In Four-dimensional space, the Levi-Civita symbol is defined as:
$$ \varepsilon_{ijkl } =$$ \begin{cases} +1 & \text{if }(i,j,k,l) \text{ is an even permutation of } (1,2,3,4) \\ -1 & \text{if }(i,j,k,l) \text{ is an odd permutation of } (1,2,3,4) \\ 0 & \text{otherwise} \end{cases}
Let's suppose that I fix the last index ( l=4 for example). I guess that the 4-indices symbol can now be replaced with a 3-indices one:
$$ \varepsilon_{ijk } =$$ \begin{cases} +1 & \text{if } (i,j,k) \text{ is } (1,2,3), (2,3,1) \text{ or } (3,1,2), \\ -1 & \text{if } (i,j,k) \text{ is } (3,2,1), (1,3,2) \text{ or } (2,1,3), \\ \;\;\,0 & \text{if }i=j \text{ or } j=k \text{ or } k=i \end{cases}
My doubt is the following: is $$ \varepsilon_{ijk4 }A^{jk} = \varepsilon_{ij4k }A^{jk } $$ true? (In the sense that the 4-indices symbols can both be replaced by the same 3-indices symbols. I'm using the Einstein notation, so multiple indices are summed) or they give two 3-indices symbols with different sign
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$\begingroup$If you fix one of the indices of $\varepsilon_{ijkl}$ to be $4$ you get $\pm\varepsilon_{ijk}$ depending on wheather you fix an odd or an even positioned index. So $\varepsilon_{ijk4}=\varepsilon_{ijk}$ but $\varepsilon_{ij4k}=-\varepsilon_{ijk}$. To see why the signs come out this way, notice that when you substitute $1,2,3$ for $i,j,k$ you get: $\varepsilon_{1234}=1=\varepsilon_{123}$, but $\varepsilon_{1243}=-1=-\varepsilon_{123}$. So in fact $$\varepsilon_{ijk4}A^{jk}=-\varepsilon_{ij4k}A^{jk}.$$
$\endgroup$ $\begingroup$Note that the permutations corresponding to $(i,j,k,4)$ and $(i,j,4,k)$ differ only by a transposition of the last two indices. Consequently, they have different parity and so their Levi-Civita symbols have opposite signs (assuming they do not vanish, of course). Hence the correct statement is $\epsilon_{ijk4}A^{jk}=-\epsilon_{ij4k}A^{jk}$.
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