Levi Civita Symbol: from 4 to 3 indices.

Question

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

Answer 1

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 $$\varepsilon_{ijk4}A^{jk}=-\varepsilon_{ij4k}A^{jk}.$$

Answer 2

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 $\epsilon_{ijk4}A^{jk}=-\epsilon_{ij4k}A^{jk}$.

Answer 3

There is a method by which we can formally verify your guess that,

$$ \epsilon^{ijk4} = \epsilon^{ijk} $$

And determine whether,

$$ \epsilon^{ijk4} = \pm \epsilon^{ij4k} $$

The $n$-indexed Levi-Civita symbol is defined in terms of the generalized Kronecker delta as:

$$ \epsilon^{i_1i_2 \dots i_n} = \delta^{i_1i_2 \dots i_n}_{12 \dots n} $$

In case of the four-indexed Levi-Civita symbol of 4D space,

$$ \epsilon^{ijkl} = \delta^{ijkl}_{1234} $$

By definition of the generalized Kronecker delta, this can be written as the determinant:

$$ \epsilon^{ijkl} = \det \begin{vmatrix} \delta^i_1 && \delta^i_2 && \delta^i_3 && \delta^i_4 \\ \delta^j_1 && \delta^j_2 && \delta^j_3 && \delta^j_4 \\ \delta^k_1 && \delta^k_2 && \delta^k_3 && \delta^k_4 \\ \delta^l_1 && \delta^l_2 && \delta^l_3 && \delta^l_4 \\ \end{vmatrix} $$

When $l = 4$,

$$ \epsilon^{ijk4} = \det \begin{vmatrix} \delta^i_1 && \delta^i_2 && \delta^i_3 && \delta^i_4 \\ \delta^j_1 && \delta^j_2 && \delta^j_3 && \delta^j_4 \\ \delta^k_1 && \delta^k_2 && \delta^k_3 && \delta^k_4 \\ \delta^4_1 && \delta^4_2 && \delta^4_3 && \delta^4_4 \\ \end{vmatrix} = \det \begin{vmatrix} \delta^i_1 && \delta^i_2 && \delta^i_3 && \delta^i_4 \\ \delta^j_1 && \delta^j_2 && \delta^j_3 && \delta^j_4 \\ \delta^k_1 && \delta^k_2 && \delta^k_3 && \delta^k_4 \\ 0 && 0 && 0 && 1 \\ \end{vmatrix} $$

Expanding this via the last row,

$$ \epsilon^{ijk4} = 0 \det \begin{vmatrix} \delta^i_2 && \delta^i_3 && \delta^i_4 \\ \delta^j_2 && \delta^j_3 && \delta^j_4 \\ \delta^k_2 && \delta^k_3 && \delta^k_4 \end{vmatrix} - 0 \det \begin{vmatrix} \delta^i_1 && \delta^i_3 && \delta^i_4 \\ \delta^j_1 && \delta^j_3 && \delta^j_4 \\ \delta^k_1 && \delta^k_3 && \delta^k_4 \end{vmatrix} + 0 \det \begin{vmatrix} \delta^i_1 && \delta^i_2 && \delta^i_4 \\ \delta^j_1 && \delta^j_2 && \delta^j_4 \\ \delta^k_1 && \delta^k_2 && \delta^k_4 \end{vmatrix} + 1 \det \begin{vmatrix} \delta^i_1 && \delta^i_2 && \delta^i_3 \\ \delta^j_1 && \delta^j_2 && \delta^j_3 \\ \delta^k_1 && \delta^k_2 && \delta^k_3 \end{vmatrix} $$

The only remaining term is,

$$ \epsilon^{ijk4} = \det \begin{vmatrix} \delta^i_1 && \delta^i_2 && \delta^i_3 \\ \delta^j_1 && \delta^j_2 && \delta^j_3 \\ \delta^k_1 && \delta^k_2 && \delta^k_3 \end{vmatrix} $$

Which by the definitions of the generalized Kronecker delta and the Levi-Civita symbol is,

$$ \epsilon^{ijk4} = \delta^{ijk}_{123} = \epsilon^{ijk} $$

When we write the Levi-Civita symbol as a determinant like we did above, the totally antisymmetric property that it possess becomes evident: swapping any two indices corresponds to interchanging their corresponding rows in the matrix due to which the determinant, which is the Levi-Civita itself, changes sign.

From this behavior, we can easily deduce what $\epsilon^{ij4k}$ would be without going through the computations once again: it is just the result of exchanging the last two indices of $\epsilon^{ijk4}$,

$$ \epsilon^{ij4k} = -\epsilon^{ijk4} = -\epsilon^{ijk} $$

So yes, depending on what index you fix, you will get a three-indexed Levi-Civita with a different sign.