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When I am asked to evaluate, $\mathbf{U^{\alpha}_{~,~~\beta}}$ for all $\alpha$ and $\beta$, what does it mean? I have not able to understand this notation.

In case of $\mathbf{g(~~,~\bar{A})}$ I understand that the blank in the metric tensor means it acts like a one-form which takes in a vector and outputs a real number. For this case it makes sense to me because it is a

$\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ tensor. How do we understand the same for a mixed tensor like the one above ?

Reference: Chapter 3, Problem 30: A First Course in General Relativity, Second Edition, B. Schutz, pg 82.

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Let us assume $U^\alpha$ is a vector, i.e. described by its contravariant components. The subscript $_{, \beta}$ means the partial derivative, that is $_{, \beta} = \partial_\beta$. Therefore $U^\alpha _{, \beta} = \partial_\beta U^\alpha$.

However in general that is not a tensor, as the partial derivative is not a tensor in arbitrary coordinates. To have a tensorial expression, you should replace the partial derivative with the covariant derivative, that is $\partial_\mu$ replaced by $\nabla_\mu = _{; \mu}$.

So the expression $U^\alpha _{; \beta} = \nabla_\beta U^\alpha$ is a tensor, contravariant in $\alpha$ and covariant in $\beta$.

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  • $\begingroup$ Thank you ! Thant makes sense. In Minkowski space-time, the partial derivative and the covariant derivative would be the same which was my case :-) $\endgroup$ – Astronomer Sep 24 '18 at 22:50

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