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$.HZ92F7.png

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,

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

3 Replies to “Notes: Simple tensor algebra I”

  1. mohamad hardanian says: Reply

    hi it not better first you define a vector space on a filde with some example then go to define tensor

    1. Hi, hardanian, thanks for your advice. I will make it better~

      1. mohamad hardanian says: Reply

        thanks so much tomorrow i wil come to study it again

Leave a Reply to mohamad hardanian Cancel reply