Skip to main content
Engineering LibreTexts

15.1: D.1 A completely antisymmetric, three dimensional array

  • Page ID
    18109
  • We will start by recalling a few important aspects of antisymmetry in ordinary matrices. The matrix shown in Figure \(\PageIndex{1}\) illustrates the idea of antisymmetry. If you flip the matrix about its main diagonal, you get the transpose. Now notice that every element in the transpose is the negative of the corresponding element in the original matrix. So \(\underset{\sim}{A}^T=-\underset{\sim}{A}\), or in index notation, \(A_{ji}=-A_{ij}\).

    clipboard_ed624024b44db639b7299565539c8f6b8.png
    Figure \(\PageIndex{1}\): (a) Example of an antisymmetric two-dimensional matrix. Transposing the matrix changes the sign of every element.

    A property shared by every antisymmetric matrix is that the elements on the main diagonal are all zero. That is obvious if you think about it: transposing the matrix does not change those elements, so if transposing the matrix changes the sign, then those elements can only be zero (because only zero is its own negative). So there are an infinite number of antisymmetric matrices, but \(A_{11}\) = \(A_{22}\) = \(A_{33}\) = 0 in all of them.

    Now imagine that there exists a three-dimensional array that has the property of complete asymmetry. We will call it \(\underset{\sim}{\varepsilon}\). Two additional properties of \(\underset{\sim}{\varepsilon}\) follow from complete antisymmetry.

    1. First, any element with two equal indices must be zero. The logic is the same as in the case of the two-dimensional matrix. Interchanging two indices changes the sign, but if the two indices are identical, the interchange makes no difference. So the value of the element equals its own negative and must therefore be zero.
    2. A second property that results from complete antisymmetry is that cyclic permutations1 of the indices have no effect. For example, starting with \(\varepsilon_{ijk}\) (where \(i\), \(j\) and \(k\) are any combination of 1,2 and 3), move the third index back to the first position and shift the other two indices one place to the right. The result is \(\varepsilon_{kij}\). Now, is the element \(\varepsilon_{kij}\) related to the original element, \(\varepsilon_{ijk}\)? Yes. We can tell this because we can recover the original ordering of the indices by making two successive interchanges, each of which changes the sign: \[\varepsilon_{kij}=-\varepsilon_{ikj}=\varepsilon_{ikj}\] The result is \(\varepsilon_{ijk}\), just what we started with. This shows that \(\varepsilon_{ijk}\) is invariant under any cyclic permutation of its indices.

    At this stage we know some things that must be true about this hypothetical array if it exists, but does it? If so what does it look like? We will now deduce the specific form of \(\underset{\sim}{\varepsilon}\) in three steps.

    • The integers 1, 2 and 3 have 27 combinations, only six of which are non-repeating: 123, 312, 231, 213, 321, and 132. The corresponding six elements of \(\underset{\sim}{\varepsilon}\) are the only ones that can be nonzero (by property 1 above).
    • The first three combinations, 123, 312 and 231, are cyclic permutations, and so are the remaining three. Because cyclic permutations make no difference (property 2 above), \[\varepsilon_{123}=\varepsilon_{312}=\varepsilon_{231},\label{eqn:1}\] and \[\varepsilon_{213}=\varepsilon_{321}=\varepsilon_{132},\label{eqn:2}\] So there are only two different nonzero values that elements of \(\underset{\sim}{\varepsilon}\) can have.
    • Finally, note that these two values must be additive inverses. For example, consider the first member of each triplet: 123 and 213. These are related by an exchange of the first and second indices, and therefore \[\varepsilon_{213}=-\varepsilon_{123}.\label{eqn:3}\] So, if we choose a value for \(\varepsilon_{123}\), we can deduce the values of all the other elements.

    To obtain the Levi-Civita tensor, we make the simplest choice \(\varepsilon_{123}\) = 1. We then have

    \[\varepsilon_{123}=\varepsilon_{312}=\varepsilon_{231}=1,\label{eqn:4}\]

    \[\varepsilon_{213}=\varepsilon_{321}=\varepsilon_{132}=-1,\label{eqn:5}\]

    and all other elements are zero. In summary:

    \[\varepsilon_{i j k}=\left\{\begin{array}{cc}
    1, & \text { if } i j k=123,312,231, \\
    -1, & \text { if } i j k=213,321,132, \\
    0, & \text { otherwise. }
    \end{array}\right.\label{eqn:6}\]

    Note that every completely antisymmetric, three-dimensional array must be proportional to \(\underset{\sim}{\varepsilon}\), i.e., equal to \(\underset{\sim}{\varepsilon}\) times some scalar. To see this, recall that we chose \(\varepsilon_{123}\) = 1 arbitrarily. What if we had chosen \(\varepsilon_{123}\) = 2? The resulting array would be exactly the same except multiplied by 2.

    1Recall that cyclic permutations are accomplished by moving the final value to the beginning, or the first value to the end, as shown in Figure 3.3.3. Interchanging any two values changes a cyclic to a non-cyclic permutation, or vice versa.