Why is rank-3 tensor in 3D with two antisymmetric indices equivalent to rank-2 tensor?

Question

I'd like to know how many irreducible representations of $SO(n)$ when it comes to rank 3 tensor. Here $n=3$. Among the rank 3 tensor components, there might be antisymmetric parts and symmetric parts and goes on. According to A.Zee's "Group Theory for Physicists in a nutshell" p.193, he mentions that we actually don't need to concern about partially antisymmetrized tensor since it's nothing new but old representation that merged in rank 2 tensor representation.

We don’t care about the antisymmetric combination $T^{[ij]k}$ , because we know that secretly it is just a 2-indexed tensor $B^{lk}$ ≡ $\epsilon_{ijl}T^{[ij]k}$, and we have already disposed of all 2-indexed tensors. Our attack is inductive, as I said.

And i don't understand the statement that $T^{[ij]k}$ is actually equivalent to $B^{lk}$ ≡ $\epsilon_{ijl}T^{[ij]k}$ so that this partially antisymmetric rank 3 tensor is nothing new but rank 2 antisymmetric tensor representation of $SO(n)$.

Is it because $B^{lk}$ here is dual tensor of $T^{[ij]k}$? I'm not sure whether dual tensors share same transformation laws. If so, is it because Levi-Civita symbol is invariant symbol under $SO(n)$?

In summary, what i want to ask is,

Why is $T^{[ij]k}$ part among the rank 3 representation of $SO(n)$ is treated as rank 2 antisymmetric tensor? While there's no such term like $B^{lk}$ = $\epsilon_{ijl}T^{[ij]k}$ in $T^{ijk}$ with additional Levi-Civita symbol put on it.

Answer 1

Yes, you can work out the detail. Btw, since your Levi Civita symbol only has $3$ indices, you are dealing with $SO(3)$.

Under a rotation $O$, a tensor transforms by definition as: $$ T_{i_1...i_n} = O_{i_1j_1}...O_{i_nj_n}T_{j_1...j_n} $$ First off, the two space have the same dimensions (first sanity check) given by the isomorphism $$ T_{ijk}\to B_{kl} = \frac{1}{2}\epsilon_{ijk}T_{ijl} \\ B_{kl}\to T_{ijl} = \epsilon_{ijk}B_{kl} $$ The two transformation are the inverse of one another thanks to the identities: $$ \epsilon_{ijk}\epsilon_{ijl} = 2\delta_{kl} \\ \epsilon_{ijk}\epsilon_{rsk} = \delta_{ir}\delta_{js}-\delta_{is}\delta_{jr} $$

Additionally, these isomorphism preserve the action of $SO(3)$. Indeed: $$ \begin{align} T_{ijk}&\to O_{ii'}O_{jj'}O_{kk'}T_{i'j'k'} \\ B_{kl}&= \frac{1}{2}\epsilon_{ijk}T_{ijl} \\ &\to \frac{1}{2}\epsilon_{ijk}O_{ii'}O_{jj'}O_{ll'}T_{i'j'l'} \\ &\to \frac{1}{2}\epsilon_{ijk'}O_{ii'}O_{jj'}O_{k'k''}O_{kk''}O_{ll'}T_{i'j'l'}\\ &\to O_{kk'}O_{ll'}\frac{1}{2}\epsilon_{i'j'k'}T_{i'j'l'} \\ &\to O_{kk'}O_{ll'}B_{k'l'} \end{align} $$

Where I used the property of the inverse of orthogonal groups: $$ O_{ik}O_{jk}=\delta_{ij} $$ and the fact that $\epsilon$ is a invariant tensor (orthogonal group elements have determinant $1$): $$ O_{ii'}O_{jj'}O_{kk'}\epsilon_{i'j'k'} = \epsilon_{ijk} $$

Thus not only are the two spaces isomorphic as vector spaces, but additionally, it preserves the group action. They are therefore considered equivalent. Furthermore, the isomorphism is natural, in the sense that it does not require the choice of an arbitrary basis.

Hope this helps.