Eddington❤︎Kronecker

以下のEddington epsilon(Levi-Civita symbol)とKronecker deltaの関係式をベクトル3重積を用いて示す．これが今のところ自分の中で一番分かりやすい．

$$
\varepsilon_{ijk} \varepsilon_{lmk} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}
$$

[証明]
まず，以下のベクトル3重積の公式を示す．

$$
\boldsymbol{u} \times (\boldsymbol{v} \times \boldsymbol{w})
=(\boldsymbol{u} \cdot \boldsymbol{w})\boldsymbol{v}
-(\boldsymbol{u} \cdot \boldsymbol{v})\boldsymbol{w}
$$

$$
\begin{aligned}
(左辺)&=
\begin{pmatrix}
u_1 \\ u_2 \\ u_3
\end{pmatrix}
\times
\Bigg[
\begin{pmatrix}
v_1 \\ v_2 \\ v_3
\end{pmatrix}
\times
\begin{pmatrix}
w_1 \\ w_2 \\ w_3
\end{pmatrix}
\Bigg] \\
&=
\begin{pmatrix}
u_1 \\ u_2 \\ u_3
\end{pmatrix}
\times
\begin{pmatrix}
v_2w_3-v_3w_2 \\
v_3w_1-v_1w_3 \\
v_1w_2-v_2w_1
\end{pmatrix} \\
&=
\begin{pmatrix}
u_2(v_1w_2-v_2w_1)-u_3(v_3w_1-v_1w_3) \\
u_3(v_2w_3-v_3w_2)-u_1(v_1w_2-v_2w_1) \\
u_1(v_3w_1-v_1w_3)-u_2(v_2w_3-v_3w_2)
\end{pmatrix} \\
\end{aligned}
$$

$$
\begin{aligned}
(右辺)&=
\Bigg[
\begin{pmatrix}
u_1 \\ u_2 \\ u_3
\end{pmatrix}
\cdot
\begin{pmatrix}
w_1 \\ w_2 \\ w_3
\end{pmatrix}
\Bigg]
\begin{pmatrix}
v_1 \\ v_2 \\ v_3
\end{pmatrix}
-
\Bigg[
\begin{pmatrix}
u_1 \\ u_2 \\ u_3
\end{pmatrix}
\cdot
\begin{pmatrix}
v_1 \\ v_2 \\ v_3
\end{pmatrix}
\Bigg]
\begin{pmatrix}
w_1 \\ w_2 \\ w_3
\end{pmatrix} \\
&=
(u_1w_1+u_2w_2+u_3w_3)
\begin{pmatrix}
v_1 \\ v_2 \\ v_3
\end{pmatrix}
-
(u_1v_1+u_2v_2+u_3v_3)
\begin{pmatrix}
w_1 \\ w_2 \\ w_3
\end{pmatrix} \\
&=
\begin{pmatrix}
(u_1w_1+u_2w_2+u_3w_3)v_1-(u_1v_1+u_2v_2+u_3v_3)w_1 \\
(u_1w_1+u_2w_2+u_3w_3)v_2-(u_1v_1+u_2v_2+u_3v_3)w_2 \\
(u_1w_1+u_2w_2+u_3w_3)v_3-(u_1v_1+u_2v_2+u_3v_3)w_3
\end{pmatrix} \\
&=
\begin{pmatrix}
u_2(v_1w_2-v_2w_1)-u_3(v_3w_1-v_1w_3) \\
u_3(v_2w_3-v_3w_2)-u_1(v_1w_2-v_2w_1) \\
u_1(v_3w_1-v_1w_3)-u_2(v_2w_3-v_3w_2)
\end{pmatrix} \\
&=(左辺)
\end{aligned}
$$

それぞれEddington epsilonとKronecker deltaを使って書けば，以下のようになる．

$$
\begin{aligned}
\boldsymbol{u} \times (\boldsymbol{v} \times \boldsymbol{w}) 
&=\varepsilon_{ijk} u_j (\varepsilon_{klm}v_lw_m) \\
&=\varepsilon_{ijk} \varepsilon_{klm}u_jv_lw_m \\
&=\varepsilon_{ijk} \varepsilon_{lmk}u_jv_lw_m
\end{aligned}
$$

$$
\begin{aligned}
(\boldsymbol{u} \cdot \boldsymbol{w})\boldsymbol{v}
-(\boldsymbol{u} \cdot \boldsymbol{v})\boldsymbol{w}
&=(u_jw_j)v_i-(u_jv_j)w_i \\
&=u_jv_iw_j-u_jv_jw_i \\
&=u_j(\delta_{il}v_l)(\delta_{jm}w_m)-u_j(\delta_{jl}v_l)(\delta_{im}w_m) \\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})u_jv_lw_m
\end{aligned}
$$

したがって，2式を比べれば求める式が得られる．