# 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

## 2 Answers

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.

This identity relates the product of the volumes spanned by two sets of three vectors (with a minus sign if the sets are oppositely oriented) in terms of the inner products of the individual vectors.

In particular, the determinant can be understood as computing the product of the volumes by projecting one set of vectors onto the other set.