In the paper [1], it is proven in Theorem 2 that any $n$-tensor $\mathcal{A}\in\mathbb{R}^{d_1\times...\times d_n}$ can be decomposed as

$$
\mathcal{A}=\mathcal{S} \times_1 U_1 ...\times_n U_n
$$
where $\times_n$ is the mode-$n$ product defined as $\mathcal{T}\times M=\sum_{i_k}\mathcal{T}_{i_1...i_k...i_n}M_{i_kj}$ and $U_n$ are appropriately shaped matrices. More interesting is the condition imposed on $\mathcal{S}$ by the theorem to ensure uniqueness of the decomposition, which the authors label *all-orthogonality* and which prescribes that

$$ \sum_{i_1,...,i_{k-1},\\i_{k+1},...,i_n}\mathcal{S}_{i_1,...,i_k,...,i_n}\mathcal{S}^{i_1,...,j_k,...,i_n}=\delta_{i_k}^{j_k}a_{k, i_k} \quad \forall k $$ Where $\delta$ is the kronecker delta and $a_k$ is a set of weights particular to each $k$, so that the right-hand side is a diagonal matrix. My question is a bit broad, as I am looking for any possible characterization of the family of all-orthogonal tensors. I would largely be happy with results for $3$-tensors, especially if it's easier to say something concrete. A parameterization or further decomposition into simpler matrix/tensor components would be the ideal, but I'd also appreciate some ideas on the space/manifold they inhabit and whether they possess any invariances. Results in $\mathbb{C}$ are also welcome.

I have been trying to attack the problem from different angles, but without much luck. By QR decomposition, it would seem that the QR of an arbitrary matricization $\mathcal{A}_{(k)}\in \mathbb{R}^{D\times d_k}$ separating one mode from the rest (where $D=\prod_{j\neq k}d_j$),

$$ \mathcal{A}_{(k)}=QR $$

has $R$ being diagonal, since the all-orthogonality condition corresponds to orthogonality of the matricizations. So all-orthogonality means that every matricization of the type above splits into an orthogonal and a diagonal matrix.

Some observations of varying utility:

- the all-orthogonality property plays a parallel role to the diagonality property in matrix SVD, which is also what makes me assume there is some underlying simplicity to the condition.
- Empirically, there does exist tensors where $R$ in the QR above is the identity for all choices of $k$, i.e. tensors where all matricizations are orthogonal.
- The Levi-Civita antisymmetric tensor is all-orthogonal after appropriate scaling, along with any transformations of it where each mode is independently transformed by orthogonal matrices.

[1] "*L. De Lathauwer, B. De Moor, and J. Vandewalle, “A Multilinear Singular Value Decomposition,” SIAM J. Matrix Anal. Appl., vol. 21, no. 4, pp. 1253–1278, Jan. 2000."*