How to show that the isotropic tensor of order $n$ is a multiple of the Kronecker delta?

Question

I have already found this question here but with the property of invariant under rotation. However, I don't have this property and I want to prove that $T_{ij} = \alpha \delta _{ij}$ where $T_{ij}$ are the components of a Cartesian tensor of order $2$ and the tensor is isotropic meaning that $T_{ij} = q_{ik}q_{jl}T_{kl}$ for all $(q_{ij}) \in \mbox{SO}(3)$ where $\mbox{SO}(3)$ is the special orthogonal group of $3 \times 3$ matrices.

Answer 1

Inspired by the answer to this related question where we do not have only ${\rm SO}(3)$, but ${\rm O}(3)$-invariance:

• First we will show, that $T$ has to be diagonal, for this consider the transform, for $1 \le j \le 3$: $$ (q^j)_{kl} = \begin{cases} 1 & k=j, l=j\\ -1 & k\ne j, l \ne j, k = l\\ 0 & \text{otherwise} \end{cases} $$ for example $q^1$ is given by $$ q^1 = \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$ Then $q^j$ is orthogonal and has determinant $1$, hence $q^j \in {\rm SO}(3)$. Now we have to compute $(q^j)^tTq^j$, but as $q^j$ is symmetric, $(q^j)^t = q^j$. We will do it for $j=1$, the other cases are analogous \begin{align*} q^1Tq^1 &= \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}T \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\\ &= \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} T_{11} & -T_{12} & -T_{13}\\ T_{21} & -T_{22} & -T_{23}\\ T_{31} & -T_{32} & -T_{33} \end{pmatrix}\\ &= \begin{pmatrix} T_{11} & -T_{12} & -T_{13}\\ -T_{21} & T_{22} & T_{23} \\ -T_{31} & T_{32} & T_{33} \end{pmatrix} \end{align*} As this equals $T$, we can read off: $$ T_{12} = -T_{12}, T_{21} = -T_{21}, T_{13} = -T_{13}, T_{31} = -T_{31} $$ hence $T_{12} = T_{21} = T_{13} = T_{31} = 0$. By looking at $q^2$, $q^3$, we see that the other off-diagonal elements of $T$ have to be 0 too. So $T$ is diagonal.

• Knowing that $T$ is diagonal, we will show, that all diagonal entries are equal, for that consider another $q^j$, namely $$ (q^j)_{kl} = \begin{cases} 1 & k=j, l=j\\ 1 & k\ne j, l \ne j, k \ne l\\ 0 & \text{otherwise} \end{cases} $$ for example $q^1$ is given by $$ q^1 = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} $$ Again, $q^j \in {\rm SO}(3)$ is symmetric, hence we have $q^jTq^j = T$, computing gives (we will use that $T$ is diagonal!): \begin{align*} q^1Tq^1 &= \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}T\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\\ &= \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} T_{11} & 0 & 0\\ 0 & 0 & T_{22} \\ 0 & T_{33} & 0 \end{pmatrix}\\ &= \begin{pmatrix} T_{11} & 0 & 0\\ 0 & T_{33} & 0 \\ 0 & 0 & T_{22} \end{pmatrix} \end{align*} Hence, as this is $T$, $T_{22} = T_{33}$. Using $q^2$, $q^3$, gives $T_{11} = T_{22} = T_{33}$, or $T_{ij} = \alpha \delta_{ij}$.