# 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 $O(N)$.

## 1 Answer

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.

I believe that the remaining invariant tensors are in fact just the combinations of Dirac deltas, but a proof doesn't spring immediately to mind.

• Thanks! I finally found a book that covers this subject. The book is called "The Classical Groups: Their Invariants and Representations" by Hermann Weyl. According to this book, isotropic tensors of O(N) are Kronecker deltas or Levi-Civitas (but not combinations if I understood it correctly). – A.Dunder Oct 13 '17 at 14:33
• This book (math.unibas.ch/uploads/x4epersdb/files/primernew.pdf) is available online and I seem to remember it's quite good and a bit more modern than Weyl (who I remember as being nearly unreadable). – Oscar Cunningham Oct 13 '17 at 14:41
• I talked with a professor at Uppsala University about these isotropic tensors. As it turns out, isotropic tensors of O(N) are a sum over products of Kronecker deltas, an even number of Levi-Civitas and combinations of those two. What differs an isotropic tensor of SO(N) with that from O(N) is that those of O(N) can only contain even numbers of Levi-Civitas, while those from SO(N) can contain any number of Levi-Civitas. Basically, uneven number of Levi-Civitas would pick up a sign under O(N) transformations, and thus they are not invariant under O(N). – A.Dunder Nov 1 '17 at 15:33
• @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
• Oh, that is actually quite useful for me. Then basically an isotropic tensor of O(N) is just sum of Kronecker deltas, where every term comes with their own constant. Thanks for the discussion. It helped a lot! – A.Dunder Nov 2 '17 at 15:41