Levi-Civita symbol, cross-products and determinants

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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.

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