# Expression for general derivative operator

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.

• instead of $i$ or $j$ under summatories, it should be $$\sum_d.$$ Jun 30, 2022 at 13:47
• 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).
– Max
Jun 30, 2022 at 20:14

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