# Why $a_i b_m c_n\left(\delta_{m k} \delta_{n i}-\delta_{m i} \delta_{n k}\right) \mathbf{e}_k=a_n b_m c_n \mathbf{e}_m-a_m b_m c_n \mathbf{e}_n$?

I am stuck in an intermediate step.

In order to evaluate the product of the Levi-Civita symbols, we use the identity $$\epsilon_{m n j} \epsilon_{i j k}=\delta_{m k} \delta_{n i}-\delta_{m i} \delta_{n k}$$ and the properties of the Kronecker delta functions. Thus, we obtain \begin{aligned} \mathbf{a} \times(\mathbf{b} \times \mathbf{c}) &=\epsilon_{m n j} \epsilon_{i j k} a_i b_m c_n \mathbf{e}_k \\ &=a_i b_m c_n\left(\delta_{m k} \delta_{n i}-\delta_{m i} \delta_{n k}\right) \mathbf{e}_k \\ &=a_n b_m c_n \mathbf{e}_m-a_m b_m c_n \mathbf{e}_n \\ &=\left(b_m \mathbf{e}_m\right)\left(c_n a_n\right)-\left(c_n \mathbf{e}_n\right)\left(a_m b_m\right) \\ &=\mathbf{b}(\mathbf{a} \cdot \mathbf{c})-\mathbf{c}(\mathbf{a} \cdot \mathbf{b}) \end{aligned}

http://people.uncw.edu/hermanr/qm/Levi_Civita.pdf

• just replace $k,i$ with $m,n$ in the first and then $k,i$ with $n,m$ in the second i.e. use the definition of $\delta_{xy}$ Commented Oct 17, 2022 at 13:52

\begin{aligned} \mathbf{a} \times(\mathbf{b} \times \mathbf{c}) &=\epsilon_{m n j} \epsilon_{i j k} a_i b_m c_n \mathbf{e}_k \\ &=a_i b_m c_n\left(\delta_{m k} \delta_{n i}-\delta_{m i} \delta_{n k}\right) \mathbf{e}_k \\ &=a_i b_m c_n\delta_{m k} \delta_{n i}\mathbf{e}_k-a_i b_m c_n\delta_{m i} \delta_{n k} \mathbf{e}_k \end{aligned} Now recall that this expression includes bunch of summations. In the the first term, $$\delta_{m k} \delta_{n i}$$ makes all other terms zero unless $$m=k$$ and $$n=i$$. So all that remains is $$a_n b_m c_n \mathbf{e}_m$$. Similarly the second term $$\delta_{m i} \delta_{n k}$$ makes all other terms zero unless $$m=i$$ and $$n=k$$. So the secomd term is $$a_m b_m c_n \mathbf{e}_n.$$ That gives you
\begin{aligned} a_i b_m c_n\delta_{m k} \delta_{n i}\mathbf{e}_k-a_i b_m c_n\delta_{m i} \delta_{n k} \mathbf{e}_k&=a_n b_m c_n \mathbf{e}_m-a_m b_m c_n \mathbf{e}_n \\ &=\left(b_m \mathbf{e}_m\right)\left(c_n a_n\right)-\left(c_n \mathbf{e}_n\right)\left(a_m b_m\right) \\ &=\mathbf{b}(\mathbf{a} \cdot \mathbf{c})-\mathbf{c}(\mathbf{a} \cdot \mathbf{b}) \end{aligned}