Consider a four-dimensional spacetime. Consider the following contraction between Levi-Civita symbols and tetrads $$\epsilon_{\alpha \beta i j}\,{\epsilon^{ij}}_k\, e^\alpha\!\wedge e^\beta\!\wedge e^0\!\wedge e^k,$$ what I have done to simplify this is the following: $$\epsilon_{\alpha \beta i j}\,{\epsilon^{ij}}_k\,e^\alpha\!\wedge e^\beta\!\wedge e^0\!\wedge e^k = 2!\,\epsilon_{mnk}\,e^0\!\wedge e^m\!\wedge e^n\!\wedge e^k = 2!\,3!\,e^0\!\wedge e^1\!\wedge e^2\!\wedge e^3$$

where the latin indices take values $\{1,2,3\}$ and Greek indices take values $\{0,1,2,3\}$.

Is this correct?


1 Answer 1


It is zero. In $\epsilon_{\alpha\beta ij}$, all indices have to be distinct. So one of them must be $0$, and therefore $e^\alpha\wedge e^\beta\wedge e^0\wedge e^k$ vanishes, since $0$ appears twice.

It also vanishes for a different reason, namely: in $\epsilon_{ijk}$, the index $k$ has to be different from $ij$, so it must equal either $\alpha$ or $\beta$. In either case, $e^\alpha\wedge e^\beta\wedge e^0\wedge e^k$ vanishes, since $e^k=e^\alpha$ or $e^k=e^\beta$.

You can do a simple Mathematica test,

Sum[Signature[{α, β, i, j}] Signature[{i, j, k}] e[α] ** e[β] ** e[0] ** e[k], {α, 0, 3}, {β, 0, 3}, {i, 1, 3}, {j, 1, 3}, {k, 1, 3}]

which yields $$ 2 e^0\wedge e^1\wedge e^0\wedge e^1+2 e^0\wedge e^2\wedge e^0\wedge e^2+2 e^0\wedge e^3\wedge e^0\wedge e^3-2 e^1\wedge e^0\wedge e^0\wedge e^1-2 e^2\wedge e^0\wedge e^0\wedge e^2-2 e^3\wedge e^0\wedge e^0\wedge e^3 $$ which clearly vanishes since all terms include repeated vectors. Indeed, $e^0$ always appears twice, and also the last vector is equal to either the first or the second, in agreement with the two observations above.


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