# The Set of Orthogonal Tensor Trains

Pardis Semnani and Elina Robeva

###### Abstract

In this paper we study the set of tensors that admit a special type of
decomposition called an orthogonal tensor train decomposition. Finding
equations defining varieties of low-rank tensors is generally a hard problem,
however, the set of orthogonally decomposable tensors is defined by appealing
quadratic equations. The tensors we consider are an extension of orthogonally
decomposable tensors. We show that they are defined by similar quadratic
equations, as well as an interesting higher-degree additional equation.

##  1 Introduction

With the emergence of big data, it is more and more often the case that
information is recorded in the form of a tensor (or multi-dimensional array).
The importance of decomposing such a tensor is (at least) twofold. First, it
provides hidden information about the data at hand, and second, having a
concise decomposition of the tensor allows us to store it much more
efficiently. One of the biggest obstacles in dealing with tensors, however, is
that decomposing them is often computationally hard. For example, finding (the
number of terms of) the CP-decomposition [7] of a general tensor is NP-hard
[6], and the set of tensors of CP rank at most $r$ is not closed for any
$r\geq 2$, making the low-rank approximation problem impossible in some
instances [9]. It is also notoriously hard to describe all the defining
equations of the set of tensors of rank at most $r$ [10].

Finding the CP-decomposition of special subclasses of tensors, however, can be
done efficiently. For example, low-rank tensors can be decomposed via
Jennrich’s algorithm and their subfamily of orthogonally decomposable tensors
can be decomposed via the slice method [8] or via the tensor power method [1].
Furthermore, the set of orthogonally decomposable tensors of bounded rank is
closed [3]. In fact, it is a real algebraic variety defined by interpretable
quadratic equations which represent symmetry in the contraction of the tensor
$T$ with itself. The eigenvectors and singular vector tuples of orthogonally
decomposable tensors can be found efficiently [12, 13], making the family of
such tensors as appealing as the set of matrices. However, such tensors are
very rare – the rank of a general $n\times n\times\cdots\times n$ ($d$-times)
tensor is $\mathcal{O}(n^{d-1})$, while that of an orthogonally decomposable
one is $n$.

Tensor networks provide many other ways of decomposing tensors. They originate
from quantum physics and are used to depict the structure of steady states of
Hamiltonians of quantum systems [2, 4]. Many types of tensor network
decompositions, like tensor trains [11], also known as matrix product states
[4], are used in machine learning to decompose data tensors in meaningful
ways. The CP-decomposition of a tensor can also be represented by a tensor
network whose underlying graph is a hypergraph [14]. While tensor network
decompositions can represent various tensor structures, they are not unique
[15].

More recently, tensor train decompositions for which the tensors at the
vertices are orthogonally decomposable have been studied [5]. It was shown
that those can be decomposed efficiently, at least in the case of length-2
tensor trains. In this work we study the equations that define the set of
tensors that can be decomposed this way. Finding the description via equations
would help in determining which tensors can be expressed in this way, and if
it is easy to approximate a given tensor by such a tensor. We compare the
equations we get to the equations defining the set of orthogonally
decomposable tensors. It turns out, the two are quite similar and structured.

The rest of this manuscript is organized as follows. In Section 2 we provide
background on orthogonally decomposable tensors as well as tensor networks. In
Section 3 we turn to orthogonally decomposable tensor trains of length 2 and
present our findings.

##  2 Background

In this section we provide the necessary background on orthogonal tensor
decomposition, as well as tensor networks.

###  2.1 Tensors

We begin by defining tensors and symmetric tensors.

A tensor $T$ is an $n_{1}\times\cdots\times n_{d}$ table with elements in a
field, which for us will be the reals. The number $d$ of sides of the table is
called the order of the tensor. The set of all tensors of size
$n_{1}\times\cdots\times n_{d}$ with real entries is denoted by
$\mathbb{R}^{n_{1}}\otimes\cdots\otimes\mathbb{R}^{n_{d}}$.

A tensor $T\in\mathbb{R}^{n_{1}}\otimes\cdots\otimes\mathbb{R}^{n_{d}}$ is a
rank-1  tensor if there exist vectors
$v^{(1)}\in\mathbb{R}^{n_{1}},\ldots,v^{(d)}\in\mathbb{R}^{n_{d}}$ such that

$T=v^{(1)}\otimes\cdots\otimes v^{(d)}.$

The CP-decomposition [7] expresses a tensor
$T\in\mathbb{R}^{n_{1}}\otimes\cdots\otimes\mathbb{R}^{n_{d}}$ as a sum of
rank-1 terms:

$T=\sum_{i=1}^{r}v_{i}^{(1)}\otimes\cdots\otimes v_{i}^{(d)}.$

An $n\times\cdots\times n$ ($d$-times) tensor $T$ is symmetric if for any
permutation $\sigma$ on $\\{1,\ldots,d\\}$, $T_{i_{1}\ldots
i_{d}}=T_{i_{\sigma(1)}\ldots i_{\sigma(d)}}$. The set of such tensors is
denoted by $S^{d}(\mathbb{R}^{n})$.

A symmetric tensor $T\in S^{d}(\mathbb{R}^{n})$ has rank-1 if it can be
expressed as

$T=\lambda v^{\otimes d},$

i.e., as a scalar times the outer product of a vector with itself $d$ times. A
symmetric CP-decomposition of a symmetric tensor $T\in S^{d}(\mathbb{R}^{n})$
expresses $T$ as a sum of symmetric rank-1 terms:

$T=\sum_{i=1}^{r}\lambda_{i}v_{i}^{\otimes d}.$

###  2.2 Tensor networks

In this subsection we provide the necessary background on tensor networks and
tensor trains. Let
$T\in\mathbb{R}^{m_{1}}\otimes\cdots\otimes\mathbb{R}^{m_{k}}$ and
$S\in\mathbb{R}^{n_{1}}\otimes\cdots\otimes\mathbb{R}^{n_{\ell}}$ be two
tensors such that $m_{i}=n_{j}$. The contraction of the tensors $T$ and $S$
along their $i$th and $j$th modes is the tensor
$R\in\mathbb{R}^{m_{1}}\otimes\cdots\otimes\mathbb{R}^{m_{i-1}}\otimes\mathbb{R}^{m_{i+1}}\otimes\cdots\otimes\mathbb{R}^{m_{k}}\otimes\mathbb{R}^{n_{1}}\otimes\cdots\otimes\mathbb{R}^{n_{j-1}}\otimes\mathbb{R}^{n_{j+1}}\otimes\cdots\otimes\mathbb{R}^{n_{\ell}}$
whose entries are

$R_{a_{1}\ldots a_{i-1}a_{i+1}\ldots a_{k}b_{1}\ldots b_{j-1}b_{j+1}\ldots
b_{\ell}}=\sum_{c=1}^{m_{i}}T_{a_{1}\ldots a_{i-1}ca_{i+1}\ldots
a_{k}}S_{b_{1}\ldots b_{j-1}cb_{j+1}\ldots b_{\ell}},$

where
$a_{1}\in\Set{1,\ldots,m_{1}},\ldots,a_{k}\in\Set{1,\ldots,m_{k}},b_{1}\in\Set{1,\ldots,n_{1}},\ldots,b_{\ell}\in\Set{1,\ldots,n_{\ell}}$.

Starting with a number of tensors, and applying repeated contractions in a
prescribed manner yields the notion of a tensor network. A tensor network can
be represented by a graph in which each node corresponds to one of the tensors
involved in forming the tensor network, and the number of the edges connected
to each node is equal to the order of the tensor corresponding to that node.

######  Definition 1.

Consider a 3-tuple $G=(V,E,D)$, where $V$ is a nonempty set, $E$ is a subset
of $\Set{\Set{v_{1},v_{2}}}{v_{1},v_{2}\in V,v_{1}\neq v_{2}}$, and $D$ is a
subset of $\Set{\Set{v}}{v\in V}$. The elements of $V$, $E\cup D$, and $D$ are
called the vertices, edges, and dangling edges of $G$ respectively, and $G$ is
called a graph. Each edge in $e\in E\cup D$ is assigned a positive integer
$n_{e}$. Each vertex $v\in V$ is assigned a tensor $T_{v}\in\otimes_{v\in e\in
E\cup D}\mathbb{R}^{n_{e}}$. The resulting tensor network state
$T\in\otimes_{e\in D}\mathbb{R}^{n_{e}}$ is obtained by contracting the
tensors $T_{v}$ along all edges $e\in E$.

For instance, in Figure 1, two order-3 tensors $T\in\mathbb{F}^{n_{1}\times
n_{2}\times n_{3}}$ and $S\in\mathbb{F}^{n_{3}\times n_{4}\times n_{5}}$ are
contracted along their 3rd and 1st modes respectively to form an $n_{1}\times
n_{2}\times n_{4}\times n_{5}$ tensor network, which is called a tensor train
of length 2.

We define a tensor train of length $l$ to be a tensor network consisting of
$l$ vertices arranged in a line, each of which has 3 adjacent edges (see
Figure 2).

Figure 1: How a tensor train of length 2 is generated Figure 2: Tensor train
of length $l$

######  Remark 2.

Matrix product states often appear in the condensed matter physics literature
with one additional vertex and edge attached to each end of the diagram, as
opposed to our diagrams as in Figure 2 [2, 4]. This format can be converted to
the format introduced here simply with a contraction of the single edges, thus
removing the vertices at each end. Conversely, given a tensor network in the
form of our diagram, a multiplication by the identity matrix at each end will
yield the matrix product state form. Thus, as the two network formats are
equivalent, we proceed with the use of the term “tensor train” to describe the
networks we study in this paper. The reason for the modified form is to better
generalize and extend the class of tensors known as orthogonally decomposable
tensors, as studied in [1, 3, 8, 12, 13].

###  2.3 Orthogonally Decomposable Tensors

A tensor $T\in\mathbb{R}^{n_{1}}\otimes\cdots\otimes\mathbb{R}^{n_{d}}$ is
orthogonally decomposable (or odeco) if it has a decomposition

$T=\sum_{i=1}^{r}v_{i}^{(1)}\otimes\cdots\otimes v_{i}^{(d)},$

where for every $j=1,\ldots,d$, the vectors
$v_{1}^{(j)},\ldots,v_{r}^{(j)}\in\mathbb{R}^{n_{j}}$ are orthogonal.

A symmetric tensor $T\in S^{d}(\mathbb{R}^{n})$ is symmetrically orthogonally
decomposable (or symmetrically odeco) if it has a decomposition

$T=\sum_{i=1}^{r}\lambda_{i}v_{i}^{\otimes d},$

where $v_{1},\ldots,v_{r}\in\mathbb{R}^{n}$ are orthonormal.

It turns out that the set of odeco tensors (both in the symmetric and ordinary
cases) is a real algebraic variety.

######  Theorem 3 ([3]).

Let $T\in\mathbb{R}^{n_{1}}\otimes\cdots\otimes\mathbb{R}^{n_{d}}$. For every
$q=1,\ldots,d$, define

$T\bullet_{q}T\in\left(\mathbb{R}^{n_{1}}\otimes\cdots\otimes\mathbb{R}^{n_{q-1}}\otimes\mathbb{R}^{n_{q+1}}\otimes\cdots\otimes\mathbb{R}^{n_{d}}\right)^{\otimes
2}$

by

$(T\bullet_{q}T)_{i_{1}\ldots i_{q-1}i_{q+1}\ldots i_{d}j_{1}\ldots
j_{q-1}j_{q+1}\ldots j_{d}}=\sum_{m=1}^{n_{q}}T_{i_{1}\ldots
i_{q-1}mi_{q+1}\ldots i_{d}}T_{j_{1}\ldots j_{q-1}mj_{q+1}\ldots j_{d}},$

i.e., it is the contraction of $T$ with itself along the $q$-th mode. Then,
$T$ is odeco if and only if for every $q=1,\ldots,d$, the tensor
$T\bullet_{q}T$ lies in $S^{2}(\mathbb{R}^{n_{1}})\otimes\cdots\otimes
S^{2}(\mathbb{R}^{n_{d}})$, which means that for all $\ell\in\Set{1,\ldots,d}$
with $\ell\neq q$, the entry $(T\bullet_{q}T)_{i_{1}\ldots
i_{q-1}i_{q+1}\ldots i_{d}j_{1}\ldots j_{q-1}j_{q+1}\ldots j_{d}}$ remains
unchanged if we permuted $i_{\ell}$ and $j_{\ell}$.

Figure 3: $T$ is odeco iff the the entry $(T\bullet_{q}T)_{i_{1}\ldots
i_{q-1}i_{q+1}\ldots i_{d}j_{1}\ldots j_{q-1}j_{q+1}\ldots j_{d}}$ remains
unchanged when any two indices corresponding to edges with the same color are
permuted.

A similar and easier to state result holds in the symmetric case.

######  Theorem 4 ([3]).

Let $T\in S^{d}(\mathbb{R}^{n})$. Define $T\bullet_{1}T\in
S^{d-1}(\mathbb{R}^{n})\otimes S^{d-1}(\mathbb{R}^{n})$ by

$(T\bullet_{1}T)_{i_{2}\ldots i_{d}j_{2}\ldots
j_{d}}=\sum_{m=1}^{n}T_{mi_{2}\ldots i_{d}}T_{mj_{2}\ldots j_{d}},$

i.e., the contraction of $T$ with itself along its first mode. Then, $T$ is
odeco if and only if

$T\bullet_{1}T\in S^{2d-2}(\mathbb{R}^{n}),$

i.e., the entry $(T\bullet_{1}T)_{i_{2}\ldots i_{d}j_{2}\ldots j_{d}}$ remains
unchanged after permuting any of its indices.

Figure 4: $T$ is odeco iff the entry $(T\bullet_{1}T)_{i_{2}\ldots
i_{d}j_{2}\ldots j_{d}}$ remains unchanged when any of the indices are
permuted.

In [12, Lemma 3.7], it has been shown that the Zariski closure in
$S^{d}(\mathbb{C}^{n})$ of the set of symmetrically orthogonally decomposable
tensors in $S^{d}(\mathbb{R}^{n})$ is an irreducible component of the variety
of the ideal defined by the symmetries (i.e., quadratic equations in the
entries of $T$) introduced in Theorem 4. So, it suffices to prove that this
ideal is prime in order to conclude that it is exactly equal to the ideal of
the Zariski closure in $S^{d}(\mathbb{C}^{n})$ of the set of symmetrically
orthogonally decomposable tensors in $S^{d}(\mathbb{R}^{n})$. This has been
proven when $n=2$ in [12, Theorem 3.6], but is still an open problem for an
arbitrary $n$.

##  3 Orthogonally Decomposable Tensor Trains

We now turn to tensors which decompose according to a tensor network, where
all the tensors at the vertices of the network are symmetric and odeco. In
particular, we confine our study to tensor trains.

######  Definition 5.

A tensor train of length $l$ (recall Figure 2) is said to be symmetrically
orthogonally decomposable, or for short, symmetrically odeco, if each of the
$l$ order-3 tensors generating the tensor train is $n\times n\times n$
symmetrically odeco. We denote the set of such symmetrically odeco tensor
trains by $SOT_{l,n}$.

It has been shown in [5] that if a tensor train of length 2 has a
symmetrically odeco decomposition, one can find the decomposition efficiently.
Our goal here is to find polynomials which vanish on the set $SOT_{l,n}$ for
certain $n,l$, and use them to determine the prime ideal of the Zariski
closure of this set. In order to derive these polynomials for $l=2$, our
approach is to consider the sets of variables

$\displaystyle P=\Set{p_{ijkl}}{i,j,k,l\in\Set{1,\ldots,n}}$ $\displaystyle
V=\Set{v_{ij}}{i,j\in\Set{1,\ldots,n}}$ $\displaystyle
W=\Set{w_{ij}}{i,j\in\Set{1,\ldots,n}}$ $\displaystyle
M=\Set{\mu_{i}}{i\in\Set{1,\ldots,n}}$ $\displaystyle
L=\Set{\lambda_{i}}{i\in\Set{1,\ldots,n}},$

define the following $\mathbb{R}$-algebra morphism

$\begin{array}[]{c}\phi:\mathbb{R}\left[P\right]\to\mathbb{R}\left[V\cup W\cup
M\cup L\right]\\\
\phi\left(p_{ijkl}\right)=\sum_{a=1}^{n}\sum_{b=1}^{n}\sum_{c=1}^{n}\mu_{a}\lambda_{b}v_{ai}v_{aj}w_{bk}w_{bl}v_{ac}w_{bc},\end{array}$

and compute $\phi^{-1}\left(I\right)$ using computer algebra software such as
Macaulay2, where

$I=\left\langle\bigcup_{k=1}^{n}\Set{\sum_{i=1}^{n}v_{ki}^{2}=1,\sum_{i=1}^{n}w_{ki}^{2}=1}\right\rangle+\left\langle\bigcup_{\begin{subarray}{c}{k,l=1}\\\
k\neq
l\end{subarray}}^{n}\Set{\sum_{i=1}^{n}v_{ki}v_{li}=0,\sum_{i=1}^{n}w_{ki}w_{li}=0}\right\rangle.$

We were unable to run an analogous program which would allow us to come up
with the polynomials in the cases where $l>2$. So, we confine our study to
length-2 tensor trains (see Figure 1) in this paper.

We now introduce one set of linear polynomials and one set of quadratic
polynomials which vanish on $SOT_{2,n}$, where $n\geq 2$.

Let $n\in\mathbb{N}$ and let $\mathcal{P}_{n}$ denote the set of polynomials

$f_{a,b,c,d,\sigma_{1},\sigma_{2}}\coloneqq
p_{abcd}-p_{\sigma_{1}(a)\sigma_{1}(b)\sigma_{2}(c)\sigma_{2}(d)},$

where $a,b,c,d\in\Set{1,\ldots,n}$, $\sigma_{1}$ is a permutation on
$\Set{a,b}$, and $\sigma_{2}$ is a permutation on $\Set{c,d}$. Moreover, let
$\mathcal{Q}_{n}$ denote the set of polynomials

$g^{(1)}_{a,b,c,d,e,f}\coloneqq\sum_{t=1}^{n}p_{abet}\ p_{cdft}-p_{abft}\
p_{cdet},$

and

$g^{(2)}_{a,b,c,d,e,f}\coloneqq\sum_{t=1}^{n}p_{etab}\ p_{ftcd}-p_{ftab}\
p_{etcd},$

where $a,b,c,d,e,f\in\Set{1,\ldots,n}$.

We now proceed to showing that the equations $\mathcal{P}_{n}$ and
$\mathcal{Q}_{n}$ vanish on $SOT_{2,n}$. In particular, these also represent
certain symmetries of the contracted tensor $T\bullet_{q}T$, similar to the
odeco case described in section 2.3.

######  Theorem 6.

Let $n\geq 2$. Then $\mathcal{P}_{n}$ and $\mathcal{Q}_{n}$ vanish on
$SOT_{2,n}$.

Figure 5: The fact that $\mathcal{P}_{n}$ vanishes on $SOT_{2,n}$ shows that
$P$ is symmetric with respect to any of the two pairs of modes indicated in
red and green.
The fact that $\mathcal{Q}_{n}$ vanishes on $SOT_{2,n}$ shows that
$P\bullet_{4}P$ and $P\bullet_{2}P$ are symmetric with respect to the modes
indicated in purple.

######  Remark 7.

The equations $\mathcal{Q}_{n}$ depict certain symmetries in the contracted
tensors $P\bullet_{2}P$ and $P\bullet_{4}P$, very similar to the odeco case
(see Theorems 3 and 4, and Figures 3 and 4). We have explained these
symmetries in Figure 5. In particular, $P\bullet_{4}P\in
S^{2}(\mathbb{R}^{n})\otimes S^{2}(\mathbb{R}^{n})\otimes
S^{2}(\mathbb{R}^{n})$, where the first two $S^{2}(\mathbb{R}^{n})$ correspond
to the red edges and the red dashed edges respectively, and the last
$S^{2}(\mathbb{R}^{n})$ corresponds to the commuting of the indices coming
from the purple edges and is the one that gives rise to half of the equations
in $\mathcal{Q}_{n}$. Similarly, $P\bullet_{2}P\in
S^{2}(\mathbb{R}^{n})\otimes S^{2}(\mathbb{R}^{n})\otimes
S^{2}(\mathbb{R}^{n})$, where the first two $S^{2}(\mathbb{R}^{n})$ correspond
to the green and green dashed edges respectively, and the last
$S^{2}(\mathbb{R}^{n})$ corresponds to the fact that the indices along the
purple edges commute and gives rise to the other half of the equations in
$\mathcal{Q}_{n}$.

###### Proof.

We begin by writing an arbitrary entry of a given tensor $P$ in $SOT_{2,n}$ in
terms of the entries of the orthogonal vectors which appear in the
decomposition of the two tensors at the nodes of $P$, and then proceed by
writing the equations in $\mathcal{P}_{n}$ in terms of the entries of the two
node tensors and the equations in $\mathcal{Q}_{n}$ in terms of the entries of
these vectors. This is helpful since it provides us with the opportunity of
applying the symmetry of the tensors and the orthogonality of the vectors.

Let $P\in SOT_{2,n}$. So, there exist vectors
$v_{1},\ldots,v_{n},w_{1},\ldots,w_{n}\in\mathbb{R}^{n}$ and scalars
$\lambda_{1},\ldots,\lambda_{n}$, $\mu_{1},\ldots,\mu_{n}\in\mathbb{R}$ such
that $\Set{v_{1},\ldots,v_{n}}$ and $\Set{w_{1},\ldots,w_{n}}$ are orthonormal
sets, $T=\sum_{i=1}^{n}\lambda_{i}v_{i}^{\otimes 3}$,
$S=\sum_{i=1}^{n}\mu_{i}w_{i}^{\otimes 3}$, and $P$ is the result of the
contraction of $T$ and $S$ along their third modes. Therefore, for all
$a,b,c,d\in\Set{1,\ldots,n}$,

$\displaystyle p_{abcd}$
$\displaystyle=\sum_{s=1}^{n}T_{abs}S_{cds}=\sum_{s=1}^{n}\left(\sum_{i=1}^{n}\lambda_{i}v_{ia}v_{ib}v_{is}\right)\left(\sum_{j=1}^{n}\mu_{j}w_{jc}w_{jd}w_{js}\right)$
$\displaystyle=\sum_{i=1}^{n}\sum_{j=1}^{n}\lambda_{i}\mu_{j}v_{ia}v_{ib}w_{jc}w_{jd}\left(\sum_{s=1}^{n}v_{is}w_{js}\right)=\sum_{i=1}^{n}\sum_{j=1}^{n}\lambda_{i}\mu_{j}v_{ia}v_{ib}w_{jc}w_{jd}\langle
v_{i},w_{j}\rangle.$

So, if $f_{a,b,c,d,\sigma_{1},\sigma_{2}}\in\mathcal{P}_{n}$, then

$\displaystyle f_{a,b,c,d,\sigma_{1},\sigma_{2}}$
$\displaystyle=p_{abcd}-p_{\sigma_{1}(a)\sigma_{1}(b)\sigma_{2}(c)\sigma_{2}(d)}$
$\displaystyle=\left(\sum_{s=1}^{n}T_{abs}S_{cds}\right)-\left(\sum_{s=1}^{n}T_{\sigma_{1}(a)\sigma_{1}(b)s}S_{\sigma_{2}(c)\sigma_{2}(d)s}\right)$
$\displaystyle=\left(\sum_{s=1}^{n}T_{abs}S_{cds}\right)-\left(\sum_{s=1}^{n}T_{abs}S_{cds}\right)$
$\displaystyle(\text{$T$ and $S$ are symmetric.})$ $\displaystyle=0.$

Thus, $\mathcal{P}_{n}$ vanishes on $SOT_{2,n}$. Now suppose that
$g^{(1)}_{a,b,c,d,e,f}\in\mathcal{Q}_{n}$. Then

$\displaystyle g^{(1)}_{a,b,c,d,e,f}=\sum_{t=1}^{n}p_{abet}\
p_{cdft}-p_{abft}\ p_{cdet}$
$\displaystyle=\sum_{t=1}^{n}\biggm{[}\left(\sum_{i=1}^{n}\sum_{j=1}^{n}\lambda_{i}\mu_{j}v_{ia}v_{ib}w_{je}w_{jt}\langle
v_{i},w_{j}\rangle\right)\left(\sum_{k=1}^{n}\sum_{l=1}^{n}\lambda_{k}\mu_{l}v_{kc}v_{kd}w_{lf}w_{lt}\langle
v_{k},w_{l}\rangle\right)$
$\displaystyle\phantom{\sum_{t=1}^{n}}-\left(\sum_{i=1}^{n}\sum_{j=1}^{n}\lambda_{i}\mu_{j}v_{ia}v_{ib}w_{jf}w_{jt}\langle
v_{i},w_{j}\rangle\right)\left(\sum_{k=1}^{n}\sum_{l=1}^{n}\lambda_{k}\mu_{l}v_{kc}v_{kd}w_{le}w_{lt}\langle
v_{k},w_{l}\rangle\right)\biggm{]}$
$\displaystyle=\sum_{t=1}^{n}\biggm{[}\left(\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}\sum_{l=1}^{n}\lambda_{i}\lambda_{k}\mu_{j}\mu_{l}v_{ia}v_{ib}v_{kc}v_{kd}w_{je}w_{jt}w_{lf}w_{lt}\langle
v_{i},w_{j}\rangle\langle v_{k},w_{l}\rangle\right)$
$\displaystyle\phantom{\sum_{t=1}^{n}}-\left(\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}\sum_{l=1}^{n}\lambda_{i}\lambda_{k}\mu_{j}\mu_{l}v_{ia}v_{ib}v_{kc}v_{kd}w_{jf}w_{jt}w_{le}w_{lt}\langle
v_{i},w_{j}\rangle\langle v_{k},w_{l}\rangle\right)\biggm{]}$
$\displaystyle=\sum_{t=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}\sum_{l=1}^{n}\lambda_{i}\lambda_{k}\mu_{j}\mu_{l}v_{ia}v_{ib}v_{kc}v_{kd}w_{jt}w_{lt}\langle
v_{i},w_{j}\rangle\langle
v_{k},w_{l}\rangle\left(w_{je}w_{lf}-w_{jf}w_{le}\right)$
$\displaystyle=\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}\sum_{l=1}^{n}\lambda_{i}\lambda_{k}\mu_{j}\mu_{l}v_{ia}v_{ib}v_{kc}v_{kd}\langle
v_{i},w_{j}\rangle\langle
v_{k},w_{l}\rangle\left(w_{je}w_{lf}-w_{jf}w_{le}\right)\left(\sum_{t=1}^{n}w_{jt}w_{lt}\right).$

If $j=l$, then $w_{je}w_{lf}-w_{jf}w_{le}=0$, and otherwise,
$\sum_{t=1}^{n}w_{jt}w_{lt}=\langle w_{j},w_{l}\rangle=0$. This shows that
$g^{(1)}_{a,b,c,d,e,f}=0$. Similarly, if
$g^{(2)}_{a,b,c,d,e,f}\in\mathcal{Q}_{n}$, then $g^{(2)}_{a,b,c,d,e,f}$
vanishes on $P$. Hence, $\mathcal{Q}_{n}$ vanishes on $SOT_{2,n}$. ∎

Now we proceed by introducing a polynomial of degree $n$ which vanishes on
$SOT_{2,n}$.

Let $n\in\mathbb{N}$. Define

$h_{n}\coloneqq\sum_{k_{1}=1}^{n}\ldots\sum_{k_{n}=1}^{n}\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}}{\rm sgn}(\sigma)\ {\rm sgn}(\gamma)\
p_{k_{1}\sigma(1)k_{n}\gamma(1)}\ p_{k_{2}\sigma(2)k_{1}\gamma(2)}\cdots
p_{k_{n}\sigma(n)k_{n-1}\gamma(n)}.$

######  Theorem 8.

Let $n\geq 2$. Then $h_{n}$ vanishes on $SOT_{2,n}$.

###### Proof.

Similar to the proof of Theorem 6, we first write the polynomial $h_{n}$ in
terms of the entries of the vectors appearing in the orthogonal decomposition
of the two tensors at the nodes of a given tensor $P$ in $SOT_{2,n}$, which
yields equation (1). When $n$ is even, we show that the terms of the right
hand side polynomial in (1) can be split into two partitions such than the
partitions cancel each other out. In the case when $n$ is odd, by manipulating
the terms of the right hand side polynomial in (1) and applying the Parseval’s
Equation, we show that $h_{n}$ vanishes on $SOT_{2,n}$.

Let $P\in SOT_{2,n}$. So, there exist vectors
$v_{1},\ldots,v_{n},w_{1},\ldots,w_{n}\in\mathbb{R}^{n}$ and scalars
$\lambda_{1},\ldots,\lambda_{n}$, $\mu_{1},\ldots,\mu_{n}\in\mathbb{R}$ such
that $\Set{v_{1},\ldots,v_{n}}$ and $\Set{w_{1},\ldots,w_{n}}$ are orthonormal
sets, $T=\sum_{i=1}^{n}\lambda_{i}v_{i}^{\otimes 3}$,
$S=\sum_{i=1}^{n}\mu_{i}w_{i}^{\otimes 3}$, and $P$ is the result of the
contraction of $T$ and $S$ along their third modes. Note that

$\displaystyle h_{n}$
$\displaystyle=\sum_{k_{1}=1}^{n}\ldots\sum_{k_{n}=1}^{n}\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}}{\rm sgn}(\sigma)\ {\rm sgn}(\gamma)\times$
$\displaystyle\Bigg{[}\left(\sum_{i_{1}=1}^{n}\sum_{j_{1}=1}^{n}\lambda_{i_{1}}\mu_{j_{1}}\langle
v_{i_{1}},w_{j_{1}}\rangle
v_{i_{1}k_{1}}v_{i_{1}\sigma(1)}w_{j_{1}k_{n}}w_{j_{1}\gamma(1)}\right)\times$
$\displaystyle\phantom{\Bigg{[}}\left(\sum_{i_{2}=1}^{n}\sum_{j_{2}=1}^{n}\lambda_{i_{2}}\mu_{j_{2}}\langle
v_{i_{2}},w_{j_{2}}\rangle
v_{i_{2}k_{2}}v_{i_{2}\sigma(2)}w_{j_{2}k_{1}}w_{j_{2}\gamma(2)}\right)\times$
$\displaystyle\phantom{\Bigg{[}\ \ \ }\vdots$
$\displaystyle\phantom{\Bigg{[}}\left(\sum_{i_{n}=1}^{n}\sum_{j_{n}=1}^{n}\lambda_{i_{n}}\mu_{j_{n}}\langle
v_{i_{n}},w_{j_{n}}\rangle
v_{i_{n}k_{n}}v_{i_{n}\sigma(n)}w_{j_{n}k_{n-1}}w_{j_{n}\gamma(n)}\right)\Bigg{]}$
$\displaystyle=\sum_{i_{1}=1}^{n}\sum_{j_{1}=1}^{n}\ldots\sum_{i_{n}=1}^{n}\sum_{j_{n}=1}^{n}\sum_{k_{1}=1}^{n}\ldots\sum_{k_{n}=1}^{n}\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}}{\rm sgn}(\sigma)\ {\rm
sgn}(\gamma)\lambda_{i_{1}}\mu_{j_{1}}\cdots\lambda_{i_{n}}\mu_{j_{n}}$
$\displaystyle\phantom{=}\langle v_{i_{1}},w_{j_{1}}\rangle\cdots\langle
v_{i_{n}},w_{j_{n}}\rangle v_{i_{1}k_{1}}v_{i_{2}k_{2}}\cdots
v_{i_{n}k_{n}}w_{j_{1}k_{n}}w_{j_{2}k_{1}}\cdots
w_{j_{n}k_{n-1}}v_{i_{1}\sigma(1)}\cdots v_{i_{n}\sigma(n)}$
$\displaystyle\phantom{=}w_{j_{1}\gamma(1)}\cdots w_{j_{n}\gamma(n)}$
$\displaystyle=\sum_{i_{1}=1}^{n}\sum_{j_{1}=1}^{n}\ldots\sum_{i_{n}=1}^{n}\sum_{j_{n}=1}^{n}\lambda_{i_{1}}\mu_{j_{1}}\cdots\lambda_{i_{n}}\mu_{j_{n}}\langle
v_{i_{1}},w_{j_{1}}\rangle\cdots\langle v_{i_{n}},w_{j_{n}}\rangle$
$\displaystyle\phantom{=}\left(\sum_{k_{1}=1}^{n}v_{i_{1}k_{1}}w_{j_{2}k_{1}}\right)\cdots\left(\sum_{k_{n-1}=1}^{n}v_{i_{n-1}k_{n-1}}w_{j_{n}k_{n-1}}\right)\left(\sum_{k_{n}=1}^{n}v_{i_{n}k_{n}}w_{j_{1}k_{n}}\right)$
$\displaystyle\phantom{=}\left(\sum_{\sigma\in S_{n}}{\rm
sgn(\sigma)}v_{i_{1}\sigma(1)}\cdots
v_{i_{n}\sigma(n)}\right)\left(\sum_{\gamma\in S_{n}}{\rm
sgn(\gamma)}w_{j_{1}\gamma(1)}\cdots w_{j_{n}\gamma(n)}\right)$
$\displaystyle=\sum_{i_{1}=1}^{n}\sum_{j_{1}=1}^{n}\ldots\sum_{i_{n}=1}^{n}\sum_{j_{n}=1}^{n}\lambda_{i_{1}}\mu_{j_{1}}\cdots\lambda_{i_{n}}\mu_{j_{n}}\langle
v_{i_{1}},w_{j_{1}}\rangle\cdots\langle v_{i_{n-1}},w_{j_{n-1}}\rangle\langle
v_{i_{n}},w_{j_{n}}\rangle$ $\displaystyle\phantom{=}\langle
v_{i_{1}},w_{j_{2}}\rangle\cdots\langle v_{i_{n-1}},w_{j_{n}}\rangle\langle
v_{i_{n}},w_{j_{1}}\rangle\det\left(V_{i_{1}\ldots
i_{n}}\right)\det\left(W_{j_{1}\ldots j_{n}}\right),$

where $V_{i_{1}\ldots
i_{n}}\coloneqq\left[\begin{array}[]{ccc}-&v_{i_{1}}^{T}-\\\ &\vdots&\\\
-&v_{i_{n}}^{T}&-\end{array}\right]$ and $W_{j_{1}\ldots
j_{n}}\coloneqq\left[\begin{array}[]{ccc}-&w_{j_{1}}^{T}-\\\ &\vdots&\\\
-&w_{j_{n}}^{T}&-\end{array}\right].$
For all $\left(i_{1},\ldots,i_{n}\right)\in\Set{1,\ldots,n}^{n}$, if there
exists $a,b\in\Set{1,\ldots,n}$ such that $a\neq b$ and $i_{a}=i_{b}$, then
$\det\left(V_{i_{1}\ldots i_{n}}\right)=0$. Similarly,
$\det\left(W_{j_{1}\ldots j_{n}}\right)=0$ if there exists a pair of equal
coordinates in $\left(j_{1},\ldots,j_{n}\right)$. Thus,

$\displaystyle h_{n}=\det\left(V_{1\ldots n}\right)\det\left(W_{1\ldots
n}\right)\sum_{\sigma,\gamma\in S_{n}}$ $\displaystyle{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ (1) $\displaystyle\langle
v_{\sigma(1)},w_{\gamma(2)}\rangle\cdots\langle
v_{\sigma(n-1)},w_{\gamma(n)}\rangle\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle.$

Now assume that $n$ is even. Define

$\displaystyle\begin{array}[]{cc}\begin{array}[]{c}F_{1}:S_{n}\to S_{n}\\\
F_{1}\left(\left[i_{1},i_{2},\ldots,i_{n}\right]\right)=\left[i_{1},i_{n},i_{n-1},\ldots,i_{2}\right]\end{array}&\begin{array}[]{c}F_{2}:S_{n}\to
S_{n}\\\
F_{2}\left(\left[j_{1},j_{2},\ldots,j_{n}\right]\right)=\left[j_{2},j_{1},j_{n},j_{n-1},j_{n-2},\ldots,j_{3}\right]\end{array}.\end{array}$

Since for all $\sigma=\left[i_{1},\ldots,i_{n}\right]\in S_{n}$,
$F_{1}(\sigma)$ is obtained from $\sigma$ by swapping $i_{2}$ and $i_{n}$,
swapping $i_{3}$ and $i_{n-1}$, …, and finally, swapping $i_{\frac{n}{2}}$ and
$i_{\frac{n+4}{2}}$, ${\rm
sgn}\left(F_{1}(\sigma)\right)=(-1)^{\frac{n}{2}-1}{\rm sgn}(\sigma)$.
Similarly, for all $\gamma=\left[j_{1},\ldots,j_{n}\right]\in S_{n}$,
$F_{2}(\gamma)$ is obtained from $\gamma$ by first swapping $j_{1}$ and
$j_{2}$, and then, swapping $j_{3}$ and $j_{n}$, swapping $j_{4}$ and
$j_{n-1}$, …, and swapping $j_{\frac{n+2}{2}}$ and $i_{\frac{n+4}{2}}$. Thus,
${\rm sgn}\left(F_{2}(\gamma)\right)=(-1)^{\frac{n}{2}}{\rm sgn}(\gamma)$.
Hence,

$\displaystyle{\rm sgn}\left(F_{1}(\sigma)F_{2}(\gamma)\right)\langle
v_{F_{1}(\sigma)(1)},w_{F_{2}(\gamma)(1)}\rangle\cdots\langle
v_{F_{1}(\sigma)(n)},w_{F_{2}(\gamma)(n)}\rangle\langle
v_{F_{1}(\sigma)(1)},w_{F_{2}(\gamma)(2)}\rangle\cdots\langle
v_{F_{1}(\sigma)(n)},w_{F_{2}(\gamma)(1)}\rangle$ $\displaystyle=$
$\displaystyle(-1)^{\frac{n}{2}+\frac{n}{2}-1}{\rm sgn}(\sigma\gamma)\langle
v_{i_{1}},w_{j_{2}}\rangle\langle v_{i_{n}},w_{j_{1}}\rangle\langle
v_{i_{n-1}},w_{j_{n}}\rangle\cdots\langle v_{i_{2}},w_{j_{3}}\rangle\langle
v_{i_{1}},w_{j_{1}}\rangle\langle v_{i_{n}},w_{j_{n}}\rangle\cdots\langle
v_{i_{3}},w_{j_{3}}\rangle\langle v_{i_{2}},w_{j_{2}}\rangle$ $\displaystyle=$
$\displaystyle-{\rm sgn}(\sigma\gamma)\langle
v_{\sigma(1)},w_{\gamma(1)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle\langle
v_{\sigma(1)},w_{\gamma(2)}\rangle\cdots\rangle\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle.$

On the other hand, for all $\sigma\in S_{n}$,
$F_{1}\left(F_{1}(\sigma)\right)=\sigma$ and
$F_{2}\left(F_{2}(\sigma)\right)=\sigma$, which implies that both $F_{1}$ and
$F_{2}$ are bijections. So,

$\displaystyle 2\left(\sum_{\sigma,\gamma\in S_{n}}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle\langle
v_{\sigma(1)},w_{\gamma(2)}\rangle\cdots\langle
v_{\sigma(n-1)},w_{\gamma(n)}\rangle\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle\right)$ $\displaystyle=$
$\displaystyle\sum_{\sigma,\gamma\in S_{n}}{\rm sgn}(\sigma\gamma)\langle
v_{\sigma(1)},w_{\gamma(1)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle\langle
v_{\sigma(1)},w_{\gamma(2)}\rangle\cdots\langle
v_{\sigma(n-1)},w_{\gamma(n)}\rangle\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle+$
$\displaystyle\sum_{\sigma,\gamma\in S_{n}}{\rm
sgn}\left(F_{1}(\sigma)F_{2}(\gamma)\right)\langle
v_{F_{1}(\sigma)(1)},w_{F_{2}(\gamma)(1)}\rangle\cdots\langle
v_{F_{1}(\sigma)(n)},w_{F_{2}(\gamma)(n)}\rangle$
$\displaystyle\phantom{\sum_{\sigma,\gamma\in S_{n}}}\langle
v_{F_{1}(\sigma)(1)},w_{F_{2}(\gamma)(2)}\rangle\cdots\langle
v_{F_{1}(\sigma)(n-1)},w_{F_{2}(\gamma)(n)}\rangle\langle
v_{F_{1}(\sigma)(n)},w_{F_{2}(\gamma)(1)}\rangle$ $\displaystyle=$
$\displaystyle\sum_{\sigma,\gamma\in S_{n}}\bigg{(}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle\langle
v_{\sigma(1)},w_{\gamma(2)}\rangle\cdots\langle
v_{\sigma(n-1)},w_{\gamma(n)}\rangle\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$
$\displaystyle\phantom{\sum_{\sigma,\gamma\in S_{n}}}+{\rm
sgn}\left(F_{1}(\sigma)F_{2}(\gamma)\right)\langle
v_{F_{1}(\sigma)(1)},w_{F_{2}(\gamma)(1)}\rangle\cdots\langle
v_{F_{1}(\sigma)(n)},w_{F_{2}(\gamma)(n)}\rangle$
$\displaystyle\phantom{\sum_{\sigma,\gamma\in S_{n}}}\langle
v_{F_{1}(\sigma)(1)},w_{F_{2}(\gamma)(2)}\rangle\cdots\langle
v_{F_{1}(\sigma)(n-1)},w_{F_{2}(\gamma)(n)}\rangle\langle
v_{F_{1}(\sigma)(n)},w_{F_{2}(\gamma)(1)}\rangle\bigg{)}$ $\displaystyle=$
$\displaystyle 0,$

and thus,

$\displaystyle\sum_{\sigma,\gamma\in S_{n}}{\rm sgn}(\sigma\gamma)\langle
v_{\sigma(1)},w_{\gamma(1)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle\langle
v_{\sigma(1)},w_{\gamma(2)}\rangle\cdots\langle
v_{\sigma(n-1)},w_{\gamma(n)}\rangle\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle=0.$

According to (1), this means that $h_{n}=0$.
We now proceed by assuming that $n$ is odd. In this case, for all
$t\in\Set{1,\ldots,n}$, define

$\displaystyle\begin{array}[]{c}F_{1}^{(t)}:S_{n}\to S_{n}\\\
F_{1}^{(t)}\left(\left[i_{1},\ldots,i_{n}\right]\right)=\left[i_{\gamma^{-1}(t)-1},i_{\gamma^{-1}(t)},i_{\gamma^{-1}(t)+1},\ldots,i_{n},i_{1},i_{2},\ldots,i_{\gamma^{-1}(t)-2}\right]\end{array}$

and

$\displaystyle\begin{array}[]{c}F_{2}^{(t)}:S_{n}\to S_{n}\\\
F_{2}^{(t)}\left(\left[j_{1},\ldots,j_{n}\right]\right)=\left[j_{\gamma^{-1}(t)-1},j_{2},j_{\gamma^{-1}(t)+1},\ldots,j_{n},j_{1},j_{\gamma^{-1}(t)},j_{3},\ldots,j_{\gamma^{-1}(t)-2}\right]\end{array},$

where if $k<0$, by $i_{k}$ and $j_{k}$ we mean $i_{l}$ and $j_{l}$
respectively, such that $k\equiv l\mod{n}$ and $l\in\Set{1,\ldots,5}$. Suppose
that for all $\sigma=\left[i_{1},\ldots,i_{n}\right]\in S_{n}$,
$F_{1}^{(t)}(\sigma)$ is obtained from $\sigma$ by swapping $m$ pairs. Then
${\rm sgn}\left(F_{1}^{(t)}(\sigma)\right)=(-1)^{m}{\rm sgn}(\sigma)$.
Moreover, in this case, according to the definition of $F_{2}^{(t)}$, for all
$\gamma=\left[j_{1},\ldots,j_{n}\right]\in S_{n}$ with $\gamma(2)\neq t$,
$F_{2}^{(t)}(\gamma)$ is obtained from $\gamma$ by first, swapping the same
$m$ pairs and then, swapping $j_{2}$ and $j_{\gamma^{-1}(t)}$ in the resulting
permutation, which means that ${\rm
sgn}\left(F_{2}^{(t)}(\gamma)\right)=(-1)^{m+1}{\rm sgn}(\gamma)$. So, if
$\gamma(2)\neq t$, then

$\displaystyle{\rm sgn}(F_{1}^{(t)}(\sigma)F_{2}^{(t)}({\gamma}))\langle
v_{F_{1}^{(t)}(\sigma)(1)},w_{F_{2}^{(t)}(\gamma)(1)}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(2)},w_{t}\rangle\langle
v_{F_{1}^{(t)}(\sigma)a(3)},w_{F_{2}^{(t)}(\gamma)(3)}\rangle\cdots\langle
v_{F_{1}^{(t)}(\sigma)(n)},w_{F_{2}^{(t)}(\gamma)(n)}\rangle$
$\displaystyle\phantom{{\rm
sgn}(F_{1}^{(t)}(\sigma)F_{1}^{(t)}(\gamma))}\langle
v_{F_{1}^{(t)}(\sigma)(1)},w_{t}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(2)},w_{F_{2}^{(t)}(\gamma)(3)}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(3)},w_{F_{2}^{(t)}(\gamma)(4)}\rangle\cdots\langle
v_{F_{1}^{(t)}(\sigma)(n)},w_{F_{2}^{(t)}(\gamma)(1)}\rangle$ $\displaystyle=$
$\displaystyle(-1)^{2m+1}{\rm sgn}(\sigma\gamma)\langle
v_{i_{\gamma^{-1}(t)-1}},w_{j_{\gamma^{-1}(t)-1}}\rangle\langle
v_{i_{\gamma^{-1}(t)}},w_{t}\rangle\cdots\langle
v_{i_{n}},w_{j_{n}}\rangle\langle v_{i_{1}},w_{j_{1}}\rangle\langle
v_{i_{2}},w_{j_{\gamma^{-1}(t)}}\rangle\langle
v_{i_{\gamma^{-1}(t)-2}},w_{j_{\gamma^{-1}(t)-2}}\rangle$
$\displaystyle\phantom{(-1)^{2m+1}{\rm sgn}(\sigma\gamma)}\langle
v_{i_{\gamma^{-1}(t)-1}},w_{t}\rangle\langle
v_{i_{\gamma^{-1}(t)}},w_{j_{\gamma^{-1}(t)+1}}\rangle\cdots\langle
v_{i_{n}},w_{j_{1}}\rangle\langle
v_{i_{1}},w_{j_{\gamma^{-1}(t)}}\rangle\langle
v_{i_{2}},w_{j_{3}}\rangle\cdots\langle
v_{i_{\gamma^{-1}(t)-2}},w_{j_{\gamma^{-1}(t)-1}}\rangle$ $\displaystyle=$
$\displaystyle-{\rm sgn}(\sigma\gamma)\langle
v_{\sigma(1)},w_{\gamma(1)}\rangle\langle v_{\sigma(2)},w_{t}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{{\rm
sgn}(\sigma\gamma)}\langle v_{\sigma(1)},w_{t}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle.$

Now let $t\in\Set{1,\ldots,n}$. Since for all $\sigma\in S_{n}$,
$F_{1}^{(t)}\left(F_{1}^{(t)}(\sigma)\right)=\sigma$ and
$F_{2}^{(t)}\left(F_{2}^{(t)}(\sigma)\right)=\sigma$, $F_{1}^{(t)}$ and
$F_{2}^{(t)}$ are bijections. Additionally, considering that for $\gamma\in
S_{n}$, if $\gamma(2)\neq t$, then $F_{2}^{(t)}(\gamma)(2)\neq t$,
$F_{2}^{(t)}$ is also a bijection on $A^{(t)}\coloneqq\Set{\gamma\in
S_{n}}{\gamma(2)\neq t}$. Hence,

$\displaystyle 2\sum_{\sigma\in S_{n}}\sum_{\gamma\in A^{(t)}}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{t}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in A^{(t)}}}\langle v_{\sigma(1)},w_{t}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle=$
$\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in A^{(t)}}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{t}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in A^{(t)}}}\langle v_{\sigma(1)},w_{t}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle+$
$\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in A^{(t)}}{\rm
sgn}(F_{1}^{(t)}(\sigma)F_{2}^{(t)})\langle
v_{F_{1}^{(t)}(\sigma)(1)},w_{F_{2}^{(t)}(1)}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(2)},w_{t}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(3)},w_{F_{2}^{(t)}(3)}\rangle\cdots\langle
v_{F_{1}^{(t)}(\sigma)(n)},w_{F_{2}^{(t)}(n)}\rangle$
$\displaystyle\phantom{\sum_{\sigma\in S_{n}}\sum_{\gamma\in A^{(t)}}}\langle
v_{F_{1}^{(t)}(\sigma)(1)},w_{t}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(2)},w_{F_{2}^{(t)}(3)}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(3)},w_{F_{2}^{(t)}(4)}\rangle\cdots\langle
v_{F_{1}^{(t)}(\sigma)(n)},w_{F_{2}^{(t)}(1)}\rangle$ $\displaystyle=$
$\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in A^{(t)}}\bigg{(}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{t}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in A^{(t)}}}\langle v_{\sigma(1)},w_{t}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in A^{(t)}}}+{\rm
sgn}(F_{1}^{(t)}(\sigma)F_{2}^{(t)})\langle
v_{F_{1}^{(t)}(\sigma)(1)},w_{F_{2}^{(t)}(1)}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(2)},w_{t}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(3)},w_{F_{2}^{(t)}(3)}\rangle\cdots\langle
v_{F_{1}^{(t)}(\sigma)(n)},w_{F_{2}^{(t)}(n)}\rangle$
$\displaystyle\phantom{\sum_{\sigma\in S_{n}}\sum_{\gamma\in A^{(t)}}}\langle
v_{F_{1}^{(t)}(\sigma)(1)},w_{t}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(2)},w_{F_{2}^{(t)}(3)}\rangle\langle
v_{F_{1}^{(t)}(\sigma)(3)},w_{F_{2}^{(t)}(4)}\rangle\cdots\langle
v_{F_{1}^{(t)}(\sigma)(n)},w_{F_{2}^{(t)}(1)}\rangle\bigg{)}$ $\displaystyle=$
$\displaystyle 0.$

Hence,

$\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in S_{n}}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{t}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}}}\langle v_{\sigma(1)},w_{t}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle=$
$\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in S_{n}\setminus
A^{(t)}}{\rm sgn}(\sigma\gamma)\langle
v_{\sigma(1)},w_{\gamma(1)}\rangle\langle v_{\sigma(2)},w_{t}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}\setminus A^{(t)}}}\langle
v_{\sigma(1)},w_{t}\rangle\langle v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle=$
$\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in S_{n}\setminus
A^{(t)}}{\rm sgn}(\sigma\gamma)\langle
v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{\gamma(2)}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}\setminus A^{(t)}}}\langle
v_{\sigma(1)},w_{\gamma(2)}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle.$

This means that

$\displaystyle\sum_{t=1}^{n}\sum_{\sigma\in S_{n}}\sum_{\gamma\in S_{n}}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{t}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$
$\displaystyle\phantom{\sum_{t=1}^{n}\sum_{\sigma\in S_{n}}\sum_{\gamma\in
S_{n}}}\langle v_{\sigma(1)},w_{t}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle=$
$\displaystyle\sum_{t=1}^{n}\sum_{\sigma\in S_{n}}\sum_{\gamma\in
S_{n}\setminus A^{(t)}}{\rm sgn}(\sigma\gamma)\langle
v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{\gamma(2)}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$
$\displaystyle\phantom{\sum_{t=1}^{n}\sum_{\sigma\in S_{n}}\sum_{\gamma\in
S_{n}\setminus A^{(t)}}}\langle v_{\sigma(1)},w_{\gamma(2)}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle=$
$\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in S_{n}}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{\gamma(2)}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}}}\langle v_{\sigma(1)},w_{\gamma(2)}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle.$

On the other hand,

$\displaystyle\sum_{t=1}^{n}\sum_{\sigma\in S_{n}}\sum_{\gamma\in S_{n}}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(2)},w_{t}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$
$\displaystyle\phantom{\sum_{t=1}^{n}\sum_{\sigma\in S_{n}}\sum_{\gamma\in
S_{n}}}\langle v_{\sigma(1)},w_{t}\rangle\langle
v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle$ $\displaystyle=$
$\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in S_{n}}{\rm
sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}}}\langle v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle\bigg{(}\sum_{t=1}^{n}\langle
v_{\sigma(2)},w_{t}\rangle\langle v_{\sigma(1)},w_{t}\rangle\bigg{)}$
$\displaystyle=$ $\displaystyle\sum_{\sigma\in S_{n}}\sum_{\gamma\in
S_{n}}{\rm sgn}(\sigma\gamma)\langle v_{\sigma(1)},w_{\gamma(1)}\rangle\langle
v_{\sigma(3)},w_{\gamma(3)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(n)}\rangle$ $\displaystyle\phantom{\sum_{\sigma\in
S_{n}}\sum_{\gamma\in S_{n}}}\langle v_{\sigma(2)},w_{\gamma(3)}\rangle\langle
v_{\sigma(3)},w_{\gamma(4)}\rangle\cdots\langle
v_{\sigma(n)},w_{\gamma(1)}\rangle\langle v_{\sigma(1)},v_{\sigma(2)}\rangle$
$\displaystyle=$ $\displaystyle 0.$

Thus, by (1), $h_{n}=0$. ∎

######  Remark 9.

Note that the second part of the above proof, which proves the theorem when
$n$ is odd, is actually applicable to the case where $n$ is even as well.
However, the approach that has been used for the case where $n$ is even
indicates that in this case, $h_{n}$ vanishes on any tensor train of length 2
whose nodes are tensors in $S^{3}\left(\mathbb{R}^{n}\right)$ with rank at
most $n$, regardless of whether or not these tensors are orthogonally
decomposable.

When $n=2$ and $n=3$, we are able to show using Maple that $h_{n}$ does not
lie in the ideal generated by $\mathcal{P}_{n}\cup\mathcal{Q}_{n}$, but
vanishes on the set $SOT_{2,n}$.

Below we state our general conjecture about $SOT_{2,n}$ and provide some
evidence for this conjecture.

######  Conjecture 10.

The Zariski closure of the set $SOT_{2,n}$ in $\mathbb{C}^{n^{4}}$ is cut out
by the vanishing of $\mathcal{P}_{n},\mathcal{Q}_{n}$, and the polynomial
$h_{n}$. The ideal defined by these equations is prime.

######  Lemma 11.

The dimension of the Zariski closure of $SOT_{2,n}$ in $\mathbb{C}^{n^{4}}$ is
$n(n+1)-1$.

###### Proof.

Define the map

$\displaystyle\begin{array}[]{c}\phi:\mathbb{R}^{n}\times
SO_{n}\times\mathbb{R}^{n}\times
SO_{n}\to\mathbb{R}^{n^{4}}\subseteq\mathbb{C}^{n^{4}}\\\
\left(\lambda_{1},\ldots,\lambda_{n}\right),V,\left(\mu_{1},\ldots,\mu_{n}\right),W\mapsto\sum_{i=1}^{n}\sum_{j=1}^{n}\lambda_{i}\mu_{j}v_{i}^{\otimes
2}\otimes w_{j}^{\otimes 2}\langle v_{i},w_{j}\rangle\end{array},$

where $v_{i}$ and $w_{i}$ are the $i$th columns of the orthogonal matrices $V$
and $W$ respectively. Then $SOT_{2,n}={\rm Im}(\phi)$. It is proved in [5]
that the decomposition of a generic tensor in $SOT_{2,n}$ can be found
uniquely up to permutations and a global rescaling of $\lambda_{i}$’s and
$\mu_{j}$’s (i.e. if we multiply all $\lambda_{i}$’s by $C$, we have to divide
all $\mu_{j}$’s by $C$.). So, the dimension of the general fibre of $\phi$ is
1. Thus,

$\displaystyle\dim\left({\rm Im}(\phi)\right)=\dim\left(\mathbb{R}^{n}\times
SO_{n}\times\mathbb{R}^{n}\times SO_{n}\right)-1=2n+2\binom{n}{2}-1=n(n+1)-1.$

So, the dimension of the Zariski closure of $SOT_{2,n}$ in
$\mathbb{C}^{n^{4}}$ is $\dim\left(\overline{{\rm Im}(\phi)}\right)=n(n+1)-1$.
∎

When $n=2$, Lemma 11 shows that the dimension of the Zariski closure of the
set $SOT_{2,2}$ in $\mathbb{C}^{16}$ equals 5. The dimension of the ideal
$I\coloneqq\langle\mathcal{P}_{2},\mathcal{Q}_{2},h_{2}\rangle$ also equals 5,
which was checked using `Groebner[HilbertDimension]` in Maple. Furthermore,
`PolynomialIdeals[PrimeDecomposition]` only generates one ideal when applied
to $I$, which means that the radical of $I$ is prime. On the other hand,
`PolynomialIdeals[IsRadical]` verifies that $I$ is radical. So, $I$ is
actually prime, and the above-mentioned equality of the dimensions proves that
$I$ is the prime ideal of the Zariski closure of $SOT_{2,2}$ in
$\mathbb{C}^{16}$. We have not been able to verify that that the ideal
generated by $\mathcal{P}_{n}$, $\mathcal{Q}_{n}$, and $h_{n}$ is prime for
higher values of $n$ using Maple due to computational limitations.

##  4 Acknowledgements

ER was supported by an NSERC Discovery Grant (DGECR-2020-00338).

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