Permutation Symbol

The permutation symbol, denoted by $\epsilon_{ijk}$, is defined by

$$\epsilon_{ijk} = \left\{\begin{array}{r} 1\\ -1\\ 0\end{array}\right\}\ \ i,j,k\ \ \left\{\begin{array}{l} \mbox{an even}\\ \mbox{an odd}\\ \mbo... ...\end{array}\right\} \begin{array}{l} \mbox{permutation of}\\ 1,2,3.\end{array}$$ (2.5.1)

That is,

$$\epsilon_{123} = \epsilon_{231} = \epsilon_{312} = 1,$$

$$\epsilon_{132} = \epsilon_{321} = \epsilon_{213} = -1,$$

$$\epsilon_{111} = \epsilon_{211} = \epsilon_{133} = \ldots = 0.$$

We note that

$$\epsilon_{ijk} = \epsilon_{jki} = \epsilon_{kij} = -\epsilon_{jik} = - \epsilon_{ikj} = - \epsilon_{kji}.$$

If $\mathbf{e}_1,\ \mathbf{e}_2,\ \mathbf{e}_3$ form a right-handed triad, then

$$\mathbf{e}_1\times \mathbf{e}_2 = \mathbf{e}_3,\ \mathbf{e}_2\times \mathbf{e}_3 = \mathbf{e}_1,\ etc.$$

which can be written as

$$\mathbf{e}_i\times \mathbf{e}_j = \epsilon_{ijk}\mathbf{e}_k.$$ (2.5.2)

Now, if $\mathbf{u} = u_i\mathbf{e}_i,\ \mathbf{v} = v_i\mathbf{e}_i$, then

$$\mathbf{u}\times \mathbf{v} =\ u_i\mathbf{e}_i \times v_j\mathbf{e}_j = u_iv_j\mathbf{e}_i\times \mathbf{e}_j$$

$$=\ u_iv_j\epsilon_{ijk}\mathbf{e}_k = \epsilon_{ijk}u_iv_j\mathbf{e}_k.$$ (2.5.3)

Exercise. Using the index notation write an expression for $\vert\sin\theta \vert$; $\theta$ being the angle between vectors $\mathbf{u}$ and $\mathbf{v}$.

Exercise. Show that

$$(1)\ \ \epsilon_{ijk}\epsilon_{jki} = 6,$$

$$(2)\ \ \epsilon_{ijk}A_jA_k = 0.$$

The following useful identity, which can be verified by long-hand calculations should be memorized.

$$\epsilon_{ijm}\epsilon_{klm} = \delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}.$$ (2.5.4)

Now by using this identity let us prove the vector identity

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

Proof: Let $\mathbf{v}\times \mathbf{w} = \mathbf{a}$. Then $\mathbf{a} = \epsilon_{ijk}v_iw_j\mathbf{e}_k$, and

$$\mathbf{u}\times \mathbf{a} = \epsilon_{ijk}u_ia_j\mathbf{e}_k,$$

$$a_j = \mathbf{a}\cdot \mathbf{e}_j = \epsilon_{ilm}v_iw_l\mathbf{e}_m\cdot \mathbf{e}_j,$$

$$\hspace*{.7in} =\epsilon_{ilm}v_iw_l\delta_{mj} = \epsilon_{ilj}v_iw_l.$$

$$\mathbf{u}\times (\mathbf{v} \times \mathbf{w})=\ \epsilon_{ijk}u_i (\epsilon_{plj}v_pw_l)\mathbf{e}_k,$$

$$=\ \epsilon_{ijk}\epsilon_{plj}u_iv_pw_l\mathbf{e}_k,$$

$$=\ \epsilon_{kij}\epsilon_{plj}u_iv_pw_l\mathbf{e}_k,$$

$$( \ (2.5.4)) =\ (\delta_{kp}\delta_{il} - \delta_{kl} \delta_{ip})u_iv_pw_l\mathbf{e}_k,$$

$$=\ \delta_{kp}\delta_{il} u_iv_pw_l\mathbf{e}_k - \delta_{kl}\delta_{ip}u_iv_pw_l\mathbf{e}_k,$$

$$=\ u_lw_lv_k\mathbf{e}_k - u_pv_pw_k\mathbf{e}_k,$$

$$( \ (2.4.2)) =\ (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u}\cdot \mathbf{v})\mathbf{w}.$$

Exercise. Show that

(a)

If $\epsilon_{ijk}T_{jk} = 0$, then $T_{ij} = T_{ji}$,

(b)

$\epsilon_{ilm}\epsilon_{jlm} = 2\delta_{ij}$, and

(c)

if $T_{ij} = -T_{ji}$, then $T_{ij}a_ia_j = 0$.

We now write ${\rm det}\, A_{ij}$ in the index notation.

$${\rm det}\, [A_{ij}] =\ {\rm det}\, \left[\begin{array}{ccc} A_{11} & A_{12} & A_{13}\\ A_{21} & A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33}\end{array}\right],$$

$$=\ A_{11} (A_{22}A_{33} - A_{32} A_{23}) - A_{21} (A_{12}A_{33} - A_{32}A_{13}) + A_{31} (A_{12} A_{23} - A_{22} A_{13}),$$

$$=\ A_{11} (\epsilon_{1jk}A_{j2}A_{k3}) - A_{21} (-\epsilon_{2jk}A_{j2}A_{k3}) + A_{31} (\epsilon_{3jk}A_{j2}A_{k3}),$$

$$=\ A_{i1} \epsilon_{ijk} A_{j2} A_{k3},$$

$$=\ \epsilon_{ijk}A_{i1}A_{j2}A_{k3}.$$

Example. Show that $\epsilon_{ijk}A_{il}A_{jm}A_{kn} = ({\rm det}\, A)\epsilon_{lmn}$.