I understood that for any $N$ the decomposition of a tensor product of two vector representations into irreducible irreps would give the singlet plus an anti-symmetric tensor plus a symmetric traceless tensor.
However, when I run compute the decomposition for for example SO(4) using lieArt I find (using the command DecomposeProduct[Irrep[D2][4], Irrep[D2][4]]:
$$1+3+\bar{3}+9$$
I assume that this is correct (otherwise sorry for bothering the math stackexchange), so it seems that the anti-symmetric tensor can be further decomposed into two 3 dimensional irreps. Why is this? This seems to be an n-dependent statement. The isomorphism with the lie algebra $su(2) \otimes su(2)$ is probably relevant as the same decomposition seems more logical there.
A similar thing happens when I try to decompose the product of two traceless symmetric tensors. Again I find that various representations split into $i$ and $\bar{i}$.
This is the decomposition of middle-degree forms in even dimensions into a self dual and an anti-self-dual part, which in dimension $4$ happens to concern two-forms and hence skew-symmetric two tensors. In general dimension $2n$, the Hodge-* operator maps $\Lambda^n\mathbb C^{2n}$. It is $\mathfrak{so}(2n)$-equivariant and squares to the identity. So there is a decomposition $ \Lambda^n\mathbb C^{2n}=\Lambda^n_+\mathbb C^{2n}\oplus\Lambda^n_-\mathbb C^{2n}$ into $\pm 1$-eigenspaces for $*$. Whether this also works in a real setting depends on the parity of the dimension modulo $4$.