Let $A$ be a super-symmetric $(n \times.....\times n)$ $m$-fold tensor, so that:
$$ A\left(i_{1},\dots,i_{m}\right) = A\left(j_{1},\dots,j_{m}\right), $$ when $\left(j_{1},\dots,j_{m}\right)$ is any permutation of $\left(i_{1},\dots,i_{m}\right)$.
I am looking to compare different such tensors to see if they are the same, up to permutations. To this end, I am looking for functions $f(A)$ (real- or complex valued) that are invariant under these permutations. The idea is that I can conclude that if $f(A) \not = f(A')$, $A$ and $A'$ are not the same up to a permutation. I'm aware that I won't be able to claim the reverse.
So far, I've come up with the following ideas, which I think are all valid:
- The total sum of all elements, i.e. $\sum_{i_{1},\dots,i_{n}}A(i_{1},\dots,i_{m})$
- The sum or product of the ordered values of the sum over all-except-one-dimensions, i.e. the length $n$ vector $v$ with entries $v_{i_{1}} = \sum_{i_{2},\dots,i_{n}}A(i_{1},i_{2},\dots,i_{m})$
- The sum or product of the eigenvalues of the matrix that you get if you sum over all-except-two-dimensions, i.e. the $n \times n$ matrix $V$ with entries $V_{(i_{1},i_{2})} = \sum_{i_{3},\dots,i_{n}}A(i_{1},i_{2},i_{3}\dots,i_{m})$.
I hope there are some other interesting approaches. Unfortunately I'm not very well versed in the theory of tensors beyond matrices. There's of course plenty of norms of matrices that are permutationally invariant, but I don't know how these generalize to higher order tensors.
N.B. I'm also aware that this will never will efficiently discern all tensors, as even for the $m = 2$ case this is a generalization of the Graph isomorphism problem.