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\myskip
How do we know if a bunch of numbers $T$\dots form
a tensor? \myskip
If $T$ is a $(p + q)$-tensor then for every
$q$-tensor $u$,
\[ v_{i_1 \cdots i_p} = T_{i_1 \cdots i_p j_1
\cdots j_q} u_{j_1 \cdots j_q} \]
is a $p$-tensor. \myskip
Conversely, if $v$ is a $p$-tensor for every
$q$-tensor $u$, then $T$ is a $(p + q)$-tensor.
This is the \emph{quotient rule}.

\begin{proof}
    Consider
    \[ u_{j_1 \cdots j_q} = c_{j_1} \cdots d_{j_q} \]
    By assumption
    \[ v_{i_1 \cdots i_p} = T_{i_1 \cdots i_p j_1
    \cdots j_q} c_{j_1} \cdots d_{j_q} \]
    is a tensor, so
    \[ a_{i_1} \cdots b_{i_p} v_{i_1 \cdots i_p} =
    T_{i_1 \cdots i_p j_1 \cdots j_q} a_{i_1}
    \cdots b_{i_p} c_{j_1} \cdots d_{j_q} \]
    is a scalar, $\forall\,\, \bf{a}, \cdots
    \bf{b}, \bf{c}, \cdots \bf{d}$ hence $T$ must
    be a $(p + q)$-tensor.
\end{proof}

\subsubsection*{(Anti)-Symmetry}

A tensor that obeys
\[ T_{ijp \cdots q} = \pm T_{jip \cdots q} \]
is said to be (anti)-symmetric in $i, j$ (anti for
$-$). This is a basis independent statement:
\begin{align*}
    T_{ijp \cdots q}'
    &= R_{ik} R_{jl} R_{pr} \cdots R_{qs} T_{klr
    \cdots s} \\
    &= \pm R_{ik} R_{jk} R_{pr} \cdots R_{qs}
    T_{lkr \cdots s} \\
    &= \pm T_{jip \cdots q}'
\end{align*}
If $T$ is (anti)-symmetric in all indices it is
said to be \emph{totally (anti)-symmetric}. A
totally anti-symmetric $p$-tensor in $\RR^n$ has
${n \choose p}$ independent components, and
vanishes in $p > n$. \myskip
In $\RR^3$, a 2-tensor $T_{ij}$ decomposes as
\[  S_{ij} = \half (T_{ij} + T_{ij}) \]
\[ A_{ij} = \half (T_{ij} - T_{ji}) \]
and $S_{ij}$ further decomposes as
\[  S_{ij} = P_{ij} + \frac{1}{3} Q \delta_{ij} \]
where $P_{ij}$ is traceless (i.e. $P_{ii} = 0$)
and the trace of $S_{ij}$ is $Q$. \myskip
In $\RR^3$ we have another invariant tensor
$\eps_{ijk}$ (see below) and we can write
\[ A_{ij} = \eps_{ijk} B_k \iff B_k \half
\eps_{klm} A_{lm} \]
So a $3 \times 3$ matrix can be written as
\[ T_{ij} = P_{ij} + \eps_{ijk} B_k \frac{1}{3} Q
\delta_{ij} \]
with $P$, $B$ and $Q$ are themselves tensors.

\subsubsection*{Invariant Tensors}

A tensor that obeys
\[ T_{i_1 \cdots i_p}' = R_{i_1j_1} \cdots
R_{i_pj_p} T_{j_1 \cdots j_p} = T_{i_1 \cdots i_p} \]
for all $R$ is called an \emph{invariant tensor}
or is said to be \emph{isotropic}. \myskip
Any rank 0 tensor is isotropic. There are no rank
1 isotropic tensors. There is a rank 2, and in
$\RR^3$, a rank 3 invariant tensor:
\begin{itemize}
    \item $\delta_{ij}$ with $\delta_{ij}' =
        R_{ik} R_{jl} \delta_{kl} = \delta_{il}$
    \item $\eps_{ijk}$ with
        \begin{align*}
            \eps_{ijk}'
            &= R_{il} R_{jm} R_{kn} \eps_{lmn} \\
            &= (\det R) \eps_{ijk} \\
            &= \eps_{ijk}
        \end{align*}
\end{itemize}

\begin{claim*}
    The only isotropic tensors in $\RR^3$ of rank
    $1 \le p \le 3$ are
    \[ T_{ij} = \alpha \delta_{ij} \]
    and
    \[ T_{ijk} = \beta \eps_{ijk} \]
    with $\alpha, \beta$ constant.
\end{claim*}

\begin{proof}
    Look for a rank 1 tensor. Must have
    \[ T_i' = R_{ij} T_j = T_i \]
    for
    \[ R_{ij} =
    \begin{pmatrix}
        -1 & 0 & 0 \\
        0 & -1 & 0 \\
        0 & 0 & 1
    \end{pmatrix}
    \]
    hence $T_1' = -T_1$ and $T_2' = -T_2$ so $T_1
    = T_2 = 0$. A similar argument gives $T_3 =
    0$. \myskip
    Look for a rank 2 tensor:
    \[ T_{ij}' = \tilde{R}_{ik} \tilde{R}_{jl}
    T_{kl} = T_{ij} \]
    with
    \[ \tilde{R}_{ij} =
    \begin{pmatrix}
        0 & +1 & 0 \\
        -1 & 0 & 0 \\
        0 & 0 & 1
    \end{pmatrix}
    \]
    ($\frac{\pi}{2}$ rotation about $z$-axis)
    This gives $T_{13}' = T_{23}$ and $T_{23}' =
    -T_{13}$ hence $T_{13} = T_{23} = 0$. Also
    $T_{11}' = T_{22}$. Similar arguments show
    that $T_{ij} = 0$ for $i \neq j$ and $T_{11} =
    T_{22} = T_{33} = \alpha$
    \[ \implies T_{ij} = \alpha \delta_{ij} \]
    For rank 3,
    \[ T_{ijk}' = R_{il} R_{jp} R_{kq} T_{lpq} =
    T_{ijk} \]
    Use
    \[ R =
    \begin{pmatrix}
        -1 & 0 & 0 \\
        0 & -1 & 0 \\
        0 & 0 & +1
    \end{pmatrix}
    \implies T_{133}' = -T_{133}, T_{111}' = -T_{111} \]
    $\implies T_{ijk} = 0$ unless $i, j, k$
    distinct. Use $R = \tilde{R}$ to show that
    $T_{123}' = -T_{213}$
    \[ \implies T_{ijk} = \beta \eps_{ijk} \]
\end{proof}

\myskip
All higher rank invariant tensors in $\RR^3$ are
built from $\eps_{ijk}$ and $\delta_{ij}$, for
example isotropic rank 4 tensor has the most
general form
\[ T_{ijkl} = \alpha \delta_{ij} \delta_{kl} +
\beta \delta_{ik} \delta_{jl} + \gamma \delta_{il}
\delta_{jk} \]
for $\alpha, \beta, \gamma$ some constants.

\subsubsection*{Invariant Integrals}

We can sometimes use this to do integrals, for
example
\[ T_{ij \cdots k} = \int_V f(r) x_i x_j \cdots x_k
\dd V \]
($V$ is a spherically symmetric region and $r =
|\bf{x}|$). Under a rotation
\begin{align*}
    T_{ij \cdots k}'
    &= R_{ip} R_{jq} \cdots R_{kr} T_{pq \cdots r}
    \\
    &= \int_V f(r) x_i' x_j' \cdots x_k' \dd V
\end{align*}
($x_i' = R_{ip} x_p$). Change variables to $x'$.
Both $r = |\bf{x}|$ and $V$ are invariant
\[ \implies T_{ij \cdots k}' = T_{ij \cdots k} \]
so must be proportional to an invariant tensor.


\begin{example*}
    Consider the integral over 3D ball of radius
    $R$:
    \[ T_{ij} = \int_V \rho(r) x_i x_j \dd V \]
    Necessarily $= \alpha \delta_{ij}$ for some
    $\alpha$. Take the trace:
    \[ \implies \int_V \rho(r) r^2 \dd V = 3\alpha \]
    ($\delta_{ii} = 3$)
    \[ \implies T_{ij} = \frac{1}{3} \delta_{ij}
    \int_V \rho(r) r^2 \dd V \]
\end{example*}
