Levi-Civita symbol, cross-products and determinants

PREVIOUS: Einstein's summation convention

The dot product of two vectors $\textbf{A}$ and $\textbf{B}$ can be written, using Einstein's convention, as $\textbf{A} \cdot \textbf{B} = A_{i}B_{i}$. The cross-product, on the other hand, is written in terms of Levi-Cevita symbol. The $i$ the component of $\textbf{A} \times \textbf{B}$ is:

$$(\textbf{A} \times \textbf{B})_{i} = \sum_{j,k=1}^{3}\epsilon_{ijk} A_{j}B_{k} = \epsilon_{ijk} A_{j}B_{k}$$

Here, $\epsilon_{ijk}$ is known as Levi-Cevita symbol or a permutation symbol. Note that the indices $j$ and $k$ are repeated or dummy, hence summed over, but the index $i$ is free. Therefore, as discussed in the previous section, the term $\epsilon_{ijk} A_{j}B_{k}$ is like the $i$th component of a vector. However, it is common to write $\textbf{A} \times \textbf{B} = \epsilon_{ijk} A_{j}B_{k}$ for notational brevity; this doesn't create any confusion as the free index $i$ makes it obvious that the term $\epsilon_{ijk} A_{j}B_{k}$ refers to the $i$th component of the cross-product and not the entire expression of the cross-products. So, keeping in mind this, henceforth we'll write $$\textbf{A} \times \textbf{B} = \epsilon_{ijk} A_{j}B_{k}$$

Now let's take a look at the properties of the Levi-Civita symbol, $\epsilon_{ijk}$. I begin by showing you what this object looks like in 2-, 3- and 4-dimensions instead of stating the definition right away.

In two dimensional space where indices can takes values in the range {1,2}, the Levi-Civita symbol has the following property:
$$\epsilon_{ij} = \left\lbrace \begin{array}{l} +1 & \mbox{if} \,\, (i,j) = (1,2) \\ -1 & \mbox{if} \,\, (i,j) = (2,1) \\ 0 & \mbox{if} \,\, i=j \end{array} \right.$$
Thus, on interchanging the values of indices $i$ and $j$, the sign of the element $\epsilon_{ij}$ changes; and when both indices take the same values, the element $\epsilon_{ii}$ or $\epsilon_{jj}$ is 0. We can arrange these elements into an array in the following way:

$$\epsilon_{ij} = \begin{pmatrix} \epsilon_{11} & \epsilon_{12} \\ \epsilon_{21} & \epsilon_{22} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$
In a three dimensional space, the Levi-Civita symbol is characterized by three indices: $\epsilon_{ijk}$, where each index takes the value in the range $\{1,2,3\}$. The sign of an element of $\epsilon_{ijk}$ depends on whether the number of steps it took to reach the specific indices of that element from $ijk=123$ is odd or even. If the number of steps is even, $\epsilon_{ijk}$ is +1, and -1 if the number of steps is odd. For instance, if we define $\epsilon_{123}=+1$, then $\epsilon_{132}=-1$ because it took 1 step (interchanging 2 and 3) to reach $ijk=132$ from $ijk=123$. $\epsilon_{312}=+1$ because it involves two steps: $ijk=123 \, \rightarrow 132 \, \rightarrow 312$; and $\epsilon_{312}=-1$ because it involves three steps: $ijk=123 \, \rightarrow 132 \, \rightarrow 312 \, \rightarrow 312$. $\epsilon_{ijk}=0$ when any two or all indices take the same values: $\epsilon_{333} = \epsilon_{112} = \epsilon_{223} = 0$Following this scheme, we can determine values of all elements of $\epsilon_{ijk}$:

$$\begin{array}{l} \epsilon_{123} &=& \epsilon_{312} &=& \epsilon_{231} &=& +1 \\ \epsilon_{132} &=& \epsilon_{321} &=& \epsilon_{213} &=& -1 \\ \epsilon_{111} &=& \epsilon_{222} &=& \epsilon_{333} &=& \epsilon_{112} &=& \epsilon_{232} = ... = 0 \end{array}$$
Another way to express the above results is to talk in terms of odd and even permutations of ${123}$.

$$\epsilon_{ijk} = \left\lbrace \begin{array}{l} +1 & \mbox{for even permutations of} \,\, (i,j,k): (1,2,3), (3,1,2), (2,3,1) \\ -1 & \mbox{for odd permutations of} \,\, (i,j,k): (1,3,2), (3,2,1), (2,1,3) \\ 0 & \mbox{if} \,\, i=j \,\, \text{or} \,\, j=k \,\, \text{or} \,\, i=k \,\, \text{or} \,\, i=j=k: (1,1,3), (2,1,2), (1,1,1), \mbox{etc.} \end{array} \right.$$

Let's now evaluate the cross-product of two vectors $\textbf{A}=(A_{1},A_{2},A_{3})$ and $\textbf{B}=(B_{1},B_{2},B_{3})$ using the index method. As stated earlier in the article, the $i$th component of the cross-product $\textbf{A} \times \textbf{B}$ is given as:
$$(\textbf{A} \times \textbf{B})_{i} = \epsilon_{ijk} A_{j}B_{k}$$ Remember that the $x,y$ and $z$ components of the cross-product correspond to the indices $1,2$ and $3$ respectively. Therefore, $$\begin{array}{l} (\textbf{A} \times \textbf{B})_{y} &=& (\textbf{A} \times \textbf{B})_{2} & \\ &=& \epsilon_{2jk} A_{j}B_{k} & (\mbox{\(j,k\) are dummy indices, hence need to be summed over}) \\ &=& \epsilon_{213} A_{1}B_{3} + \epsilon_{231} A_{3}B_{1} & (\mbox{the only values \(j,k\) can take; \(\epsilon_{ijk}\) vanished for the rest}) \\ &=& - A_{1}B_{3} + A_{3}B_{1} & \\ &=& - A_{x}B_{z} + A_{z}B_{x} & \\ &=& - (A_{x}B_{z} - A_{z}B_{x}) & \end{array}$$ Similarly, we can easily show that
$$\begin{array}{l} (\textbf{A} \times \textbf{B})_{x} &=& A_{y}B_{z} - A_{z}B_{y} \\ (\textbf{A} \times \textbf{B})_{z} &=& A_{x}B_{y} - A_{y}B_{x} \end{array}$$

The Levi-Civita symbol also comes in handy when evaluating a determinant. For example:
$$\begin{array}{l} D &=& \left| \begin{array}{l} A_{1} & A_{2} & A_{3} \\ B_{1} & B_{2} & B_{3} \\ C_{1} & C_{2} & C_{3} \end{array} \right| \\ &=& \epsilon_{ijk} A_{i}B_{j}C_{k} \\ &=& \epsilon_{123} A_{1}B_{2}C_{3} + \epsilon_{132} A_{1}B_{3}C_{2} + \\ & & \epsilon_{231} A_{2}B_{3}C_{1} + \epsilon_{213} A_{2}B_{1}C_{3} + \\ & & \epsilon_{312} A_{3}B_{1}C_{1} + \epsilon_{321} A_{3}B_{1}C_{1} \\ &=& A_{1}B_{2}C_{3} - A_{1}B_{3}C_{2} + \\ & & A_{2}B_{3}C_{1} - A_{2}B_{1}C_{3} + \\ & & A_{3}B_{1}C_{2} - A_{3}B_{2}C_{1} \\ &=& A_{1}(B_{2}C_{3} - B_{3}C_{2}) - A_{2}(B_{1}C_{3} - B_{3}C_{1}) + A_{3}(B_{1}C_{2} - B_{2}C_{1}) \,\,\,\, \mbox{in more familiar form} \end{array}$$
In general, in case of an $N \times N$ determinant:
$$\left| \begin{array}{l} A_{1} & A_{2} & A_{3} & \cdot \cdot \cdot & A_{N} \\ B_{1} & B_{2} & B_{3} & \cdot \cdot \cdot & B_{N} \\ \cdot & & & & \cdot \\ \cdot & & & & \cdot \\ \cdot & & & & \cdot\\ X_{1} & X_{2} & X_{3} & \cdot \cdot \cdot & X_{N} \end{array} \right| = \epsilon_{ijk...n} \, A_{i}B_{j}C_{k}...X_{n}$$
where each index $(i,j,k,...,n)$ takes values in the range $\{1,2,3,...,N\}$. You will appreciate the importance and usefulness of the Levi-Civita symbol when you attempt to solve a 4- or higher-dimensional determinants or cross-products.To solve problems in higher dimensions, all one has to do is to keep track of indices. As an exercise, try to solve a $4 \times 4$ determinant using this method.

NEXT: Using the method of index notation