# Isotropic Tensors of $O(N)$

Recently I have come across something I am not particularly familiar with, namely tensor invariants of $O(N)$, or isotropic tensors of $O(N)$ as I believe they are also called. What I would like to know are whether there exist any isotropic tensors of $O(N)$ that are not Kronecker deltas, Levi-Civita or a combinations of them both.

I have been looking at articles like https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500006832 , and it says that it is true for $SO(N)$, i.e. $O(N)$ with determinant one. Unfortunately, this article only covers tensor invariants of $SO(N)$ and not the full $O(N)$.

So you know every tensor invariant of $SO(N)$ is a combination of Kronecker deltas and Levi-Civitas. And $SO(N)$ is a subgroup of $O(N)$, so its invariant tensors are a superset of those of $O(N)$. This answers your main question: all invariant tensors of $O(N)$ can be written as a combination of Kronecker deltas and Levi-Civitas.
But the converse needn't be true. Not every combination of Kronecker deltas and Levi-Civitas is $O(N)$ invariant. For example the Levi-Civita symbol itself isn't.
• @A.Dunder That's what I suspected was true. By the way, any tensor containing deltas and an even number of Levi-Civitas can also be written using just deltas: $\varepsilon_{ab\dots c}\varepsilon_{ij\dots k}=\sum_\pi \mathrm{sign}(\pi)\delta_{a\pi(i)}\delta_{b\pi(j)}\dots\delta_{c\pi(k)}$ where $\pi$ ranges over all permutations of $N$ elements. – Oscar Cunningham Nov 1 '17 at 21:04