# Levi-Civita Properties

Prove the following

a) Let $(\hat e_1,\hat e_2. \hat e_3)$ be the unit vectors of a right handed, orthogonal coordinate system. Demonstrate that Levi-Civita symbol satisfies  $\epsilon_{ijk} =\hat e_i(\hat e_j \times \hat e_k)$

b) Prove that $\epsilon_{ijk}\epsilon_{ist}=\delta_{js}\delta_{kt}-\delta_{jt}\delta_{ks}$

c) Prove that $a \times b=\epsilon_{ijk}\hat e_i a_j b_k$

a)

$ϵ_{ijk}$  is defined as

$det\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{vmatrix}=∑_{i=1}^3 ∑_{j=1}^3 ∑_{k=1}^3 ϵ_{ijk} a_{1i} a_{2j} a_{3k}=ϵ_{ijk} a_{1i} a_{2j} a_{3k}$

Taking

$a_i=|\hat e_i |$  $a_j=|\hat e_j|$, $a_k=|\hat e_k|$   we obtain $ϵ_{ijk}=det⁡ |\hat e_i,\hat e_j ,\hat e_k |=(\hat e_i )(\hat e_j × \hat e_k)$

b)

In general

$ϵ_{ijk} ϵ_{lmn}=δ_{il} δ_{jm} δ_{kn}+δ_{im} δ_{jl} δ_{kl}+δ_{in} δ_{jl} δ_{km}-δ_{im} δ_{jl} δ_{kn}-δ_{il} δ_{jn} δ_{km}-δ_{in} δ_{jm} δ_{kl}$

Taking

$i=l$, $m=s$ and  $n=t$   one has

$ϵ_{ijk} ϵ_{ist}=δ_{ii} δ_{js} δ_{kt}-δ_{ii} δ_{jt} δ_{ks}=δ_{js} δ_{kt}-δ_{jt} δ_{ks}$

c)

$(a×b)=det⁡ \begin {vmatrix}\hat e_1&\hat e_2&\hat e_3 \\a_1&a_2&a_3\\b_1&b_2&b_3 \end{vmatrix}=ϵ_{ijk} (\hat e_i ) a_j b_k$    and thus $(a×b)_i=ϵ_{ijk} a_j b_k$

Evaluate the following expressions

a) $\delta_{ii}$

b) $\delta_{ij}\epsilon_{ijk}$

c) $\epsilon_{ijk}\epsilon_{ljk}$

d) $\epsilon{ijk}\epsilon_{ijk}$

a)

$δ_{ii}=1$

b)

$δ_{ij} ϵ_{ijk}=ϵ_{iik}=0$    because $i=j$

c)

$ϵ_{ijk} ϵ_{lmn}=δ_{il} δ_{jm} δ_{kn}+δ_{im} δ_{jl} δ_{kl}+δ_{in} δ_{jl} δ_{km}-δ_{im} δ_{jl} δ_{kn}-δ_{il} δ_{jn} δ_{km}-δ_{in} δ_{jm} δ_{kl}$

therefore

$ϵ_{ijk} ϵ_{ljk}=δ_{il} δ_{jj} δ_{kk}+δ_{ij} δ_{jl} δ_{kk}+δ_{ik} δ_{jl} δ_{kj}-δ_{ij} δ_{jl} δ_{kk}-δ_{il} δ_{jk} δ_{kj}-δ_{ik} δ_{jj} δ_{ki}$

$ϵ_{ijk} ϵ_{ljk}=1+δ_{ik} δ_{jl} δ_{kj}-δ_{ij} δ_{jk} δ_{kj}-δ_{ik}^2$

d)

$ϵ_{ijk} ϵ_{ijk}=δ_{ii} δ_{jj} δ_{kk}+δ_{ij} δ_{ji} δ_{kk}+δ_{ij} δ_{jk} δ_{kk}-δ_{ij} δ_{jk} δ_{kk}-δ_{ii} δ_{jk} δ_{kj}-δ_{ik} δ_{jj} δ_{ki}$

$ϵ_{ikj} ϵ_{ijk}=1+δ_{ij}^2+δ_{ij} δ_{jk}-δ_{ik} δ_{jk}-δ_{jk}^2-δ_{ik}^2=1+δ_{ij}^2-δ_{jk}^2-δ_{ik}^2$