chapter7.tex

\chapter{Tensors in Coordinates}

\begin{flushright}
 "The introduction of numbers as coordinates is an act of violence." \\ Hermann Weyl.

\end{flushright}




\section{Index notation for tensors\label{sub:Index-notation}}

So far we have used a coordinate-free formalism to define and
describe tensors. However, in
many calculations a basis in $V$ is fixed, and one needs to compute
the components of tensors in that basis. 
In this cases the \textbf{index notation} makes such calculations easier.

Suppose a basis $\left\{ \vector{e}_{1},...,\vector{e}_{n}\right\} $
in $V$ is fixed; then the dual basis $\left\{ \vector{e}^{k}\right\} $
is also fixed. Any vector $\vector{v}\in V$ is decomposed as $\vector{v}=\dsum_{k}v^{k}\vector{e}_{k}$
and any covector as $\vector{f}^{*}=\dsum_{k}f_{k}\vector{e}^{k}$.

Any tensor from $V\otimes V$ is decomposed as\[
A=\dsum_{j,k}A^{jk}\vector{e}_{j}\otimes\vector{e}_{k}\in V\otimes V\]
and so on. The action of a covector on a vector is $\vector{f}^{*}\left(\vector{v}\right)=\dsum_{k}f_{k}v_{k}$,
and the action of an operator on a vector is $\dsum_{j,k}\tsor{A}_{jk}v_{k}\vector{e}_{k}$.
However, it is cumbersome to keep writing these sums. In the index
notation, one writes \emph{only} the components $v_{k}$ or $\tsor{A}_{jk}$
of vectors and tensors.


\begin{df}
 Given $T \in \tsor{T}^r_s(V)$:
\[T=\dsum_{j_1=1}^{n} \cdots \dsum_{j_n=1}^{n}  T_{j_1\cdots j_r} ^{j_{r+1}\cdots j_n} \vector{e}^{j_1}\otimes \vector{e}^{j_r} \otimes \vector{e}_{j_{r+1}} \cdots \otimes \vector{e}_{j_{r+s}}
\]

The index notation of this tensor is
\[ T_{j_1\cdots j_r} ^{j_{r+1}\cdots j_n} \]
\end{df}




\subsection{Definition of  index notation}

The rules for expressing tensors in the index notations are as follows:
\begin{dinglist}{111}
 \item Basis vectors $\vector{e}_{k}$ and basis tensors (e.g. $\vector{e}_{k}\otimes\vector{e}_{l}^{*}$)
are never written explicitly. (It is assumed that the basis is fixed
and known.)
\item Instead of a vector $\vector{v}\in V$, one writes its array of components
$v^{k}$ with the \emph{superscript} index.
Covectors $\vector{f}^{*}\in V^{*}$
are written $f_{k}$ with the \emph{subscript} index. The index $k$
runs over integers from $1$ to $N$. Components of vectors and tensors
may be thought of as numbers.
\item Tensors are written as multidimen\-sion\-al arrays of components
with superscript or subscript indices as necessary, for example $\tsor{A}_{jk}\in V^{*}\otimes V^{*}$
or $\tsor{B}_{k}^{lm}\in V\otimes V\otimes V^{*}$. Thus e.g.~the Kronecker
delta symbol is written as $\delta_{k}^{j}$ when it represents the
identity operator $\hat{1}_{V}$. 


\item Tensors with subscript indices, like $\tsor{A}_{ij}$, are called
covariant, while tensors with superscript indices, like $A^{k}$,
are called contravariant. Tensors with both types of indices, like
$\tsor{A}_{lk}^{lmn}$, are called mixed type.

\item Subscript indices, rather than subscripted tensors, are
also dubbed ``covariant'' and superscript indices are dubbed ``contravariant''.

\item For tensor invariance, a pair of dummy indices should in
general be complementary in their variance type, i.e. one covariant
and the other contravariant. 

\item As indicated earlier, tensor order is equal to the number
of its indices while tensor rank is equal to the number of its free
indices; hence vectors (terms, expressions and equalities) are represented
by a single free index and rank-2 tensors are represented by two free
indices. The dimension of a tensor is determined by the range taken
by its indices.

\item The choice of indices must be consistent; each index corresponds to
a particular copy of $V$ or $V^{*}$. Thus it is wrong to write $v_{j}=u_{k}$
or $v_{i}+u^{i}=0$. Correct equations are $v_{j}=u_{j}$ and $v^{i}+u^{i}=0$.
This disallows meaningless expressions such as $\vector{v}^{*}+\vector{u}$
(one cannot add vectors from different spaces).
\item Sums over indices such as $\dsum_{k=1}^{n}a_{k}b_{k}$ are not written
explicitly, the $\dsum$ symbol is omitted, and the \textbf{Einstein
summation convention} is used instead: Summation over all values of
an index is \emph{always implied} when that index letter appears once
as a subscript and once as a superscript. In this case the letter
is called a \textbf{dummy}\index{dummy index} (or \textbf{mute})
\textbf{index}. Thus one writes $f_{k}v^{k}$ instead of $\dsum_{k}f_{k}v_{k}$
and $\tsor{A}_{k}^{j}v^{k}$ instead of $\dsum_{k}\tsor{A}_{jk}v_{k}$. 
\item Summation is allowed \emph{only} over one subscript and one superscript
but never over two subscripts or two superscripts and never over three
or more coincident indices. This corresponds to requiring that we
are only allowed to compute the canonical pairing of $V$ and $V^{*}$
but no other pairing. The expression
$v^{k}v^{k}$ is not allowed because there is no canonical pairing
of $V$ and $V$, so, for instance, the sum $\dsum_{k=1}^{n}v^{k}v^{k}$
depends on the choice of the basis. For the same reason (dependence
on the basis), expressions such as $u^{i}v^{i}w^{i}$ or $\tsor{A}_{ii}\tsor{B}^{ii}$
are not allowed. Correct expressions are $u_{i}v^{i}w_{k}$ and $\tsor{A}_{ik}\tsor{B}^{ik}$.
\item One needs to pay close attention to the choice and the position of
the letters such as $j,k,l$,...~used as indices. Indices that are
not repeated are \textbf{free}\index{free index} indices. The rank
of a tensor expression is equal to the number of free subscript and
superscript indices. Thus $\tsor{A}_{k}^{j}v^{k}$ is a rank $1$ tensor
(i.e.~a vector) because the expression $\tsor{A}_{k}^{j}v^{k}$ has a single
free index, $j$, and a summation over $k$ is implied. 
\item The tensor product symbol $\otimes$ is never written. For example,
if $\vector{v}\otimes\vector{f}^{*}=\dsum_{jk}v_{j}f_{k}^{*}\vector{e}_{j}\otimes\vector{e}^{k}$,
one writes $v^{k}f_{j}$ to represent the tensor $\vector{v}\otimes\vector{f}^{*}$.
The index letters in the expression $v^{k}f_{j}$ are intentionally
chosen to be \emph{different} (in this case, $k$ and $j$) so that
no summation would be implied. In other words, a tensor product is
written simply as a product of components, and the index letters are
chosen appropriately. Then one can interpret $v^{k}f_{j}$ as simply
the product of \emph{numbers}. In particular, it makes no difference
whether one writes $f_{j}v^{k}$ or $v^{k}f_{j}$. The \emph{position
of the indices} (rather than the ordering of vectors) shows in every
case how the tensor product is formed. Note that it is not possible
to distinguish $V\otimes V^{*}$ from $V^{*}\otimes V$ in the index
notation.

\end{dinglist}

\begin{exa}
 It follows from the definition of $\delta_{j}^{i}$ that $\delta_{j}^{i}v^{j}=v^{i}$.
This is the index representation of the identity transformation $\hat{1}\vector{v}=\vector{v}$. 
\end{exa}





\begin{exa}
Suppose $\vector{w}$, $\vector{x}$, $\vector{y}$, and $\vector{z}$
are vectors from $V$ whose components are $w^{i}$, $x^{i}$, $y^{i}$,
$z^{i}$. What are the components of the tensor $\vector{w}\otimes\vector{x}+2\vector{y}\otimes\vector{z}\in V\otimes V$?
\end{exa}


\begin{solu}
 $w^{i}x^{k}+2y^{i}z^{k}$. (We need to choose another letter for the
second free index, $k$, which corresponds to the second copy of $V$
in $V\otimes V$.)
\end{solu}


\begin{exa}
The operator $\hat{A}\equiv\hat{1}_{V}+\lambda\vector{v}\otimes\vector{u}^{*}\in V\otimes V^{*}$
acts on a vector $\vector{x}\in V$. Calculate the resulting vector
$\vector{y}\equiv\hat{A}\vector{x}$.

In the index-free notation, the calculation is\[
\vector{y}=\hat{A}\vector{x}=\left(\hat{1}_{V}+\lambda\vector{v}\otimes\vector{u}^{*}\right)\vector{x}=\vector{x}+\lambda\vector{u}^{*}\left(\vector{x}\right)\vector{v}.\]
In the index notation, the calculation looks like this:\[
y^{k}=\left(\delta_{j}^{k}+\lambda v^{k}u_{j}\right)x^{j}=x^{k}+\lambda v^{k}u_{j}x^{j}.\]
In this formula, $j$ is a dummy index and $k$ is a free index. We
could have also written $\lambda x^{j}v^{k}u_{j}$ instead of $\lambda v^{k}u_{j}x^{j}$
since the ordering of components makes no difference in the index
notation. 
\end{exa}


\begin{exa}
In a physics book you find the following formula, \[
H_{\mu\nu}^{\alpha}=\dfrac{1}{2}\left(h_{\beta\mu\nu}+h_{\beta\nu\mu}-h_{\mu\nu\beta}\right)g^{\alpha\beta}.\]
To what spaces do the tensors $H$, $g$, $h$ belong (assuming these
quantities represent tensors)? Rewrite this formula in the coordinate-free
notation.
\end{exa}


\begin{solu}
$H\in V\otimes V^{*}\otimes V^{*}$, $h\in V^{*}\otimes V^{*}\otimes V^{*}$,
$g\in V\otimes V$. Assuming the simplest case,\[
h=\vector{h}_{1}^{*}\otimes\vector{h}_{2}^{*}\otimes\vector{h}_{3}^{*},\; g=\vector{g}_{1}\otimes\vector{g}_{2},\]
the coordinate-free formula is\[
H=\dfrac{1}{2}\vector{g}_{1}\otimes\left(\vector{h}_{1}^{*}\left(\vector{g}_{2}\right)\vector{h}_{2}^{*}\otimes\vector{h}_{3}^{*}+\vector{h}_{1}^{*}\left(\vector{g}_{2}\right)\vector{h}_{3}^{*}\otimes\vector{h}_{2}^{*}-\vector{h}_{3}^{*}\left(\vector{g}_{2}\right)\vector{h}_{1}^{*}\otimes\vector{h}_{2}^{*}\right).\]
\end{solu}








\subsection{Advantages and disadvantages of index notation}

Index notation is conceptually easier than the index-free notation
because one can imagine manipulating ``merely'' some tables of
numbers, rather than ``abstract vectors.'' In other words, we
are working with \textit{less abstract objects}. The price is that we obscure
the geometric interpretation of what we are doing, and proofs of general
theorems become more difficult to understand.

The main advantage of the index notation is that it makes computations
 with complicated tensors quicker. 

Some \emph{disadvantages} of the index notation are: 

\begin{dinglist}{111}
 \item If the basis is changed, all components need to be recomputed. In
textbooks that use the index notation, quite some time is spent studying
the transformation laws of tensor components under a change of basis.
If different basis are used simultaneously, confusion may result.


\item  The geometrical meaning of many calculations appears hidden behind
a mass of indices. It is sometimes unclear whether a long expression
with indices can be simplified and how to proceed with calculations.

\end{dinglist}

Despite these disadvantages, the index notation enables one to perform
practical calculations with high-rank tensor spaces, such as those
required in field theory and in general relativity. For this reason,
and also for historical reasons (Einstein used the index notation
when developing the theory of relativity), most physics textbooks
use the index notation. In some cases, calculations can be performed
equally quickly using index and index-free notations. In other cases,
especially when deriving general properties of tensors, the index-free
notation is superior.%


\section{Tensor Revisited: Change of Coordinate}

Vectors, covectors, linear operators, and bilinear forms
are examples of tensors.  They are multilinear maps that are
represented numerically when some basis in the space is chosen.

This numeric representation is specific to each of them: vectors
and covectors are represented by one-dimensional arrays, linear
operators and quadratic forms are represented by two-dimensional
arrays. Apart from the number of indices, their position does
matter. The coordinates of a vector are numerated by one upper 
index, which is called the contravariant index. The coordinates of
a covector are numerated by one lower index, which is called the 
covariant index. In a matrix of bilinear form we
use two lower indices; therefore bilinear forms are called
\textbf{twice-covariant tensors}. Linear operators are tensors
of \textbf{mixed type}; their components are numerated by one upper
and one lower index. The number of indices and their positions
determine the transformation rules, i\.\,e\. the way the components
of each particular tensor behave under a change of basis. In the
general case, any tensor is represented by a multidimensional
array with a definite number of upper indices and a definite number
of lower indices. Let's denote these numbers by $r$ and $s$.
Then we have \textbf{a tensor of the type $(r,s)$}, or sometimes the
term \textbf{valency} is used. A tensor of type $(r,s)$, or of valency
$(r,s)$ is called \textbf{an $r$-times contravariant} and
\textbf{an $s$-times covariant} tensor. This is terminology; now let's
proceed to the exact definition. It is based on the following general
transformation formulas:
\begin{align}
&\hskip -2em
\tsor{X}^{i_1\dots\,i_r}_{j_1\dots\,j_s}=\multsum \tsor{S}^{i_1}_{h_1}\dots\,\tsor{S}^{i_r}_{h_r}\tsor{T}^{k_1}_{j_1}
\dots\,\tsor{T}^{k_s}_{j_s}\,\tilde{ \tsor{X}}^{h_1\dots\,h_r}_{k_1\dots\,k_s},
\label{12.1}\\
&\hskip -2em
\tilde{\tsor{X}}^{i_1\dots\,i_r}_{j_1\dots\,j_s}=\multsum \tsor{T}^{i_1}_{h_1}\dots\,\tsor{T}^{i_r}_{h_r}\tsor{S}^{k_1}_{j_1}
\dots\,\tsor{S}^{k_s}_{j_s}\,\tsor{X}^{h_1\dots\,h_r}_{k_1\dots\,k_s}.
\label{12.2}
\end{align}

\begin{definition}[Tensor Definition in Coordinate] A  $(r+s)$-dimensional array $\tsor{X}^{i_1\dots\,i_r}_{j_1
\dots\,j_s}$ of real numbers and such that the components of this
array obey the transformation rules
\begin{align}
&\hskip -2em
\tsor{X}^{i_1\dots\,i_r}_{j_1\dots\,j_s}=\multsum \tsor{S}^{i_1}_{h_1}\dots\,\tsor{S}^{i_r}_{h_r}\tsor{T}^{k_1}_{j_1}
\dots\,\tsor{T}^{k_s}_{j_s}\,\tilde{ \tsor{X}}^{h_1\dots\,h_r}_{k_1\dots\,k_s},
\label{12.1}\\
&\hskip -2em
\tilde{\tsor{X}}^{i_1\dots\,i_r}_{j_1\dots\,j_s}=\multsum \tsor{T}^{i_1}_{h_1}\dots\,\tsor{T}^{i_r}_{h_r}\tsor{S}^{k_1}_{j_1}
\dots\,\tsor{S}^{k_s}_{j_s}\,\tsor{X}^{h_1\dots\,h_r}_{k_1\dots\,k_s}.
\label{12.2}
\end{align}
under a change of basis is called \textbf{tensor} of type $(r,s)$, or
of valency $(r,s)$.
\end{definition}
    Formula \ref{12.2} is derived from \ref{12.1}, so it is
sufficient to remember only one of them. Let it be the formula
\ref{12.1}. Though huge, formula \ref{12.1} is easy to
remember. 

    Indices $i_1,\,\dots,\,i_r$ and $j_1,\,\dots,\,j_s$
are free indices. In right hand side of the equality \ref{12.1}
they are distributed in $S$-s and $T$-s, each having only one
entry and each keeping its position, i\.\,e\. upper indices
$i_1,\,\dots,\,i_r$ remain upper and lower indices $j_1,\,
\dots,\,j_s$ remain lower in right hand side of the equality
\ref{12.1}.\par
    Other indices $h_1,\,\dots,\,h_r$ and $k_1,\,\dots,\,k_s$
are summation indices; they enter the right hand side of
\ref{12.1} pairwise: once as an upper index and once as a
lower index, once in $S$-s or $T$-s and once in components of
array $\tilde{ \tsor{X}}^{h_1\dots\,h_r}_{k_1\dots\,k_s}$.\par
    When expressing $\tsor{X}^{i_1\dots\,i_r}_{j_1\dots\,j_s}$ through
$\tilde{ \tsor{X}}^{h_1\dots\,h_r}_{k_1\dots\,k_s}$ each upper index is
served by direct transition matrix $S$ and produces one summation
in \ref{12.1}:

\begin{equation}
\tsor{X}^{\dots\,\textcolor{Maroon}{i_{\!\scriptscriptstyle\alpha}}\,\dots}_{\dots\,\dots\,
\dots}=\dsum\dots\blue{\dsum}^n_{h_{\!\scriptscriptstyle\alpha}=1}
\dots\dsum\dots\,\tsor{S}^{\,\textcolor{Maroon}{i_{\!\scriptscriptstyle\alpha}}}_{\blue{h_{\!
\scriptscriptstyle\alpha}}}\,\dots\,\tilde{ \tsor{X}}^{\dots\,\blue{h_{\!\scriptscriptstyle\alpha}}\,
\dots}_{\dots\,\dots\,\dots}.
\label{12.3}
\end{equation}
In a similar way, each lower index is served by inverse transition
matrix $T$ and also produces one summation in formula \ref{12.1}:
\begin{equation}
\hskip -2em
\tsor{X}^{\dots\,\dots\,\dots}_{\dots\,\textcolor{Maroon}{j_{\!\scriptscriptstyle\alpha}}\,
\dots}=\dsum\dots\blue{\dsum}^n_{k_{\!\scriptscriptstyle\alpha}=1}
\dots\dsum\dots\,\tsor{T}^{\blue{k_{\!\scriptscriptstyle\alpha}}}_{\,\textcolor{Maroon}{j_{\!
\scriptscriptstyle\alpha}}}\,\dots\,\tilde{ \tsor{X}}^{\dots\,\dots\,\dots}_{\dots\,
\blue{k_{\!\scriptscriptstyle\alpha}}\,\dots}.
\label{12.4}
\end{equation}
Formulas \ref{12.3} and \ref{12.4} are the same as \ref{12.1}
and used to highlight how \ref{12.1} is written. So tensors are
defined. Further we shall consider more examples showing that many
well-known objects undergo the definition~12.1.


\begin{exa}\label{ex:12.1} Verify that formulas  for change of basis of vectors, covectors, linear transformation and bilinear forms are special cases
of formula \ref{12.1}. What are the valencies of vectors, covectors,
linear operators, and bilinear forms when they are considered as
tensors.
\end{exa}


\begin{exa}
 The $\delta_{i}^ j$ is a tensor.
\end{exa}

\begin{solu}
\[ \delta'^j_i=A^j_k(A^{-1})^l_i\delta^k_l=A^j_k(A^{-1})^k_i =\delta^j_i
 \]
\end{solu}


\todoin{terminar os exemplos}

\begin{exa}
  The $\epsilon_{ijk}$ is a pseudo-tensor.
\end{exa}


\begin{exa}\label{ex:12.2} Let $a_{ij}$ be the matrix of some bilinear
form $a$. Let's denote by $b^{ij}$ components of inverse matrix for
$a_{ij}$. Prove that matrix $b^{ij}$ under a change of basis
transforms like matrix of twice-contravariant tensor. Hence it
determines tensor $b$ of valency $(2,0)$. Tensor $b$ is called \negrito{a dual bilinear form} for $a$.
\end{exa}




\subsection{Rank}
The order of a tensor is identified by the number of its
indices (e.g. $\tsor{A}_{jk}^{i}$ is a tensor of order 3) which normally
identifies the tensor rank as well. However, when contraction (see
$\S$ \ref{subsubContraction}) takes place once or more, the order
of the tensor is not affected but its rank is reduced by two for each
contraction operation.\footnote{In the literature of tensor calculus, rank and order of tensors are
generally used interchangeably; however some authors differentiate
between the two as they assign order to the total number of indices,
including repetitive indices, while they keep rank to the number of
free indices. We think the latter is better and hence we follow this
convention in the present text.}

\begin{itemize}
\item ``Zero tensor'' is a tensor whose all components are
zero.
\item ``Unit tensor'' or ``unity tensor'', which is usually
defined for rank-2 tensors, is a tensor whose all elements are zero
except the ones with identical values of all indices which are assigned
the value 1.
\item  While tensors of rank-0 are generally represented in a
common form of light face non-indexed symbols, tensors of rank $\ge1$
are represented in several forms and notations, the main ones are
the index-free notation, which may also be called direct or symbolic
or Gibbs notation, and the indicial notation which is also called
index or component or tensor notation. The first is a geometrically
oriented notation with no reference to a particular reference frame
and hence it is intrinsically invariant to the choice of coordinate
systems, whereas the second takes an algebraic form based on components
identified by indices and hence the notation is suggestive of an underlying
coordinate system, although being a tensor makes it form-invariant
under certain coordinate transformations and therefore it possesses
certain invariant properties. The index-free notation is usually identified
by using bold face symbols, like $\vector{a}$ and $\vector{B}$,
while the indicial notation is identified by using light face indexed
symbols such as $a^{i}$ and $\tsor{B}_{ij}$.
\end{itemize}



\subsection{Examples of Tensors of Different Ranks}
\begin{dinglist}{111}
 \item Examples of rank-0 tensors (scalars) are energy, mass,
temperature, volume and density. These are totally identified by a
single number regardless of any coordinate system and hence they are
invariant under coordinate transformations.

 \item Examples of rank-1 tensors (vectors) are displacement,
force, electric field, velocity and acceleration. These need for their
complete identification a number, representing their magnitude, and
a direction representing their geometric orientation within their
space. Alternatively, they can be uniquely identified by a set of
numbers, equal to the number of dimensions of the underlying space,
in reference to a particular coordinate system and hence this identification
is system-dependent although they still have system-invariant properties
such as length.

 \item Examples of rank-2 tensors are Kronecker delta (see $\S$
\ref{subKronecker}), stress, strain, rate of strain and inertia tensors.
These require for their full identification a set of numbers each
of which is associated with two directions.

 \item Examples of rank-3 tensors are the Levi-Civita tensor (see
$\S$ \ref{subPermutation}) and the tensor of piezoelectric moduli.

 \item Examples of rank-4 tensors are the elasticity or stiffness
tensor, the compliance tensor and the fourth-order moment of inertia
tensor.

 \item Tensors of high ranks are relatively rare in science.
\end{dinglist}


\section{Tensor Operations in Coordinates}

 There are many operations that can be performed on tensors
to produce other tensors in general. Some examples of these operations
are addition/subtraction, multiplication by a scalar (rank-0 tensor),
multiplication of tensors (each of rank $>0$), contraction and permutation.
Some of these operations, such as addition and multiplication, involve
more than one tensor while others are performed on a single tensor,
such as contraction and permutation.

 In tensor algebra, division is allowed only for scalars,
hence if the components of an indexed tensor should appear in a denominator,
the tensor should be redefined to avoid this, e.g. $\tsor{B}_{i}=\dfrac{1}{\tsor{A}_{i}}$.


\subsection{Addition and Subtraction}

 Tensors of the same rank and type can be added algebraically to produce a tensor of
the same rank and type, e.g.
\begin{equation}
a=b+c
\end{equation}
\begin{equation}
\tsor{A}_{i}=\tsor{B}_{i}-\tsor{C}_{i}
\end{equation}
\begin{equation}
\tsor{A}_{j}^{i}=\tsor{B}_{j}^{i}+\tsor{C}_{j}^{i}
\end{equation}

\begin{df}
 Given two tensors $\tsor{Y}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}$ and $\tsor{Z}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}$ of the same type then we define their sum as
 \[\tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}+
\tsor{Y}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}=
\tsor{Z}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}.\]
\end{df}



\begin{theorem} \label{sum-tensors} Given two tensors $\tsor{Y}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}$ and $\tsor{Z}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}$ of type $(r,s)$ then their sum
\[
\tsor{Z}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}=
\tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}+
\tsor{Y}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}.
\]
is also a tensor of type $(r,s)$.
\end{theorem}

\begin{proof}
\[ \tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}=\multsum  \tsor{S}^{i_1}_{h_1}\ldots\,\tsor{S}^{i_r}_{h_r}\tsor{T}^{k_1}_{j_1}
\ldots\,\tsor{T}^{k_s}_{j_s}\,\tilde{ \tsor{X}}^{h_1\ldots\,h_r}_{k_1\ldots\,k_s},\]

\[\tsor{Y}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}=\multsum  \tsor{S}^{i_1}_{h_1}\ldots\,\tsor{S}^{i_r}_{h_r}\tsor{T}^{k_1}_{j_1}
\ldots\,\tsor{T}^{k_s}_{j_s}\,\tilde{ \tsor{Y}}^{h_1\ldots\,h_r}_{k_1\ldots\,k_s},\]

Then
\[\tsor{Z}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}=\multsum  \tsor{S}^{i_1}_{h_1}\ldots\,\tsor{S}^{i_r}_{h_r}\tsor{T}^{k_1}_{j_1}
\ldots\,\tsor{T}^{k_s}_{j_s}\,\tilde{ \tsor{X}}^{h_1\ldots\,h_r}_{k_1\ldots\,k_s}\]
\[+\multsum  \tsor{S}^{i_1}_{h_1}\ldots\,\tsor{S}^{i_r}_{h_r}\tsor{T}^{k_1}_{j_1}
\ldots\,\tsor{T}^{k_s}_{j_s}\,\tilde{ \tsor{Y}}^{h_1\ldots\,h_r}_{k_1\ldots\,k_s}\]

\[\tsor{Z}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}=\multsum  \tsor{S}^{i_1}_{h_1}\ldots\,\tsor{S}^{i_r}_{h_r}\tsor{T}^{k_1}_{j_1}
\ldots\,\tsor{T}^{k_s}_{j_s}\,\left( \tilde{ \tsor{X}}^{h_1\ldots\,h_r}_{k_1\ldots\,k_s} +\tilde{ \tsor{Y}}^{h_1\ldots\,h_r}_{k_1\ldots\,k_s}\right)
\]

\end{proof}




 Addition of tensors is associative and commutative:
\begin{equation}
\left(\vector{A}+\vector{B}\right)+\vector{C}=\vector{A}+\left(\vector{B}+\vector{C}\right)
\end{equation}
\begin{equation}
\vector{A}+\vector{B}=\vector{B}+\vector{A}
\end{equation}



\subsection{Multiplication by Scalar\label{subMultiplicationbyScalar}}

 A tensor can be multiplied by a scalar, which generally
should not be zero, to produce a tensor of the same variance type
and rank, e.g.
\begin{equation}
\tsor{A}_{ik}^{j}=a\tsor{B}_{ik}^{j}
\end{equation}
where $a$ is a non-zero scalar.

\begin{df} Given $\tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}$ a tensor of type $(r,s)$  and $\alpha$ a scalar we define the  multiplication of $\tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}$ by $\alpha$ as:
$$
\tsor{Y}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}=\alpha\,
\tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}.
$$
\end{df}

\begin{theorem}
 Given $\tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}$ a tensor of type $(r,s)$  and $\alpha$ a scalar then 
$$
\tsor{Y}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}=\alpha\,
\tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}.
$$
is also a tensor of type $(r,s)$
\end{theorem}
The proof of this Theorem is very similar to the proof of the Theorem \ref{sum-tensors} and the proof is left as an exercise to the reader.



 As indicated above, multiplying a tensor by a scalar means
multiplying each component of the tensor by that scalar.

 Multiplication by a scalar is commutative, and associative
when more than two factors are involved.


\subsection{Tensor Product \label{subTensorMultiplication} }

 This may also be called outer or exterior or direct or
dyadic multiplication, although some of these names may be reserved
for operations on vectors.

   The tensor product is defined by a more tricky formula. Suppose
we have tensor $\tsor{X}$ of type $(r,s)$ and tensor $\tsor{Y}$
of type $(p,q)$, then we can write:
$$
\hskip -2em
\tsor{Z}^{i_1\ldots\,i_{r+p}}_{j_1\ldots\,j_{s+q}}=
\tsor{X}^{i_1\ldots\,i_r}_{j_1\ldots\,j_s}\,
\tsor{Y}^{i_{r+1}\ldots\,i_{r+p}}_{j_{s+1}\ldots\,j_{s+q}}.
\label{product-tensor}
$$
Formula \ref{product-tensor} produces new tensor $\bold Z$
of the type $(r+p,s+q)$. It is called \negrito{ the tensor product}
of $\tsor{X}$ and $\tsor{Y}$ and denoted $\bold Z=\tsor{X}
\otimes\tsor{Y}$. 


\begin{exa}
 \begin{equation}
\tsor{A}_{i}\tsor{B}_{j}=\tsor{C}_{ij}
\end{equation}
\begin{equation}
A^{ij}\tsor{B}_{kl}=\tsor{C}_{\,\,\,kl}^{ij}
\end{equation}
\end{exa}



 Direct multiplication of tensors is not commutative.

 
 \begin{exa}[Outer Product of Vectors]
  The outer product of two vectors is equivalent to a matrix multiplication
$\vector{uv}^T$, provided that $\vector{u}$ is represented
as a column vector and $\vector{v}$ as a column
vector. And so $\vector{v}^T$ is a row
vector.

\begin{align}\vector{u} \otimes \vector{v} = \vector{u} \vector{v}^\mathrm{T}
= \begin{bmatrix}u_1 \\ u_2 \\ u_3 \\ u_4\end{bmatrix}
\begin{bmatrix}v_1 & v_2 & v_3\end{bmatrix}
= \begin{bmatrix}u_1v_1 & u_1v_2 & u_1v_3 \\ u_2v_1 & u_2v_2 & u_2v_3 \\ u_3v_1 & u_3v_2 & u_3v_3 \\ u_4v_1 & u_4v_2 & u_4v_3\end{bmatrix}.\end{align}

In index notation:

\[(\vector{u} \vector{v}^\mathrm{T})_{ij}=u_iv_j\]

 \end{exa}

 
 
 The outer product operation is distributive with respect
to the algebraic sum of tensors:
\begin{equation}
\vector{A}\left(\vector{B}\pm\vector{C}\right)=\vector{A}\vector{B}\pm\vector{A}\vector{C}\,\,\,\,\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,\,\,\,\,\,\left(\vector{B}\pm\vector{C}\right)\vector{A}=\vector{B}\vector{A}\pm\vector{C\vector{A}}
\end{equation}


 Multiplication of a tensor by a scalar (refer to $\S$
\ref{subMultiplicationbyScalar}) may be regarded as a special case
of direct multiplication.

 The rank-2 tensor constructed as a result of the direct
multiplication of two vectors is commonly called dyad.

 Tensors may be expressed as an outer product of vectors
where the rank of the resultant product is equal to the number of
the vectors involved (e.g. 2 for dyads and 3 for triads).

 Not every tensor can be synthesized as a product of lower
rank tensors.



\subsection{Contraction\label{subsubContraction}}

 Contraction of a tensor of rank $>1$ is to make two free
indices identical, by unifying their symbols, and perform summation
over these repeated indices, e.g.
\begin{equation}
\tsor{A}_{i}^{j}\,\,\,\,\,\,\,\,\,\,\underrightarrow{\mathrm{contraction\,\,}}\,\,\,\,\,\,\,\,\,\,\tsor{A}_{i}^{i}
\end{equation}
\begin{equation}
\tsor{A}_{il}^{jk}\,\,\,\,\,\,\,\,\,\,\underrightarrow{\mathrm{contraction\,\,on}\,\,jl\,\,}\,\,\,\,\,\,\,\,\,\,\tsor{A}_{im}^{mk}
\end{equation}


 Contraction results in a reduction of the rank by 2 since
it implies the annihilation of two free indices. Therefore, the contraction
of a rank-2 tensor is a scalar, the contraction of a rank-3 tensor
is a vector, the contraction of a rank-4 tensor is a rank-2 tensor,
and so on.

 For  non-Cartesian coordinate systems, the pair
of contracted indices should be different in their variance type,
i.e. one upper and one lower. Hence, contraction of a mixed tensor
of type ($m,n$) will, in general, produce a tensor of type ($m-1,n-1$).

 A tensor of type ($p,q$) can have $p\times q$ possible
contractions, i.e. one contraction for each pair of lower and upper
indices.




 \begin{exa}[Trace]
   In matrix algebra, taking the trace (summing the diagonal
elements) can also be considered as contraction of the matrix, which
under certain conditions can represent a rank-2 tensor, and hence
it yields the trace which is a scalar.

 \end{exa}



\todoin{Exemplo}

\subsection{Inner Product\label{secInnerProduct}}

 On taking the outer product 
of two tensors of rank $\ge1$ followed by a contraction on two indices
of the product, an inner product of the two tensors is formed. Hence
if one of the original tensors is of rank-$m$ and the other is of
rank-$n$, the inner product will be of rank-($m+n-2$).

 The inner product operation is usually symbolized by a
single dot between the two tensors, e.g. $\vector{A}\cdot\vector{B}$,
to indicate contraction following outer multiplication.

 In general, the inner product is not commutative. When
one or both of the tensors involved in the inner product are of rank
$>1$ the order of the multiplicands does matter.

 The inner product operation is distributive with respect
to the algebraic sum of tensors:
\begin{equation}
\vector{A}\cdot\left(\vector{B}\pm\vector{C}\right)=\vector{A}\cdot\vector{B}\pm\vector{A}\cdot\vector{C}\,\,\,\,\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,\,\,\,\,\,\left(\vector{B}\pm\vector{C}\right)\cdot\vector{A}=\vector{B}\cdot\vector{A}\pm\vector{C}\cdot\vector{A}
\end{equation}


 \begin{exa}[Dot Product]
 A common example of contraction is the dot product operation
on vectors which can be regarded as a direct multiplication (refer
to $\S$ \ref{subTensorMultiplication}) of the two vectors, which
results in a rank-2 tensor, followed by a contraction.
  
 \end{exa}


 \begin{exa}[Matrix acting on vectors]
 Another common example (from linear algebra) of inner product
is the multiplication of a matrix (representing a rank-2 tensor)  by a vector (rank-1 tensor) to produce a vector,
e.g.
\begin{equation}
\left[\vector{A}\vector{b}\right]_{ij}^{\,\,\,k}=\tsor{A}_{ij}b^{k}\,\,\,\,\,\,\,\,\,\,\underrightarrow{\mathrm{contraction\,\,on}\,\,jk\,\,}\,\,\,\,\,\,\,\,\,\,\left[\vector{A}\cdot\vector{b}\right]_{i}=\tsor{A}_{ij}b^{j}
\end{equation}
\end{exa}


The multiplication of two $n\times n$ matrices is another example
of inner product (see Eq. \ref{eqMatrixMultiplication}).

 For tensors whose outer product produces a tensor of rank
$>2$, various contraction operations between different sets of indices
can occur and hence more than one inner product, which are different
in general, can be defined. Moreover, when the outer product produces
a tensor of rank $>3$ more than one contraction can take place simultaneously.


\subsection{Permutation}

 A tensor may be obtained by exchanging the indices of another
tensor, e.g. transposition of rank-2 tensors.

 Obviously, tensor permutation applies only to tensors of rank $\ge2$.

 The collection of tensors obtained by permuting the indices
of a basic tensor may be called \negrito{isomers}.


\section{Tensor Test: Quotient Rule}

 Sometimes a tensor-like object may be suspected for being
a tensor; in such cases a test based on the ``quotient rule'' can
be used to clarify the situation. According to this rule, if the inner
product of a suspected tensor with a known tensor is a tensor then
the suspect is a tensor. In more formal terms, if it is not known
if $\vector{A}$ is a tensor but it is known that $\vector{B}$ and
$\vector{C}$ are tensors; moreover it is known that the following
relation holds true in all rotated (properly-transformed) coordinate
frames:
\begin{equation}
\tsor{A}_{pq\ldots k\ldots m}\tsor{B}_{ij\ldots k\ldots n}=\tsor{C}_{pq\ldots mij\ldots n}\label{eqQuotientRule}
\end{equation}
then $\vector{A}$ is a tensor. Here, $\vector{A}$, $\vector{B}$
and $\vector{C}$ are respectively of ranks $m,\,n$ and ($m+n-2$),
due to the contraction on $k$ which can be any index of $\vector{A}$
and $\vector{B}$ independently.

 Testing for being a tensor can also be done by applying
first principles through direct substitution in the transformation
equations. However, using the quotient rule is generally more convenient
and requires less work.

 The quotient rule may be considered as a replacement for
the division operation which is not defined for tensors.




\section{Kronecker and Levi-Civita Tensors}

 These tensors are of particular importance in tensor calculus
due to their distinctive properties and unique transformation attributes.
They are numerical tensors with fixed components in all coordinate
systems. The first is called Kronecker delta or unit tensor, while
the second is called Levi-Civita

 The $\delta$ and $\epsilon$ tensors are conserved under
coordinate transformations and hence they are the same for all systems
of coordinate.\footnote{For the permutation tensor, the statement applies to proper coordinate
transformations.}


\subsection{Kronecker $\delta$\label{subKronecker}}

 This is a rank-2 symmetric tensor in all dimensions, i.e.
\begin{equation}
\delta_{ij}=\delta_{ji}\,\,\,\,\,\,\,\,\,\,\,\,\,\left(i,j=1,2,\ldots,n\right)
\end{equation}
Similar identities apply to the contravariant and mixed types of this
tensor.

 It is invariant in all coordinate systems, and hence it
is an isotropic tensor.\footnote{In fact it is more general than isotropic as it is invariant even
under improper coordinate transformations.}

 It is defined as:
\begin{equation}
\delta_{ij}=\begin{cases}
1 & (i=j)\\
0\,\,\,\,\,\,\,\,\,\,\,\,\,\, & (i\neq j)
\end{cases}\label{eqKroneckerDefinitionNormal}
\end{equation}
and hence it can be considered as the identity matrix, e.g. for 3D
\begin{equation}
\left[\delta_{ij}\right]=\left[\begin{array}{ccc}
\delta_{11} & \delta_{12} & \delta_{13}\\
\delta_{21} & \delta_{22} & \delta_{23}\\
\delta_{31} & \delta_{32} & \delta_{33}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]
\end{equation}


 Covariant, contravariant and mixed type of this tensor
are the same, that is
\begin{equation}
\delta_{\,j}^{i}=\delta_{i}^{\,j}=\delta^{ij}=\delta_{ij}
\end{equation}



\subsection{Permutation $\epsilon$\label{subPermutation}}

 This is an isotropic tensor. It has a rank equal to the
number of dimensions; hence, a rank-$n$ permutation tensor has $n^{n}$
components.

 It is totally anti-symmetric in each pair of its indices,
i.e. it changes sign on swapping any two of its indices, that is
\begin{equation}
\epsilon_{i_{1}\ldots i_{k}\ldots i_{l}\ldots i_{n}}=-\epsilon_{i_{1}\ldots i_{l}\ldots i_{k}\ldots i_{n}}
\end{equation}
The reason is that any exchange of two indices requires an even/odd
number of single-step shifts to the right of the first index plus
an odd/even number of single-step shifts to the left of the second
index, so the total number of shifts is odd and hence it is an odd
permutation of the original arrangement.

 It is a pseudo tensor since it acquires a minus sign under
improper orthogonal transformation of coordinates (inversion of axes
with possible superposition of rotation).

 Definition of rank-2 $\epsilon$ ($\epsilon_{ij}$):
\begin{equation}
\epsilon_{12}=1,\,\,\,\,\,\,\,\,\,\,\epsilon_{21}=-1\,\,\,\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,\,\,\,\epsilon_{11}=\epsilon_{22}=0
\end{equation}


 Definition of rank-3 $\epsilon$ ($\epsilon_{ijk}$):
\begin{equation}
\epsilon_{ijk}=\begin{cases}
\,\,\,\,\,1 & (i,j,k\text{ is even permutation of 1,2,3})\\
-1 & (i,j,k\text{ is odd permutation of 1,2,3})\\
\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\, & (\text{repeated\,index})
\end{cases}\label{eqEpsilon3Definition}
\end{equation}


 The definition of rank-$n$ $\epsilon$ ($\epsilon_{i_{1}i_{2}\ldots i_{n}}$)
is similar to the definition of rank-3 $\epsilon$ considering index
repetition and even or odd permutations of its indices $\left(i_{1},i_{2},\cdots,i_{n}\right)$
corresponding to $\left(1,2,\cdots,n\right)$, that is
\begin{equation}
\epsilon_{i_{1}i_{2}\ldots i_{n}}=\begin{cases}
\,\,\,\,\,1 & \left[\left(i_{1},i_{2},\ldots,i_{n}\right)\text{ is even permutation of (\ensuremath{1,2,\ldots,n})}\right]\\
-1 & \left[\left(i_{1},i_{2},\ldots,i_{n}\right)\text{ is odd permutation of (\ensuremath{1,2,\ldots,n})}\right]\\
\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\, & \text{\ensuremath{\left[\mathrm{repeated\,index}\right]}}
\end{cases}\label{eqEpsilonnDefinition}
\end{equation}


 $\epsilon$ may be considered a contravariant relative
tensor of weight $+1$ or a covariant relative tensor of weight $-1$.
Hence, in 2, 3 and $n$ dimensional spaces respectively we have:
\begin{eqnarray}
\epsilon_{ij} & = & \epsilon^{ij}\\
\epsilon_{ijk} & = & \epsilon^{ijk}\\
\epsilon_{i_{1}i_{2}\ldots i_{n}} & = & \epsilon^{i_{1}i_{2}\ldots i_{n}}
\end{eqnarray}



\subsection{Useful Identities Involving $\delta$ or/and $\epsilon$}


\subsubsection{Identities Involving $\delta$}

 When an index of the Kronecker delta is involved in a contraction
operation by repeating an index in another tensor in its own term,
the effect of this is to replace the shared index in the other tensor
by the other index of the Kronecker delta, that is
\begin{equation}
\delta_{ij}\tsor{A}_{j}=\tsor{A}_{i}\label{EqIndexReplace}
\end{equation}
In such cases the Kronecker delta is described as the substitution
or index replacement operator. Hence,
\begin{equation}
\delta_{ij}\delta_{jk}=\delta_{ik}
\end{equation}
Similarly,
\begin{equation}
\delta_{ij}\delta_{jk}\delta_{ki}=\delta_{ik}\delta_{ki}=\delta_{ii}=n\label{eqdeltas}
\end{equation}
where $n$ is the space dimension.

 Because the coordinates are independent of each other:
\begin{equation}
\dfrac{\partial x_{i}}{\partial x_{j}}=\partial_{j}x_{i}=x_{i,j}=\delta_{ij}\label{eqdxdelta}
\end{equation}
Hence, in an $n$ dimensional space we have
\begin{equation}
\partial_{i}x_{i}=\delta_{ii}=n\label{eqdxn}
\end{equation}


 For orthonormal Cartesian systems:
\begin{equation}
\dfrac{\partial x^{i}}{\partial x^{j}}=\dfrac{\partial x^{j}}{\partial x^{i}}=\delta_{ij}=\delta^{ij}\label{eqdxdxdelta}
\end{equation}


 For a set of orthonormal basis vectors in orthonormal Cartesian
systems:
\begin{equation}
\vector{e}_{i}\cdot\vector{e}_{j}=\delta_{ij}
\end{equation}


 The double inner product of two dyads formed by orthonormal
basis vectors of an orthonormal Cartesian system is given by:
\begin{equation}
\vector{e}_{i}\vector{e}_{j}\colon\vector{e}_{k}\vector{e}_{l}=\delta_{ik}\delta_{jl}
\end{equation}



\subsubsection{Identities Involving $\epsilon$}

 For rank-3 $\epsilon$:
\begin{equation}
\epsilon_{ijk}=\epsilon_{kij}=\epsilon_{jki}=-\epsilon_{ikj}=-\epsilon_{jik}=-\epsilon_{kji}\,\,\,\,\,\,\,\,\,\,\text{(sense of cyclic order)}\label{EqEpsilonCycle}
\end{equation}
These equations demonstrate the fact that rank-3 $\epsilon$ is totally
anti-symmetric in all of its indices since a shift of any two indices
reverses the sign. This also reflects the fact that the above tensor
system has only one independent component.

 For rank-2 $\epsilon$:
\begin{equation}
\epsilon_{ij}=\left(j-i\right)
\end{equation}


 For rank-3 $\epsilon$:
\begin{equation}
\epsilon_{ijk}=\dfrac{1}{2}\left(j-i\right)\left(k-i\right)\left(k-j\right)
\end{equation}


 For rank-4 $\epsilon$:
\begin{equation}
\epsilon_{ijkl}=\dfrac{1}{12}\left(j-i\right)\left(k-i\right)\left(l-i\right)\left(k-j\right)\left(l-j\right)\left(l-k\right)
\end{equation}


 For rank-$n$ $\epsilon$:
\begin{equation}
\epsilon_{a_{1}a_{2}\cdots a_{n}}=\prod_{i=1}^{n-1}\left[\dfrac{1}{i!}\prod_{j=i+1}^{n}\left(a_{j}-a_{i}\right)\right]=\dfrac{1}{S(n-1)}\prod_{1\le i<j\le n}\left(a_{j}-a_{i}\right)
\end{equation}
where $S(n-1)$ is the super-factorial function of $(n-1)$ which
is defined as
\begin{equation}
S(k)=\prod_{i=1}^{k}i!=1!\cdot2!\cdot\ldots\cdot k!
\end{equation}
A simpler formula for rank-$n$ $\epsilon$ can be obtained from the
previous one by ignoring the magnitude of the multiplication factors
and taking only their signs, that is
\begin{equation}
\epsilon_{a_{1}a_{2}\cdots a_{n}}=\prod_{1\le i<j\le n}\sigma\left(a_{j}-a_{i}\right)=\sigma\left(\prod_{1\le i<j\le n}\left(a_{j}-a_{i}\right)\right)
\end{equation}
where
\begin{equation}
\sigma(k)=\begin{cases}
+1 & (k>0)\\
-1 & (k<0)\\
\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\, & (k=0)
\end{cases}
\end{equation}


 For rank-$n$ $\epsilon$:
\begin{equation}
\epsilon_{i_{1}i_{2}\cdots i_{n}}\,\epsilon_{i_{1}i_{2}\cdots i_{n}}=n!
\end{equation}
because this is the sum of the squares of $\epsilon_{i_{1}i_{2}\cdots i_{n}}$
over all the permutations of $n$ different indices which is equal
to $n!$ where the value of $\epsilon$ of each one of these permutations
is either $+1$ or $-1$ and hence in both cases their square is 1.

 For a symmetric tensor $\tsor{A}_{jk}$:
\begin{equation}
\epsilon_{ijk}\tsor{A}_{jk}=0
\end{equation}
because an exchange of the two indices of $\tsor{A}_{jk}$ does not affect
its value due to the symmetry whereas a similar exchange in these
indices in $\epsilon_{ijk}$ results in a sign change; hence each
term in the sum has its own negative and therefore the total sum will
vanish.


\begin{equation}
\epsilon_{ijk}\tsor{A}_{i}\tsor{A}_{j}=\epsilon_{ijk}\tsor{A}_{i}\tsor{A}_{k}=\epsilon_{ijk}\tsor{A}_{j}\tsor{A}_{k}=0\label{eqPermutingTwoFactors}
\end{equation}
because, due to the commutativity of multiplication, an exchange of
the indices in $A$'s will not affect the value but a similar exchange
in the corresponding indices of $\epsilon_{ijk}$ will cause a change
in sign; hence each term in the sum has its own negative and therefore
the total sum will be zero.

 For a set of orthonormal basis vectors in a 3D space with
a right-handed orthonormal Cartesian coordinate system:
\begin{equation}
\vector{e}_{i}\times\vector{e}_{j}=\epsilon_{ijk}\vector{e}_{k}
\end{equation}
\begin{equation}
\vector{e}_{i}\cdot\left(\vector{e}_{j}\times\vector{e}_{k}\right)=\epsilon_{ijk}
\end{equation}



\subsubsection{Identities Involving $\delta$ and $\epsilon$}


\begin{equation}
\epsilon_{ijk}\delta_{1i}\delta_{2j}\delta_{3k}=\epsilon_{123}=1
\end{equation}


 For rank-2 $\epsilon$:
\begin{equation}
\epsilon_{ij}\epsilon_{kl}=\begin{vmatrix}\begin{array}{cc}
\delta_{ik} & \delta_{il}\\
\delta_{jk} & \delta_{jl}
\end{array}\end{vmatrix}=\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}
\end{equation}
\begin{equation}
\epsilon_{il}\epsilon_{kl}=\delta_{ik}
\end{equation}
\begin{equation}
\epsilon_{ij}\epsilon_{ij}=2
\end{equation}


 For rank-3 $\epsilon$:
\begin{equation}
\epsilon_{ijk}\epsilon_{lmn}=\begin{vmatrix}\begin{array}{ccc}
\delta_{il} & \delta_{im} & \delta_{in}\\
\delta_{jl} & \delta_{jm} & \delta_{jn}\\
\delta_{kl} & \delta_{km} & \delta_{kn}
\end{array}\end{vmatrix}=\delta_{il}\delta_{jm}\delta_{kn}+\delta_{im}\delta_{jn}\delta_{kl}+\delta_{in}\delta_{jl}\delta_{km}-\delta_{il}\delta_{jn}\delta_{km}-\delta_{im}\delta_{jl}\delta_{kn}-\delta_{in}\delta_{jm}\delta_{kl}
\end{equation}
\begin{equation}
\epsilon_{ijk}\epsilon_{lmk}=\begin{vmatrix}\begin{array}{cc}
\delta_{il} & \delta_{im}\\
\delta_{jl} & \delta_{jm}
\end{array}\end{vmatrix}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\label{EqEpsilonDelta}
\end{equation}
The last identity is very useful in manipulating and simplifying tensor
expressions and proving vector and tensor identities.
\begin{equation}
\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}
\end{equation}
\begin{equation}
\epsilon_{ijk}\epsilon_{ijk}=2\delta_{ii}=6
\end{equation}
since the rank and dimension of $\epsilon$ are the same, which is
3 in this case.

 For rank-$n$ $\epsilon$:
\begin{equation}
\epsilon_{i_{1}i_{2}\cdots i_{n}}\,\epsilon_{j_{1}j_{2}\cdots j_{n}}=\begin{vmatrix}\begin{array}{cccc}
\delta_{i_{1}j_{1}} & \delta_{i_{1}j_{2}} & \cdots & \delta_{i_{1}j_{n}}\\
\delta_{i_{2}j_{1}} & \delta_{i_{2}j_{2}} & \cdots & \delta_{i_{2}j_{n}}\\
\vdots & \vdots & \ddots & \vdots\\
\delta_{i_{n}j_{1}} & \delta_{i_{n}j_{2}} & \cdots & \delta_{i_{n}j_{n}}
\end{array}\end{vmatrix}\label{eqEpsilon2}
\end{equation}


 According to Eqs. \ref{eqEpsilon3Definition} and \ref{EqIndexReplace}:
\begin{equation}
\epsilon_{ijk}\delta_{ij}=\epsilon_{ijk}\delta_{ik}=\epsilon_{ijk}\delta_{jk}=0
\end{equation}




\subsection{$\star$ Generalized Kronecker delta}

 The generalized Kronecker delta is defined by:
\begin{equation}
\delta_{j_{1}\ldots j_{n}}^{i_{1}\ldots i_{n}}=\begin{cases}
\,\,\,\,\,1 & \left[(j_{1}\ldots j_{n})\text{ is even permutation of (\ensuremath{i_{1}\ldots i_{n})}}\right]\\
-1 & \left[(j_{1}\ldots j_{n})\text{ is odd permutation of (\ensuremath{i_{1}\ldots i_{n})}}\right]\\
\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\, & \left[\text{repeated\,\,\ensuremath{j}'s}\right]
\end{cases}\label{eqGeneralizedKronecker}
\end{equation}
It can also be defined by the following $n\times n$ determinant:
\begin{equation}
\delta_{j_{1}\ldots j_{n}}^{i_{1}\ldots i_{n}}=\begin{vmatrix}\begin{array}{cccc}
\delta_{j_{1}}^{i_{1}} & \delta_{j_{2}}^{i_{1}} & \cdots & \delta_{j_{n}}^{i_{1}}\\
\delta_{j_{1}}^{i_{2}} & \delta_{j_{2}}^{i_{2}} & \cdots & \delta_{j_{n}}^{i_{2}}\\
\vdots & \vdots & \ddots & \vdots\\
\delta_{j_{1}}^{i_{n}} & \delta_{j_{2}}^{i_{n}} & \cdots & \delta_{j_{n}}^{i_{n}}
\end{array}\end{vmatrix}\label{eqGeneralizedKronecker2}
\end{equation}
where the $\delta_{j}^{i}$ entries in the determinant are the normal
Kronecker delta as defined by Eq. \ref{eqKroneckerDefinitionNormal}.

 Accordingly, the relation between the rank-$n$ $\epsilon$
and the generalized Kronecker delta in an $n$ dimensional space is given by:
\begin{equation}
\epsilon_{i_{1}i_{2}\ldots i_{n}}=\delta_{i_{1}i_{2}\ldots i_{n}}^{1\,2\ldots n}\,\,\,\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,\,\,\,\epsilon^{i_{1}i_{2}\ldots i_{n}}=\delta_{1\,2\ldots n}^{i_{1}i_{2}\ldots i_{n}}
\end{equation}
Hence, the permutation tensor $\epsilon$ may be considered as a special
case of the generalized Kronecker delta. Consequently the permutation
symbol can be written as an $n\times n$ determinant consisting of
the normal Kronecker deltas.

 If we define
\begin{equation}
\delta_{lm}^{ij}=\delta_{lmk}^{ijk}
\end{equation}
then Eq. \ref{EqEpsilonDelta} will take the following form:
\begin{equation}
\delta_{lm}^{ij}=\delta_{l}^{i}\delta_{m}^{j}-\delta_{m}^{i}\delta_{l}^{j}
\end{equation}
Other identities involving $\delta$ and $\epsilon$ can also be formulated
in terms of the generalized Kronecker delta.

 On comparing Eq. \ref{eqEpsilon2} with Eq. \ref{eqGeneralizedKronecker2}
we conclude
\begin{equation}
\delta_{j_{1}\ldots j_{n}}^{i_{1}\ldots i_{n}}=\epsilon^{i_{1}\ldots i_{n}}\,\epsilon_{j_{1}\ldots j_{n}}
\end{equation}



\section{Types of Tensors Fields}

 In the following subsections we introduce a number of tensor
types and categories and highlight their main characteristics and
differences. These types and categories are not mutually exclusive
and hence they overlap in general; moreover they may not be exhaustive
in their classes as some tensors may not instantiate any one of a
complementary set of types such as being symmetric or anti-symmetric.


\subsection{Isotropic and Anisotropic Tensors}

 Isotropic tensors are characterized by the property that
the values of their components are invariant under coordinate transformation
by proper rotation of axes. In contrast, the values of the components
of anisotropic tensors are dependent on the orientation of the coordinate
axes. Notable examples of isotropic tensors are scalars (rank-0),
the vector $\vector{0}$ (rank-1), Kronecker delta $\delta_{ij}$
(rank-2) and Levi-Civita tensor $\epsilon_{ijk}$ (rank-3). Many tensors
describing physical properties of materials, such as stress and magnetic
susceptibility, are anisotropic.

 Direct and inner products of isotropic tensors are isotropic
tensors.

 The zero tensor of any rank is isotropic; therefore if
the components of a tensor vanish in a particular coordinate system
they will vanish in all properly and improperly rotated coordinate
systems.\footnote{For improper rotation, this is more general than being isotropic.}
Consequently, if the components of two tensors are identical in a
particular coordinate system they are identical in all transformed
coordinate systems.

 As indicated, all rank-0 tensors (scalars) are isotropic.
Also, the zero vector, $\vector{0}$, of any dimension is isotropic;
in fact it is the only rank-1 isotropic tensor.


\begin{theorem}
Any isotropic second order tensor $T_{ij}$ we can be written as 
\[T_{ij} = \lambda \delta_{ij}\] for some scalar $\lambda$.
\end{theorem}


\begin{proof}
First we will prove that $T$ is diagonal. Let
$R$ be the reflection in the hyperplane perpendicular to the $j$-th vector in the standard ordered basis.
$$R_{kl}=\begin{cases}
 -1 & \text{if }k=l=j\\
 \delta_{kl} & \text{otherwise}\\
\end{cases}$$ 
therefore 
$$R=R^T\land R^2=I\Rightarrow R^TR=RR^T=I$$
Therefore:
$$T_{ij}=\sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\\
i\neq j\Rightarrow T_{ij}=-T_{ij}\Rightarrow T_{ij}=0$$

Now we will prove that  $T_{jj} = T_{11}$. Let  $P$ be the permutation matrix that interchanges the 1st and j-th rows when acrting by  left multiplication.
$$P_{kl}=\begin{cases}
 \delta_{jl} & \text{if } k=1\\
 \delta_{1l} & \text{if } k=j\\
 \delta_{kl} & \text{otherwise} 
\end{cases}$$
$$(P^TP)_{kl}=\sum_{m}P^T_{km}P_{ml}=\sum_{m}P_{mk}P_{ml}=\sum_{m\neq1,j}P_{mk}P_{ml}+\sum_{m=1,j}P_{mk}P_{ml}\\
=\sum_{m\neq1,j}\delta_{mk}\delta_{ml}+\delta_{jk}\delta_{jl}+\delta_{1k}\delta_{1l}=\sum_{m}\delta_{mk}\delta_{ml}=\delta_{kl}$$
Therefore:
$$T_{jj}=\sum_{p,q}P_{jp}P_{jq}T_{pq}=\sum_{q}P_{jq}^2T_{qq}=\sum_{q}\delta_{1q}^2T_{qq}=\sum_{q}\delta_{1q}T_{qq}=T_{11}$$
\end{proof}



\subsection{Symmetric and Anti-symmetric Tensors}

 These types of tensor apply to high ranks only (rank $\ge2$).
Moreover, these types are not exhaustive, even for tensors of ranks
$\ge2$, as there are high-rank tensors which are neither symmetric
nor anti-symmetric.

 A rank-2 tensor $\tsor{A}_{ij}$ is symmetric \textit{iff }for
all $i$ and $j$
\begin{equation}
\tsor{A}_{ji}=\tsor{A}_{ij}
\end{equation}
and anti-symmetric or skew-symmetric \textit{iff}
\begin{equation}
\tsor{A}_{ji}=-\tsor{A}_{ij}
\end{equation}
Similar conditions apply to contravariant type tensors (refer also
to the following).

 A rank-$n$ tensor $\tsor{A}_{i_{1}\ldots i_{n}}$ is symmetric
in its two indices $i_{j}$ and $i_{l}$ \textit{iff}
\begin{equation}
\tsor{A}_{i_{1}\ldots i_{l}\ldots i_{j}\ldots i_{n}}=\tsor{A}_{i_{1}\ldots i_{j}\ldots i_{l}\ldots i_{n}}
\end{equation}
and anti-symmetric or skew-symmetric in its two indices $i_{j}$ and
$i_{l}$ \textit{iff}
\begin{equation}
\tsor{A}_{i_{1}\ldots i_{l}\ldots i_{j}\ldots i_{n}}=-\tsor{A}_{i_{1}\ldots i_{j}\ldots i_{l}\ldots i_{n}}
\end{equation}


 Any rank-2 tensor $\tsor{A}_{ij}$ can be synthesized from (or
decomposed into) a symmetric part $\tsor{A}_{(ij)}$ (marked with round brackets
enclosing the indices) and an anti-symmetric part $\tsor{A}_{[ij]}$ (marked
with square brackets) where
\begin{equation}
\tsor{A}_{ij}=\tsor{A}_{(ij)}+\tsor{A}_{[ij]},\,\,\,\,\,\,\,\,\,\,\tsor{A}_{(ij)}=\dfrac{1}{2}\left(\tsor{A}_{ij}+\tsor{A}_{ji}\right)\,\,\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,\,\,\tsor{A}_{[ij]}=\dfrac{1}{2}\left(\tsor{A}_{ij}-\tsor{A}_{ji}\right)
\end{equation}


 A rank-3 tensor $\tsor{A}_{ijk}$ can be symmetrized by
\begin{equation}
\tsor{A}_{(ijk)}=\dfrac{1}{3!}\left(\tsor{A}_{ijk}+\tsor{A}_{kij}+\tsor{A}_{jki}+\tsor{A}_{ikj}+\tsor{A}_{jik}+\tsor{A}_{kji}\right)
\end{equation}
and anti-symmetrized by
\begin{equation}
\tsor{A}_{[ijk]}=\dfrac{1}{3!}\left(\tsor{A}_{ijk}+\tsor{A}_{kij}+\tsor{A}_{jki}-\tsor{A}_{ikj}-\tsor{A}_{jik}-\tsor{A}_{kji}\right)
\end{equation}


 A rank-$n$ tensor $\tsor{A}_{i_{1}\ldots i_{n}}$ can be symmetrized
by
\begin{equation}
\tsor{A}_{(i_{1}\ldots i_{n})}=\dfrac{1}{n!}\left(\text{sum of all even \& odd permutations of indices \ensuremath{i}'s}\right)
\end{equation}
and anti-symmetrized by
\begin{equation}
\tsor{A}_{\left[i_{1}\ldots i_{n}\right]}=\dfrac{1}{n!}\left(\text{sum of all even permutations minus sum of all odd permutations}\right)
\end{equation}


 For a symmetric tensor $\ensuremath{\tsor{A}_{ij}}$ and an anti-symmetric
tensor $\ensuremath{\tsor{B}^{ij}}$ (or the other way around) we have
\begin{equation}
\tsor{A}_{ij}\tsor{B}^{ij}=0
\end{equation}


 The indices whose exchange defines the symmetry and anti-symmetry
relations should be of the same variance type, i.e. both upper or
both lower.

 The symmetry and anti-symmetry characteristic of a tensor
is invariant under coordinate transformation.

 A tensor of high rank ($>2$) may be symmetrized or anti-symmetrized
with respect to only some of its indices instead of all of its indices,
e.g.
\begin{equation}
\tsor{A}_{(ij)k}=\dfrac{1}{2}\left(\tsor{A}_{ijk}+\tsor{A}_{jik}\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tsor{A}_{[ij]k}=\dfrac{1}{2}\left(\tsor{A}_{ijk}-\tsor{A}_{jik}\right)
\end{equation}


 A tensor is totally symmetric \textit{iff}
\begin{equation}
\tsor{A}_{i_{1}\ldots i_{n}}=\tsor{A}_{(i_{1}\ldots i_{n})}
\end{equation}
and totally anti-symmetric \textit{iff}
\begin{equation}
\tsor{A}_{i_{1}\ldots i_{n}}=\tsor{A}_{[i_{1}\ldots i_{n}]}
\end{equation}


 For a totally skew-symmetric tensor (i.e. anti-symmetric
in all of its indices), nonzero entries can occur only when all the
indices are different.