It is often claimed that the only tensors invariant under the orthogonal transformations (rotations) are the Kronecker delta $\delta_{ij}$, the LeviCivita 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.

1It is not clear to me exactly what this question means; the way mathematicians and physicists use the word "tensor" is slightly different, and I can't figure out what the LeviCivita symbol actually describes as a tensor in purely mathematical language. Are you familiar with the purely mathematical language? My interpretation of the question is the following: let $V$ be a finitedimensional real inner product space. Say that an element of the tensor product $V^{\otimes n}$ is an invariant tensor if it is invariant under the action of $\text{O}(V)$. Then you are asking whether... – Qiaochu Yuan Dec 23 '12 at 02:08

...invariant tensors are generated under tensor product and contraction by the inner product, which is an invariant tensor in $V^{\otimes 2}$, and a second tensor, perhaps the "determinant" in $V^{\dim V}$? Or are you only interested in the $3$dimensional case? – Qiaochu Yuan Dec 23 '12 at 02:10

(It looks to me from a quick google search like the LeviCivita symbol is not actually a tensor, but I can't really make heads or tails of this.) – Qiaochu Yuan Dec 23 '12 at 02:13

3@QiaochuYuan In physicist's notation, I would like to construct all tensors $T_{i_1 i_2 \ldots i_n}$ that satisfy the following equation: $O_{ij} T_j = T_i$ for a rank one tensor $T_i$, $O_{ij} O_{kl} T_{jl} = T_{ik}$ for a rank two tensor $T_{ij}$ etc. Here $O_{ij}$ is an arbitrary orthogonal matrix. Now, the claim is that for the case of three dimensional rotations all such tensors can be expressed as a combination of the Kronecker delta and LeviCivita tensor (the totally antisymmetric symbol). I have never seen a really convincing proof of this. – Dec 23 '12 at 10:20

You might like: http://mathworld.wolfram.com/IsotropicTensor.html – jinawee Jan 20 '14 at 22:38
3 Answers
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 nontrivial, and given in
Weyl, H., The Classical Groups, Princeton University Press, 1939
Constructions for higherorder isotropic Cartesian tensors are also given there.
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Harold Jeffreys (1973). On isotropic tensors. Mathematical Proceedings of the Cambridge Philosophical Society, 73, pp 173176.
The proof given is a lot more concrete and "hands on" than Weyl's proof linked to by user_of_math.
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$\mathtt{Definition:}$ $T$ 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 skewsymmetricness. 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 skewsymmetric in its $1$st two indices. Symmetry of this argument shows that $T$ is fully skewsymmetric. Therefore $T$ is a multiple of the LeviCivita tensor.
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