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I am working my way through Carroll's text on GR and am having trouble understanding what it means when an index is inside/outside parentheses. For example, in his discussion of geodesic deviation, he defines the relative velocity of geodesics as:

$$V^\mu=(\nabla_TS)^\mu$$

Can someone help me understand this notation?

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    $\begingroup$ Your question would be clearer if you explained what $S$ and $T$ are. Please don’t assume that readers have a copy of the book you’re using. $\endgroup$
    – Ghoster
    Commented Dec 25, 2022 at 20:46
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    $\begingroup$ It's short for $\nabla^\mu=T^\nu\nabla_\nu S^\mu$. $\endgroup$
    – J.G.
    Commented Dec 25, 2022 at 20:54

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The idea is that $\nabla_T S$ is a tensor field (in this case, a vector field), and $\big(\nabla_T S\big)^\mu$ is its $\mu^{th}$ component. The parentheses are meant to distinguish this from $\nabla_TS^\mu \overset{?}=\nabla_T \big(S^\mu\big)$, which in the present context does not actually make sense, but which could nevertheless be confusing to students learning about tensors for the first time.

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There is (at least) two equivalent ways of performing such calculations on tensors.

One would have the covariant derivative acting on the whole tensor field, for example on a (1,1)-valence one:

$$T = T^{\mu}_{\;\;\;\nu}\partial_{\mu} \otimes dx^{\nu} $$

$$\nabla_{\alpha}T = \nabla_{\alpha}(T^{\mu}_{\;\;\;\nu}\partial_{\mu} \otimes dx^{\nu}) = \nabla_{\alpha}(T^{\mu}_{\;\;\;\nu}) \partial_{\mu} \otimes dx^{\nu} + T^{\mu}_{\;\;\;\nu} \nabla_{\alpha} \partial_{\mu} \otimes dx^{\nu} + T^{\mu}_{\;\;\;\nu} \partial_{\mu} \otimes \nabla_{\alpha} dx^{\nu},$$

where $\nabla_{\alpha}(T^{\mu}_{\;\;\;\nu}) = \partial_{\alpha}T^{\mu}_{\;\;\;\nu} $ because tensor field components are functions, and the covariant derivatives of coordinate vector fields and 1-forms are:

$$ \nabla_{\alpha}\partial_{\mu} = \Gamma^{\lambda}_{\alpha\mu}\partial_{\lambda}$$

$$ \nabla_{\alpha} dx^{\nu} = - \Gamma^{\nu}_{\alpha\lambda} dx^{\lambda}$$

If you change the dummy (summation) indices appropriately, you will get:

$$ \nabla_{\alpha }T = \big{[} \partial_{\alpha}T^{\mu}_{\;\;\;\nu} + T^{\lambda}_{\;\;\;\nu}\Gamma^{\mu}_{\alpha\lambda} - T^{\mu}_{\;\;\;\lambda} \Gamma^{\lambda}_{\alpha\nu}\big{]} \partial_{\mu}\otimes dx^{\nu} \equiv (\nabla_{\alpha}T)^{\mu}_{\;\;\;\nu}\partial_{\mu}\otimes dx^{\nu}.$$

In the above, the final expression for the tensor components of the derived field is present: $(\nabla_{\alpha}T)^{\mu}_{\;\;\;\nu}$. In the community one would typically write it as $\nabla_{\alpha}T^{\mu}_{\;\;\;\nu}$ as it is clear from the context that we do mean first the derivative, then reading off its components. You can also use the semicolon notation $(\nabla_{\alpha}T)^{\mu}_{\;\;\;\nu}=T^{\mu}_{\;\;\;\nu;\alpha}$, where it is again understood that first the covariant derivative along $\partial_{\alpha}$ is taken, and then the tensor components are read off.

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