Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 1 / 1C (Algebra and Geometry)

The two free indices are the row and column indices of a matrix. The Kronecker delta has value one on its diagonal and zero off it, so the answer is the $3\times 3$ identity matrix $I$.

This is the incorrectly formed expression. In the Einstein summation convention, an index in a term is either free and occurs once, or summed and occurs twice. Here $i$ occurs three times in the same product, so it cannot consistently be classified as either. Renaming a dummy index before multiplying would give a valid different expression, such as ${\delta }_{kk}{\delta }_{ij}=3{\delta }_{ij}$, but does not make the printed expression valid.

The contracted Kronecker delta is ${\delta }_{ll}=3$. The other scalar contraction is $a_i{C}_{ij}{b}_j={\mathbf{a}}^{T}C\mathbf{b}$, while $C_{ik}{d}_i$ is the $k$th component of the matrix transpose applied to $\mathbf{d}$. Hence the resulting vector isBoth terms have the same single free index $k$, as required by the Einstein summation convention.

Compare with $(\mathbf{b}\times \mathbf{a}{)}_i={ϵ}_{ijk}{b}_j{a}_k$. The scalar factors commute, so the answer isThe order of the contracted indices matters because the Levi-Civita symbol is antisymmetric.

Interchange the two dummy indices $j,k$. The product $a_j{a}_k$ is symmetric under this interchange, while the Levi-Civita symbol changes sign. Thus the contraction equals its negative and is the zero vector, equivalently $\mathbf{a}\times \mathbf{a}=0$.

For the unheaded continuation, write the vector triple product asThe Levi-Civita symbol contraction is ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$. One way to check its sign is to fix distinct $i,j$: the only nonzero term has $k$ the remaining index, and the pairs $(l,m)=(i,j),(j,i)$ contribute $+1,-1$ respectively. When $i=j$, both sides vanish. Contracting with the vectors now gives both requested notations:If $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, then $\mathbf{a}\cdot \mathbf{b}=0$. With the specified $\mathbf{c}$, the perpendicularity of the cross product gives $\mathbf{a}\cdot \mathbf{c}=|\mathbf{a}{|}^2$. ThereforeThe same answer covers the degenerate case in which one of these two vectors is zero.