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

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.

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 $$1$$): $$O_{ii'}O_{jj'}O_{kk'}\epsilon_{i'j'k'} = \epsilon_{ijk}$$