Notes: Simple tensor algebra I

The vector space $\mathbb{V}$ with the operation (+) over a field of real number $\mathbb{R}$ is an abelian group with a scalar multiplication. So, suppose $\boldsymbol{x}$, $\boldsymbol{y}$ and $\boldsymbol{z} \in \mathbb{V}$, they satisfy the following conditions:

Abelian group:

• Closure:$\boldsymbol{x} + \boldsymbol{y} \in \mathbb{V}$,
• Associativity: $( \boldsymbol{x} + \boldsymbol{y} )+ \boldsymbol{z}= \boldsymbol{y} + ( \boldsymbol{x}+ \boldsymbol{z} )$,
• Identity: $\exists \boldsymbol{0} \in \mathbb{V}, \forall \boldsymbol{x} \in \mathbb{V}$ such that $\boldsymbol{0} + \boldsymbol{x} = \boldsymbol{x}$ and $\boldsymbol{x} + \boldsymbol{0} = \boldsymbol{x}$,
• Invertibility $\exists ! ( -\boldsymbol{x} ) \in \mathbb{V}, \forall \boldsymbol{x} \in \mathbb{V}$ such that $( -\boldsymbol{x} ) + \boldsymbol{x} = \boldsymbol{0}$ and $\boldsymbol{x} + ( -\boldsymbol{x} ) = \boldsymbol{0}$,
• Commutativity: $\boldsymbol{x} + \boldsymbol{y} = \boldsymbol{y} + \boldsymbol{x}$.

Scalar multiplication:

• $\forall \alpha, \beta \in \mathbb{R}$, $\forall \boldsymbol{x} \in \mathbb{V}$ such that $( \alpha \beta ) \boldsymbol{x} = \alpha ( \beta \boldsymbol{x} )$,
• $\forall \alpha, \beta \in \mathbb{R}$, $\forall \boldsymbol{x}, \boldsymbol{y} \in \mathbb{V}$ such that $\alpha ( \boldsymbol{x} + \boldsymbol{y} ) = \alpha \boldsymbol{x} + \alpha \boldsymbol{y}$ and $( \alpha + \beta ) \boldsymbol{x} = \alpha \boldsymbol{x} + \beta \boldsymbol{x}$,
• $1 \boldsymbol{x} = \boldsymbol{x}$.

This is not a tutorial, please refer to the books for tensor algebra.

Linearly independent:

For a set of non-zero vectors $\{\boldsymbol{x}_1, \boldsymbol{x}_2, \boldsymbol{x}_3, \dots, \boldsymbol{x}_n, \}$,  $\exists \alpha_1, \alpha_2, \alpha_3, \dots, \alpha_n \in \mathbb{R}$, such that,

$$\displaystyle \sum_{i=1}^n \alpha_i \boldsymbol{x}_i = \boldsymbol{0}$$

if and only if  $\alpha_1, \alpha_2, \alpha_3, \dots, \alpha_n = 0$.

Such set of the vectors $\{\boldsymbol{x}_1, \boldsymbol{x}_2, \boldsymbol{x}_3, \dots, \boldsymbol{x}_n, \}$ is called linearly independent.

Vector space dimension:

Let $\mathcal{G} = \{\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3,\dots,\boldsymbol{g}_n \}$ be a subset of a vector space $\mathbb{V}$ and $latex \alpha_i \in \mathbb{R}$, the vector,

$$\displaystyle \boldsymbol{x} = \sum_{i=1}^n \alpha_i \boldsymbol{g}_i$$

is the linear combination of the vectors $\{\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3,\dots,\boldsymbol{g}_n \}$.

With the definition of the linear combination, the vector space dimension can be defined: A set $\mathcal{G} \subset \mathbb{V}$) of linearly independent vectors is a basis of a vector space $\mathbb{V}$ if every vector in $\mathbb{V}$ is the linear combination of the elements in $( \mathcal{G}$.

Theorem: All the basis of a finite-dimensional vector space $\mathbb{V}$ contains the same number of vectors.

Theorem: For a n-dimension vector space $\mathbb{V}$, every set contains n linearly independent vectors is a basis of $\mathbb{V}$, every set contains more than n vectors is linearly dependent.

Summation convention:

Let $\mathcal{G} = \{\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3,\dots,\boldsymbol{g}_n \}$ be the basis of the vector space $\mathbb{V}$, any vector in such space can be represented with the linear combination of the basis,

$$\displaystyle \boldsymbol{x} = \sum_{i=1}^{n} x^i \boldsymbol{g}_i = x^i \boldsymbol{g}_i$$

The expression is shorten without the summation symbol. This is called the Einstein summation. As shown on the mathworld, there are three rules of this notational convention:

• Repeated indices are implicitly summed over,
• Each index can appear at most twice in any term,
• Each term must contain identical non-repeated indices.

Theorem: The representation of the vector in the vector space $\mathbb{V}$ with respect to a given basis $\mathcal{G}$ is unique.

Euclidean space:

Suppose $\boldsymbol{x}$, $\boldsymbol{y}$ and $\boldsymbol{z} \in \mathbb{V}$, then the vector space $\mathbb{V}$with inner product of the vectors satisfying the following conditions is called Euclidean space $\mathbb{E}^n$ :

• Commutativity: $\boldsymbol{x} \cdot \boldsymbol{y} = \boldsymbol{y} \cdot \boldsymbol{x}$,
• Distributivity: $\boldsymbol{x} \cdot ( \boldsymbol{y} + \boldsymbol{z}) = \boldsymbol{x} \cdot \boldsymbol{y} + \boldsymbol{x} \cdot \boldsymbol{z}$,
• Associativity with a scalar: $\alpha (\boldsymbol{x} \cdot \boldsymbol{y}) = (\alpha \boldsymbol{x}) \cdot \boldsymbol{y} = \boldsymbol{x} \cdot (\alpha \boldsymbol{y})$,
• $\boldsymbol{x} \cdot \boldsymbol{x} \ge 0, \forall \boldsymbol{x} \in \mathbb{V}$, $\boldsymbol{x} \cdot \boldsymbol{x} = 0$ if and only if $\boldsymbol{x} = 0$.

Since the inner product of the vectors in $\mathbb{E}^n$ is non-negative, the Euclidean distance or length (Norm) is defined by,

$$\displaystyle \|x\|=\sqrt{\boldsymbol{x} \cdot \boldsymbol{x}}$$

Orthonormal basis:

• Orthogonal: $\boldsymbol{x} \cdot \boldsymbol{y} = 0$,
• Normal: $\|x\|=1$.

Let $\mathcal{E} = \{\boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3, \dots, \boldsymbol{e}_n \}$ be a basis of $\mathbb{V}$, if $\forall \boldsymbol{e}_i, \boldsymbol{e}_j \in \mathcal{E}$ satisfy the preceding conditions, then the set $\mathcal{E}^n$ is called orthonormal basis, i.e.

$$\displaystyle \boldsymbol{e}_i \cdot \boldsymbol{e}_j = \delta_{ij} ,i,j = 1, 2, 3, \dots, n$$

The symbol is called Kronecker delta,

$$\displaystyle \delta_{ij}=\delta^{ij}=\delta^i_j=\begin{cases} 1 & \text{if}\ i = j\\ 0 & \text{if}\ i \ne j \end{cases}$$

Every set of linearly independent vectors in Euclidean space $\mathbb{E}^n$ can be orthogonalized and normalized with the Gram-Schmidt procedure.

Dual basis:

The dual basis of vectors is a biorthogonal system of the original basis of vectors in vector space $\mathbb{V}$, the detail information can be found in wikipedia. It is a little bit abstract, the following figure shows the original basis $\{ \boldsymbol{g}_1, \boldsymbol{g}_2 \}$ and the dual basis $\{ \boldsymbol{g}^1, \boldsymbol{g}^2 \}$ in 2D Euclidean space $\mathbb{E}^2$.

In this figure, $\boldsymbol{g}_i \cdot \boldsymbol{g}^j = \delta_i^j$. Now, extend the 2D Euclidean space $\mathbb{E}^2$ to nD Euclidean space $\mathbb{E}^n$, the basis $\mathcal{G}^ \prime = \{\boldsymbol{g}^1, \boldsymbol{g}^2, \boldsymbol{g}^3, \dots,\boldsymbol{g}^n \}$ is the dual basis to the basis $\mathcal{G} = \{\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3, \dots,\boldsymbol{g}_n \}$, if that,

$$\displaystyle \boldsymbol{g}_i \cdot \boldsymbol{g}^j = \delta_i^j \quad i,j = 1, 2, 3, \dots, n$$

Theorem: There exists a unique dual basis of each basis in an Euclidean space $\mathbb{E}^n$.

Proof: Let $\mathcal{G}^ \prime = \{\boldsymbol{g}^1, \boldsymbol{g}^2, \boldsymbol{g}^3, \dots,\boldsymbol{g}^n \}$ and $\mathcal{H}^ \prime = \{\boldsymbol{h}^1, \boldsymbol{h}^2, \boldsymbol{h}^3, \dots,\boldsymbol{g}^n \}$ be the two dual bases of the original basis $\mathcal{G} = \{\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3, \dots,\boldsymbol{g}_n \}$, recall the properties of the dual basis,

$$\displaystyle \boldsymbol{g}_i \cdot \boldsymbol{g}^j = \delta^j_i$$

$$\displaystyle \boldsymbol{g}_i \cdot \boldsymbol{h}^j = \delta^j_i$$

Expand $\boldsymbol{h}^j = h^j_k \boldsymbol{g}^k$ we get,

$$\displaystyle \delta^j_i = \boldsymbol{g}_i \cdot \boldsymbol{h}^j = \boldsymbol{g}_i \cdot (h^j_k \boldsymbol{g}^k) = h^j_k \delta_i^k = h^j_i$$

Then,

$$\displaystyle \boldsymbol{h}^j = h^j_k \boldsymbol{g}^k = \delta^j_i \boldsymbol{g}^k = \boldsymbol{g}^j$$

Now, let $\boldsymbol{g}_i$ be the basis of $\boldsymbol{g}^i$, then,

$$\displaystyle \boldsymbol{g}^i = g^{ij} \boldsymbol{g}_j$$

Expend the $\boldsymbol{g}_j$ with respect to the $\boldsymbol{g}^j$, then,

$$\displaystyle \boldsymbol{g}_j = g_{jk} \boldsymbol{g}^k$$

Thus,

$$\displaystyle \boldsymbol{g}^i = g^{ij} g_{jk} \boldsymbol{g}^k$$

$$\displaystyle \boldsymbol{0} = g^{ij} g_{jk} \boldsymbol{g}^k – \boldsymbol{g}^i = (g^{ij} g_{jk} – \delta_k^i) \boldsymbol{g}^k$$

we get,

$$\displaystyle g^{ij} g_{jk} = \delta_k^i$$

Then, the matrices $g^{ij}$ and $g_{jk}$ (i.e. $g_{ji}$) are inverse of each other.

Dual basis Properties:

For orthonormal basis of $\mathbb{E}^n$, it is a self dual vector space,

$$\displaystyle \boldsymbol{e}_i = \boldsymbol{e}^i$$

$$\displaystyle \boldsymbol{e}_i \cdot \boldsymbol{e}^j = \delta_i^j = \delta_{ij} = \boldsymbol{e}_i \cdot \boldsymbol{e}_j = \delta^{ij} = \boldsymbol{e}^i \cdot \boldsymbol{e}^j$$

For non-orthonormal basis of $\mathbb{V}$,

$$\displaystyle \boldsymbol{g}^i \cdot \boldsymbol{g}^j = \boldsymbol{g}^j \cdot \boldsymbol{g}^i = g^{ij} = g^{ji}$$

$$\displaystyle \boldsymbol{g}_i \cdot \boldsymbol{g}_j = \boldsymbol{g}_j \cdot \boldsymbol{g}_i = g_{ij} = g_{ji}$$

The inner product of vector $\boldsymbol{x}$ and $\boldsymbol{y}$,

$$\displaystyle \boldsymbol{x} \cdot \boldsymbol{y} = x^i \boldsymbol{g}_i \cdot y_j \boldsymbol{y}^j = x^i y_j \delta^j_i = x^i y_i = x^j y_j$$

Levi-Civita symbol:

Levi-Civita symbol is the permutation symbol,

$$\displaystyle \varepsilon_{ijk} = \varepsilon^{ijk} = \begin{cases} 1 & \text{ if } ijk \text{ is an even permutation }\\ -1 & \text{ if } ijk \text{ is an odd permutation } \\ 0& \text{ if otherwise } \end{cases}$$

• Even permutation: $123, 231, 312$,
• Odd permutation: $132, 321, 213$,
• Otherwise: more than two symbols are the same e.g. $112, 223, 333$.

Let $\{\boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3 \}$ be the orthonormal vectors in $\mathbb{E}^3$ and [ ] be the symbol of mixed product of three vectors,

$$\displaystyle \varepsilon_{ijk} = [\boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k], \quad i, j, k = 1,2,3$$

Denote g as the result for mixed product of basis vectors $\{\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3\}$ in $\mathbb{E}^3$,

$$\displaystyle g = [ \boldsymbol{g}_1 \boldsymbol{g}_2 \boldsymbol{g}_3] = [ \beta^i_1 \beta^j_2 \beta^k_3 \boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k] = \beta^i_1 \beta^j_2 \beta^k_3 [ \boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k] = \beta^i_1 \beta^j_2 \beta^k_3 \varepsilon_{ijk}$$

The last term $\beta^i_1 \beta^j_2 \beta^k_3 \varepsilon_{ijk}$ is the determinant of the matrix i.e. $g = \textbf{Det}[\beta^i_j]$,

Since we get,

$$\displaystyle g_{ij} = \boldsymbol{g}_i \cdot \boldsymbol{g}_j = \beta_i^k \boldsymbol{e}_k \cdot \beta_j^l \boldsymbol{e}_l = \beta_i^k \beta_j^l \delta_{kl} = \beta_i^k \beta_j^k$$

Thus,

$$\displaystyle |g_{ij}| = \textbf{Det}[g_{ij}] = \|[\beta_i^k][\beta_j^k]\| = \textbf{Det}[\beta^i_j]^2 = g^2$$

Levi-Civita symbol properties:

$$\displaystyle \varepsilon_{ijk} \varepsilon^{imn}=\delta_j^{m}\delta_k^n – \delta_j^n\delta_k^m$$

$$\displaystyle \varepsilon_{jmn} \varepsilon^{imn}=2 \delta_j^i$$

$$\displaystyle \varepsilon_{ijk} \varepsilon^{ijk}= 6$$

$$\varepsilon_{ijk} \varepsilon_{lmn} = \delta_{il} ( \delta_{jm}\delta_{kn} – \delta_{jn} \delta_{km}) – \delta_{im} ( \delta_{jl} \delta_{kn} – \delta_{jn} \delta_{kl} ) + \delta_{in} ( \delta_{jl} \delta_{km} – \delta_{jm} \delta_{kl})$$

Mixed product (triple product):

With the permutation symbol, the mixed product can be expressed by,

$$\displaystyle \varepsilon_{ijk}g = [\boldsymbol{g}_i \boldsymbol{g}_j \boldsymbol{g}_k] = \boldsymbol{g}_i \cdot (\boldsymbol{g}_j \times \boldsymbol{g}_k)$$

then multiplying both sides of this relation by the vector $\boldsymbol{g}^i$

$$\displaystyle \varepsilon_{ijk}g \boldsymbol{g}^i = \boldsymbol{g}_j \times \boldsymbol{g}_k$$

Since $\textbf{Det}[\boldsymbol{g}_i \cdot \boldsymbol{g}^j] = \textbf{Det}[\delta_i^j]=1$,

$$\displaystyle 1= \textbf{Det}[\boldsymbol{g}_i \cdot \boldsymbol{g}^j] = \textbf{Det}[\boldsymbol{g}_i \cdot g^{jk} \boldsymbol{g}_k] = \textbf{Det}[g_{jk}] \textbf{Det}[g^{ik}]$$

Then we get,

$$\displaystyle |g^{ik}| = \frac{1} {|g_{jk}|} \leftrightarrow |g^{ij}| = \frac{1} {|g_{ij}|} = \frac{1}{g^2} = \frac{1}{[\boldsymbol{g}_1 \boldsymbol{g}_2 \boldsymbol{g}_3]^2}$$

Thus,

$$\displaystyle \frac{\varepsilon_{ijk}}{g} \boldsymbol{g}_i = \boldsymbol{g}^j \times \boldsymbol{g}^k$$

Then,

$$\displaystyle \boldsymbol{a} \times \boldsymbol{b} = (a^i \boldsymbol{g}_i)\times(a^j \boldsymbol{g}_j) = \varepsilon_{ijk} a^i b^j g \boldsymbol{g}^k=g\begin{vmatrix} a^1 &a^2 &a^3 \\ b^1 &b^2 &b^3 \\ \boldsymbol{g}^1 &\boldsymbol{g}^2 &\boldsymbol{g}^3 \end{vmatrix}$$

$$\displaystyle \boldsymbol{a} \times \boldsymbol{b} = (a_i \boldsymbol{g}^i)\times(a_j \boldsymbol{g}^j) = \varepsilon^{ijk} a_i b_j \frac{1}{g} \boldsymbol{g}_k=\frac{1}{g} \begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 &b_2 &b_3 \\ \boldsymbol{g}_1 &\boldsymbol{g}_2 &\boldsymbol{g}_3 \end{vmatrix}$$

For 3D Euclidean space $\mathbb{E}^3$, $g = 1$, the cross product is,

$$\displaystyle \boldsymbol{a} \times \boldsymbol{b} = \begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 &b_2 &b_3 \\ \boldsymbol{e}_1 &\boldsymbol{e}_2 &\boldsymbol{e}_3 \end{vmatrix}$$