# 15.2: D.2 The $$\epsilon$$-$$\delta$$ relation
$\varepsilon_{i j k} \varepsilon_{k l m}=\delta_{i l} \delta_{j m}-\delta_{i m} \delta_{j l}.\label{eqn:1}$
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.