Reordering Terms in Abstract Index Notation — General Relativity

Question

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

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

Context:

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?

Answer 1

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.

Answer 2

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.