15.2: D.2 The ε-δ relation

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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.

15.2: D.2 The ε-δ relation is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.