7. Tensor Algebra#

This section introduces tensor algebra

Warning

This introduction is meant for spaces with inner product that allows to introduce quite naturally the useful concept of the reciprocal basis of a given basis of \(\mathscr{V}\), belonging to the same space \(\mathscr{V}\), somehow avoiding the complications coming from the introduction of dual space \(\mathscr{V}^*\) and basis, and everything is required for that.

7.1. Vector space \(\mathscr{V}\)#

A vector space is an algebraic structure with:

  • a set \(\mathscr{V}\), whose elements are called vectors, here indicated with bold symbols \(\mathbf{v} \in \mathscr{V}\)

  • a field \(K\) (usually \(\mathbb{R}\) or \(\mathbb{C}\)), whose elements are calles scalars,

  • 2 operations, closed w.r.t. the set \(\mathscr{V}\):

    • vector sum, \(\mathbf{u} + \mathbf{v} \in \mathscr{V}\) for \(\forall \mathbf{u}, \ \mathbf{v} \in \mathscr{V}\)

    • multiplication of a scalar and a vector, \(a \mathbf{v} \in \mathscr{V}\) for \(\forall a \in K, \ \mathbf{u} \in \mathscr{V}\)

    with properties discussed below todo …

7.1.1. Operations (I)#

7.1.1.1. Sum#

…

7.1.1.2. Multiplication by a scalar#

…

Definition 7.1 (Linear combination)

A linear combination of \(n\) vectors \(\{ \mathbf{v}_i \}_{i=1:n}\), \(\mathbf{v}_i \in \mathscr{V}\), is the weighted sum

\[a^1 \mathbf{v}_1 + a^2 \mathbf{v}_2 + \dots + a^n \mathbf{v}_n = a^i \mathbf{v}_i \in \mathscr{V} \ ,\]

having used Einstein’s summation convention over repeated indices. Here the position of the indices has no particular meaning, but it’ll have soon in the following sections.

7.1.1.3. Inner product#

The existence of an inner product is not a requirement of a vector space

…

An inner product in a vectors space \(\mathscr{V}\) over field \(K\) is an operation \(\cdot: \mathscr{V} \times \mathscr{V} \rightarrow K\),

\[\mathbf{u} \cdot \mathbf{v} \ ,\]

with the following properties: todo

7.1.2. Basis of a vector space \(\mathscr{V}\)#

Definition 7.2 (Basis)

A basis is a minimal set of vectors of \(\mathscr{V}\) that can represent all the elements of \(\mathscr{V}\).

Definition 7.3 (Reciprocal basis)

In a inner product space, the reciprocal basis of a given basis \(\{ \mathbf{b}_a \}_{a=1:d}\) is the set of vectors \(\{ \mathbf{b}^{b} \}_{b=1:d}\), s.t.

\[\mathbf{b}^b \cdot \mathbf{b}_a = \delta_a^b \ .\]

Definition 7.4 (“Metric tensor”)

\[\begin{split}\begin{aligned} g_{ab} & := \mathbf{b}_a \cdot \mathbf{b}_b \\ g^{ab} & := \mathbf{b}^a \cdot \mathbf{b}^b \\ \end{aligned}\end{split}\]

The following holds

(7.1)#\[\begin{split}\begin{aligned} \mathbf{b}_a & = g_{ab} \mathbf{b}^b & (1) \\ \mathbf{b}^a & = g^{ab} \mathbf{b}_b & (2) \\ \end{aligned}\end{split}\]
“Proof”

Taking the dot product with \(\mathbf{b}_c\) of the relation (7.1)(1),

\[\mathbf{b}_c \cdot \mathbf{b}_a = g_{ab} \underbrace{\mathbf{b}_c \cdot \mathbf{b}^b}_{=\delta_c^b} = g_{ac} \ .\]

7.1.3. Change of basis#

Let \(T_{a}^b\) the elements of the matrix of change of basis, representing the vectors of the basis \(\{ \widetilde{b}_a \}\) as linear combinations of the vectors of the basis \(\{ \mathbf{b}_b \}\),

\[\widetilde{\mathbf{b}}_b = T_b^a \mathbf{b}_a \ .\]

Inverse transformation. Let \(\widetilde{T}\) be the elements of the inverse transformation,

\[\mathbf{b}_c = \widetilde{T}_{c}^{b} \widetilde{\mathbf{b}}_b = \widetilde{T}_{c}^{b} T_b^a \mathbf{b}_a \ ,\]

and thus

\[\widetilde{T}_{c}^{b} T_b^a = \delta_{c}^a \ .\]

Transformation of the reciprocal basis. todo

7.1.3.1. Transformation of components#

A vector \(\mathbf{v}\) can be represented in different basis, as different linear combinations of the elements of those bases,

\[\mathbf{v} = v^{a} \mathbf{b}_a = \widetilde{v}^b \widetilde{\mathbf{b}}_b \ .\]

Given the rules of change of basis, the rule of transformation of components immediately follwos

\[\begin{split}\begin{aligned} & \widetilde{\mathbf{b}}_b = T_b^a \mathbf{b}_a && \widetilde{v}^b = \widetilde{T}^b_a v^a \\ & \mathbf{b}_b = \widetilde{T}_{b}^{a} \widetilde{\mathbf{b}}_a && {v}^b = T^{b}_a \widetilde{v}^a \end{aligned}\end{split}\]
Proof
\[\mathbf{v} = v^a \mathbf{b}_a = \underbrace{v^a \widetilde{T}^{b}_a}_{\widetilde{v}^b} \widetilde{\mathbf{b}}_b = \widetilde{v}^b \widetilde{\mathbf{b}}_b \]

or

\[\mathbf{v} = \widetilde{v}^b \widetilde{\mathbf{b}}_b = \underbrace{\widetilde{v}^b \widetilde{T}^{b}_c}_{v^c} \mathbf{b}_c = v^c \mathbf{b}_c\]

It’s clear that the vectors of the bases and the components follow inverse transformations to preserve the invariance of vector w.r.t. a change of basis: the vector \(\mathbf{v}\) doesn’t change if we change our description of it, by changing a basis.

Matrix of change of basis as a tensor (todo maybe later? Tensor not introduced yet here)

The rule of transformation between different basis can be interpreted using dot product between tensors and vectors

\[\begin{split}\begin{aligned} \widetilde{\mathbf{b}}_b = T^{a}_b \mathbf{b}_a & = T^{a}_c \mathbf{b}_a \delta^c_b = T^{a}_c \mathbf{b}_a \otimes \mathbf{b}^c \cdot \mathbf{b}_b = \left( T^{a}_c \mathbf{b}_a \otimes \mathbf{b}^c \right) \cdot \mathbf{b}_b = \mathbb{T} \cdot \mathbf{b}_b \\ \end{aligned}\end{split}\]

Inverse and transpose

Interpreting the indices of the transformation matrix as the indices of rows and columns of a 2x2 matrix, transformation of components involves the transpose of the inverse matrix, as the indices \(a\), \(b\) are swapped.

Example 7.1

Example 7.2 (Rotation)

As the inverse of a rotation is its transpose, \(\widetilde{T}^{b}_{a} = T^{a}_b\). So the rule of transformation of components follows the same rule of change of basis. As an example, let the transformation between two Cartesian bases be the rotation

\[\begin{split} \begin{aligned} \widetilde{\mathbf{x}} & = \ \ \cos \theta \, \mathbf{x} + \sin \theta \, \mathbf{y} \\ \widetilde{\mathbf{y}} & = - \sin \theta \, \mathbf{x} + \cos \theta \, \mathbf{y} \end{aligned} \quad , \quad \begin{aligned} \mathbf{x} & = \cos \theta \, \widetilde{\mathbf{x}} - \sin \theta \, \widetilde{\mathbf{y}} \\ \mathbf{y} & = \sin \theta \, \widetilde{\mathbf{x}} + \cos \theta \, \widetilde{\mathbf{y}} \end{aligned} \end{split}\]

Let a vector

\[\begin{split}\begin{aligned} \mathbf{v} & = v_x \mathbf{x} + v_y \mathbf{y} = \\ & = v_x \left( \cos \theta \, \widetilde{\mathbf{x}} - \sin \theta \, \widetilde{\mathbf{y}} \right) + v_y \left( \sin \theta \, \widetilde{\mathbf{x}} + \cos \theta \, \widetilde{\mathbf{y}} \right) = \\ & = \widetilde{v}_x \widetilde{\mathbf{x}} + \widetilde{v}_y \widetilde{\mathbf{y}} = \\ & = \widetilde{v}_x \left( \cos \theta \, \mathbf{x} + \sin \theta \, \mathbf{y} \right) + \widetilde{v}_y \left( - \sin \theta \, \mathbf{x} + \cos \theta \, \mathbf{y} \right) \end{aligned}\end{split}\]

and thus

\[\begin{split} \begin{aligned} \widetilde{v}_x & = \ \ \cos \theta \, v_x + \sin \theta \, v_y \\ \widetilde{v}_y & = - \sin \theta \, v_x + \sin \theta \, v_y \\ \end{aligned} \quad , \quad \begin{aligned} v_x & = \ \ \cos \theta \, \widetilde{v}_x - \sin \theta \, \widetilde{v}_y \\ v_y & = \ \ \sin \theta \, \widetilde{v}_x + \sin \theta \, \widetilde{v}_y \\ \end{aligned} \end{split}\]

7.1.4. Operations#

7.1.4.1. Tensor product of vectors#

\[\mathbf{v}_{(1)} \otimes \mathbf{v}_{(2)} \otimes \dots \otimes \mathbf{v}_{(r)}\]

or writing each vector as a linear combination of the elements of a basis \(\mathbf{b}_a\),

\[\mathbf{v}_{(i)} = v_{(i)}^{a_i} \mathbf{b}_{a_i} \ , \dots\]
\[\begin{split}\begin{aligned} \mathbf{v}_{(1)} \otimes \mathbf{v}_{(2)} \otimes \dots \otimes \mathbf{v}_{(3)} & = \left( v_{(1)}^{a_1} \mathbf{b}_{a_1} \right) \otimes \left( v_{(2)}^{a_2} \mathbf{b}_{a_2} \right) \otimes \dots \otimes \left( v_{(r)}^{a_r} \mathbf{b}_{a_r} \right) = \\ & = v_{(1)}^{a_1} \, v_{(2)}^{a_2} \, \dots \, v_{(r)}^{a_r} \mathbf{b}_{a_1} \otimes \mathbf{b}_{a_2} \otimes \dots \otimes \mathbf{b}_{a_r} \end{aligned}\end{split}\]

The result of the tensor product of \(r\) vectors is a rank-\(r\) tensor, \(\mathbb{T}\), as it will be clear below, with components

\[T^{a_1 a_2 \dots a_r} = v_{(1)}^{a_1} \, v_{(2)}^{a_2} \, \dots \, v_{(r)}^{a_r} \ .\]

7.2. Space of tensors#

Definition 7.5

7.2.1. Operations (I)#

7.2.1.1. Sum#

7.2.1.2. Multiplication by a scalar#

Vector space of tensors

The set of tensors of a given rank with the operations of sum and multiplication by a scalar defined above forms a vector space.

7.2.2. Basis#

7.2.2.1. Change of basis and rule of transformation of components - classical definition of a tensor#

\[\begin{split}\begin{aligned} \widetilde{\mathbf{b}}_{i_a} & = T_{i_b}^{i_a} \mathbf{b}_{i_b} \\ \mathbf{b }_{i_a} & = \widetilde{T}_{i_a}^{i_b} \widetilde{\mathbf{b}}_{i_b} \\ \end{aligned}\end{split}\]
\[\begin{split}\begin{aligned} \mathbf{A} & = A^{i_1 \dots i_p} \mathbf{b}_{i_1} \dots \mathbf{b}_{i_p} = \\ & = A^{i_1 \dots i_p} \left( T_{i_1}^{j_1} \widetilde{\mathbf{b}}_{j_1} \right) \dots \left( T_{i_p}^{j_p} \widetilde{\mathbf{b}}_{j_p} \right) = \\ & = A^{i_1 \dots i_p} T_{i_1}^{j_1} \dots T_{i_p}^{j_p} \widetilde{\mathbf{b}}_{j_1} \dots \widetilde{\mathbf{b}}_{j_p} = \\ & = \widetilde{A}^{j_1 \dots j_p} \widetilde{\mathbf{b}}_{j_1} \dots \widetilde{\mathbf{b}}_{j_p} \ , \end{aligned}\end{split}\]

with

(7.2)#\[\widetilde{A}^{j_1 \dots j_p} = A^{i_1 \dots i_p} T_{i_1}^{j_1} \dots T_{i_p}^{j_p} \ .\]

Definition 7.6 (Classical definition of a tensor)

Relation (7.2) is the “historical” definition of a tensor, through the law of transformation of its components following a change of basis.

7.2.3. Operations (II)#

7.2.3.1. Tensor product#

The tensor product of a \(p\)-rank tensor \(\mathbf{A}\) and a \(q\)-rank tensor \(\mathbf{B}\) is the \((p+q)\)-rank tensor \(\mathbf{A} \otimes \mathbf{B} = \mathbf{A} \mathbf{B}\), that can be defined using component representation in a given basis,

\[\begin{split}\begin{aligned} \mathbf{A} \otimes \mathbf{B} & = \left( A^{a_1 \dots a_p} \mathbf{b}_{a_1} \dots \mathbf{b}_{a_p} \right) \otimes \left( B^{b_1 \dots b_q} \mathbf{b}_{b_1} \dots \mathbf{b}_{b_q} \right) = \\ & = A^{a_1 \dots a_{p-1} a_p} \, B^{b_1 \, b_2 \dots b_q} \mathbf{b}_{a_1} \dots \mathbf{b}_{a_p} \mathbf{b}_{b_1} \dots\mathbf{b}_{b_q} \end{aligned}\end{split}\]

7.2.3.2. Dot product#

The dot product of a \(p\)-rank tensor \(\mathbf{A}\) and a \(q\)-rank tensor \(\mathbf{B}\) is the \((p+q-2)\)-rank tensor \(\mathbf{A} \cdot \mathbf{B}\), that can be defined using component representation in a given basis,

\[\begin{split}\begin{aligned} \mathbf{A} \cdot \mathbf{B} & = \left( A^{a_1 \dots a_p} \mathbf{b}_{a_1} \dots \mathbf{b}_{a_p} \right) \cdot \left( B_{b_1 \dots b_q} \mathbf{b}^{b_1} \dots \mathbf{b}^{b_q} \right) = \\ & = A^{a_1 \dots a_p} \, B_{b_1 \dots b_q} \mathbf{b}_{a_1} \dots \underbrace{ \mathbf{b}_{a_p} \cdot \mathbf{b}^{b_1} }_{= \delta_{a_p}^{b_1} } \dots \mathbf{b}^{b_q} = \\ & = A^{a_1 \dots a_{p-1} k} \, B_{k \, b_2 \dots b_q} \mathbf{b}_{a_1} \dots \mathbf{b}_{a_{p-1}} \mathbf{b}^{b_2} \dots \mathbf{b}^{b_q} \end{aligned}\end{split}\]

7.2.3.3. Contraction#

Contraction of a pair of index of a \(p\)-rank tensor \(\mathbf{A}\) returns a \(p-2\)-rank tensor defined as

\[\begin{split}\begin{aligned} C_{a}^b \left( \mathbf{A} \right) & = \dots \\ \end{aligned}\end{split}\]

7.2.3.4. Exterior product#

todo see exterior algebra

7.2.4. Invariants#

7.3. Exterior algebra#

\[\land\]

7.3.1. Exterior product#

Generalization of the vector product