingroup$ @DanielC by $3$-rd order tensor, you mean tensor of type $(0,3)$? $\endgroup$– JBuckCommented Feb 26, 2021 at 9:59
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$\begingroup$ Yes. Exactly. .. $\endgroup$– DanielCCommented Feb 26, 2021 at 10:00
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1 Answer
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As the comments mention, $\partial_{i} g_{jk}$ isn't a tensor in general. You can confirm this by calculating how the object transforms.
If you take the covariant derivative of a general type $(0,2)$ tensor $t_{ij}$ then you'll get a type $(0,3)$ tensor $\nabla_{k} t_{ij}$. Although if by $g$ you mean the metric, then this obviously vanishes for the Levi-Civita connection. For a different connection $\nabla_{i} g_{jk}$ is a non-zero type $(0,3)$ tensor.
If you're only concerned with special relativity, then the covariant derivative reduces to the partial derivative and the metric to the Minkowski metric $g_{\mu \nu} = \eta_{\mu \nu}$ (whose partial derivative also vanishes).
