In Wald's General Relativity, the covariant derivative is an operator acting on $(k,l)$ type tensors. But referring to the second point in the yellow block in this post, the covariant derivative acts on a smooth function $M \rightarrow \mathbb{R}$, which really confuses me. Can anyone explaim this? Thank you.
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5$\begingroup$ Well, smooth functions are (0,0) tensors, aren't they? $\endgroup$– DanielCAug 17, 2018 at 12:37
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$\begingroup$ Why can we see it as (0,0) tensors? By definition, $(k,l)$ tensor is a multi-linear map $\underbrace{V^* \times ... \times V^*}_{k \text{ times}} \times \underbrace{V \times ... \times V}_{l \text{ times}} \rightarrow \mathbb{R}$. So a (0,0) tensor is a map from the empty set to $\mathbb{R}$. (Since an empty product is the empty set) $\endgroup$– Loafy LoaferAug 17, 2018 at 14:25
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1 Answer
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It's part of the axioms for the covariant derivative:
- $\nabla_X\equiv X^\mu \nabla_\mu$ is a derivation.
- $\nabla_X$ acts on a basis vector field as $\nabla_X {\bf e}_a = {\bf e}_b{\omega^b}_{a\mu}X^\mu$ where $\omega$ is the matrix-valued connection form.
- On functions $\nabla_X f= X^\mu \partial_\mu f$.
From these axioms one can find the action of $\nabla_X$ on any tensor.
answered Aug 17, 2018 at 12:47