# If $g_{ij}$ is a tensor of type $(0,2)$, what is kind of tensor is $\partial_{i}g_{jk}$?

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?

• Only in flat spacetime. It is a 3-rd order tensor, called the gradient of g. – DanielC Feb 26 at 9:54
• 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). – fresh Feb 26 at 9:56
• @DanielC by $3$-rd order tensor, you mean tensor of type $(0,3)$? – JBuck Feb 26 at 9:59
• Yes. Exactly. .. – DanielC Feb 26 at 10:00

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).