# 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.

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It 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 Levi-Civita 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 finite-dimensional 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 2:08
...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 2:10
(It looks to me from a quick google search like the Levi-Civita symbol is not actually a tensor, but I can't really make heads or tails of this.) – Qiaochu Yuan Dec 23 '12 at 2:13
@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 Levi-Civita tensor (the totally antisymmetric symbol). I have never seen a really convincing proof of this. – LBO Dec 23 '12 at 10:20
You might like: mathworld.wolfram.com/IsotropicTensor.html – jinawee Jan 20 '14 at 22:38

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.

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By "invariant" I mean invariance of the components, of course. – user_of_math Jul 1 '14 at 11:54

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.

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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 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.

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