Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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: – jinawee Jan 20 '14 at 22:38
up vote 3 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

$\mathtt{Definition:}$ $T$ 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.