Is this a correct identity for the Kronecker delta and the Alternating Tensor?

Question

If $\varepsilon_{ijk}$ is the alternating tensor and $\delta_{in}$ is the Kronecker delta, am I correct in thinking that $$ \delta_{in}\varepsilon_{ijk} = \varepsilon_{ink} $$

If not, what is the correct evaluation of this expression?

Many thanks

Answer 1

The correct formula is $$\delta_{in}\epsilon_{ijk} = \epsilon_{njk}$$

One way to know that your proposed formula is wrong is to consider what type of index each of the four are. $i$ appears twice so it is a summing index. Thus it shouldn't appear on the other side (unless it happens to be a summing index on the other side of the equation as well). But $n$, $j$, and $k$ only appear once each. Thus they are free indices. That means that must appear on both sides of the equation.

Here's something interesting: That product comes up when deriving the relationship between the triple scalar product and the $3\times 3$ determinant: $$\begin{align}a\cdot (b\times c) &= a_n\mathbf e_n\cdot \epsilon_{ijk}b_jc_k\mathbf e_i \\ &= a_n\epsilon_{ijk}b_jc_k\mathbf e_n\cdot\mathbf e_i \\ &= a_n\epsilon_{ijk}b_jc_k\delta_{ni} \\ &= \delta_{ni}\epsilon_{ijk}a_nb_jc_k \\ &= \epsilon_{njk}a_nb_jc_k \\ &= \det(a,b,c)\end{align}$$