# A combinatory thing! [closed]

I've always been terrible at combinatorial problems, and thus I'd thank you a lot if you can help me with this.

Background

I have a tensorial current $J^{ABC}$, which is antisymmetric in all their indices. Therefore, it is written as

$$J^{ABC}=\frac{1}{6}\left(T^{ABC}+\text{5 permutations} \right),$$

where the $T$ is not necessary symmetric not antisymmetric.

Problem

Assume that $A$ takes values over $(a,\xi)$, with $a$ a set of numbers (think in extra dimensions).

How does $J^{ABC}$ decompose into $j$'s with lower-case indices?

NOTE 1: The index $\xi$ anticommutes with any value of $a$

NOTE 2: The decompositions should be expressed in term of antisymmetric currents.

## My try

$$J^{ABC}= J^{abc}+J^{ab\xi}+J^{a\xi b}+J^{\xi ab}.$$

Now, $J^{abc}=j^{abc}$ because they have the same combinatorial normalization $\frac{1}{6}$.

However, $$J^{ab\xi} = \frac{1}{3}\left( 3j^{ab\xi} \right)=j^{ab\xi},$$ where $j^{ab\xi}$ is antisymmetric only on the $a,b$ indices.

Is it Ok? or Am I wrong? If so, where is my mistake?

Finally, How does $J^{ABC}J_{ABC}$ decompose in lower-case indices? I have tried but I am more doubtfully than in the previous case.

Thank you!

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that's not a proper title. –  Peter Sheldrick Nov 22 '12 at 15:50
I suggest you add some tags here. This might help you get the attention of someone who, unlike me, knows what a "tensorial current" is. :-) –  Hew Wolff Nov 22 '12 at 15:50
You leave out too much information to make this question answerable. Is $A$ a set or a number or rather a list of numbers? What does "takes vales over ..." mean? Specifically, what does it mean for $B$? –  Phira Nov 28 '12 at 13:07