# Product of Levi-Civita symbol is determinant?

I am confused with how can one write product of Levi-Civita symbol as determinant? I want to prove 'epsilon-delta' identity and found this questions answers it. But I am stuck at product of Levi-Civita symbol

Proof relation between Levi-Civita symbol and Kronecker deltas in Group Theory

$$\varepsilon_{ijk}\varepsilon_{lmn} = \begin{vmatrix} \delta_{il} & \delta_{im} & \delta_{in} \\ \delta_{jl} & \delta_{jm} & \delta_{jn} \\ \delta_{kl} & \delta_{km} & \delta_{kn} \\ \end{vmatrix}$$

This Wikipedia article gives the above relation. I am confused how Product of Levi-Civita symbol is a determinant. Can someone explain?

• This chapter in Pavel Grinfeld's book I think explains it fairly well. – user137731 Sep 29 '16 at 13:57

The most simple way (i.e. without lengthy calculations) is to prove it on the basis of the symmetry properties which fully charaterize the two members. A levi-Civita tensor $$\epsilon_{ijk}$$ is fully charaterized in terms of its complete skew-symmetry with respect to transposition of two indices and the fact that $$\epsilon_{123}=1$$. On the other hand, the determinant is fully charaterized by: 1) the skew-symmetry with respect to transposition of rows (and columns), 2) the determinant of identity is one and 3) the multilinearity. Note that here multilinearity is not relevant, since here we are restricted to consider only matrices with entries equal to 0 or 1.
It is important to understand that usually the determinant is a scalar but in this case it is not. Recall the cross product $$A \times B$$, can be computed as a determinant and it is a vector. We will interpret as determinant to the regular expansion of determinants but this time instead of a scalar entries we have 2-rank tensor entries.
We expand the determinant as follows: $$\begin{eqnarray*} \left | \begin{array}{ccc} \delta_m^r & \delta_n^r & \delta_p^r \\ \delta_m^s & \delta_n^s & \delta_p^s \\ \delta_m^t & \delta_n^t & \delta_p^t \end{array} \right | &=& \delta_m^r \delta_n^s \delta_p^t + \delta_n^r \delta_p^s \delta_m^t + \delta_p^r \delta_m^s \delta_n^t - \delta_p^r \delta_n^s \delta_m^t - \delta_n^r \delta_m^s \delta_p^t - \delta_m^r \delta_p^s \delta_n^t . \end{eqnarray*}$$ We define $$\begin{eqnarray*} a_{mnp}^{rst} = &=& \delta_m^r \delta_n^s \delta_p^t + \delta_m^r \delta_n^s \delta_p^t + \delta_m^r \delta_n^s \delta_p^t - \delta_m^r \delta_n^s \delta_p^t - \delta_m^r \delta_n^s \delta_p^t -\delta_m^r \delta_n^s \delta_p^t \end{eqnarray*}$$ where we commuted factors so $$m,n,p$$ are in that order in all subsscripts. We recognize that the sequence of delta operations $$a_{mnp}^{rst}$$ is a completely skew-symmetri system in $$r,s,t$$. That is, for example, let us reverse the order or $$s$$ and $$t$$ and find $$\begin{eqnarray*} a^{rts}_{mnp} &=& \delta_m^r \delta_n^t \delta_p^s + \delta_m^s \delta_n^r \delta_p^t + \delta_m^t \delta_n^s \delta_p^r - \delta_m^s \delta_n^t \delta_p^r - \delta_m^t \delta_n^r \delta_p^s -\delta_m^r \delta_n^s \delta_p^t \\ &=& -\delta_m^r \delta_n^s \delta_p^t -\delta_m^t \delta_n^r \delta_p^s -\delta_m^s \delta_n^t \delta_p^r + \delta_m^t \delta_n^s \delta_p^r + \delta_m^s \delta_n^r \delta_p^t + \delta_m^r \delta_n^t \delta_p^s \\ &=& -a^{rst}_{mnp} \end{eqnarray*}$$ where we fixed the order by using the commutative law of the sum (with sign) operation . Likewise we can interchange the order of $$r,s$$ and $$r,t$$ an get $$\begin{eqnarray*} a_{mnp}^{srt} = -a_{mnp}^{rst} \quad \text{and} \quad a_{mnp}^{tsr} = -a_{mnp}^{rst}. \end{eqnarray*}$$ Hence $$a_{mnp}^{rst}$$ ($$m,n,p$$ fixed) is completely skew-symmetric in $$r,s,t$$ and so $$\begin{eqnarray*} a_{mnp}^{rst} = \epsilon^{rst} a_{mnp}^{123} \end{eqnarray*}$$ with $$\begin{eqnarray*} a_{mnp}^{123} = \delta_m^1 \delta_n^2 \delta_p^3 + \delta_m^3 \delta_n^1 \delta_p^2 + \delta_m^2 \delta_n^3 \delta_p^1 - \delta_m^3 \delta_n^2 \delta_p^1 - \delta_m^2 \delta_n^1 \delta_p^3 -\delta_m^1 \delta_n^3 \delta_p^2 . \end{eqnarray*}$$ This is equal to $$1$$ if $$mpn$$ is an even permutation of $$(1 \; 2 \; 3)$$ or $$-1$$ if $$mpn$$ is an odd permutation of $$(1 \; 2 \; 3)$$. That is, $$\begin{eqnarray*} a_{mnp}^{rst} = \epsilon^{rst} \epsilon_{mnp} \end{eqnarray*}$$ if $$m=n$$ or $$m=p$$ we would get 0. Then
$$\begin{eqnarray*} \left | \begin{array}{ccc} \delta_m^r & \delta_n^r & \delta_p^r \\ \delta_m^s & \delta_n^s & \delta_p^s \\ \delta_m^t & \delta_n^t & \delta_p^t \end{array} \right | &=& \epsilon^{rst} \epsilon_{mnp} . \end{eqnarray*}$$