Tensors 1 | bearish….

Notations

Einstein summation convention is used here. A matrix M is denoted as $[M]$ and its ij-th element is referred to by $[M]_{ij}$ . Quantities or coefficients are indexed as for example $u^i$ , $A_{ij}$ or $A_i^j$ . These indices do not automatically pertain to row and column indices of a matrix, but the quantities can be presented by matrices through isomorphisms once their indices are freely interpreted as rows and columns of matrices.

Coordinates of a vector

Let $V$ be a n-dimensional vector space and $\mathcal B=\{e_1,\cdots, e_n\}$ with $e_i\in V$ be a basis for $V$ . Then, we define the coordinate function as,

$[\cdot]_{\mathcal B}:V\to \mathbb M^{n\times 1}$

such that for a vector $v\in V$ written by its components (with respect to $\mathcal B$ ) as $v=v_ie_i$ the function acts as,

$[v]_{\mathcal B}=\begin{bmatrix}v_1 \\ \vdots \\ v_n \end{bmatrix}$

The coordinate function is a linear map.

Change of basis for vectors

Let $\mathcal B$ and $\tilde {\mathcal B}$ be two basis for $V$ , then,

$\tilde e_i=F_{ji}e_j$ and $e_i=B_{ji}\tilde e_j$

where the indices of the scalar terms $F_{ji}$ and $B_{ji}$ are intentionally set this way. So, if all $F_{mn}$ are collected into a matrix $[F]$ , then the sum $F_{ji}e_j$ is over the rows of the matrix for a particular column. In other words, we can utilize the rule of matrix multiplication and write,

$\tilde e_i=F_{ji}e_j=[F]^{\rm T}\begin{bmatrix} e_1\\ \vdots \\ e_n \end{bmatrix}$

The same is true for $[B]:=B_{ji}$ . In above formulations, note that $j$ is a dummy index (i.e. we can equivalently write $\tilde e_j=F_{ij}e_i=F_{ki}e_k$ )

Setting $\mathcal B$ as the initial (old) basis and writing the current (new) basis $\tilde {\mathcal B}$ in terms of $\mathcal B$ is referred to as forward transform denoted by $F_{ij}$ . Relatively, $B_{ij}$ is called backward transform.

The relation between that forward and backward transforms is obtained as follows,

$\begin{split}e_i &= B_{ji} \tilde e_j=B_{ji}F_{kj}e_k\\&\implies B_{ji}F_{kj}=\delta_{ik}\\&\therefore [F]=[B]^{-1} \ , [B] = [F]^{-1}\end{split}$

We now find how vector coordinates are transformed relative to different bases. A particular $v\in V$ can be expressed by its components according to any of $\mathcal B$ or $\tilde{\mathcal B}$ basis, therefore,

$v=v_ie_i=\tilde v_i \tilde e_i$

To find the relation between $[v]_{\tilde{\mathcal B}}$ and $[v]_{\mathcal B}$ we write,

$\begin{split}v&=v_ie_i=\tilde v_i\tilde e_i \implies v_ie_i = v_i B_{ji} \tilde e_j\equiv C_j\tilde e_j\implies C_j=\tilde v_j\\&\therefore \tilde v_i = B_{ij}v_j\equiv [B][v]_{\mathcal B}\\&\implies v_i = F_{ij}\tilde v_j\equiv [F][v]_{\tilde {\mathcal B}}\end{split}$

As it can be observed, the old basis to new basis is transformed by the forward transform $F_{ij}$ while the old coordinates $v_i$ are transformed to the new ones, $\tilde v_i$ , by the backward transform $B_{ij}$ . Because the coordinates of $v$ behave contrary to the basis vectors in transformation, the coordinates or the scalar components are said to be contravariant. A vector can be called a contravariant object because its scalar components (coordinates) transforms differently from the basis vectors whose their linearly combination equals to the vector. Briefly,

Proposition: Let $v=v_ie_i$ . Then, the scalar components/coordinates $v_i$ are transformed by $B_{ij}$ if and only if the basis vectors $e_i$ are transformed by $F_{ij}$ , such that $B_{ji}F_{kj}=\delta_{ik}$ .

Later, a vector is called a contravariant tensor. For the sake of notation and to distinguish between the transformations of the basis and the coordinates of a vector, in index of a coordinate is written as superscript to show it is contravariant. Therefore,

$v = v^ie_i=\tilde v^i\tilde e_i$

Linear maps and linear functionals

Definition: $\mathcal L(V, W)$ is defined as the space of all linear maps $V\to W$ where the domain and codomain are vectors spaces.

It can be proved that $\mathcal L$ is a vector space $(\mathcal L, +, \cdot)$ , hence, for $T_1, T_2\in \mathcal L(V,W)$ and $\alpha\in \mathbb R$

$\alpha\cdot T_1=\alpha T_1\quad , (T_1+T_2)()=T_1()+T_2()$

Note that the addition on the LHS is an operator in $\mathcal L$ and the addition on the RHS is an operator in $W$ .

Proposition 1: Let $T\in \mathcal L(V, W)$ , i.e a linear map from a vector space $V$ to another one $W$ . If $\mathcal B=\{e_1, \cdots, e_n\}$ is a basis for $V$ , and $T(e_i)=w_i$ for $w_i\in W$ and $i=1,\cdots , n$ , then $T$ is uniquely defined over $V$ .

This proposition says a linear map over a space is uniquely determined by its action on the basis vectors of that space. In other words, if $T(e_i)=w_i$ and $T^*(e_i)=w_i$ then $\forall v\in V, \ T(v)=T^*(v)$ . proof: let $T(e_i)=w_i$ (given by the nature of $T$ ), then for $v\in V$ such that $v=v^ie_i$ , we can write $v^iT(e_i)=v^iw_i$ , therefore, $T(v^ie_i)=T(v)=v^iw_i$ . Because, $v^i$ ‘s are unique for (a particular) $v$ then $v^iw_i$ is unique for $v$ and hence $T(v)$ must be unique for any $v\in V$ . In other word, there is only one unique $T$ over $V$ such that $T(e_i)=w_i$ .

As a side remark, if $\mathcal B=\{e_1, \cdots, e_n\}$ is a basis for $V$ , hence spanning $V$ , then $\{T(e_i)| i=1,\cdots n\}$ spans the range of $T$ ; The range of $T$ is a subset of $W$ .

By this proposition, a matrix completely determining a linear can be obtained for the linear map. let $V$ be n-dimensional with a basis $\mathcal B=\{e_i\}_1^n$ , and $W$ be m-dimensional with a basis $\mathcal B'=\{e_i'\}_1^m$ . Then there are coefficients $T_i^j$ such that,

$T(e_i)= T_i^j e_j'$

In the notation $T_i^j$ , the index $j$ is superscript because for a fixed $e_i$ and hence a fixed $i$ , the term $T_i^j$ is the coordinate of $T(e_i)\in W$ and it is a contravariant (e.g $T(e_3)=T_3^j e_j'\equiv v^je_j'$ ).

For $v\in V$ , and $w=T(v)$ , with the coordinates $[v]_{\mathcal B}$ and $[w]_{\mathcal B'}$ , we write,

$\begin{split}w=T(v^ie_i)&=v^iT(e_i)=v^iT_i^je_j' \\&\implies w^j = v^iT_i^j \equiv T_i^j v^i \end{split}$

This expression can be written as a matrix multiplication of $[w]_{\mathcal B'}=[M][v]_{\mathcal B}$ , where $[T]:=\mathcal M(T)\in \mathbb M^{m\times n}$ presented by its elements as,

$\begin{bmatrix} T_1^1 && T_2^1 && \cdots && T_n^1 \\T_1^2 && T_2^2 && \cdots && T_n^2 \\\vdots && \vdots && \cdots && \vdots\\T_1^m && T_2^m && \cdots && T_n^m \end{bmatrix}$

As a remark, above can be viewed as columns of the matrix and written as,

$[T]=\begin{bmatrix} [T(e_1)]_{\mathcal B'} && [T(e_2)]_{\mathcal B'} && \cdots && [T(e_n)]_{\mathcal B'} \end{bmatrix}$

Linear functional (linear form or covector)

Definition: a linear functional on $V$ is a linear map $f\in V^* :=\mathcal L(V,\mathbb F)$ . The space $V^*$ is called the dual space of $V$ .

Proposition: Let $\mathcal B=\{e_1, \cdots, e_n\}$ and $\varepsilon_i \in V^*$ be defined as $\varepsilon_i(e_j):=\delta_{ij}$ . Then, $\{\varepsilon_i\}_1^n$ called dual basis of $\mathcal B$ , is a basis of $V^*$ , and hence $\dim V = \dim V^*$ .

Proof: first we show that $\varepsilon_i$ ‘s are linearly independent, i.e. $c_i\varepsilon_i=0 \implies c_i=0 \forall i=1, \cdots, n$ . Note that on the RHS, $0\in V^*$ . For a $v\in V$ we can write $c_i\varepsilon_i(v)$ and assume $c_i\varepsilon_i(v)=0$ . Then,

$c_i\varepsilon_i(v)=c_i\varepsilon_i(v^je_j)=0\implies c_iv^j\varepsilon_i(e_j)=0\implies c_iv^j\delta_{ij}=0\implies c_iv_i=0$

Since $v$ is arbitrary, $c_i=0$ ■ .

Now we prove that $\{\varepsilon_i\}_1^n$ spans $V^*$ . I.e $\forall f \in V^* \exists \{c_1, \cdots, c_n\}$ such that $f=c_i\varepsilon_i$ . To this end, we apply both sides to a basis vector of $V$ and write $f(e_j)=c_i\varepsilon_i(e_j)$ which implies $f(e_j)=c_j$ or explicitly $c_j$ is found as $c_j=f(e_j)$ . Consequently, $f=f(e_i)\varepsilon_i$ ■.

Consider $V$ and $\mathcal B$ . If $f\in V^*$ , then the matrix of the linear functional/map $f$ is

$[M]=\mathcal M(f)=\begin{bmatrix} f(e_1) && \cdots && f(e_n)\end{bmatrix}\in \mathbb M^{1\times n}$

So, for $v\in V$ as $v=v^ie_i$ we can write,

$f(v)=[M][v]_\mathcal B\quad \in \mathbb R$

Result: if the coordinates of a vector is shown by a column vector or single-column matrix (which is a vector in the space of $\mathbb M^{n\times 1}$ ), then a row vector or a single-row matrix represents the matrix of a linear functional.

Definition: a linear functional $f\in V^*$ , which can be identified with a row vector as its matrix, is also called a covector.

Like vectors, a covector (and any linear map) is a mathematical object that is independent of a basis (i.e. invariant). The geometric representation of a vector in (or by an isomorphism in) $\mathbb R^3$ is an arrow in $\mathbb E^3$ . For a covector isomorphic to $\mathbb R^2$ , the representation is a set (stack) of planes in $\mathbb E^3$ that can be represented by iso lines in $\mathbb E^2$ . A covector that is isomorphic to $\mathbb R^3$ can be represented by iso surfaces in $\mathbb E^3$ .

Example: Let $\mathcal B = \{e_1, e_2\}$ be a basis of $V$ and $[2,1]$ be the matrix of a covector $f$ in some $V^*$ . Then, if $[x]_{\mathcal B} = [x_1,x_2]^{\rm T}$ , we can write,

$y=[2,1]\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \implies y = 2x_1 + x_2$

which, for different values of $y$ , is a set of (iso) lines in a Cartesian CS defined by two axes $x_1$ and $x_2$ along $e_{1g}$ and $e_{2g}$ that are the geometric representations of $e_1$ and $e_2$ . The Cartesian axes are not necessarily orthogonal.

If we chose any other basis $\tilde {\mathcal B} = \{\tilde e_1, \tilde e_2\}$ for $V$ , then the matrix of the covector $f$ changes. Also, the geometric representations of $\{\tilde e_1, \tilde e_2\}$ are different from $e_{1g}$ and $e_{2g}$ and hence the geometric representation of the covector stays the same shape.

Example: Let $\mathcal B = \{e_1, e_2\}$ be a basis of $V$ and $\mathcal B^* = \{\varepsilon_1, \varepsilon_2\}$ be a basis for $V^*$ . This means $\varepsilon_i\in V^*$ and $\varepsilon_i(e_j):=\delta_{ij}$ . Then, the matrix of each dual basis vector is as,

$\mathcal M(\varepsilon_i)=\begin{bmatrix}\varepsilon_i(e_1) && \varepsilon_i(e_2)\end{bmatrix} = \begin{bmatrix}\delta_{i1} && \delta_{i2}\end{bmatrix}$

Change of basis for covectors

Let $\mathcal B = \{e_i\}_1^n$ and $\tilde{\mathcal B} = \{\tilde e_i\}_1^n$ be two bases for $V$ , and hence, $\mathcal B^* = \{\varepsilon_i\}_1^n$ and $\tilde{\mathcal B^*} = \{\tilde \varepsilon_i\}_1^n$ be two bases for $V^*$ . Each dual basis vector $\tilde \epsilon_i$ can be written in terms of the (old) dual basis vectors by using a linear transformation as $\tilde \varepsilon_i = Q_{ij}\varepsilon_j$ . Now, the coefficients $Q_{ij}$ are to be determined as follows,

$\begin{split}\tilde \varepsilon_i(e_k) &= Q_{ij}\varepsilon_j(e_k)=Q_{ij}\delta_{jk}=Q_{ik}\\&\implies \tilde \varepsilon_i(e_k) = Q_{ik}\\&\therefore Q_{ij}=\tilde \varepsilon_i(e_j)\end{split}$

Using the formula $e_i=B_{ji}\tilde e_j$ regarding the change of basis of vectors, the above continues as,

$\begin{split}Q_{ij}&=\tilde \varepsilon_i(e_j)=\tilde \varepsilon_i(B_{kj}\tilde e_k)\\&\text{by linearity of covectors}= B_{kj}\tilde \varepsilon_i(\tilde e_k)=B_{kj}\delta_{ik}=B_{ij}\\\therefore Q_{ij}&=B_{ij}\end{split}$

This indicates that the dual basis are transformed by the backward transformation. Referring to the index convention, we use superscript for components that are transformed trough a backward transformation. Therefore,

$\tilde \varepsilon^i=B_{ij}\varepsilon^j$

meaning that dual basis vectors are contravariant because they behave contrary to the basis vectors in transformation from $e_i$ to $\tilde e_i$ .

Now let $f\in V^*$ . Writing $f=c_i\varepsilon^i=\tilde c_j \tilde \varepsilon^j$ and using the above relation, we get,

$\begin{split}c_i\varepsilon^i&=\tilde c_j \tilde \varepsilon^j \implies c_i F_{ij}\tilde \varepsilon^j=\tilde c_j \tilde \varepsilon^j\\\tilde c_j &= F_{ij}c_i\end{split}$

meaning that they are transforming in a covariant manner when the basis of the vector space changes from $e_i$ to $\tilde e_i$ .

Briefly the following relations have been shown.

Basis and change of basis for the space of linear maps $\mathcal L(V, W)$

As can be proved $\mathcal L(V,W)$ is a linear vector space and any linear map is a vector. Therefore, we should be able to find a basis for this space. If $V$ is n-dimensional and $W$ is m-dimensional, the $\mathcal L(V,W)$ is mn-dimensional and hence its basis should have $m\times n$ vectors, i.e. linear maps. Let’s enumerate the basis vectors of $\mathcal L$ as $\varphi_{ij}\in \mathcal L (V,W)$ for $i=1, \cdots , m$ and $j=1, \cdots , n$ , then any linear map $T$ can be written as,

$T = c_{ij}\varphi_{ij}$

By proposition 1, any linear map is uniquely determined by its action on the basis vectors of its codomain. If $\mathcal B = \{e_i\}_1^n$ be a basis for $V$ , then for any basis vector $e_k$ ,

$T(e_k)=c_{ij}\varphi_{ij}(e_k)$

Setting a basis for $W$ as $\mathcal B' = \{e_i'\}_1^m$ , the above equation becomes,

$a_{ik}e_i'=c_{ij}\varphi_{ij}(e_k)$

Note that $i$ s are dummy indices. This equation holds if,

$\begin{matrix}c_{ij}=a_{ik} \text{ and } \varphi_{ij}(e_k)=e_i' && \text{ if } k=j \\c_{ij} = 0 \text{ and } \varphi_{ij}(e_k)=0 && \text{ if } k\ne j\end{matrix}$

Therefore, we can choose a set of $m\times n$ basis vectors $\varphi_{ij}$ for $\mathcal L (V,W)$ as,

$\varphi_{ij}(e_k) = \begin{cases}e_i'\text{ if }k=j\\0\text{ if } k\neq j\end{cases}$

By recruiting the basis of $V^*$ , the above can be written as,

$\{\varphi_i^j}=e_i'\varepsilon^j | i=1,\cdots, m \text{ and } j=1,\cdots, n\}$

The term $e_i'\varepsilon^j$ is obviously a linear map $V\to W$ . It can be readily shown that $c_{ij}\varphi_i^j=c_{ij}e_i'\varepsilon^j$ being a linear combinations of the derived basis vectors is linearly independent, i.e. $c_{ij}e'_i\varepsilon^j(v)=0(v)$ for any $v\in V$ (here, note that $0\in \mathcal L$ ).

So, a linear map $T$ can be written as a linear combination $T = c_{ij}e_i'\varepsilon^j$ . Here, it is necessary to use the index level convention. To this end, we observe that for a fixed $i$ the term $c_{ij}$ couples with $\varepsilon^j$ and represents the coordinates of a covector. As coordinates of a covector are covariant, index $j$ is written as subscript. For a fixed $j$ though, the term $c_{ij}$ couples with $e_i'$ and represents the coordinates of a vector. As coordinates of a vector are contravariant, index $i$ should rise. Therefore, we write,

$T = c_j^ie_i'\varepsilon^j$

The coefficients $c_j^i$ can be determined as,

$\forall e_k\in \mathcal B\quad T(e_k) = c_j^ie_i'\varepsilon^j(e_k)=c_k^ie_i'$

Stating that $c_k^i$ are the coordinates of $T(e_k)$ with respect to the basis of $W$ , i.e. $\mathcal B'$ . Comparing with what was derived as $T(e_k)=T_k^ie_i'$ , we can conclude that $c_k^i = T_k^i$ . Therefore,

$T=T_j^i e_i'\varepsilon^j$

The above result can also be derived from $w^i = T_j^iv^j$ as follows.

$\begin{split} w^i &= T_j^iv^j \implies w = (T_j^iv^j)e_i' = T_j^i\varepsilon_j(v)e_i'\\&\therefore T(v) = T_j^i\varepsilon_j(v)e_i' \text{ or } T = T_j^i e_i'\varepsilon^j \end{split}$

Change of basis of $\mathcal L(V,W)$ is as follows.

For $\mathcal L(V,W)$ , let $\mathcal B=\{e_i\}_1^n$ and $\tilde {\mathcal B}=\{\tilde e_i\}_1^n$ be bases for $V$ , and $\mathcal B'=\{e_i'\}_1^m$ and $\tilde {\mathcal B}'=\{\tilde e_i'\}_1^m$ be bases for $W$ . Also, $\mathcal B^*=\{\varepsilon^i\}_1^n$ and $\tilde {\mathcal B}^*=\{\tilde \varepsilon^i\}_1^n$ are corresponding bases of $V^*$ . Forward and backward transformation pairs in $V$ and $W$ are denoted as $(F_{ij}, B_{ij})$ and $(F'_{ij}, B'_{ij})$ .

$\begin{split} T&=T_j^ie_i'\varepsilon^j = \tilde {T_j^i} \tilde e_i'\tilde \varepsilon^j \\&\implies T_j^ie_i'\varepsilon^j = \tilde {T_j^i} F'_{ki} e_k' B_{js}\varepsilon^s \implies T_s^k = \tilde {T_j^i} F'_{ki} B_{js}\\&(\text{ by } B_{nl}F_{lm}=\delta_{nm}) \ \implies B'_{lk}F_{sr}T_s^k = \tilde {T_j^i} \delta_{li}\delta_{jr}=\tilde T_r^l\\&\therefore \tilde {T_j^i} = B'_{ik}F_{sj}T_s^k\end {split}$

Note that the coordinates $T_s^k$ of a linear map need two transformations such that the covariant index $s$ of $T_s^k$ pertains to the forward transformation and the contravariant index $k$ pertains to the backward transformation.

Example: let $T\in \mathcal L(V,V)$ , then,

$T = T_i^je_j\varepsilon^i \quad \tilde {T_s^t} = B_{tj}F_{is}T_i^j$

If the matrices, $[F]$ , $[F]^{-1}=[B]$ , and $[T]$ are considered, we can write,

$[\tilde T] = [F]^{-1}[T][F]$

Bilinear forms

A bilinear form is a bilinear map defined as $T:V\times V\to \mathbb R$ . Setting a basis for $V$ , a bilinear form can be represented by matrix multiplications on the coordinates of the input vectors. If $\{e_i\}_1^n$ is a basis for $V$ , then

$\begin{split} T(u,v)&=T(u^ie_i,v^je_j)=u^iv^jT(e_i, e_j)\\&\implies B(u,v)=u^iv^jT_{ij} \quad \text{ with } T_{ij}:=T(e_i, e_j) \end{split}$

which can be written as,

$[u]^{\rm T}[T][v]$

where $[T]\in\mathbb M^{n\times n}$ with $[T]_{ij}=T_{ij}$ .

The expression $u^iv^jT(e_i, e_j)$ indicates that a bilinear form is uniquely defined by its action on the basis vectors. This is the same as what was shown for linear maps by proposition 1. This comes from the fact that a bilinear form is a linear map with respect to one of its arguments at a time.

Now we seek a basis for the space of bilinear forms, i.e. $\mathcal L_b(V\times V, \mathbb R)$ . This is a vector space with the following defined operations.

$\begin{split}\forall B_1, B_2 \in \mathcal B\quad (B_1+B_2)(u,v) &= B_1(u,v) + B_2(u,v)\\\forall \alpha\in \mathbb R\quad \alpha B(u,v) &= B(\alpha u, v)= B(u, \alpha v)\end{split}$

The dimension of this space is $n\times n$ , therefore, for any bilinear form $T$ there are bilinear forms $\rho_{ij}\in \mathcal B_b$ such that,

$T=c_{ij}\rho_{ij}$

From the result $T(u,v)=u^iv^jT(e_i, e_j)=u^iv^jT_{i,j}$ we can conclude that

$\begin{split}T(u,v)&=u^iv^jT_{ij}= \varepsilon^i(u)\varepsilon^j(v)T_{ij}\implies T(.,.) = T_{ij}\varepsilon^i(.)\varepsilon^j(.)\\\therefore c_{ij}&=T_{ij}, \quad \rho_{ij}= \varepsilon^i\varepsilon^j\quad \text {or } \rho_{ij}(e_s,e_t)=\begin{cases} 1& s=i \text { and } t=j\ 0& \text {otherwise}\end{cases}\end{split}$

Following the index level convention, the indices of $T_{ij}$ should stay as subscripts because each index pertains to the covariant coordinates of a covector after fixing the other index.

If $\mathcal B$ and $\tilde {\mathcal B}$ are two bases for $V$ , then the change of basis of the space of bilinear forms are as follows.

$\begin{split} T&=T_{ij}\varepsilon^i\varepsilon^j = \tilde T_{ij} \tilde \varepsilon^i\tilde \varepsilon^j = \tilde T_{ij}B_{is}\varepsilon^sB_{jt}\varepsilon^t\\&\implies T_{st}=\tilde T_{ij}B_{is}B_{jt}\\&\therefore \tilde T_{kl} = F_{sk}F_{tl}T_{st}\end {split}$

Example: the metric bilinear map (metric tensor)

Dot/inner product on the vector space $V$ over $\mathbb R$ is defined as a bilinear map $\langle \cdot , \cdot \rangle : V\times V \to \mathbb R$ such that, $\langle u , v \rangle = \langle v , u \rangle$ and $v\ne 0 \iff \langle v , v \rangle > 0$ . With this regard two objects (that can have geometric interpretations for Euclidean spaces) are defined as,

1- Length of a vector $\|v\|^2:=\langle v,v\rangle$
2- Angle between two vectors $\cos \theta :=\langle u/\|u\|,v/|v\|\rangle$

Let see how the dot product is expressed through the coordinates of vectors. With $\{e_i\}_1^n$ being a basis for $V$ , we can write,

$u\cdot v :=g\langle u, v \rangle = u^iv^jg_{ij} \quad \text {s.t}\quad g_{ij}=e_i\cdot e_j$

The term $g_{ij}$ is called the metric tensor and its components can be presented by an n-by-n matrix as $[g]_{ij}=e_i\cdot e_j$ .

If the basis is an orthonormal basis, i.e. $e_i\cdot e_j=0 \forall i\ne j$ , then $g_{ij}=\delta_{ij}$ and $[g]$ is the identity matrix. Therefore, $v\cdot u= u^iv^i$ and $\|v\|^2 = v^iv^i$ .

Multilinear forms

A general multilinear form is a multilinear map defined as $T:V_1\times V_2\times \cdots \times V_n\to \mathbb R$ , where $V_i$ is a vector space. Particularly setting $V_i=V$ leads to a simpler multilinear form as $T:V\times V\times \cdots \times V\to \mathbb R$ .

Following the same steps as shown for a bilinear map, a multilinear form $T:V\times V\times \cdots \times V\to \mathbb R$ can be written as,

$\begin{split} T(u,v,\cdots , z)&=T(u^ie_i,v^je_j, \cdots, z^ke_k)= u^iv^j\cdots z^k T(e_i,e_j, \cdots, e_k)\\&\implies T(u,v,\cdots , z)=u^iv^j\cdots z^kT_{ij\cdots k} \quad \text{with}\quad T_{ij\cdots k}:=T(e_i, e_j, \cdots e_k) \end{split}$

which implies,

$T = T_{ij\cdots k}\varepsilon^i \varepsilon^j\cdots \varepsilon^k$

showing that a multilinear form can be written as a linear combination of covectors.

Multilinear maps

A multilinear map is a map $T:V_1\times \cdots \times V_n \to W$ . A multilinear map can be written in terms of vector and covector basis. For example, consider $T:V\times V \to W$ as $T(u,v)=w$ with $\{e_i\}_1^n$ and $\{e'_i\}_1^m$ being bases for $V$ and $W$ . We can write,

$T(u,v)=T(u^ie_i,v^je_j)=u^iv^jT(e_i,e_j)$

Because for each $i$ and $j$ we have $T(e_i,e_j)\in W$ , we can write,

$T(e_i,e_j)=T_{ij}^ke'_k$

We write the indices $i$ and $j$ in $T_{ij}^k$ as subscripts in accordance with their position on the LHS; however, we’ll see that $T_{ij}^k$ is a coordinate of a covector for each $i$ or $j$ when $k$ is fixed. Combining above, we can write,

$\begin{split}T(u,v)&=u^iv^jT(e_i,e_j)=u^iv^jT_{ij}^ke'_k=\varepsilon^i(u)\varepsilon^j(v)T_{ij}^ke'_k\\&\therefore T = \varepsilon^i\varepsilon^jT_{ij}^ke'_k\end{split}$

The term $T_{ij}^k$ collects $n\times n \times m$ coefficients and uniqully defines the multilinear map. We can imagine this term as a 3-dimensional array/matrix. Above also shows that the multilinear map can be written as a linear combination of basis covectors and basis vectors.

Definition of a tensor

Defining the following terms,

Vector space $V$ and basis $\{e_i\}_1^n$ and another basis $\{\tilde e_i\}_1^n$ .
Basis transformation as $\tilde e_j=F_{ij}e_{i}$ , and therefore $e_j=B_{ij}\tilde e_{i}$ .
The dual vector space of $V$ as $V^*$ .
Vector space $V'$ and basis $\{e_i'\}_1^m$ and another basis $\{\tilde e_i'\}_1^m$ .
Basis transformation as $\tilde e_j'=F'_{ij}e'_i$ , and therefore $e_j'=B'_{ij}\tilde e'_i$
Linear map $T\in \mathcal L(V,V')$ .
Bilinear form $\mathfrak B \in \mathcal L(V,V; \mathbb R)$ .

we concluded that,

$\begin{split}v=v^ie_i=\tilde v^i\tilde e_i &\implies \tilde v^i = B_{ij}v^j\quad \text{contravariantly}\\\varepsilon^i, \tilde \varepsilon^i \in V^*, \varepsilon^i(e_j)=\tilde \varepsilon^i(\tilde e_j)=\delta_{ij} &\implies \tilde \varepsilon^i=B_{ij}\varepsilon^j \quad \text{contravariantly}\\f=c_i\varepsilon^i=\tilde c_j \tilde \varepsilon^j&\implies \tilde c_j = F_{ij}c_{i}\quad \text{covariantly}\\T= T_j^ie_i'\varepsilon^j = \tilde {T_j^i} \tilde e_i'\tilde \varepsilon^j &\implies \tilde {T_j^i} = B'_{il}F_{kj}T_k^l\quad \text{contravariant- and covariantly}\\\mathfrak B=\mathfrak B_{ij}\varepsilon^i\varepsilon^j = \tilde {\mathfrak B}_{ij} \tilde \varepsilon^i\tilde \varepsilon^j &\implies \tilde {\mathfrak B}_{ij} = F_{ki}F_{lj}{\mathfrak B}_{kl} \quad \text{covariantly and covariantly}\end{split}$

It is observed that if a vector $v\in V$ is written in terms a single sum/linear combination of basis vectors of $V$ , then the components of the vectors change contravariantely with respect to a change of basis. Then, the covectors are considered and it is observed that their components change covariently upon change of basis of $V^*$ or $V$ . A linear map can be written as a linear combination of vectors and covectors. The coefficients of this combination is seen to change both contra- and covariantely when the bases (of $V$ and $V'$ ) change. A bilinear form though can be written in terms of a linear combination of covectors. The corresponding coefficients change covariantly with change of basis. These results can be generalized toward an abstract definition of a mathematical object called a tensor. There are two following approaches for algebraically denfining a tensor.

Tensor as a multilinear form

Motivated by how linear maps, bilinear forms, and multilinear forms and maps can be written by combining basis vectors and covectors, a generalized combination of these vectors can considered. For example,

$\mathcal T:=\mathcal T^{ij}_k^l_s^t e_i e_j\varepsilon^k e_l\varepsilon^s e_t$

This object $\mathcal T$ consists of a linear combination of a unified (merged) set of basis vector and covectors $e_i e_j\varepsilon^k e_l\varepsilon^s e_t$ (of $V$ and $V^*$ ) by scalar coefficients $\mathcal T^{ij}_k^l_s^t$ . According to the type of the basis vectors, the indices become sub- or superscript, and hence it determines the type of the transformation regarding that index. By reordering the basis vectors and covectors, we can write,

$\mathcal T:=\mathcal T_{ks}^{ijlt} \varepsilon^k \varepsilon^s e_i e_j e_l e_t$

Recalling that vector components can be written as $v^i=\varepsilon^i (v)$ implines that there is a map $T_{(v)}$ for a particular vector $v$ such that,

$v^i=T_{(v)}(\varepsilon^i) \quad\text{s.t}\quad T_{(v)}:V^* \to \mathbb R$

And also the components of a covector $f\in V^*$ determined as,

$f_i=f(e_i)\equiv T_{(f)}(e_i)$

motivates defining $\mathcal T_{ks}^{ijlt}$ as the array (collection of the coefficients) of a multilinear form,

$T:V\times V\times V^*\times V^*\times V^*\times V^*\to \mathbb R \iff \mathcal T_{ks}^{ijlt}\equiv T_{ks}^{ijlt}=T(e_k,e_s,\varepsilon^i,\varepsilon^j,\varepsilon^l,\varepsilon^t)$

Therefore, the object or the multi-dimensional array $\mathcal T_{ks}^{ijlt}$ which is dependent on chosen bases of $V$ and $V^*$ and its transformation rules are based on the types of the beases (or indices) can be intrinsically related to an underlying multilineat map $T$ . By virtue of this observation, an object called a tensor is defined as the following.

Definition: A tensor of type (r,s) on a vector space $V$ over a field $\mathbb R$ (or $\mathbb C$ ) is a multilinear map as,

$\mathcal T: \underbrace{V\times \cdots\times V}_{\text{r times}}\times \underbrace{V^*\times \cdots\times V^*}_{\text{s times}} \to \mathbb R \text{ or } \mathbb C$

The coordinates or the (scalar) components of a tensor $\mathcal T$ can then be determined once a basis $\{e_i\}_1^n$ for $V$ and a basis $\{\varepsilon^i\}_1^n$ for $V^*$ are fixed. Therefore,

$\mathcal T_{i_1\cdots i_r}^{j_1\cdots j_s}\equiv \mathcal T (e_{i_1}, \cdots, e_{i_r}, \varepsilon^{j_1}, \cdots, \varepsilon^{j_s})$

Note that r is the number of covariant indices and s is the number of contravariant indices. A tensor of type (r,s) can be imagined as an r+s-dimensional array of data containing $(\text {dim} V)^{r+s}$ elements. Each index corresponds to a dimention of the data array.

By this definition, a vector $v=v^ie_i$ is a (0,1) tensor as it can be viewed as,

$\begin{split}v&=v^ie_i = \varepsilon^i(v)e_i \implies v^i = \varepsilon^i(v)\\\mathcal T^i&:=v^i\implies \exists \mathcal T:V^*\to \mathbb R\quad \text {s.t}\quad \mathcal T(\varepsilon^i) = v^i\end{split}$

This implies that for each $v\in V$ there is a (multilinear with one input) map $\mathcal T$ receiving a basis covector and returning the corresponding scalar component of the vector ( $\mathcal T:V^*\to \mathbb R$ ). This corresponding map is unique for each vector and it is called a tensor.

A covector (dual vector or a linear form) $f\in V^*$ is a (1,0) tensor because a covector $f$ is a linear map $\mathcal T:V\to \mathbb R$

َ A linear map $T:V\toV$ (or $V\to W$ ) is a (1,1) tensor because the term $T_j^i$ in $T=T_j^ie_i\varepsilon^j$ pertains to a covector’s components for a fixed $i$ and to a vector’s components for a fixed $j$ . Therefore,

$T_j^i=\mathcal T_j^i=\mathcal T(e_j, \varepsilon^i)$

in other words, a linear map is a tensor viewed as a multilinear map $\mathcal T: V\times V^* \to \mahbb R$ . Here, the multilibear form can be considered as $\mathcal T(v,f) = f\circ T(v)$ by which gives $T_j^i=\mathcal T_j^i=\mathcal(e_j, \varepsilon^i)$ . Note that $T(e_i)$ returns a vector and $f$ being $\varepsilon^j$ extract its j-th coordinate, and hencem the array/matrix of the linear map is retrieved.

A bilinear form $\mathfrak B=\mathfrak B_{ij}\varepsilon^i\varepsilon^j$ is then a (2,0) tensor, where $\mathfrak B_{ij}=\mathfrak B(e_i,e_j)$

A multilinear form $T = T_{ij}^k \varepsilon^i\varepsilon^j e'_k$ is a (2,1) tensor where $T_{ij}^k=\mathcal T_{ij}^k=\mathcal T(e_i,e_j,\varepsilon^k)$ where $\mathcal T:V\times V \times V^* \to \mathbb R$ can be considered as $\mathcal T (u,v,f)= f\circ T(u,v)$ .

As an example, the cross product of two vectors $u,v\in \mathbb R^3$ defined as $w=u\times v$ is a multilinear map $\mathbb R^3 \times \mathbb R^3 \to \mathbb R^3$ is a (2,1) tensor.

By convension, scalars are (0,0) tensors.

Remark: for a tensor $\mathcal T_{i_1\cdots i_r}^{j_1\cdots j_s}$ we can write,

$\mathcal T = \mathcal T_{i_1\cdots i_r}^{j_1\cdots j_s} \varepsilon^{i_1}\cdots \varepsilon^{i_r} e_{j_1}\cdots e_{j_s} = \mathcal T (e_{i_1}, \cdots, e_{i_r}, \varepsilon^{j_1}, \cdots, \varepsilon^{j_s})\varepsilon^{i_1}\cdots \varepsilon^{i_r} e_{j_1}\cdots e_{j_s}$

Example: Stress tensor. The Cauchy stress tensor in mechanics is a linear map and hence a (1,1) tensor.

Rank of a tensor

The rank of a (r,s)-type tensor is defined as r+s. In this regard, tensors of different types can have the same rank. For example tensors of types (1,1), (2,0), (0,2) have the same rank being 2. Here we compare these tensors with eachother.

A (1,1) tensor representing a linear map is $\mathcal T_j^i$ where $\mathcal T_j^i=\mathcal T(e_j, \varepsilon^i)$ with $\mathcal T:V\times V^* \to \mathbb R$ .

A (2,0) tensor representing a bilinear form is $\mathcal T_{ij}$ where $\mathcal T_{ij}=\mathcal T(e_i, e_j)$ with $\mathcal T:V\times V \to \mathbb R$ .

A (0,2) tensor is $\mathcal T^{ij}$ where $\mathcal T^{ij}=\mathcal T(\varepsilon^i, \varepsilon^j)$ with $\mathcal T:V^*\times V^* \to \mathbb R$ .

The coeffieints of the above tensor are collected in 2-dimensional array/matrix; however, they follow different transformation rules based on their types.