## Notes: Simple tensor algebra III

Previously, the vector space $\mathbb{V}$ and three vector product operations were considered in Simple tensor algebra I and Simple tensor algebra II. But we do not talk much about the tensor. Here, the tensor will be introduced.

So, what is a tensor? The following figure shows a tensor in a 2D Cartesian coordinate system.

The vector is a tensor, and it is a first-order (1st-order) tensor. Actually, the scalar is a tensor as well, we can consider the scalar as a zeroth-order (0th-order) tensor.

Let $a, b$ be two scalars in $\mathbb{R}$, we put these two scalars (0th-order tensor) together with $\{ a, b \}$ to form a vector (1st-order tensor). So, if we can put two vectors $\boldsymbol g, \boldsymbol f$ together, then we will get a 2nd-order tensor. The tensor product $\otimes$ plays such role here ($\boldsymbol g \otimes \boldsymbol f$).

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

## Notes: Simple tensor algebra II

The vector space $\mathbb{V}$ is introduced in Simple tensor algrbra I, several rules are provided in it. To reduce the abstract, we represent the vectors $\boldsymbol{x}, \boldsymbol{y}$ with the scalar coefficients $\{{x_1},{x_2},{x_3} \}$ and $\{{y_1},{y_2},{y_3} \}$ in $\mathbb{E}^3$, respectively. The composition of two vectors $\boldsymbol{x}, \boldsymbol{y} \in\mathbb{E}^3$ is expanded by,

Dot product ($\cdot$):

$${\boldsymbol{x}} \cdot {\boldsymbol{y}} = {x_i}{{\boldsymbol{g}}^i} \cdot {y_j}{{\boldsymbol{g}}^j} = {x_i}{y_j}{{\boldsymbol{g}}^i} \cdot {{\boldsymbol{g}}^j} = {x_i}{y_j}{g^{ij}} = {x^i}{y_i} = \left[ {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}}&{{x_3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ {{y_3}} \end{array}} \right] = \sum\limits_{i = 1}^3 {{x_i}} {y_i}$$

The dot product is also called scalar product or inner product under Cartesian coordinates and it provides the Euclidean magnitudes of these two vectors and the cosine of the angle between them.

Cross product ($\times$):

$${\boldsymbol{x}} \times {\boldsymbol{y}} = {x_i}{{\boldsymbol{g}}^i} \times {y_j}{{\boldsymbol{g}}^j} = {x_i}{y_j}{{\boldsymbol{g}}^i} \times {{\boldsymbol{g}}^j} = {x_i}{y_j}{\varepsilon ^{ijk}}g{{\boldsymbol{g}}_k} = \left| {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}}&{{x_3}}\\ {{y_1}}&{{y_2}}&{{y_3}}\\ {{{\boldsymbol{e}}_1}}&{{{\boldsymbol{e}}_2}}&{{{\boldsymbol{e}}_3}} \end{array}} \right|$$

The cross product is also called vector product, and it represents the area of a parallelogram with the sides of these two vectors. Now, a new composition for two vectors $\boldsymbol{x}, \boldsymbol{y} \in\mathbb{E}^3$ is introduced here.

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

## 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.