Exercise 1

At first we prove \[\left( \vec{V} \times \vec{A} \right)_i = \sum_j \sum_k \epsilon_{ijk} V_j A_k \]

For \(i=1\) all expressions are equal to zero, except the expressions with \(\epsilon_{123} = +1\) and \(\epsilon_{132} = -1\): \[ \left( \vec{V} \times \vec{A} \right)_1 = V_2 A_3 - V_3 A_2 \]

Similarly, for \(i=2\), only the expressions with \(\epsilon_{213} = -1\) and \(\epsilon_{231} = +1\) are not equal to zero: \[ \left( \vec{V} \times \vec{A} \right)_2 = - V_1 A_3 + V_3 A_1 \]

And for \(i=3\): \[ \left( \vec{V} \times \vec{A} \right)_3 = V_1 A_2 - V_2 A_1 \]

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Now we prove the equation: \[ V_i A_j - V_j A_i = \sum_k \epsilon_{ijk} \left( \vec{V} \times \vec{A} \right)_k \]

It is easy to see that for \(i=j\) the left-hand side of the equation is equal to zero. The right-hand side equals zero, too, because the indices in the Levi-Civita symbol are doubled.

For \(i \neq j\), only the term with \(k \ne i\) and \(k \ne j\) survives in the sum on the right-hand side, other terms are equal to zero due to double indices of the Levi-Civita symbol. For \(i= 1\) and \(j = 2\) you get the \(k=3\) component of the cross product. If \(i\) and \(j\) are swapped, the sign of \(\epsilon_{ijk}\) is reversed.

Analogous considerations also apply to the cases \(i=1\) and \(j= 3\) or \(i=2\) and \(j = 3\).