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
-\delta_{jm} & \text{if } l=i\\
\delta_{im} & \text{if } l=j\\
\delta_{lm} & \text{otherwise}
(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.