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In 3 dimensions the product of the Levi-Cevita tensor with itself is just,

$$ \epsilon_{ijk} \epsilon^{ijk} = 6$$

My question is how does this apply to 4 dimensions?, i.e.

$$ \epsilon^{\mu \nu \alpha \beta}\epsilon_{\mu \nu \alpha \beta} = ?$$

Thanks

1 Answers 1

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Notice $\epsilon^{\mu\nu\alpha\beta}\epsilon_{\mu\nu\alpha\beta}$ is a sum of squares of $0$s and $\pm1$s. Since $1$ and $-1$ square to $1$, the resulting sum will simply count the number of "entries" of $\epsilon$ which are nonzero. It is nonzero precisely when the indices $\mu\nu\alpha\beta$ are a permutation of $1234$, and there are $4!=24$ permutations.

Notice the $n=3$ case had a sum of $3!=6$, and in the general case the sum will be $n!$.

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    That's perfect thanks a bunch! I also posted a similar question here: http://math.stackexchange.com/questions/2135276/tensor-manipulation-involving-levi-cevita-tensor that you might have some insight on? – 2017-02-09
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    Also does the order of the indices matter? i.e. what about $\epsilon^{\mu \nu \alpha \beta} \epsilon_{\beta \nu \alpha \mu}$ ? – 2017-02-09
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    @user1887919 You mean does it matter that both factors have the same order? Yes. We have $\epsilon_{\alpha\beta\gamma\delta}=\pm\epsilon_{\alpha'\beta'\gamma'\delta}$, with sign depending on whether $\alpha\beta\gamma\delta$ is an even or an odd permutation of $\alpha'\beta'\gamma'\delta'$. Do you see how this will affect the value of $\epsilon^{\alpha\beta\gamma\delta}\epsilon_{\alpha'\beta'\gamma'\delta'}$? – 2017-02-09