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Let $T^{b_1 b_2 \dots b_k}_{c_1 c_2 \dots c_l}$ be a $(k,l)$-tensor and $\nabla_a: \tau(k,l) \mapsto \tau(k, l+1)$ be a derivative operator ($\tau$ denotes the space of tensors of specified rank). In Wald's book on General Relativity, the general expression for the difference between two derivative operators $\nabla$ and $\tilde{\nabla}$ is expressed as:

$$ \nabla_a T^{b_1 b_2 \dots b_k}_{c_1 c_2 \dots c_l} = \tilde{\nabla_a} T^{b_1 b_2 \dots b_k}_{c_1 c_2 \dots c_l} + \sum_i C^{b_i}_{ad} T^{b_1 \dots d \dots b_k}_{c_1 \dots c_l} - \sum_j C^{d}_{ac_j} T^{b_1 \dots b_k}_{c_1 \dots d \dots c_l} $$

The derivation of this formula was convincing at first, but the more I looked at it, the less sense it made to me. Basically, I'm wondering:

  1. In the summations, the tensors $T$ and $C$ are contracted with respect to $d$, but I don't understand where this index is located in the $T$-tensor. It says "$\dots$", but I think this makes it unclear from the circumstance. Or is it somehow assumed that $d$ is shifted into every possible slot then summed over? In that case, wouldn't it require some sort of notation describing just that?
  2. It says that $b_i$ is summed over for different $i$:s. However, since each $b_i$ is repeated (let's ignore the impact of $d$ at this point), should it be contracted? To me it doesn't make sense to contract based on $b_i$, since both indices are contravariant (and the same problem arises with the $c_j$-indices in the second sum). How does this work?

I'm kind of stuck at this formula and can't find it online. Any help is gladly appreciated.

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  • $\begingroup$ instead of $i$ or $j$ under summatories, it should be $$\sum_d.$$ $\endgroup$
    – janmarqz
    Jun 30, 2022 at 13:47
  • $\begingroup$ The contraction with respect to $d$ is assumed by repeated indices (or summation, if you think of it in terms of components). What is meant by the sum is that $b_i$ ranges over $b_1$, $b_2$ and so on, and for each such case, the term is further contracted with respect to $d$. So because of this, I think the index is right, even if it isn't explicitly obvious that $d$ shifts place for each such term (see answer below). $\endgroup$
    – Max
    Jun 30, 2022 at 20:14

1 Answer 1

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I finally figured it out: For each term in the sum, the $d$ is in the place where $b_i$ would normally be. In other words:

$$ \sum_i C^{b_i}_{ad} T^{b_1 \dots d \dots b_k}_{c_1 \dots c_l} = C^{b_1}_{ad}T^{d b_2 \dots b_k}_{c_1 \dots c_l} + C^{b_2}_{ad}T^{b_1 d \dots b_k}_{c_1 \dots c_l} + \dots $$

This solves both problem 1 and 2 simultaneously in the following manner:

  1. Index $d$ is placed wherever $b_i$ would be, for each term.
  2. Because of this, there will never be any repeated $b_i$ or $c_i$ indices.
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