# Help understanding the proof of a vector identity using Levi-Civita and Kronecker notation.

I am looking at the proof of the following identity:

a x (b x c) = (a.c)b - (a.b)c

I have only just been introduced to Levi-Civita notation and the Kronecker delta, so could you please break down your answer using summations where possible.

My issue is with a specific line in the proof, so I will write out the proof and then state which line I am struggling to understand.

a x (b x c) $=\epsilon_{ijk}a_j$(b x c)$_k=\epsilon_{ijk}\epsilon_{klm}a_jb_lc_m$

Then applying the '$\epsilon-\delta$ Identity',

$\epsilon_{ijk}\epsilon_{klm}a_jb_lc_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})a_jb_lc_m=a_jb_ic_j-a_jb_jc_j.$

The last line of this proof is the part I struggle with. I have tried expanding the bracket and changing the middle to

$(\delta_{il}b_l)\delta_{jm}a_jc_m-(\delta_{im}c_m)\delta_{jl}a_jb_l$

and then doing the summation of each separate bit but it doesn't seem to be getting me any closer to the final part.

• See the answer given by @anomaly below. If your trying to understand why the $\epsilon-\delta$ Identity is true there are other links here in the math.se but I think for the final step you have just to see that $\vec{a}\cdot \vec{b} = a_ib_i$. – R.W Feb 20 '17 at 2:40

The result should be a vector; you should either look at the $k$th component of $a\times (b\times c)$ or else add in an extra $e_k$ term: \begin{align*} u\times v = \epsilon_{ijk} u_j v_k e_i, \end{align*} taking the usual basis $e_1, e_2, e_3$ of $\mathbb{R}^3$. For fixed $i$, the $i$th component of $a\times (b\times c)$ is \begin{align*} a\times (b\times c) = \epsilon_{ijk} a_j (b\times c)_k = \epsilon_{ijk}\epsilon_{klm}\, a_j b_l c_m \end{align*} as stated. Furthermore, \begin{align*} \epsilon_{ijk}\epsilon_{klm} = \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}, \end{align*} so \begin{align*} \epsilon_{ijk}\epsilon_{klm}\, a_j b_l c_m &= \delta_{il}\delta_{jm} a_j b_l c_m - \delta_{im}\delta_{jl} a_j b_l c_m = a_j b_i c_j - a_j b_j c_i, \end{align*} using the fact that $\delta_{ij} u_i v_j = u_i v_i$. (This is more or less the definition of $\delta$.) But \begin{align*} a_j b_i c_j - a_j b_j c_i = (a.c)b_i - (a.b)c_i, \end{align*} from which the result follows.
$$\delta_{ij}v_j = \delta_{ji} v_j = v_i$$ because $$\delta_{ij} = \delta_{ji} = \begin{cases} 0, & i\ne j \\ 1, & i=j\end{cases}$$ so the sum $\sum_j \delta_{ij}v_j$ is ONLY NONZERO when $i=j$. That is $$\sum_j \delta_{ij}v_j = \delta_{i1}v_1 + \delta_{i2}v_2 + \cdots + \delta_{ii}v_i + \cdots + \delta_{in}v_n = 0v_1 + 0v_2 + \cdots + 1v_i + \cdots + 0v_n = v_i$$
Knowing that, we see that \begin{align}(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})a_jb_lc_m &= \delta_{il}\delta_{jm}a_jb_lc_m-\delta_{im}\delta_{jl}a_jb_lc_m \\ &= a_j(\delta_{il}b_l)(\delta_{jm}c_m) - a_j(\delta_{jl}b_l)(\delta_{im}c_m) \\ &= a_j(b_i)(c_j)-a_j(b_j)(c_i)\end{align}