In R. Wald's book, this condition is written as:

$$\nabla_a \nabla_b f=\nabla_b \nabla_a f.$$

But aren't $\nabla_a$ and $\nabla_b$, like, the exact same derivative operator with a re-naming of index? Then isn't this condition trivially true?

By derivative operator, I mean a derivative operator for tensor fields.

EDIT- I think it's saying that the tensor is symmetric. If $\nabla_a \nabla _b f=T(a,b)$, then $T(a,b)=T(b,a)$. Is this it? $a$ and $b$ are input vectors to the tensor $T$


1 Answer 1


Your edit is correct: it means that the tensor $\nabla_a \nabla_b f$ is symmetric when acting on vectors. This need not be the case, as Problem 1 of Chapter 3 shows. As an example, let me open up the expression $\nabla_a \nabla_b f$ in terms of Christoffel symbols. These symbols are given on Wald's Eq. (3.1.15), the key point being that they allow us to write a generic derivative operator in terms of ordinary derivatives plus an extra term. For a dual vector, it is given by (cf. Eq. (3.1.7)) $$\nabla_a \omega_b = \partial_a \omega_b - \Gamma^c{}_{ab}\omega_c. \tag{1}$$

Notice then that $$\begin{align} \nabla_a \nabla_b f &= \nabla_a (\partial_b f), \\ &= \partial_a\partial_b f - \Gamma^c{}_{ab}\nabla_c f , \end{align}$$ while the same calculation yields $$\nabla_b \nabla_a f = \partial_b\partial_a f - \Gamma^c{}_{ba}\nabla_c f.$$

Hence, the torsionless condition translates on $\Gamma^c{}_{ab} = \Gamma^c{}_{ba}$, which is equivalent to what your edit stated, since the partial derivatives do commute.

In summary, you should notice that you are not applying the very same object twice, although notation seems to point otherwise. The first derivative operator acts like $\nabla_a f = \partial_a f$, while the second one acts according to Eq. (1) I've written above. Furthermore, yes, you can understand the torsionless condition in terms of the symmetry properties of the tensor $\nabla_a \nabla_b f$. I also suggest working out Problem 1 of Chapter 3, since it might shed some light on the concept of torsion.


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