Does double antisymmetrisation always introduce a factor 2?

In theoretical electrodynamics, I came across terms with double antisymmetrisation, one with brackets, the other with a Levi-Civita-Tensor ($\epsilon$).

The particular example was $\epsilon^{\mu\nu\alpha\beta} \left( \partial_{[\mu} A_{\nu]} \right) \left( \partial_{[\alpha} A_{\beta]} \right)$.

I tried a couple simple examples and came up with this:

• $\epsilon_{ij} a_{[i} b_{j]} = 2 \epsilon_{ij} a_{i} b_{j}$

• $\epsilon_{ijk} a_{[i} b_{j]} c_k = 2 \epsilon_{ijk} a_{i} b_{j} c_k$

Penrose-Tensor-Diagram:

Does it generally work that way, that double antisymmetrisation in one of the factors simply causes a factor 2?

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I think you'll have to explain those diagrams, since there's no factor $2$ anywhere in your equations. –  joriki Jan 21 '13 at 9:01
I just forgot to add the “2” after I copied it. Fixed now. –  queueoverflow Jan 21 '13 at 9:40
The notation you're using comes from physics, not mathematics, so you're going to have to explain it. I, for one, have never seen it before. –  Qiaochu Yuan Jan 21 '13 at 9:54

There are two conventions for antisymmetrizing a function of two variables:

$$A_1f(x,y) = \frac{1}{2}\left( f(x,y) - f(y,x) \right)$$ $$A_2f(x,y) = f(x,y) - f(y,x)$$

The first convention is idempotent:

$$A_1 A_1 f(x,y) = \frac{1}{2}\left( A_1 f(x,y) - A_2 f(y,x) \right) = \frac{1}{4} \left( (f(x,y) - f(y,x)) - (f(y,x) - f(x,y)) \right) = \frac{1}{2}\left( f(x,y) - f(y,x) \right) = A_1 f(x,y)$$

and the second picks up a factor of $2$:

$$A_2 A_2 f(x,y) = A_2 f(x,y) - A_2 f(y,x) = (f(x,y) - f(y,x)) - (f(y,x) - f(x,y)) = 2(f(x,y) - f(y,x)) = 2 A_2 f(x,y)$$

(note that an expression like $X_{ij}$ can be viewed as a function in the two variables $i$ and $j$)

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