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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. –  Little Brown One Dec 23 '12 at 10:20
You might like: mathworld.wolfram.com/IsotropicTensor.html –  jinawee Jan 20 at 22:38

2 Answers 2

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 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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