In General Relativity by Wald, the author makes a claim that I am trying to understand. The crux of my question comes down to understanding why $\eqref{eq1}$ is true, but I have included the context below in case that is helpful or if the proof must use a different method. But in short, can someone explain why

$$t^ag_{bc}v^b\nabla_a(w^c) = g_{bc}v^bt^a \nabla_a (w^c) \tag{1}\label{eq1}$$


The definition of a geodesic $v^b$ (in abstract index notation) is that $$t^a\nabla_av^b = 0$$ Now, assume we have two geodesics $v^b$ and $w^c$ and a covariant derivative such that $$t^a \nabla_a (g_{bc} v^bw^c) = 0$$ The textbook claims it follows from the Leibnitz rule that $$t^av^bw^c\nabla_ag_{bc} = 0 \tag{2}\label{eq2}$$ This is the result I am trying to replicate. It is clear to me that the Leibnitz rule implies $$t^a \nabla_a (g_{bc} v^bw^c) = t^av^bw^c\nabla_a(g_{bc}) + t^a g_{bc} \nabla_a (v^bw^c)$$ Hence the statement is equivalent to the claim $$t^a g_{bc} \nabla_a (v^bw^c) = 0 \tag{3}\label{eq3}$$ If we apply the Leibnitz rule again, it is immediate that this quantity equals $$t^a g_{bc} v^b \nabla_a (w^c) + t^a g_{bc} w^c \nabla_a (v^b)$$ from which \eqref{eq1} would easily allow us to apply the definition of a geodesic and conclude the proof. But why is \eqref{eq1} true? Can someone explain why it seems this tensor product is taken to commute for some reason?

  • $\begingroup$ So you're only asking about (1)? There isn't any need for the rest of the question body then if that's the case, your issue is then about Einstein summation notation, if you write out the summation symbols you can quickly see it doesn't matter the order you put the individual items as they get summed by index $\endgroup$
    – Triatticus
    Feb 8, 2022 at 21:58
  • $\begingroup$ @Triatticus Yes my question is just about (1), but a priori I didn't know if the statement was true on it's own or if more context was needed. In general $t^ag_{bc}v^b \neq g_{bc}v^bt^a$, since tensor products are not commutative. But if I understand you right, it doesn't matter because we are contracting these indices? $\endgroup$ Feb 8, 2022 at 22:06

2 Answers 2


This is just a feature of abstract index notation. The order that a tensor product is written out doesn't actually matter, because the abstract indices keep track of "which slot is which".

Suppose we have two one-forms $m_a$ and $n_b$. The tensor $T_{ab} \equiv m_a n_b$ stands for the tensor $T: V \times V \to \mathbb{R}$ whose actions on the elements $\mathbf{e}$ of some basis of $V$ are given by $$ T(\mathbf{e}_\alpha, \mathbf{e}_\beta) = m(\mathbf{e}_\alpha) n(\mathbf{e}_\beta) $$ while the tensor $T_{ab} \equiv n_b m_a$ is the tensor whose action is given by $$ T(\mathbf{e}_\alpha, \mathbf{e}_\beta) = n(\mathbf{e}_\beta) m(\mathbf{e}_\alpha) $$ which we can see is the same action.

You are correct that the tensor product does not commute in the sense that $\mathbf{m} \otimes \mathbf{n} \neq \mathbf{n} \otimes \mathbf{m}$ in "mathematicians' notation"; but the latter expression in abstract index notation would not be $n_b m_a $ but rather $T'_{ab} \equiv n_a m_b$, and its action on basis elements would be $$ T'(\mathbf{e}_\alpha, \mathbf{e}_\beta) = n(\mathbf{e}_\alpha) m(\mathbf{e}_\beta), $$ which is of course a different tensor.

To apply this to your case: the tensor $T^{a} {}_c \equiv t^a g_{bc} v^b$ stands for the tensor $T : V^* \times V \to \mathbb{R}$ defined by $$ T(\underbrace{\pmb{\omega}^\alpha}_\text{"a" slot}, \underbrace{\mathbf{e}_\gamma}_\text{"c" slot}) \equiv \sum_\beta t(\underbrace{\pmb{\omega}^\alpha}_\text{"a" slot}) g(\mathbf{e}_\beta, \underbrace{\mathbf{e}_\gamma}_\text{"c" slot}) v(\pmb{\omega}^\beta) $$ where $\mathbf{e}$ and $\pmb{\omega}$ belong to some basis on $V$ and its dual basis, respectively. We could then look at the tensor $\tilde{T}^{a} {}_c \equiv g_{bc} v^b t^a $; it would stands for the tensor $\tilde{T} : V^* \times V \to \mathbb{R}$ defined by $$ \tilde{T}(\underbrace{\pmb{\omega}^\alpha}_\text{"a" slot}, \underbrace{\mathbf{e}_\gamma}_\text{"c" slot}) \equiv \sum_\beta g(\mathbf{e}_\beta, \underbrace{\mathbf{e}_\gamma}_\text{"c" slot}) v(\pmb{\omega}^\beta) t(\underbrace{\pmb{\omega}^\alpha}_\text{"a" slot}) $$ Hopefully it is obvious from these expressions that the two tensors $T$ and $\tilde{T}$ are equal.

  • $\begingroup$ I have probably played a little too fast & loose with the formal mathematical background here; the whole point of abstract index notation is to ensure that you don't have to think about this stuff! As always, I welcome corrections or requests for clarification. $\endgroup$ Feb 8, 2022 at 22:23
  • $\begingroup$ Out of curiousity... does the tensor theory have a different name if the scalars are noncommutative. $\endgroup$
    – Emil
    Feb 8, 2022 at 22:35
  • $\begingroup$ @MichaelSeifert Thank you, I think I understand now! I really dislike the abstract index notation for exactly this reason, but hopefully this helps me get used to it $\endgroup$ Feb 8, 2022 at 23:53
  • $\begingroup$ @user2233816: Funny, I couldn't imagine doing calculations in GR without it. Not infrequently in my research, I'll need to construct a tensor of rank 4 or higher via products of lower-rank tensors. I can't imagine having to keep track of all the various contractions and permutations of such a tensor using "mathematicians' notation." (But then, I may be biased, since Wald was my doctoral advisor.) $\endgroup$ Feb 9, 2022 at 3:05

The expression (1) is only a sum of products. And the order of factors doesn't change each product.

For each value of a, b, and c there is a term: $t^ag_{bc}v^b\nabla_a(w^c)$. And all this terms are added. There is no issue in changing the order of the functions.


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