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

Expand \boldsymbol{h}^j = h^j_k \boldsymbol{g}^kwe 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 and \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}  and \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}

4 thoughts on “Notes: Simple tensor algebra I

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

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