# 15.2: D.2 The \(\epsilon\)-\(\delta\) relation

- Page ID
- 18110

As was stated without proof in section 3.3.7, the alternating tensor is related to the 2nd-order identity tensor by

\[\varepsilon_{i j k} \varepsilon_{k l m}=\delta_{i l} \delta_{j m}-\delta_{i m} \delta_{j l}.\label{eqn:1}\]

The easiest way to convince yourself of this is to try a few tests. First, set \(i\) = \(j\) and verify that the right-hand side is zero, as it should be. Then try interchanging \(i\) and \(j\) and check that the right-hand side changes sign, as it should. The same tests work with \(l\) and \(m\). To remember Equation \(\ref{eqn:1}\), note that the first \(\delta\) on the right-hand side has subscripts \(i\) and \(l\); these are the first free indices of the two \(\varepsilon\)’s on the left-hand side. After this, the remaining pairs of indices fall into place naturally.