Levi Civita and Kronecker Delta I've been working on some quantum mechanics problems and arrived to this one where I have to deal with subscripts. I got stuck doing this:
I have $\epsilon_{imk}\epsilon_{ikn}=\delta_{mk}\delta_{kn}-\delta_{mn}\delta_{kk}$. But then I went to check and $\delta_{mk}\delta_{kn}-\delta_{mn}\delta_{kk}$ is equal to $\delta_{mn}-3\delta_{mn}$. Why is that so?
Thank you in advance.
 A: Levi Civita symbol $\epsilon_{ikl}$ is defined as it follows
$$\epsilon_{ikl}  = \left\{
\begin{array}{cl}
1 & if\quad i\neq k\neq l\quad and \quad even \quad permutation\\
-1& if\quad i\neq k\neq l\quad and \quad odd\quad permutation\\
0 & otherwise
\end{array}\right.$$
From this definition, let's start with the contraction of $\epsilon_{ijk}$ in its first index:
$$\epsilon_{ikl}\epsilon_{imn}=\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm}\tag1$$
Where $\delta_{ik}$ is the Kronecker delta (identity matrix), a symmetric isotropic tensor and it is defined as it follows
$$\delta_{ik}  = \left\{
\begin{array}{cl}
1 & if\quad i= k\\
0 & otherwise
\end{array}\right.$$
Contracting $(1)$ once more, $\textit{i.e.}$ multiplying it by $\delta_{km}$ we have
$$\delta_{km}\epsilon_{ikl}\epsilon_{imn}=\epsilon_{ikl}\epsilon_{ikn}=\delta_{km}(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})=\delta_{kk}\delta_{ln}-\delta_{kn}\delta_{kl}=\delta_{kk}\delta_{ln}-\delta_{ln}\tag2$$
Recall that $\delta_{lk} = \delta_{kl}$ due to symmetry properties and $\delta_{km}\delta_{km}=\delta_{kk}=\delta_{mm}$ since they are dummy indices (repeated indices indicated summation over this index).
Now comes the term $\delta_{ii}$, this quantity is a scalar, and represents the trace of the identity matrix in a n-dimensional space, therefore $\delta_{ii}=n$ and finally $(2)$ is written as it follows
$$\epsilon_{ikl}\epsilon_{ikn}=(\delta_{kk}-1)\delta_{ln}=(n-1)\delta_{ln}$$
If one contracts again, the following identity
$$\epsilon_{ikl}\epsilon_{ikl}=n(n-1)$$
In your case $n=3$.
Hope this helps you
A: Another approach is to write $$\epsilon_{ijk}=\hat x_i\cdot(\hat x_j\times \hat x_k)$$Then, we have
$$\begin{align}
\epsilon_{imk}\epsilon_{ikn}&=\hat x_i\cdot(\hat x_m\times \hat x_k)x_i\cdot(\hat x_k\times \hat x_n)\\\\
&=(\hat x_m\times \hat x_k)\cdot(\hat x_k\times \hat x_n)\\\\
&=\hat x_m \cdot (\hat x_k\times(\hat x_k\times \hat x_n))\\\\
&=\hat x_m \cdot(\delta_{nk}\hat x_k-\delta_{kk}\hat x_n)\\\\
&=\delta_{nk}\delta_{mk}-3\delta_{mn}
\end{align}$$
