Riemann curvature tensor notation in Wald

Question

This question is entirely on tensorial notation in Wald's General Relativity. When specifying the properties of the Riemann tensor on pg39, he states:

$R_{[abc]}^{\quad \ \ \ d} = 0$

and

For the derivative operator $\nabla_a$ naturally associated with the metric, $\nabla_a g_{bc}=0$, we have $R_{abcd} = -R_{abdc}$.

and

The Bianchi identity holds: $\nabla_{[a}R_{bc]d}^{\quad \ \ e} = 0$

Questions:

1. What do the square brackets around "abc" mean?

2. Why does $R_{abc}^{\quad d}$ become $R_{abcd}$? What is the relation between the two?

3. What does having $R$ in the square brackets mean?

Thank you.

Answer 1

1. The square brackets mean antisymmetrization. That is: $$ X_{[a_1a_2\dots a_n]} = \frac{1}{n!}\sum_{P\in S(n)} \text{Sign}(P) X_{a_{P(1)}a_{P(2)}\dots a_{P(n}} $$ where $S(n)$ is the set of permutations of $n$ elements, and $\text{Sign}(P)$ is the sign of the permutation $P$, that is, $\text{Sign}(P)=-1$ if you need an odd number of element exchanges, and $\text{Sign}(P)=+1$ if you need an even number of element exchanges.

   In particular,

   • $R_{[abc]}{}^d = \frac{1}{6}\left(R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d-R_{bac}{}^d-R_{acb}{}^d-R_{cba}{}^d\right)$
   • $\nabla_{[a}R_{bc]d}{}^{e} = \frac{1}{6}\left(\nabla_{a}R_{bcd}{}^{e}+\nabla_{b}R_{cad}{}^{e}+\nabla_{c}R_{abd}{}^{e}-\nabla_{b}R_{ac d}{}^{e}-\nabla_{a}R_{cbd}{}^{e}-\nabla_{c}R_{bad}{}^{e}\right)$

2. You can "move" indices up and down using the metric tensor. That is, $$R_{abcd} = g_{de}R_{abc}{}^e,\quad R_{abc}{}^d = g^{de}R_{abce}.$$

3. The square brackets just affects the indices; the $R$ is inside because the antisymmetrization affects indices both from $\nabla$ and from $R$.