Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 3 / 5E / Solution

Using Einstein summation convention and the contraction of two Levi-Civita symbols,This proves the vector triple product identity directly from the Levi-Civita symbol and Kronecker delta.

Define the scalar triple product by $[a,b,c]=a\cdot (b\times c)$. Its expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by cyclically relabeling the three indices: a three-cycle has positive sign. ThusIn particular, let $\Delta =[e_1,e_2,e_3]$. Since the three vectors form a basis, their determinant $\Delta$ is nonzero, so their reciprocal basis vectors are well defined. Cyclic invariance and vanishing of a scalar triple product with repeated vectors giveTo calculate the new scalar triple product explicitly, apply the already proved vector triple product identity with $A=e_3\times e_1$:Consequentlyso the reciprocal vectors also form a basis. Taking the reciprocal basis of that basis now givesThe same cyclic calculation gives ${\hat{\hat{e}}}_2=e_2$ and ${\hat{\hat{e}}}_3=e_3$, proving reciprocal basis involution.

Finally, write $V=V_1{e}_1+V_2{e}_2+V_3{e}_3$. Taking a dot product with ${\hat{e}}_j$ isolates $V_j$. Applying the same argument to the reciprocal basis, whose reciprocal is the original basis, gives the two sets of coordinates:No orthonormality assumption is needed; the reciprocal basis is what makes these coordinate-extraction formulas work.