Levi Civita Tensor in 4D

[div curl F= 0]

Is there an identity for a product of 2 LC Tensors in 4D if one sums over 3 of the indicies?

i.e.

\epsilon^{\mu \beta \gamma \delta} \epsilon_{\nu \beta \gamma \delta} = ?


What if gamma is constrained to be 0? Does this reduce things?


Best Regards

[fantispug]

There is a general formula for the product of multidimensional Levi-Civita symbols:

\epsilon_{i_1 i_2 i_3 \ldots i_n}\epsilon_{j_1 j_2 j_3 \ldots j_n}= \det A
where A is the matrix with elements
(A)_{mn}=\delta_{i_m j_n}.

Using this you could push out an identity with a bit of work.

A more direct way is to look at symmetry consider expressions of the form:
\epsilon^{\mu i_1 \ldots i_n} \epsilon_{\nu i_1 \ldots i_n}. (In your case n=3).

The Levi-Civita symbol is zero unless all the terms are different, and there are only n+1 different choices for the indicies; thus for any given choice of i_1,\ldots,i_n there is only one choice of mu such that the first term doesn't vanish, and only one choice of nu such that the second term doesn't vanish. Consequently the whole expression is proportional to \delta^{\mu}_{\nu}.

Now to find the constant of proportionality just work with any case: to get a non-vanishing term we require all the indicies to be different. If we choose mu=nu, then we have n ways of choosing i_1, (n-1) ways of choosing i_2, ..., 1 way of choosing i_n; and so we get exactly n! non-vanishing terms.
Now clearly each term is either 0 or 1, and so we conclude
\epsilon^{\mu i_1 \ldots i_n} \epsilon_{\nu i_1 \ldots i_n}=n! \delta^{\mu}_{\nu}.

So

\epsilon^{\mu \beta \gamma \delta} \epsilon_{\nu \beta \gamma \delta} = 3! \delta^{\mu}_{\nu}


Hope I convinced you!