The Levi-Civita tensor is $\epsilon_{ijkl}$. It goes by many names Levi-Civita epsilon, permutation symbol or completely antisymmetric tensor. The four indices tell us that this is a rank four tensor. When we work in four dimensions each index can take four values 1, 2, 3 or 4. Thus the number of components is 4$\times$4$\times$4$\times$4=256. Note that the number of components equals four to the fourth power or dimension to the power of rank. To list some of these components for illustration we mention $\epsilon_{1234}$, $\epsilon_{3241}$, $\epsilon_{1244}$ and $\epsilon_{1111}$. The 256 components are classified into three mutually exclusive groups:

(i) The four indices are different and are an even permutation of 1, 2, 3, 4 e.g. $\epsilon_{2143}$

(ii) The four indices are different and are an odd permutation of 1, 2, 3, 4 e.g. $\epsilon_{2134}$

(iii) The four indices are not all different e.g. $\epsilon_{1133}$

Now there are 4!=24 permutations of 1, 2, 3, 4 and 12 of them are even and twelve odd. So there are 12 components in the first group, 12 in the second and 234 in the third. How are we to determine the value of so many components? Amazingly all these values are set by a simple and elegant rule namely that the tensor is completely antisymmetric with respect to all indices. This just mean that if we swap any two indices the value of the tensor is negated for example

$$\epsilon_{1234}=\mathbin{-}\epsilon_{1243}=\epsilon_{2143}=\mathbin{-}\epsilon_{2413}=\epsilon_{4213}\cdots$$

This means all the components in the first group have the same value which we call $a$. This then means that all the components in the second group take the value $\mathbin{-}a$. Complete asymmetry means that all components in the third group are zero. To see this consider $\epsilon_{2114}$. If we swap the ones the tensor is negated by asymmetry but of course the tensor still remains $\epsilon_{2114}$ after swapping. Thus

$$\epsilon_{2114}=\mathbin{-}\epsilon_{2114}$$

$$\epsilon_{2114}+\epsilon_{2114}=0$$

$$2\epsilon_{2114}=0$$

$$\epsilon_{2114}=0$$

All that remains is to choose a value of $a$. The choice that is almost always made is $a=1$ and with this we can state the formal definition of the Levi Civita Tensor.

$$\epsilon_{ijkl}=\begin{cases}1&\text{if indices are an even permutation of 1, 2, 3, 4}\\\mathbin{-}1&\text{if indices are an odd permutation of 1, 2, 3, 4}\\0&\text{if the indices are not all different}\end{cases}$$

Now what to make of the expression 

$$\epsilon_{ijkl}\epsilon_{ijkl}$$

This expression has four pairs for dummy indices and is thus hiding four summations. Writing out the sum over $i$ we get

$$\epsilon_{1jkl}\epsilon_{1jkl}+\epsilon_{2jkl}\epsilon_{2jkl}+\epsilon_{3jkl}\epsilon_{3jkl}+\epsilon_{4jkl}\epsilon_{4jkl}$$

So we have four terms at this point but if write out the sum over $j$ each of these terms balloons into four terms giving us 16 in all. Thus we see when we go on to write out the sum over $k$ we would have 64 terms and when we end with $l$ we have a 256 term summation. Fortunately in this very long summation 234 terms are $0\times0=0$, 12 terms are $1\times1=1$ and 12 are $\mathbin{-}1\times\mathbin{-}1=1$. Thus we see that 

$$\epsilon_{ijkl}\epsilon_{ijkl}=24$$

We now prove this by contracting four corresponding pairs of indices in the expression $\epsilon_{ijkl}\epsilon_{pqrs}$ i.e. we contract $l$ with $s$, $k$ with $r$, $j$ with $q$ and $i$ with $p$. We state without proof the following 

$$\epsilon_{ijkl}\epsilon_{pqrs}=\begin{vmatrix}\delta_{ip}&\delta_{iq}&\delta_{ir}&\delta_{is}\\\delta_{jp}&\delta_{jq}&\delta_{jr}&\delta_{js}\\\delta_{kp}&\delta_{kq}&\delta_{kr}&\delta_{ks}\\\delta_{lp}&\delta_{lq}&\delta{lr}&\delta_{ls}\end{vmatrix}$$

Doing a Laplace development from the first row gives (we won’t write out the entire thing because 4×4 determinants are very lengthy)

$$\epsilon_{ijkl}\epsilon_{pqrs}=\delta_{ip}[\delta_{jq}(\delta_{kr}\delta_{ls}\mathbin{-}\delta_{ks}\delta_{lr})\mathbin{-}\delta_{jr}(\delta_{kq}\delta_{ls}\mathbin{-}\delta_{ks}\delta_{lq})+\delta_{jq}(\delta_{kq}\delta_{lr}\mathbin{-}\delta_{kr}\delta_{lq})]\mathbin{-}\delta_{iq}[\cdots]+\delta_{ir}[\cdots]\mathbin{-}\delta_{is}[\cdots]$$

Now we contract $l$ and $s$ i.e. we set $l=s$ and sum. Keeping in mind two rules i.e. that $\delta_{ls}=\delta_{ll}=\delta_{11}+\delta_{22}+\delta_{33}+\delta_{44}=1+1+1+1=4$ and $\delta_{ks}\delta_{lr}=\delta_{kl}\delta_{lr}=\delta_{kr}$ we make use of these rules repeatedly to get

$$\epsilon_{ijkl}\epsilon_{pqrl}=\delta_{ip}(\delta_{jq}\delta_{kr}\mathbin{-}\delta_{kq}\delta_{jr})\mathbin{-}\delta_{iq}(\delta_{jp}\delta_{kr}\mathbin{-}\delta_{jr}\delta_{kp})+\delta_{ir}(\delta_{jp}\delta_{kq}\mathbin{-}\delta_{jq}\delta_{kp})=\begin{vmatrix}\delta_{ip}&\delta_{iq}&\delta_{ir}\\\delta_{jp}&\delta_{jq}&\delta_{jk}\\\delta_{kp}&\delta_{kq}&\delta_{kr}\end{vmatrix}$$

Next we contract $k$ with $r$ and thus replace every $r$ with $k$ to get

$$\epsilon_{ijkl}\epsilon_{pqkl}=\delta_{ip}(\delta_{jq}\delta_{kk}\mathbin{-}\delta_{jk}\delta_{kq})\mathbin{-}\delta_{iq}(\delta_{jp}\delta_{kk}\mathbin{-}\delta_{jk}\delta_{kp})+\delta_{jp}\delta_{ik}\delta_{kq}\mathbin{-}\delta_{jq}\delta_{ik}\delta_{kp}$$

Using our two rules we get

$$\epsilon_{ijkl}\epsilon_{pqkl}=\delta_{ip}(4\delta_{jq}\mathbin{-}\delta_{jq})\mathbin{-}\delta_{iq}(4\delta_{jp}\mathbin{-}\delta_{jp})+\delta_{jp}\delta_{iq}\mathbin{-}\delta_{jq}\delta_{ip}=2(\delta_{ip}\delta_{jq}\mathbin{-}\delta_{iq}\delta_{jq})=2\begin{vmatrix}\delta_{ip}&\delta_{iq}\\\delta_{jp}&\delta_{jq}\end{vmatrix}$$

Repeating the process we contract $j$ with $q$ and replace every instance of $q$ with $j$

$$\epsilon_{ijkl}\epsilon_{pjkl}=2(\delta_{ip}\delta_{jj}\mathbin{-}\delta_{ij}\delta_{jp})=2(4\delta_{ip}\mathbin{-}\delta_{ip})=2\cdot3\delta_{ip}=6\delta_{ip}$$

Finally we contract $i$ with $p$ and by setting $p=i$ to get

$$\epsilon_{ijkl}\epsilon_{ijkl}=6\delta_{ii}=6\times4=24$$

Next we note that in three dimensions two rank three Levi Civita tensor is written as the following determinant

$$\epsilon_{ijk}\epsilon_{pqr}=\begin{vmatrix}\delta_{ip}&\delta_{iq}&\delta_{ir}\\\delta_{jp}&\delta_{jq}&\delta_{jr}\\\delta_{kp}&\delta_{kq}&\delta_{kr}\end{vmatrix}=\delta_{ip}(\delta_{jq}\delta_{kr}\mathbin{-}\delta_{jr}\delta_{kq})\mathbin{-}\delta_{iq}(\delta_{jp}\delta_{kr}\mathbin{-}\delta_{jr}\delta_{kp})+\delta_{ir}(\delta_{jp}\delta_{kq}\mathbin{-}\delta_{jq}\delta_{kp})$$

We now contract $p$ with $i$ remembering that $\delta_{ip}=\delta_{ii}=3$ and $\delta_{iq}\delta_{jr}\delta_{kp}=\delta_{ki}\delta_{iq}\delta_{jr}=\delta_{kq}{jr}$ etc

$$\epsilon_{ijk}\epsilon_{iqr}=3(\delta_{jq}\delta_{kr}\mathbin{-}\delta_{jr}\delta_{kq})\mathbin{-}\delta_{kr}\delta_{jq}+\delta_{kq}\delta_{jr}+\delta_{kq}\delta_{jr}\mathbin{-}\delta_{jq}\delta_{kr}$$

Hence we get the very important formula which we will put to great use

$$\epsilon_{ijk}\epsilon_{iqr}=\delta_{jq}\delta_{kr}\mathbin{-}\delta_{jr}\delta_{kq}$$