6
$\begingroup$

I know about the () symmetrisation and [] anti-symmetrisation brackets on tensor indices so long as they appear on their own, such as :

$$V_{[\alpha \beta ]}=\frac{1}{2}\left ( V_{\alpha \beta }-V_{\beta \alpha } \right )$$

However, I am somewhat confused as to what happens when these brackets span indices on more than one tensor at a time, and especially if there is more than one bracket within such a term. To see what I mean, consider the following expression for the Weyl tensor, as given in chapter 13 of MTW :

$$C{^{\alpha \beta }}_{\gamma \delta }=R{^{\alpha \beta }}_{\gamma \delta }-2\delta {^{[\alpha}}_{[\gamma }R{^{\beta ]}}_{\delta ]}+\frac{1}{3}\delta {^{[\alpha}}_{[\gamma }\delta{^{\beta ]}}_{\delta ]}R$$

How do you expand bracketed expressions such as the ones in the second and third terms above in index notation without brackets ? I find this notation really confusing.

$\endgroup$
10
$\begingroup$

The (anti)symmetrization simply acts on all the enclosed indices (at the same "height" which are really enclosed between the brackets), regardless of their belonging to the same tensor or different tensors. For example, $$ \delta^{[\alpha}{}_{[\gamma} R^{\beta]}{}_{\delta]} = \frac 12 \left(\delta^{[\alpha}{}_{\gamma} R^{\beta]}{}_{\delta} - \delta^{[\alpha}{}_{\delta} R^{\beta]}{}_{\gamma} \right) = \dots $$ and each term on the right hand side may still be expanded because it is an antisymmetrization of the $\alpha\beta$ indices: $$\dots = \frac 14 \left( \delta^{\alpha}{}_{\gamma} R^{\beta}{}_{\delta} - \delta^{\alpha}{}_{\delta} R^{\beta}{}_{\gamma} - \delta^{\beta}{}_{\gamma} R^{\alpha}{}_{\delta} + \delta^{\beta}{}_{\delta} R^{\alpha}{}_{\gamma} \right) $$ Similarly for all your other terms (the $\delta-\delta$ term is computed in the other answer that I endorse), or any other tensors. The number of terms in these expansions is increasing "exponentially" if there are many (anti)symmetrizations so it becomes really tiresome to write all of the terms – this tiresomeness is the reason why use the (anti)symmetrization symbols, after all.

As long as the several groups of indices that are (anti)symmetrized are disjoint (and groups of upper indices never have any intersection with a group of lower indices), it doesn't matter in which order you expand the (anti)symmetrization. You are guaranteed to get the same result.

$\endgroup$
3
$\begingroup$

You are not supposed to consider the $\gamma$ to be part of the upper bracket in your example. You are just anti-symmetrizing over the pair $\alpha$ and $\beta$ and the pair $\gamma$ and $\delta$. The rule would be $V{^{[\alpha}}_{[\gamma }{^{\beta ]}}_{\delta ]} = \frac{1}{2}(V{^{\alpha}}_{[\gamma }{^{\beta }}_{\delta ]} - V{^{\beta}}_{[\gamma }{^{\alpha }}_{\delta ]}) = \frac{1}{2}(V{^{[\alpha}}_{\gamma }{^{\beta ]}}_{\delta } - V{^{[\alpha}}_{\delta }{^{\beta ]}}_{\gamma }) = \frac{1}{4}(V{^{\alpha}}_{\gamma }{^{\beta }}_{\delta } -V{^{\alpha}}_{\delta }{^{\beta }}_{\gamma } - V{^{\beta}}_{\gamma }{^{\alpha }}_{\delta } + V{^{\beta}}_{\delta }{^{\alpha }}_{\gamma })$.

Notice that the order that you do the two antisymmetrizations doesn't matter.

As an example, $\delta{^{[\alpha}}_{[\gamma }\delta{^{\beta ]}}_{\delta ]} = \frac{1}{4}(\delta{^{\alpha}}_{\gamma }\delta{^{\beta }}_{\delta } -\delta{^{\alpha}}_{\delta }\delta{^{\beta }}_{\gamma } - \delta{^{\beta}}_{\gamma }\delta{^{\alpha }}_{\delta } + \delta{^{\beta}}_{\delta }\delta{^{\alpha }}_{\gamma })$.

The one with the $\delta$ and the $R$ gives the same answer except the second $\delta$ becomes an $R$.

$\endgroup$
  • $\begingroup$ LOL, keep it, so that the OP sees some consensus - even though I reproduced his or her factors of $1/2$ and you didn't. ;-) $\endgroup$ – Luboš Motl Jan 24 '14 at 18:50
  • $\begingroup$ oops just edited. I thought I remembered it without the 1/2. Its been a while since I have done this stuff. $\endgroup$ – Brian Moths Jan 24 '14 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.