Suffix Notation
Suffix notation is a concise way for expressing vectors (and tensors). It is generally easier to manipulate vectors with that, together with Einstein’s summation convention.
Kronecker Delta
Definition. The Kronecker delta $\delta_{ij}$ is a tensor, defined by, in matrix form
\[\begin{pmatrix} \delta_{11} & \delta_{12} & \delta_{13} \\ \delta_{21} & \delta_{22} & \delta_{23} \\ \delta_{31} & \delta_{32} & \delta_{33} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]
Property. $\delta_{ij}$ is symmetric, i.e.
\[\delta_{ij} = \delta_{ji}\]
Property. $\delta_{ii} = 3 (= n)$.
Property. The delta function behaves like a suffix switching quantity, i.e.
\[\begin{align*} \delta_{ij} a_j &= a_i \\ \delta_{ij} \delta_{jk} &= \delta_{ik} \\ \delta_{ij} A_{jk} &= A_{ik} \\ \delta_{ij} \varepsilon_{jkl} &= \varepsilon_{ikl} \\ \delta_{ij} \sigma_{jklmno...} &= \sigma_{iklmno...} \end{align*}\]Hence,
\[\delta_{ij} \delta_{ji} = \delta_{ii} = 3 (= n)\]
Levi-Civita Symbol / Alternating Tensor
Definition. The Levi-Civita symbol $\varepsilon_{ijk}$ or alternating tensor is the set of $27$ quantities such that
\[\varepsilon_{ijk} = \begin{cases} +1 & \text{if } ijk \text{ is an even permutation} \\ -1 & \text{if } ijk \text{ is an odd permutation} \\ 0 & \text{otherwise (i.e. repeated suffices)} \end{cases}\]Therefore, the non-zero components are
\[\begin{align*} \varepsilon_{123} = \varepsilon_{231} = \varepsilon_{312} &= 1 \\ \varepsilon_{213} = \varepsilon_{132} = \varepsilon_{321} &= -1 \\ \end{align*}\]Further
\[\varepsilon_{ijk} = \varepsilon_{jki} = \varepsilon_{kij} = -\varepsilon_{jik} = -\varepsilon_{ikj} = -\varepsilon_{jki}\]
Proposition. For a symmetric tensor $s_{ij}$,
\[\varepsilon_{ijk} s_{ij} = 0\]Hence,
\[\varepsilon_{ijk} \delta_{ij} = 0\]Proof.
By relabelling the dummy suffices, we have
\[\varepsilon_{ijk} s_{ij} = \varepsilon_{jik} s_{ji}\]As $\varepsilon_{jik} = -\varepsilon_{ijk}$ and $s_{ji} = s_{ij}$, we have
\[\varepsilon_{jik} s_{ji} = -\varepsilon_{ijk} s_{ij}\]As $\varepsilon_{ijk} s_{ij} = -\varepsilon_{ijk} s_{ij}$, we conclude
\[\varepsilon_{ijk} s_{ij} = 0\]and so as
\[\varepsilon_{ijk} \delta_{ij} = 0\]
Proposition. For no repeated indices,
\[\varepsilon_{ijk} \varepsilon_{pqr} = \begin{vmatrix} \delta_{ip} & \delta_{iq} & \delta_{ir} \\ \delta_{jp} & \delta_{jq} & \delta_{jr} \\ \delta_{kp} & \delta_{kq} & \delta_{kr} \\ \end{vmatrix}\]
Proposition. For one repeated index,
\[\epsilon_{ijk}\epsilon_{ipq} = \delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp}\]
Proposition. For two repeated indices,
\[\varepsilon_{ijk} \varepsilon_{ijp} = 2 \delta_{kp}\]Proof.
When $k \not = p$, both side vanishs. When $k = p$, there are two ways left to choose $i, j$.
Proposition. For three repeated indices,
\[\varepsilon_{ijk} \varepsilon_{ijk} = 6 (= n!)\]Proof.
As for any distinct values of $i, j, k$, the “product” is $1$, and there are $3!$ choices.
Basis Vectors
Let $\mathbf{e_1} = \mathbf{i}$, $\mathbf{e_2} = \mathbf{j}$ and $\mathbf{e_3} = \mathbf{k}$ be the standard basis vectors of $\mathbb{R}^3$. Be noted that $\mathbf{e}_i$ means one of the basis vector for $i = 1, 2, 3$ instead of $\mathbf{i}$.
Proposition. For any two of the basis vector,
\[\mathbf{e}_i \cdot \mathbf{e}_j = \delta_{ij}\]Proof.
$\mathbf{e}_i \cdot \mathbf{e}_j = 1$ iff $i = j$, matching the definition of $\delta_{ij}$.
Proposition. In terms of suffix notation,
\[\mathbf{a} \cdot \mathbf{e}_i = a_i\]
Proposition. The suffix label of basis vector is interchangeable, i.e.
\[(\mathbf{e}_j)_i = (\mathbf{e}_i)_j = \delta_{ij}\]Proof.
The $i$-th component of $\mathbf{e}_j$ is given by
\[(\mathbf{e}_j)_i = \mathbf{e}_j \cdot \mathbf{e}_i = \delta_{ij}\]Similarily,
\[(\mathbf{e}_i)_j = \mathbf{e}_i \cdot \mathbf{e}_j = \delta_{ij}\]
Vector Algebra
Proposition. The scalar/dot product is given by
\[\mathbf{a} \cdot \mathbf{b} = \delta_{ij} a_i b_j = a_i b_i = a_j b_j\]
Proposition. The vector/cross product is given by
\[(\mathbf{a} \times \mathbf{b})_i = \varepsilon_{ijk} a_j b_k\]
Proposition. The scalar triple product is given by
\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \varepsilon_{ijk} a_i b_j c_k\]Proof.
\[\begin{align*} \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) &= a_i (\mathbf{b} \times \mathbf{c})_i \\ &= \varepsilon_{ijk} a_i b_j c_k \end{align*}\]
Proposition. The vector triple product is given by
\[(\mathbf{a} \times (\mathbf{b} \times \mathbf{c}))_i = a_jb_ic_j - a_jb_jc_i\]Proof.
\[\begin{align*} \mathbf{a} \times (\mathbf{b} \times \mathbf{c})_i &= \epsilon_{ijk} a_j (\mathbf{b} \times \mathbf{c})_k \\ &= \epsilon_{ijk} a_j \epsilon_{klm} b_l c_m \\ &= -\epsilon_{kji} \epsilon_{klm} a_j b_l c_m \\ &= -(\delta_{jl}\delta_{im} - \delta_{jm}\delta_{il}) a_j b_l c_m \\ &= a_j b_i c_j - a_j b_j c_i \\ &= ((\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c})_i \end{align*}\]
Proposition. Cauthy-Schwarz inequality can be proved by suffix notation.
Proof.
\[\begin{align*} \vert \mathbf{x} \vert^2 \vert \mathbf{y} \vert^2 - \vert x \cdot y \vert^2 &= x_ix_iy_jy_j - x_iy_ix_jy_j \\ &= {1 \over 2}x_ix_iy_jy_j + {1 \over 2}x_jx_jy_iy_i - x_iy_ix_jy_j \\ &= {1 \over 2}(x_iy_j - x_jy_i)(x_iy_j - x_jy_i) \\ &\ge 0 \end{align*}\]