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 $latex \mathbb{V} $ with respect to a given basis $latex \mathcal{G}$ 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

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