# Rotation invariant tensors

It is often claimed that the only tensors invariant under the orthogonal transformations (rotations) are the Kronecker delta $\delta_{ij}$, the Levi-Civita epsilon $\epsilon_{ijk}$ and various combinations of their tensor products. While it is easy to check that $\delta_{ij}$ and $\epsilon_{ijk}$ are indeed invariant under rotations, I would like to know if there exist any proof by construction that they are the only (irreducible) tensors with this property.

#### Solutions Collecting From Web of "Rotation invariant tensors"

This is somewhat late in the day / year, but I suspect the author is asking about representations of isotropic Cartesian tensors, where “isotropic” means “invariant under the action of proper orthogonal transformations” and “Cartesian” means the underlying space is Euclidean $R^n$ ($R^3$ is a case of great practical interest).

The proofs for the two cases asked here are non-trivial, and given in

Weyl, H., The Classical Groups, Princeton University Press, 1939

Constructions for higher-order isotropic Cartesian tensors are also given there.

Harold Jeffreys (1973). On isotropic tensors. Mathematical Proceedings of the Cambridge Philosophical Society, 73, pp 173-176.

The proof given is a lot more concrete and “hands on” than Weyl’s proof linked to by user_of_math.

$\mathtt{Definition:}$ $T$ is an isotropic tensor of type $(0,n)$ if $\;T_{i_1i_2…i_n}=R_{i_1j_1}R_{i_2j_2}…R_{i_nj_n}T_{j_1j_2…j_n}$ whenever $R$ is an orthogonal matrix i.e $R^TR=RR^T=I$.

$\mathtt{n=2:}$See my answer here.

$\mathtt{n=3:}$For tensors of type $(0,3)$ we can mimick the proof for $n=2$ to deduce skew-symmetricness. Suppose $T_{pqr}$ is an isotropic tensor. Let $R$ be a diagonal matrix whose diagonal entries are $1$ except for $R_{ii}$ and $R_{ii}=-1$. $R$ is diagonal and its own inverse hence it’s orthogonal.
$$T_{ijj}=\sum_{p,q,r}R_{ip}R_{jq}R_{kj}T_{pqr}=R_{ii}R_{jj}R_{jj}T_{ijj}\text{( using the fact that R is diagonal)}\\ \Rightarrow T_{ijj}=-T_{ijj}=0$$
Using the symmetry of this argument we can show that the only nonzero components of $T$ are those whose indices are a permutation of $(1,2,3)$. Suppose $i\neq j$. Define
$$R_{lm}=\begin{cases} -\delta_{jm} & \text{if } l=i\\ \delta_{im} & \text{if } l=j\\ \delta_{lm} & \text{otherwise} \end{cases}\\ (R^TR)_{lm}=\sum_{n}R_{nl}R_{nm}=\sum_{n\neq i,j}R_{nl}R_{nm}+(-\delta_{jl})(-\delta_{jm})+\delta_{il}\delta_{im}\\ =\sum_{n\neq i,j}\delta_{nl}\delta_{nm}+\delta_{jl}\delta_{jm}+\delta_{il}\delta_{im}=\sum_{n}\delta_{nl}\delta_{nm}=\delta_{lm}\\$$
So $R$ is orthogonal. Suppose $k\neq i,j$.
$$T_{ijk}=\sum_{p,q,r}R_{ip}R_{jq}R_{kr}T_{pqr}=\sum_{p,q,r}-\delta_{jp}\delta_{iq}\delta_{kr}T_{pqr}=-T_{jik}$$ So $T$ is skew-symmetric in its $1$st two indices. Symmetry of this argument shows that $T$ is fully skew-symmetric. Therefore $T$ is a multiple of the Levi-Civita tensor.