# Quick question on notation.

• cristo
In summary, the conversation discusses the meaning of \nabla_{[a}F_{bc]}, with one person providing an explanation that involves permuting indices with even and odd permutations. They also mention that this equation is related to the electromagnetic field tensor and the Bianchi identity. Another person adds that this equation is equivalent to the statement d\mathbf{F}=0, which is guaranteed for any p-form field \mathbf{A}.

#### cristo

Staff Emeritus
Does anyone know what this means: $$\nabla_{[a}F_{bc]}$$? I know that $$F_{(ab;c)}=\frac{1}{3}(F_{ab;c}+F_{bc;a}+F_{ca;b})$$, and presume that the first expression can be written thus $$F_{[bc;a]}$$, but am not sure what it means!

Can anyone help?

cristo said:
Does anyone know what this means: $$\nabla_{[a}F_{bc]}$$? I know that $$F_{(ab;c)}=\frac{1}{3}(F_{ab;c}+F_{bc;a}+F_{ca;b})$$, and presume that the first expression can be written thus $$F_{[bc;a]}$$, but am not sure what it means!

Can anyone help?

Any time you see square brackets around indices it means that you permute the indices, with even permutations receiving a plus sign and odd permutations receiving a minus sign. Thus,

$$\nabla_{[a}F_{bc]} = \frac{1}{3!}\sum_{\pi\in S(3)}\textrm{sign}(\pi)\nabla_{\pi(a)}F_{\pi(b)\pi(c)}$$

where $S(3)$ is the symmetric group of order three, $\pi$ is a permutation, and sign$(\pi)$ equals one for an even permutation of the elements and minus one for an odd permutation of the elements. In your case you can expand out the above definition to obtain

$$\nabla_{[a}F_{bc]} = \frac{1}{3!}(\nabla_aF_{bc} + \nabla_bF_{ca} + \nabla_cF_{ab} - \nabla_aF_{cb} - \nabla_bF_{ac} - \nabla_cF_{ba})$$

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shoehorn said:
Any time you see square brackets around indices it means that you permute the indices, with even permutations receiving a plus sign and odd permutations receiving a minus sign. Thus,

$$\nabla_{[a}F_{bc]} = \nabla_aF_{bc} + \nabla_bF_{ca} + \nabla_cF_{ab} - \nabla_aF_{cb} - \nabla_bF_{ac} - \nabla_cF_{ba}$$

Ahh ok, that makes sense. Thanks for the quick reply! In this case F is the electromagnetic field tensor, and so is antisymmetric. Would I be right in assuming that in this case the equation becomes $$\nabla_{[a}F_{bc]} = \frac{1}{3}\left(\nabla_aF_{bc} + \nabla_bF_{ca} + \nabla_cF_{ab}\right)$$

cristo said:
Ahh ok, that makes sense. Thanks for the quick reply! In this case F is the electromagnetic field tensor, and so is antisymmetric. Would I be right in assuming that in this case the equation becomes $$\nabla_{[a}F_{bc]} = \frac{1}{3}\left(\nabla_aF_{bc} + \nabla_bF_{ca} + \nabla_cF_{ab}\right)$$

Yes. If $F_{ab}$ are taken as the components of the Maxwell tensor then $\nabla_{[a}F_{bc]}=0$ is essentially the Bianchi identity for the electromagnetic field.

Another way to think about it is to notice that $\nabla_{[a}F_{bc]}=0$ is precisely the same statement as $d\mathbf{F}=0$ where $\mathbf{F}=d\mathbf{A}$ is the Maxwell two-form. The identity $d\mathbf{F}=0$ is guaranteed since for any $p$-form field $\mathbf{A}$ one has $d\cdot(d\mathbf{A})=0$.

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That's a good way to think about it. Thanks a lot for your help, shoehorn!