# On the Structure Tensor of $\mathfrak{sl}_{n}$

Kashif K. Bari<EMAIL_ADDRESS>
###### Abstract.

The structure tensor of $\mathfrak{sl}_{n}$, denoted $T_{\mathfrak{sl}_{n}}$,
is the tensor arising from the Lie bracket bilinear operation on the set of
traceless $n\times n$ matrices over $\mathbb{C}$. This tensor is intimately
related to the well studied matrix multiplication tensor. Studying the
structure tensor of $\mathfrak{sl}_{n}$ may provide further insight into the
complexity of matrix multiplication and the “hay in a haystack” problem of
finding explicit sequences tensors with high rank or border rank. We aim to
find new bounds on the rank and border rank of this structure tensor in the
case of $\mathfrak{sl}_{3}$ and $\mathfrak{sl}_{4}$. We additionally provide
bounds in the case of the lie algebras $\mathfrak{so}_{4}$ and
$\mathfrak{so}_{5}$. The lower bounds on the border ranks were obtained via
various recent techniques, namely Koszul flattenings, border substitution, and
border apolarity. Upper bounds on the rank of $T_{\mathfrak{sl}_{3}}$ are
obtained via numerical methods that allowed us to find an explicit rank
decomposition.

###### Key words and phrases:

bilinear complexity, structure tensor, rank, border rank, Lie algebra

###### 2010 Mathematics Subject Classification:

15A69, 14L35, 68Q15

Author was supported by the Mathematics Department at Texas A&M University and
NSF grant AF-1814254.

##  1\. Introduction

In 1969, Strassen presented a novel algorithm for matrix multiplication of
$n\times n$ matrices. Strassen’s algorithm used fewer than the $O(n^{3})$
arithmetic operations needed for the standard algorithm. This led to the
question: what is the minimal number of arithmetic operations required to
multiply $n\times n$ matrices, or in other words, what is the complexity of
matrix multiplication [Str69] [Str83]. An asymptotic version of the problem is
to determine the exponent of matrix multiplication, $\omega$, which is the
infimum of all values such that for all $\epsilon>0$, multiplying $n\times n$
matrices can be performed in $O(n^{\omega+\epsilon})$ arithmetic operations.
Any bilinear operation, including matrix multiplication, may be thought of as
tensor in the following way: Let $A,B,$ and $C$ denote vector spaces over
$\mathbb{C}$. Given a bilinear map $A^{*}\times B^{*}\to C$, the universal
property of tensor products induces a linear map $A^{*}\otimes B^{*}\to C$.
Since $\text{Hom}_{\mathbb{C}}(A^{*}\otimes B^{*},C)\simeq A\otimes B\otimes
C$, then we can take our bilinear map to be a tensor in $A\otimes B\otimes C$.
Let $M_{\langle n\rangle}$ denote the matrix multiplication tensor arising
from the bilinear operation of multiplying $n\times n$ matrices.

An important invariant of a tensor is its rank. For a tensor $T\in A\otimes
B\otimes C$ the rank, denoted $\textbf{R}(T)$, is the minimal $r$ such that
$T=\sum_{i=1}^{r}a_{i}\otimes b_{i}\otimes c_{i}$ with $a_{i}\in A,b_{i}\in
B,c_{i}\in C$ for $1\leq i\leq r$. Given precise $T_{i}=a_{i}\otimes
b_{i}\otimes c_{i}$, then we call $T=\sum_{i=1}^{r}T_{i}$ a rank decomposition
of $T$. Strassen also showed that the rank of the matrix multiplication tensor
is a valid measure of its complexity; in particular, he proved
$\omega=\inf\\{\tau\in\mathbb{R}\ |\ \textbf{R}(M_{\langle
n\rangle})=O(n^{\tau})\\}$ [Str69].

For a tensor $T\in A\otimes B\otimes C$, the border rank of $T$, denoted
$\textbf{\text@underline{R}}(T)$, is another invariant of interest and defined
to be the minimal $r$ such that $T=\displaystyle\lim_{\epsilon\to
0}T_{\epsilon}$ where for all $\epsilon>0$, $T_{\epsilon}$ has rank $r$. Given
rank decompositions of $\displaystyle
T_{\epsilon}=\sum_{i=1}^{r}T_{i}(\epsilon)$, we then call $\lim_{\epsilon\to
0}\sum_{i=1}^{r}T_{i}(\epsilon)$ a border rank decomposition of $T$. Later, in
1980, it was shown by Bini that the border rank of matrix multiplication is
also a valid measure of its complexity by proving that
$\omega=\inf\\{\tau\in\mathbb{R}\ |\ \textbf{\text@underline{R}}(M_{\langle
n\rangle})=O(n^{\tau})\\}$ [Bin80].

Intimately related to the matrix multiplication tensor is the structure tensor
of the Lie algebra $\mathfrak{sl}_{n}$, the set of traceless $n\times n$
matrices over $\mathbb{C}$ equipped with the Lie bracket $[x,y]=xy-yx$. The
structure tensor of $\mathfrak{sl}_{n}$ is defined as the tensor arising from
the Lie bracket bilinear operation, and we denote it by
$T_{\mathfrak{sl}_{n}}$. One example of how matrix multiplication is related
to $T_{\mathfrak{sl}_{n}}$ is by closer examination of a skew-symmetric
version of the matrix multiplication tensor; consider the tensor arising from
the Lie bracket bilinear operation on $\mathfrak{gl}_{n}$, (which is just
$M_{n}$, but considered as a Lie algebra) [CHI+18]. Since
$\mathfrak{gl}_{n}=\mathfrak{sl}_{n}\oplus z$, where $z$ indicates the scalar
matrices which are central in $\mathfrak{gl}_{n}$, $T_{\mathfrak{sl}_{n}}$
determines the commutator action on all of $\mathfrak{gl}_{n}$. While the
matrix multiplication tensor has been well studied [CHI+18] [CHL19], the
structure tensor of $\mathfrak{sl}_{n}$ has not been studied to the same
extent. Currently, the only known non-trivial results are lower bounds on the
rank of the structure tensor of $\mathfrak{sl}_{n}$. In [dGH86] it was shown
$\textbf{R}(T_{\mathfrak{sl}_{n}})\geq 2n^{2}-n-1$. Studying the structure
tensor of $\mathfrak{sl}_{n}$ may provide some further insight into two
central problems in complexity theory.

In complexity theory, it is of interest to find explicit objects that behave
generically. This type of problem is known as a “hay in a haystack” problem.
Algebraic geometry tells us that a “random” tensor $T$ in
$\mathbb{C}^{m}\otimes\mathbb{C}^{m}\otimes\mathbb{C}^{m}$ will have
rank/border rank $\lceil\frac{m^{3}}{3m-2}\rceil$. By an explicit sequence of
tensors, we will mean a collection of tensors
$T_{m}\in\mathbb{C}^{m}\otimes\mathbb{C}^{m}\otimes\mathbb{C}^{m}$ such that
the coefficients of $T_{m}$ are computable in polynomial time in m. The “hay
in a haystack” problem for tensors is to find an example of an explicit
sequence of tensors of high rank or border rank, asymptotically in $m$.
Currently, there exists an explicit sequence of tensors, $S_{m}$ over
$\mathbb{C}$, such that $\textbf{R}(S_{m})\geq 3m-o(\log(m))$[AFT11] and a
different explicit sequence of tensors over $\mathbb{C}$, $T_{m}$, such that
$\textbf{\text@underline{R}}(T_{m})\geq 2.02m-o(m)$ [LM19]. One should note
that the sequence $T_{m}$ of [LM19] has border rank equal to $2m$ when $m=13$
and has been shown to exceed $2m$ for $m>364175$. It would be of interest to
find sequences of tensors for which the border rank exceeds $2m$ for smaller
values of $m$.

The second problem is Strassen’s problem of computing of the complexity of
matrix multiplication. The exponent of $\mathfrak{sl}_{n}$ is defined as
$\displaystyle\omega(\mathfrak{sl}_{n}):=\liminf_{n\to\infty}\log_{n}(\textbf{R}(T_{\mathfrak{sl}_{n}}))$.
By Theorem 4.1 from [LY20], the exponent of matrix multiplication is equal to
the exponent of $\mathfrak{sl}_{n}$. Consequently, upper bounds on the rank
and even the border rank of $T_{\mathfrak{sl}_{n}}$ provide upper bounds on
$\omega$.

These two problems motivate our study of the border rank of
$T_{\mathfrak{sl}_{n}}$. We prove new bounds in the case of
$\mathfrak{sl}_{3}$ and $\mathfrak{sl}_{4}$.

######  Theorem 1.

$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{3}})\geq 16$

######  Theorem 2.

$\textbf{R}(T_{\mathfrak{sl}_{3}})\leq 20$

######  Theorem 3.

$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{4}})\geq 28$

Additionally, we obtain lower bounds on the border rank of the structure
tensors of $\mathfrak{so}_{4}$ and $\mathfrak{so}_{5}$. We note that
$\mathfrak{so}_{4}\simeq\mathfrak{sl}_{2}\times\mathfrak{sl}_{2}$ and
$\mathfrak{so}_{5}\simeq\mathfrak{sp}_{4}$.

######  Theorem 4.

$\textbf{\text@underline{R}}(T_{\mathfrak{so}_{4}})\geq 9$

######  Theorem 5.

$\textbf{\text@underline{R}}(T_{\mathfrak{so}_{5}})\geq 19$

##  2\. Preliminaries

The above definition of $T_{\mathfrak{sl}_{n}}$ is independent of choice of
basis, but we may also write the tensor in terms of bases. Let
$\displaystyle\\{a_{i}\\}_{i=1}^{n^{2}-1}$ be a basis of $\mathfrak{sl}_{n}$
and $\displaystyle\\{\alpha^{i}\\}_{i=1}^{n^{2}-1}$ a dual basis. Recall that
$\mathfrak{sl}_{n}$ has a bilinear operation called the Lie bracket, given by
$\displaystyle[a_{i},a_{j}]=a_{i}a_{j}-a_{j}a_{i}=\sum_{k=1}^{n^{2}-1}A_{ij}^{k}a_{k}$.
The structure tensor of $\mathfrak{sl}_{n}$ in this basis is
$T_{\mathfrak{sl}_{n}}=\sum_{i,j,k}A_{ij}^{k}\alpha^{i}\otimes\alpha^{j}\otimes
a_{k}\in\mathfrak{sl}_{n}^{*}\otimes\mathfrak{sl}_{n}^{*}\otimes\mathfrak{sl}_{n}$.
Let $e_{i}^{j}$ denote the $n\times n$ matrix with $1$ in the $(i,j)$th entry.
If we take the standard weight basis of $\mathfrak{sl}_{n}$, namely the
matrices $e_{i}^{j}$ for $1\leq i\neq j\leq n$ and $e_{i}^{i}-e_{i+1}^{i+1}$
for $1\leq i\leq n-1$, then we can express the tensor $T_{\mathfrak{sl}_{3}}$
as follows.

$\begin{bmatrix}0&0&0&0&0&0&0&0\\\ 0&2&0&0&0&0&0&0\\\ 0&0&1&0&0&0&0&0\\\
0&0&0&-2&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&-1&0&0\\\ 0&0&0&0&0&0&-1&0\\\
0&0&0&0&0&0&0&1\end{bmatrix},\begin{bmatrix}0&-2&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 1&0&0&0&0&0&0&0\\\ 0&1&0&0&0&0&0&0\\\
0&0&1&0&0&0&0&0\\\ 0&0&0&0&0&0&0&-1\\\ 0&0&0&0&0&0&0&0\end{bmatrix}$

$\begin{bmatrix}0&0&-1&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\
0&0&0&0&0&-1&0&0\\\ 0&0&-1&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 1&0&0&0&1&0&0&0\\\
0&1&0&0&0&0&0&0\end{bmatrix},\begin{bmatrix}0&0&0&2&0&0&0&0\\\
-1&0&0&0&0&0&0&0\\\ 0&0&0&0&0&1&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&0&0&-1&0&0&0&0\\\
0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&-1&0\end{bmatrix}$

$\begin{bmatrix}0&0&0&0&0&0&0&0\\\ 0&-1&0&0&0&0&0&0\\\ 0&0&1&0&0&0&0&0\\\
0&0&0&1&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&2&0&0\\\ 0&0&0&0&0&0&-1&0\\\
0&0&0&0&0&0&0&-2\end{bmatrix},\begin{bmatrix}0&0&0&0&0&1&0&0\\\
0&0&-1&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&-2&0&0\\\
0&0&0&0&0&0&0&0\\\ 0&0&0&1&0&0&0&0\\\ 0&0&0&0&1&0&0&0\end{bmatrix}$

$\begin{bmatrix}0&0&0&0&0&0&1&0\\\ 0&0&0&0&0&0&0&1\\\ -1&0&0&0&-1&0&0&0\\\
0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&1&0\\\ 0&0&0&-1&0&0&0&0\\\ 0&0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0&0\end{bmatrix},\begin{bmatrix}0&0&0&0&0&0&0&-1\\\
0&0&0&0&0&0&0&0\\\ 0&-1&0&0&0&0&0&0\\\ 0&0&0&0&0&0&1&0\\\ 0&0&0&0&0&0&0&2\\\
0&0&0&0&-1&0&0&0\\\ 0&0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0&0\end{bmatrix}$

We establish some basic definitions of algebraic geometry and representation
theory that will be used in our study of the structure tensor of
$\mathfrak{sl}_{n}$.

###  2.1. Algebraic Geometry

The language of algebraic geometry will prove useful in studying our problems.
Throughout, let $A,B,C,$ and $V$ be complex vector spaces. Let
$\pi:V\backslash\\{0\\}\to\mathbb{P}V$ be the projection map from $V$ to its
projectivization. Denote $[v]:=\pi(v)\in\mathbb{P}V$ for nonzero $v\in V$. For
$v_{1},\cdots,v_{k}\in V$, let $\langle v_{1},\cdots,v_{k}\rangle$ denote the
projectivization of the linear span of $v_{1},\cdots,v_{k}$.

See [Sha13] for the definitions of the Zariski topology, projective variety,
and the dimension of a projective variety. For a projective variety,
$X\subset\mathbb{P}V$, let S[X] denote its homogeneous coordinate ring and let
$I(X)\subset Sym(A^{*})$ denote the ideal of $X$. Given $Y\subset\mathbb{P}V$
a nondegenerate projective variety, we define the $r$th secant variety of $Y$,
denoted $\sigma_{r}(Y)\subset\mathbb{P}V$.

######  Definition 6.

The $r$th secant variety of $Y$ is
$\sigma_{r}(Y)=\displaystyle\overline{\bigcup_{y_{i}\in Y}\langle
y_{1},\cdots,y_{r}\rangle}$

The closure operation indicated in Definition 1 is the Zariski closure. As
polynomials are continuous, this closure also contains the Euclidean closure
of the set. In this case, we have that the Euclidean closure is in fact equal
to the Zariski closure (see Theorem 3.1.6.1 in [Lan17] or Theorem 2.3.3 in
[Mum95]).

Let $A,B,C$ be vector spaces over $\mathbb{C}$. The Segre embedding is defined
as map
$Seg:\mathbb{P}A\times\mathbb{P}B\times\mathbb{P}C\to\mathbb{P}(A\otimes
B\otimes C)$ given by $([a],[b],[c])\mapsto[a\otimes b\otimes c]$. Let
$X=Seg(\mathbb{P}A\times\mathbb{P}B\times\mathbb{P}C)$ denote the image of the
Segre embedding, which is a projective variety of dimension $\dim A+\dim
B+\dim C-3$. Note that $X$ is the space of rank one tensors in
$\mathbb{P}(A\otimes B\otimes C)$. Consequently, one can redefine the border
rank, $\textbf{\text@underline{R}}(T)$, for $T\in A\otimes B\otimes C$ as the
$r$ such that $T\in\sigma_{r}(X)$ and $T\notin\sigma_{r-1}(X)$. We make a
remark here that $\textbf{R}(T)\geq\textbf{\text@underline{R}}(T)$, trivially,
as a tensor rank decomposition can be considered as a border rank
decomposition by taking the tensor rank decomposition as a constant sequence.

For a secant variety $\sigma_{r}(Y)\subset\mathbb{P}V$ in general, the
expected dimension will be $\min\\{r\dim Y+(r-1),\dim\mathbb{P}V\\}$, since we
expect to choose $r$ points from $Y$ and have an additional $r-1$ parameters
to span the $r$-plane generated by those points. The dimension of
$\displaystyle\sigma_{r}(Seg(\mathbb{P}V\times\mathbb{P}V\times\mathbb{P}V))$
is the expected dimension, except when $r=4$ and $\dim V=3$ [Lic85].

Another variety we will make use of is the Grassmannian, denoted $G(k,V)$. Let
$\Lambda^{k}V$ denote the $k$th exterior power of the vector space $V$. The
Grassmannian variety is $G(k,V):=\mathbb{P}\\{T\in\Lambda^{k}V\ |\ \exists
v_{1},\cdots,v_{k}\in V\text{ such that }T=v_{1}\wedge\cdots\wedge v_{k}\\}$.

###  2.2. Representation Theory

Recall that $A,B,C,$ and $V$ are complex vector spaces. A guiding principle in
geometric complexity theory is to use symmetry to reduce the problem of
testing a space of tensors to testing particular representatives of families
of tensors. To describe the symmetry of our tensor, we use the language of the
representation theory of linear algebraic groups and Lie algebras. See [FH91]
for the definitions of linear algebraic groups, semisimple Lie algebras,
representations, orbits of group actions, $G$-modules, irreducibility of a
representation/module, Borel subgroups, a maximal torus, and the
correspondence between Lie groups and Lie algebras.

The group $GL(A)\times GL(B)\times GL(C)$ acts on $A\otimes B\otimes C$ by the
product of the natural actions of $GL(A)$ on $A$, etc. Identify
$(\mathbb{C}^{*})^{\times 2}$ with the subgroup
$\\{(aId_{A},bId_{B},cId_{C})\in GL(A)\times GL(B)\times GL(C)\ |\ abc=1\\}$
of $GL(A)\times GL(B)\times GL(C)$. Note that $(\mathbb{C}^{*})^{\times 2}$
acts trivially on $A\otimes B\otimes C$.

######  Definition 7.

For a tensor $T\in A\otimes B\otimes C$, define the symmetry group of $T$ to
be the group $\displaystyle G_{T}:=\\{g\in GL(A)\times GL(B)\times
GL(C)/(\mathbb{C}^{*})^{\times 2}\ |\ gT=T\\}$.

In the case of $T_{\mathfrak{sl}_{n}}$, our symmetry group,
$G_{T_{\mathfrak{sl}_{n}}}$, is in fact isomorphic to $SL_{n}$. For any
element $g\in SL_{n}$, we have the element $g^{*}\otimes g^{*}\otimes g$ acts
on $\mathfrak{sl}_{n}^{*}\otimes\mathfrak{sl}_{n}^{*}\otimes\mathfrak{sl}_{n}$
and leave $T_{\mathfrak{sl}_{n}}$ invariant. It is always the case that for
any automorphism of $\mathfrak{sl}_{n}$, we will have an automorphism of
$\mathfrak{sl}_{n}^{*}\otimes\mathfrak{sl}_{n}^{*}\otimes\mathfrak{sl}_{n}$.
See [Mir85] for a proof that these are all elements of the symmetry group for
$T_{\mathfrak{sl}_{n}}$.

Let $B_{T}\subset G_{T}$ denote a Borel subgroup. In the case of
$T_{\mathfrak{sl}_{n}}$, where our symmetry group is isomorphic to $SL_{n}$,
we take $B_{T}$ to be the Borel subgroup of upper triangular matrices of
determinant 1. We note that Borel subgroups are not unique, but are all
conjugate. For this Borel subgroup, let $N\subset B_{T}$ denote the group of
upper triangular matrices with diagonal entries equal to 1, called the maximal
unipotent group, and let $\mathbb{T}$ denote the subgroup of diagonal
matrices, also called the maximal torus.

######  Definition 8.

A vector $v\in V^{\otimes k}$ is a weight vector if $\mathbb{T}[v]=[v]$. In
particular, for $t\in\mathbb{T}$, where
$t=\text{diag}\\{t_{1},\cdots,t_{n}\\}$, if $tv=t_{1}^{p_{1}}\cdots
t_{n}^{p_{n}}v$, then $v$ is said to have weight
$(p_{1},\cdots,p_{n})\in\mathbb{Z}^{n}$.

Furthermore, call $v$ a highest weight vector if $B_{T}[v]=[v]$, i.e. if
$Nv=v$.

In our case where $G_{T}\simeq SL_{n}$, every irreducible $G_{T}$-module will
have a highest weight line and addtionally will be be uniquely determined by
this highest weight line.

Given a symmetry group, $G_{T}$, we also have a symmetry Lie algebra, denoted
$\mathfrak{g}_{T}$, which will be more convenient to work with. Let
$\mathfrak{b}_{T}$ denote the Borel subalgebra and $\mathfrak{h}$ denote the
Cartan subalgebra, which will be the Lie algebras of the Borel subgroup,
$B_{T}$, and maximal torus, $\mathbb{T}$, respectively. In the case of our
symmetry Lie algebra, $\mathfrak{sl}_{n}$, we take $\mathfrak{h}$ to be the
subalgebra of traceless diagonal matrices. Additionally, for $N\subset B_{T}$,
we have the corresponding Lie subalgebra
$\mathfrak{n}\subset\mathfrak{b}_{T}$, which will consist of the strictly
upper triangular elements of $\mathfrak{sl}_{n}$. We will refer to elements of
$\mathfrak{n}$ as raising operators.

For a Lie algebra representation, a vector is a weight vector, if
$\mathfrak{h}[v]=[v]$. One may regard the weight of a vector $v$ as a linear
functional $\lambda\in\mathfrak{h}^{*}$, such that for all $H\in\mathfrak{h}$,
$Hv=\lambda(H)v$. Analogous to the above definition, $[v]$ is a highest weight
vector if and only if $\mathfrak{n}[v]=0$. Consider the weights
$\omega_{i}\in\mathfrak{h}^{*}$ satisfying $\omega_{i}(h_{j})=\delta_{ij}$ for
all $1\leq i,j\leq n-1$, where $h_{j}=e_{j}^{j}-e_{j+1}^{j+1}\in\mathfrak{h}$.
Call these weights $\omega_{1},\cdots,\omega_{n-1}\in\mathfrak{h}^{*}$ the
fundamental weights of $\mathfrak{sl}_{n}$. It is well known that the highest
weight $\lambda$ of an irreducible representation can be represented as an
integral linear combination of fundamental weights. For notational
convenience, we denote the irreducible representation with highest weight
$\lambda=a_{1}\omega_{1}+\cdots+a_{n-1}\omega_{n-1}$ by $[a_{1}\ \cdots\
a_{n-1}]$ and denote its highest weight vector by $v_{[a_{1}\ \cdots\
a_{n-1}]}$. We note that this notation is different from the widely used
diagram notation of Weyl [FH91]

######  Definition 9.

A variety $X\subset\mathbb{P}V$ is $G$-homogeneous if it is a closed orbit of
some point $x\in\mathbb{P}V$ under the action of some group $G\subset GL(V)$.
If $P\subset G$ is the subgroup fixing $x$, write $X=G/P$.

The purpose for introducing the language of homogeneous varieties is to
introduce the following Normal Form Lemma. Recall that $V$ is a complex vector
space.

######  Lemma 10 (Normal Form Lemma).

Let $X=G/P\subset\mathbb{P}V$ be a homogeneous variety and $v\in V$ such that
$G_{v}=\\{g\in G\ |\ g[v]=[v]\\}$ has a single closed orbit
$\mathcal{O}_{min}$ in $X$. Then any border rank $r$ decomposition of $v$ may
be modified using $G_{v}$ to a border rank $r$ decomposition
$\lim_{\epsilon\to 0}x_{1}(\epsilon)+\cdots+x_{r}(\epsilon)$ such that there
is a stationary point $x_{1}(t)\equiv x_{1}$ (i.e. $x_{1}$ is independent of
$t$) lying in $\mathcal{O}_{min}$.

If, moreover, every orbit of $G_{v}\cap G_{x_{1}}$ contains $x_{1}$ in its
closure, we may further assume that for all $j\neq 1$, $\lim_{\epsilon\to
0}x_{j}(\epsilon)=x_{1}$.

See Lemma 3.1 in [LM17] for the proof. This Lemma allows one describe the
interactions between the different $G$-orbits. It can be thought of as a
consequence of the Lie’s theorem. Lie’s Theorem states that for a solvable
group, $H$, an $H$-module, $W$, and a point $[w]\in\mathbb{P}W$, the closed
orbit $\overline{H[w]}$ contains an $H$-fixed point [FH91].

##  3\. Methodology

The most fruitful current techniques for finding lower bounds on the border
rank of a tensor are Koszul flattenings, the border substitution method, and
border apolarity. We review each of these techniques.

###  3.1. Koszul flattenings

For $T\in A\otimes B\otimes C$, we may consider it as a linear map
$T_{B}:B^{*}\to A\otimes C$. We have analogous maps $T_{A},T_{B},T_{C}$, which
are called the coordinate flattenings of $T$. Consider the linear map obtained
by composing the map $T_{B}\otimes
Id_{\Lambda^{p}A}:B^{*}\otimes\Lambda^{p}A\to\Lambda^{p}A\otimes A\otimes C$
with the map $\pi\otimes Id_{C}:\Lambda^{p}A\otimes A\otimes
C\to\Lambda^{p+1}A\otimes C$. Note that $\pi\otimes Id_{C}$ is the tensor
product of the exterior multiplication map with the identity on $C$. Denote
this composition by $T_{A}^{\wedge p}$. Let $rank$ denote the rank of a linear
map.

######  Proposition 11 (Landsberg-Ottaviani).

Let $T\in A\otimes B\otimes C$ and $t=a\otimes b\otimes c\in A\otimes B\otimes
C$, then

$\displaystyle\displaystyle\textbf{\text@underline{R}}(T)\geq\frac{\text{rank}(T_{A}^{\wedge
p})}{\text{rank}(t_{A}^{\wedge p})}=\frac{\text{rank}(T_{A}^{\wedge p})}{{\dim
A-1\choose p}}$

See [LO15] for the proof. One should note that we achieve the best bounds when
$\dim A=2p+1$. Thus, if $\dim A>2p+1$, we may restrict $T$ to subspaces
$A^{\prime}\subset A$ of dimension $2p+1$, since border rank is upper semi-
continuous with respect to restriction i.e. for a restriction $T^{\prime}$ of
$T$,
$\textbf{\text@underline{R}}(T^{\prime})\leq\textbf{\text@underline{R}}(T)$.
Koszul flattenings alone are insufficient to prove
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{n}})\geq 2(n^{2}-1)$, as the
limit of the method for
$T\in\mathbb{C}^{m}\otimes\mathbb{C}^{m}\otimes\mathbb{C}^{m}$ is below
$2m-3$(m even) and $2m-5$(m odd). See [LO15] for more on this method.

###  3.2. Border Substitution

The only known technique for computing lower bounds on the rank of a tensor is
the substitution method. A tensor $T$ is A-concise if the coordinate
flattening map $T_{A}$ is injective, and define similarly for $B$-concise and
$C$-concise. If a tensor is $A$-concise, $B$-concise, and $C$-concise, then we
simply call it concise. We remark that $T_{\mathfrak{sl}_{n}}$ is in fact a
concise tensor, since $\mathfrak{sl}_{n}$ is a simple Lie algebra and the
coordinate flattening maps do not send everything to 0.

The substitution method is a standard technique for obtaining lower bounds on
the rank of a tensor [AFT11]. This technique can be extended to border rank.

######  Proposition 12 (Landsberg-Michałek).

Let $T\in A\otimes B\otimes C$ be $A$-concise. Let $\dim A=m$ and let $k<m$.
Then

$\displaystyle\textbf{\text@underline{R}}(T)\geq\min_{\displaystyle
A^{\prime}\in
G(k,A^{*})}\textbf{\text@underline{R}}(T\big{|}_{A^{\prime}\otimes
B^{*}\otimes C^{*}})+(m-k)$

See [LM19] for proof. Note that the notation $T\big{|}_{A^{\prime}\otimes
B^{*}\otimes C^{*}}$ is a restriction of $T$ when considering $T$ as a
trilinear form $T:A^{*}\otimes B^{*}\otimes C^{*}\to\mathbb{C}$. If we let
$\tilde{A}=A/(A^{\prime})^{\perp}$ then our restricted tensor will be an
element of $\tilde{A}\otimes B\otimes C$.

Also note that in the border substitution proposition, we are minimizing over
all elements in the Grassmannian. In practice, applying border substitution
uses tensors with large symmetry groups $G_{T}$. The utility is that one may
restrict to looking at representatives of closed $G_{T}$-orbits in the
Grassmannian, rather than by examining all elements of the Grassmannian. One
often achieves the best results on the rank of a tensor by using border
substitution in conjunction with Koszul flattenings. Naively, the largest
lower bound obtainable by the method, i.e. the limit of the method, is at most
$\dim A+\dim B+\dim C-3$, however, the limit is in fact slightly less. For
tensors $T\in\mathbb{C}^{m}\otimes\mathbb{C}^{m}\otimes\mathbb{C}^{m}$, the
limit of the method is $3m-3\sqrt{3m+\frac{9}{4}}+\frac{9}{2}$. See [LM17] for
a proof of this and more on this method. The best lower bound achieved on the
border rank that is mentioned in the introduction is achieved using this
method [LM19].

###  3.3. Border Apolarity

In order to establish larger lower bounds on
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{3}})$ than can be achieved by
Koszul flattenings and border substitution for $T_{\mathfrak{sl}_{3}}$, we
will use border apolarity, as developed in [BB20] and [CHL19].

Suppose $T$ has a border rank $r$ decomposition, $T=\lim_{\epsilon\to
0}T_{\epsilon}$, where $T_{\epsilon}=\sum_{i=1}^{r}T_{i}(\epsilon)$. If the
rank summands $T_{i}(\epsilon)$ are in general position in $A\otimes B\otimes
C$, then we may identify the border rank decomposition with a curve
$E_{\epsilon}$ in the Grassmannian variety $G(r,A\otimes B\otimes C)$, by
taking the exterior product of the $T_{i}(\epsilon)$, i.e.
$E_{\epsilon}=[T_{1}(\epsilon)\wedge\cdots\wedge T_{r}(\epsilon)]$.

Now we define a $\mathbb{Z}^{3}$-grading on ideals of subsets of
$\mathbb{P}A\times\mathbb{P}B\times\mathbb{P}C$ (i.e. the Segre variety) from
the natural $\mathbb{Z}^{3}$-grading of $Sym(A\oplus B\oplus C)^{*}$. Let
$Irrel:=\\{0\oplus B\oplus C\\}\cup\\{A\oplus 0\oplus C\\}\cup\\{A\oplus
B\oplus 0\\}\subset A\oplus B\oplus C$. Since
$\mathbb{P}A\times\mathbb{P}B\times\mathbb{P}C\simeq(A\oplus B\oplus
C\backslash Irrel)/(\mathbb{C}^{*})^{\times 3}$, then we may consider the
quotient map $q:(A\oplus B\oplus C)\backslash
Irrel\to\mathbb{P}A\times\mathbb{P}B\times\mathbb{P}C$, which will be
invariant under the action of $(\mathbb{C}^{*})^{\times 3}$. Therefore, for a
set $Z\subset\mathbb{P}A\times\mathbb{P}B\times\mathbb{P}C)$, the ideal of
this set $I(Z)=I(q^{-1}(Z))\subset Sym(A\oplus B\oplus C)^{*}$ will have a
$\mathbb{Z}^{3}$ grading. In particular, for a single point
$([a],[b],[c])\in\mathbb{P}A\times\mathbb{P}A\times\mathbb{P}C$, corresponding
to a rank one tensor ($[a\otimes b\otimes c]\in\mathbb{P}(A\otimes B\otimes
C)$), we are considering the ideal in $Sym(A\oplus B\oplus C)^{*}$ of
polynomials vanishing along the lines $a,b,$ and $c$.

Let $I_{\epsilon}$ denote the $\mathbb{Z}^{3}$-graded ideal of the set of the
r points $[T_{i}(\epsilon)]$, i.e. $I_{ijk,\epsilon}\subset S^{i}A^{*}\otimes
S^{j}B^{*}\otimes S^{k}C^{*}$. Since the $r$ points are in general position,
then $\operatorname{codim}I_{ijk,\epsilon}=r$. Define $\displaystyle
I_{ijk}:=\lim_{\epsilon\to 0}I_{ijk,\epsilon}$ as the limit of points in the
Grassmannian $G(\dim(S^{i}A^{*}\otimes S^{j}B^{*}\otimes
S^{k}C^{*})-r,\dim(S^{i}A^{*}\otimes S^{j}B^{*}\otimes S^{k}C^{*}))$.
$I_{ijk}$ will exist, since the Grassmannian is compact, however, the
resulting ideal $I$ may not be saturated. See [BB20] for further discussion on
this.

Recall that a tensor $T$ is concise if all the coordinate flattening maps
$T_{A},T_{B},T_{C}$ are injective. For a subspace $U\subset V$, define
$U^{\perp}:=\\{\alpha\in V^{*}\ |\ \alpha(u)=0\ \forall u\in U\\}$. In a
nutshell, border apolarity gives us some necessary conditions on the possible
limiting ideals, $I$, that can arise from a border rank decomposition.

######  Theorem 13.

(Weak Border Apolarity) Let $X=\mathbb{P}A\times\mathbb{P}B\times\mathbb{P}C$
and $S[X]$ be its coordinate ring. Suppose a tensor $T$ has
$\textbf{\text@underline{R}}(T)\leq r$. Then there exists a (multi)homogeneous
ideal $I\subset S[X]$ such that

  * •

$I\subset Ann(T)$

  * •

For each multidegree $ijk$, the $ijk$th graded piece, $I_{ijk}$, of $I$ has
$\operatorname{codim}I_{ijk}=\min\\{r,\dim S[X]_{ijk}\\}$

In addition, if $G_{T}$ is a group acting on $X$ and preserving $T$, then
there exists an $I$ as above which in addition is invariant under a Borel
subgroup of $G_{T}$.

In [BB20], see Theorems 3.15 (Border Apolarity) for proof of the first part
and see Theorem 4.3 (Fixed Ideal Theorem) for a proof of the second part.
Theorem 3.15 and 4.3 of [BB20] are not stated here as they are stated in
greater generality than we require using the language of schemes. We remark
that the Weak Border Apolarity Theorem provides sufficiency that if a border
rank $r$ decomposition exists, then there will exist another border rank $r$
decomposition satisfying the given conditions.

We note that the second condition says that we may in fact take the $r$ points
of the border rank decomposition to be in general position, and so our initial
supposition that the $r$ points are in general position is justified. Lie’s
Theorem and the Normal Form Lemma allow us to take $I_{111}$ to be
$B_{T}$-fixed. The Fixed Ideal Theorem of [BB20] uses the same reasoning to
generalize this to prove $B_{T}$-invariance for all multigraded components
$I_{ijk}$, not just a finite number of multigraded components.

In [CHL19], using Weak Border Apolarity Theorem, they assert that for $T$ a
concise tensor with a border rank $r$ decomposition, there will exist an ideal
$I$ satisfying the following:

  1. (1)

$I_{ijk}$ is $B_{T}$-stable ($I_{ijk}$ is a Borel fixed weight space)

  2. (2)

$I\subset Ann(T)$ i.e. $I_{110}\subset T(C^{*})^{\perp}\subset A^{*}\otimes
B^{*}$, etc. and $I_{111}\subset T^{\perp}\subset A^{*}\otimes B^{*}\otimes C$

  3. (3)

For all $i,j,k$ such that $i+j+k>1$, then $\operatorname{codim}I_{ijk}=r$,
i.e. the condition that we may take the $r$ points of the border rank
decomposition to be in general position.

  4. (4)

Since $I$ is an ideal, the image of the multiplication map $I_{i-1,j,k}\otimes
A^{*}\oplus I_{i,j-1,k}\otimes B^{*}\oplus I_{i,j,k-1}\otimes C^{*}\to
S^{i}A^{*}\otimes S^{j}B^{*}\otimes S^{k}C^{*}$ is contained in $I_{ijk}$

We note that the last condition is simply the condition that $I$ is an ideal
where the multiplication respects the grading. The border apolarity algorithm
presented in [CHL19] makes use of these conditions and attempts to iteratively
construct all possible ideals, $I$, in each multidegree. In particular, if at
any multidegree $ijk$, there does not exist an $I_{ijk}$ satisfying the above,
then we may conclude that $\textbf{\text@underline{R}}(T)>r$. We remark that
the candidate ideals, $I$, do not necessarily correspond to an actual border
rank decomposition of the tensor. We describe the algorithm of [CHL19]
precisely below:

Algorithm 1 Border Apolarity Algorithm of Conner, Harper, Landsberg

Input: $T\in A\otimes B\otimes C$, $r$

Output: Candidate ideals or $\textbf{\text@underline{R}}(T)>r$

  1. (1)

For each $B_{T}$-fixed space $F_{110}$ of codim $r-\dim C$ in
$T(C^{*})^{\perp}$ (i.e. codim $r$ in $A^{*}\otimes B^{*}$) compute ranks of
maps $F_{110}\otimes A^{*}\to S^{2}A^{*}\otimes B^{*}$ and $F_{110}\otimes
B^{*}\to A^{*}\otimes S^{2}B^{*}$

If both have images of codim at least $r$, then $F_{110}$ is possible
$I_{110}$. These are called the (210) and (120) tests.

  2. (2)

Perform analogously for possible $I_{101}\subset T(B^{*})^{\perp}$ and
$I_{011}\subset T(A^{*})^{\perp}$ for candidate $F_{101}$ and $F_{011}$

  3. (3)

For each triple $F_{110},F_{101},F_{011}$, compute rank of map
$(F_{110}\otimes C^{*})\oplus(F_{101}\otimes B^{*})\oplus(F_{011}\otimes
A^{*})\to A^{*}\otimes B^{*}\otimes C^{*}$

If codim of image is at least $r$, then have a candidate triple. $F_{111}$ is
candidate for $I_{111}$ if it is codim $r$, it is contained in $T^{\perp}$ and
contains image of above map.

  4. (4)

Analogous higher degree tests

  5. (5)

If at any point there are no such candidates
$\textbf{\text@underline{R}}(T)>r$, otherwise stabilization of candidate
ideals will occur at worst multi-degree $(r,r,r)$

The condition that $I_{ijk}$ is $B_{T}$-fixed allows us to greatly reduce the
search for possible candidate ideals. The $B_{T}$-fixed spaces are easier to
list than trying to list all possible $I_{ijk}$. This condition allows the
algorithm of [CHL19] to be feasible for tensors with large symmetry groups.
Then, using the assumption that our points are in general position, one has
rank conditions on the multiplication maps, as the images must have codim at
least $r$.

####  3.3.1. Implementation of Border Apolarity for $T_{\mathfrak{sl}_{3}}$

We show how to implement the algorithm of [CHL19] for $T_{\mathfrak{sl}_{n}}$
by describing how to compute all possible $B_{T}$-fixed $I_{110}$.
Additionally, we can leverage the skew-symmetry of $T_{\mathfrak{sl}_{n}}$ to
reduce the amount of computation involved for determining potential
$I_{110},I_{101},I_{011}$.

Let
$T=T_{\mathfrak{sl}_{3}}\in\mathfrak{sl}_{3}^{*}\otimes\mathfrak{sl}_{3}^{*}\otimes\mathfrak{sl}_{3}=A\otimes
B\otimes C$. The first and third condition from border apolarity tells us to
compute all $B_{T}$-fixed weight subspaces $F_{110}\subset A^{*}\otimes B^{*}$
of codimension $r$; however, since $T$ is concise with $\dim
T(C^{*})=\dim\mathfrak{sl}_{3}=8$ and $F_{110}\subset T(C^{*})^{\perp}$ by the
second condition of border apolarity, we compute all $B_{T}$-fixed weight
spaces $F_{110}\subset T(C^{*})^{\perp}$ of codimension $r-8$.

Standard computational methods, see [FH91], yield that the irreducible
decomposition of $A^{*}\otimes B^{*}$ as $\mathfrak{sl}_{n}$-modules is as
follows.

$\displaystyle A^{*}\otimes B^{*}=\mathfrak{sl}_{n}\otimes\mathfrak{sl}_{n}$
$\displaystyle\simeq[2\ 0\ \cdots\ 0\ 2]\oplus[2\ 0\cdots 0\ 1\ 0]\oplus[0\ 1\
0\cdots\ 0\ 2]\oplus[0\ 1\ 0\cdots\ 0\ 1\ 0]\oplus 2[1\ 0\ \cdots 0\
1]\oplus[0\ \cdots\ 0]$

In particular, for $\mathfrak{sl}_{3}$, we have the following decomposition
into $\mathfrak{sl}_{3}$-modules.

$\displaystyle\mathfrak{sl}_{3}\otimes\mathfrak{sl}_{3}\simeq[2\ 2]\oplus[3\
0]\oplus[0\ 3]\oplus 2[1\ 1]\oplus[0\ 0]$

Using this decomposition and the conciseness of $T$, we have the
$\mathfrak{sl}_{3}$-module decomposition $T(C^{*})^{\perp}\simeq[2\
2]\oplus[3\ 0]\oplus[0\ 3]\oplus[1\ 1]\oplus[0\ 0]$ (Note that $\dim
T(C^{*})^{\perp}=56$). Using this $\mathfrak{sl}_{3}$-module decomposition, we
can obtain a decomposition of $T(C^{*})^{\perp}$ into weight spaces
(decomposition into $\mathfrak{h}$-modules) by combining the weight space
decompositions of each $\mathfrak{sl}_{3}$-module into one poset. The result
is the Figure 1. Each node represents a weight space of weight $\lambda$
labeled by $(\lambda,$ dimension of weight space in $T(C^{*})^{\perp})$. Also
note that arrows go from lower weights to higher weights, so the highest
weight occuring in $T(C^{*})^{\perp}$ is $[2\ 2]$. The weight space $[3\ 0]$
is of dimension 2, where one of the basis elements for this weight space is
the highest weight vector of the $\mathfrak{sl}_{3}$-module $[3\ 0]$ and the
other basis element for this weight space is the vector arising from lowering
the highest weight vector of the $\mathfrak{sl}_{3}$-module $[2\ 2]$.

Figure 1. Weight Decomposition of $T(C^{*})^{\perp}$

The utility of this decomposition is that we can generate all $B_{T}$-fixed
weight subspaces $F_{110}$ by taking a collection of weight vectors $v_{i}$
from the poset such that $v_{1}\wedge v_{2}\wedge\cdots\wedge v_{56-(r-8)}$ is
a highest weight vector in $G(56-(r-8),T(C^{*})^{\perp})$ and consequently is
closed under the raising operators. One should note that if $v_{i}$ comes from
a weight space of dimension greater than 1 then one needs to include linear
combinations of basis vectors of that weight space.

######  Example 14.

We show a few small examples of possible $F_{110}$ $B_{T}$-fixed subspaces to
provide intuition of how these spaces are computed.

For $r=63$, we have that $F_{110}$ will be of the form $v_{1}$. Since it must
be closed under raising operators, it will necessarily be a highest weight
vector. Therefore, our choices will be $v_{1}=v_{\lambda}$ where of
$v_{\lambda}$ is a highest weight vector of weight $\lambda=[2\ 2]$, $[3\ 0]$,
$[0\ 3]$, $[1\ 1]$, or $[0\ 0]$.

For $r=62$, $F_{110}$ will be of the form $v_{1}\wedge v_{2}$. Necessarily, we
must have that $v_{1}$ must be a highest weight vector. The second vector
$v_{2}$ may either be another highest weight vector, a weight vector that can
be raised to $v_{1}$, or a linear combination of the two previous cases if
they are vectors of the same weight. For example, if we take $v_{1}=v_{[2\
2]}$ to be the highest weight vector of $[2\ 2]$. Let $v_{[3\ 0]}$ and $u_{[3\
0]}$ be a weight basis for $[3\ 0]$ with $v$ being a highest weight vector.
Assume similarly for $[0\ 3]$ weight space. The possible choices for $v_{2}$
are weight vectors of the following types: $v_{[3\ 0]},sv_{[3\ 0]}+tu_{[3\
0]},v_{[0\ 3]},sv_{[0\ 3]}+tu_{[0\ 3]},v_{[1\ 1]},v_{[0\ 0]}$ with $s,t$
parameters. Applying all possible raising operators to $v_{1}\wedge v_{2}$
where $v_{2}$ has parameters will provide equations for what values of $s,t$
give us a highest weight vector.

For smaller values of $r$, the number of possible Borel fixed spaces is much
larger and more difficult to list by hand without the aide of a computer. The
computationally difficult step in this algorithm lies in computing the ranks
of the multiplication maps such as $F_{110}\otimes A^{*}\to S^{2}A^{*}\otimes
B^{*}$. In some cases there are many parameters which arise from choosing
weight vectors from high dimensional weight spaces, such as the $[0\ 0]$
weight space in Figure 1. Recall that a linear map has rank at most $k$ is the
$k+1$ minors all vanish. In order to determine whether the multiplication map
has image codimension $r$, we look at the appropriate minors of this linear
map. When there are no parameters involved, this is a simple linear algebra
calculation. However, in some cases, the entries of the multiplication map are
linear polynomials in the parameters coming from choosing a linear combination
of weight vectors. In order to determine whether the multiplication map has
image of codimension $r$, one needs to look at the ideal of these minors, as
well as some polynomial equations in the parameters that are needed for the
space to be Borel fixed. One must do a Groebner Basis computation on this
ideal to determine whether all the minors vanish or not. This can become an
unfeasible computation if there are too many parameters and/or equations.

##  4\. New Bounds

It is known that $\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{2}})=5$
[dGH86], so we aim to find bounds on
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{n}})$ for $n=3$ and $4$ using
the above techniques. Additionally, we provide new lower bounds on
$\textbf{\text@underline{R}}(T_{\mathfrak{so}_{n}})$ for $n=4$ and $5$.

###  4.1. Koszul flattenings

In the case of $T_{\mathfrak{sl}_{3}}$, we achieve the best results when $p=3$
and we restrict to a generic $7$ dimensional subspace of $\mathfrak{sl}_{3}$,
since $\dim\mathfrak{sl}_{3}=8$. The best bound achieved is
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{3}})\geq 14$ (See Table 2).

Table 1. $T_{\mathfrak{sl}_{3}}$ Results p | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---
1 | (64,224) | 0 | 10
2 | (224,448) | 1 | 11
3 | (448,560) | 8 | 13
Table 2. $T_{\mathfrak{sl}_{3}}$ Restriction to a generic subspace of dim $k$ Results p | k | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---|---
1 | 3 | (24,24) | 0 | 12
2 | 5 | (80,80) | 4 | 13
3 | 7 | (280,280) | 7 | 14

In the case of $T_{\mathfrak{sl}_{4}}$, we achieve the best lower bound of
$27$ when $p=4$ or $5$ while restricting to a subspace (See Table 4).

Table 3. $T_{\mathfrak{sl}_{4}}$ Results p | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---
1 | (225,1575) | 0 | 17
2 | (1575,6825) | 1 | 18
3 | (6825,20475) | 15 | 19
4 | (20475,45045) | 106 | 21
5 | (45045,75075) | 470 | 23
6 | (75075,96525) | 2680 | 25
7 | (96525,96525) | 11039 | 25
Table 4. $T_{\mathfrak{sl}_{4}}$ Restriction to a generic subspace of dim $k$ Results  p | k | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---|---
1 | 3 | (45,45) | 0 | 23
2 | 5 | (150,150) | 2 | 25
3 | 7 | (525,525) | 7 | 26
4 | 9 | (1890,1890) | 38 | 27
5 | 11 | (6930,6930) | 176 | 27
6 | 13 | (25740,25740) | 2254 | 26

As stated above, Koszul flattenings alone are insufficient to obtain border
rank lower bounds exceeding $2m$, i.e. Koszul flattenings will not prove
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{3}})\geq 16$ and
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{4}})\geq 30$.

In addition to lower bounds on these tensor, we computed lower bounds on
$T_{\mathfrak{so}_{4}}$ and $T_{\mathfrak{so}_{5}}$.

Table 5. $T_{\mathfrak{so}_{4}}$ Results p | Dimensions of Linear Map | Rank | Koszul Bound
---|---|---|---
1 | (90,36) | 36 | 8
2 | (120,90) | 78 | 8
3 | (90,120) | 78 | 8
Table 6. $T_{\mathfrak{so}_{4}}$ Restriction to a generic subspace of dim $k$ Results  p | k | Dimensions of Linear Map | Rank | Koszul Bound
---|---|---|---|---
1 | 3 | (18,18) | 18 | 9
2 | 5 | (60,60) | 48 | 8

From [Mir85], it is shown that for the Lie algebras
$\mathfrak{g}_{n}=\mathfrak{sl}_{2}^{\times n}$ that
$\textbf{R}(T_{\mathfrak{g}_{n}})=5n$, so
$\textbf{R}(T_{\mathfrak{so}_{4}})=10$. Therefore,
$\textbf{\text@underline{R}}(T_{\mathfrak{so}_{4}})$ will either be 9 or 10.

Table 7. $T_{\mathfrak{so}_{5}}$ Results p | Dimensions of Linear Map | Rank | Koszul Bound
---|---|---|---
1 | (450,100) | 100 | 12
2 | (1200,450) | 449 | 13
3 | (2100,1200) | 1190 | 15
4 | (2520,2100) | 1971 | 16
5 | (2100,2520) | 1971 | 16
Table 8. $T_{\mathfrak{so}_{5}}$ Restriction to a generic subspace of dim $k$ Results  p | k | Dimensions of Linear Map | Rank | Koszul Bound
---|---|---|---|---
1 | 3 | (30,30) | 30 | 15
2 | 5 | (100,100) | 98 | 17
3 | 7 | (350,350) | 343 | 18
4 | 9 | (1260,1260) | 1136 | 17

###  4.2. Border Substitution

For
$T_{\mathfrak{sl}_{n}}\in\mathfrak{sl}_{n}^{*}\otimes\mathfrak{sl}_{n}^{*}\otimes\mathfrak{sl}_{n}$,
we may identify the space $\mathfrak{sl}_{n}^{*}$ with $\mathfrak{sl}_{n}$ (by
sending an element to its negative transpose). Therefore, we may identify
$T_{\mathfrak{sl}_{n}}$ as an element of
$\mathfrak{sl}_{n}\otimes\mathfrak{sl}_{n}\otimes\mathfrak{sl}_{n}$. As a
first step in applying border substitution, we restrict
$T_{\mathfrak{sl}_{n}}\in A\otimes B\otimes C$ in the $A$ tensor factor. Since
we may restrict to looking at representatives of closed
$G_{T_{\mathfrak{sl}_{n}}}$-orbits, then the only planes we need to check are
the highest weight planes in $G(k,\mathfrak{sl}_{n})$. In order to compute the
border rank of the restricted tensor, we use Koszul flattenings on the
restricted tensor. Once again, let $v_{\lambda}$ denote the unique weight
vector in weight space $\lambda$.

For $T_{\mathfrak{sl}_{3}}$, border substitution did not generate a better
lower bound than the Koszul flattenings. However, we were able to obtain a
better lower bound for $T_{\mathfrak{sl}_{4}}$. Let $A^{\prime}$, as in
Proposition 15, be $\tilde{A}^{\perp}$ where we take $\tilde{A}$ to be a space
of dimension $m-k$. If we restrict our tensor by a one dimensional subspace,
then the only choice for $\tilde{A}$ will be the space spanned by $v_{[1\ 0\
1]}$, which is the highest weight vector of $\mathfrak{sl}_{n}$.

Table 9. $T_{\mathfrak{sl}_{4}}$ with Restriction $\tilde{A}=v_{[1\ 0\ 1]}$ p | k | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---|---
1 | 3 | (45,45) | 0 | 23
2 | 5 | (150,150) | 2 | 25
3 | 7 | (525,525) | 7 | 26
4 | 9 | (1890,1890) | 38 | 27
5 | 11 | (6930,6930) | 248 | 27
6 | 13 | (25740,25740) | 2254 | 26

Restricting by a two dimensional subspace, we have one choice for $\tilde{A}$
up to symmetry in the weight space decomposition for $\mathfrak{sl}_{4}$,
namely $v_{[1\ 0\ 1]}\wedge v_{[-1\ 1\ 1]}$.

Table 10. $T_{\mathfrak{sl}_{4}}$ with Restriction $\tilde{A}=v_{[1\ 0\ 1]}\wedge v_{[-1\ 1\ 1]}$ p | k | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---|---
1 | 3 | (45,45) | 0 | 23
2 | 5 | (150,150) | 2 | 25
3 | 7 | (525,525) | 7 | 26
4 | 9 | (1890,1890) | 78 | 26
5 | 11 | (6930,6930) | 498 | 26

Restricting by a three dimensional subspace, we have three choices for
$\tilde{A}$ up to symmetry. $\tilde{A}$ may be $v_{[1\ 0\ 1]}\wedge v_{[1\ 1\
-1]}\wedge v_{[-1\ 1\ 1]}$, $v_{[1\ 0\ 1]}\wedge v_{[1\ 1\ -1]}\wedge v_{[2\
-1\ 0]}$, or $v_{[1\ 0\ 1]}\wedge v_{[1\ 1\ -1]}\wedge v_{[-1\ 2\ -1]}$.

Table 11. $T_{\mathfrak{sl}_{4}}$ with Restriction $\tilde{A}=v_{[1\ 0\ 1]}\wedge v_{[1\ 1\ -1]}\wedge v_{[-1\ 1\ 1]}$ p | k | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---|---
1 | 3 | (45,45) | 0 | 23
2 | 5 | (150,150) | 2 | 25
3 | 7 | (525,525) | 31 | 25
4 | 9 | (1890,1890) | 168 | 25
5 | 11 | (6930,6930) | 755 | 25
Table 12. $T_{\mathfrak{sl}_{4}}$ with Restriction $\tilde{A}=v_{[1\ 0\ 1]}\wedge v_{[1\ 1\ -1]}\wedge v_{[2\ -1\ 0]}$ p | k | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---|---
1 | 3 | (45,45) | 0 | 23
2 | 5 | (150,150) | 2 | 25
3 | 7 | (525,525) | 7 | 25
4 | 9 | (1890,1890) | 72 | 25
5 | 11 | (6930,6930) | 498 | 25
Table 13. $T_{\mathfrak{sl}_{4}}$ with Restriction $\tilde{A}=v_{[1\ 0\ 1]}\wedge v_{[1\ 1\ -1]}\wedge v_{[-1\ 2\ -1]}$ p | k | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---|---
1 | 3 | (45,45) | 3 | 21
2 | 5 | (150,150) | 14 | 23
3 | 7 | (525,525) | 42 | 25
4 | 9 | (1890,1890) | 254 | 24
5 | 11 | (6930,6930) | 1072 | 24

The best bound we obtain is
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{4}}\big{|}_{(v_{[1\ 0\
1]})^{\perp}\otimes\mathfrak{sl}_{n}\otimes\mathfrak{sl}_{n}})\geq 27$ (See
Table 9). By Proposition 15, this proves Theorem 3,
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{4}})\geq 28$. After restricting
in the $A$ tensor factor, we cannot restrict by the highest weight vector of
$\mathfrak{sl}_{n}$ in the $B$ or $C$ factor as the symmetry group of the
restricted tensor will have a different symmetry group.

Additionally, border substitution gives a better lower bound on
$\textbf{\text@underline{R}}(T_{\mathfrak{so}_{5}})$. Restricting by a two
dimension subspace (See Table 14) and using Proposition 15 proves Theorem 5.

Table 14. $T_{\mathfrak{so}_{5}}$ with Restriction $\tilde{A}=v_{[0\ 1]}\wedge v_{[1\ 0]}$ p | k | Dimensions of Linear Map | Dimension of Kernel | Koszul Bound
---|---|---|---|---
1 | 3 | (30,30) | 0 | 15
2 | 5 | (100,100) | 4 | 16
3 | 7 | (350,350) | 28 | 17
4 | 9 | (1260,1260) | 194 | 16

###  4.3. Border Apolarity

We use border apolarity to disprove that $T_{\mathfrak{sl}_{3}}$ has rank
$r=15$. We first compute candidate $F_{110}$ spaces which passed the
$(210)$-test. There were a total of 5 candidate $F_{110}$ subspaces out of a
total of more than 1245 possible $F_{110}$ spaces. The candidate $F_{110}$
spaces came in three types of weight space decompositions:

Table 15. Two candidate $F_{110}$ planes have the following weight decomposition Weight | Dimension of weight space in $A^{*}\otimes B^{*}$ | Dimension of weight space in $F_{110}$
---|---|---
$[2,2]$ | 1 | 1
$[3,0]$ | 2 | 2
$[4,-2]$ | 1 | 1
$[0,3]$ | 2 | 2
$[1,1]$ | 5 | 5
$[2,-1]$ | 5 | 5
$[3,-3]$ | 2 | 2
$[-2,4]$ | 1 | 1
$[-1,2]$ | 5 | 5
$[0,0]$ | 8 | 8
$[1,-2]$ | 5 | 5
$[2,-4]$ | 1 | 1
$[-3,3]$ | 2 | 2
$[-2,1]$ | 5 | 4
$[-1,-1]$ | 5 | 4
$[0,-3]$ | 2 | 1
Table 16. One candidate $F_{110}$ plane has the following weight decomposition Weight | Dimension of weight space in $A^{*}\otimes B^{*}$ | Dimension of weight space in $F_{110}$
---|---|---
$[2,2]$ | 1 | 1
$[3,0]$ | 2 | 2
$[4,-2]$ | 1 | 1
$[0,3]$ | 2 | 2
$[1,1]$ | 5 | 5
$[2,-1]$ | 5 | 5
$[3,-3]$ | 2 | 2
$[-2,4]$ | 1 | 1
$[-1,2]$ | 5 | 5
$[0,0]$ | 8 | 8
$[1,-2]$ | 5 | 5
$[2,-4]$ | 1 | 1
$[-3,3]$ | 2 | 2
$[-2,1]$ | 5 | 5
$[-1,-1]$ | 5 | 3
$[-4,2]$ | 1 | 1
Table 17. Two candidate $F_{110}$ planes have the following weight decomposition Weight | Dimension of weight space in $A^{*}\otimes B^{*}$ | Dimension of weight space in $F_{110}$
---|---|---
$[2,2]$ | 1 | 1
$[3,0]$ | 2 | 2
$[4,-2]$ | 1 | 1
$[0,3]$ | 2 | 2
$[1,1]$ | 5 | 5
$[2,-1]$ | 5 | 5
$[3,-3]$ | 2 | 2
$[-2,4]$ | 1 | 1
$[-1,2]$ | 5 | 5
$[0,0]$ | 8 | 8
$[1,-2]$ | 5 | 5
$[2,-4]$ | 1 | 1
$[-3,3]$ | 2 | 2
$[-2,1]$ | 5 | 5
$[-1,-1]$ | 5 | 4
$[-4,2]$ | 1 | 1
$[-3,0]$ | 2 | 1

The computation to produce these 5 candidate $F_{110}$ planes took extensive
time in some cases, due to the parameters creating a difficult groebner basis
computation when determining whether an $F_{110}$ plane passes $(210)$-test.
The large number of candidates was not as much of a computational issue as all
the $(210)$-tests can be parallelized. Some of these computations were done on
Texas A&M’s High Performance Research Cluster as well as Texas A&M’s Math
Department Cluster.

Using the skew-symmetry of $T_{\mathfrak{sl}_{n}}$, we are able to produce
$F_{011}$ and $F_{101}$ candidate weight spaces from the candidate $F_{110}$
spaces. A computer calculation verified that for each candidate triple
$F_{110},F_{011},F_{101}$, the rank condition is not met for the $(111)$-test
and consequently, there are no candidate $F_{111}$ spaces. Therefore, the rank
of $T_{\mathfrak{sl}_{3}}$ is greater than 15 and this proves Theorem 1,
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{3}})\geq 16$. This result is
significant as it is the first example of an explicit tensor such that the
border rank is at least $2m$ when $m<13$.

###  4.4. Upper Bounds

A numerical computer search has given a rank 20 decomposition of
$T_{\mathfrak{sl}_{3}}$. The technique used was a combination of Newton’s
Method and Lenstra–Lenstra–Lovász Algorithm to find rational approximations
[CLGV20]. This technique formulated the problem as a nonlinear optimization
problem that was solved to machine precision and then subsequently modified
using the Lenstra-Lenstra-Lovász Algorithm to generate a precise solution with
algebraic numbers given the numerical solution. As
$T_{\mathfrak{sl}_{3}}\in\mathbb{C}^{8}\otimes\mathbb{C}^{8}\otimes\mathbb{C}^{8}$,
a rank 20 decomposition consists of finding
$a_{i},b_{i},c_{i}\in\mathbb{C}^{8}$ such that $T=\sum_{i=1}^{20}a_{i}\otimes
b_{i}\otimes c_{i}$. We take each vector $a_{i},b_{i},c_{i}$ to be a vector 8
variables, and using properties of elements of tensor products, we can
multiply out the right hand side and have a system of equations for each entry
of the tensor. This amounts to solving a system of 512 polynomial equations of
degree 3 in 480 variables. We then use Newton’s method to find roots to this
system of equations. If it appears to converge to a solution, then we compute
it to machine precision and use Lenstra-Lenstra-Lovász to find an algebraic
solution that satisfies the initial polynomial conditions. Therefore, this
decomposition proves Theorem 2, $\textbf{R}(T_{\mathfrak{sl}_{3}})\leq 20$.

Let $\zeta_{6}$ denote a primitive 6th root of unity. The following is the
rank 20 decomposition of $T_{\mathfrak{sl}_{3}}$. One may verify that this is
in fact a rank decomposition by showing that it satisfies the polynomial
equations described above. We note that the below decomposition has some
localy symmetry. For example, the first three terms share
$\begin{bmatrix}0&\zeta_{6}&0\\\ 0&0&0\\\ 0&1&0\end{bmatrix}$ as a common
factor with standard cyclic $\mathbb{Z}_{3}$-symmetry, i.e. this element
occurs once in each tensor factor in the first three terms. We have grouped
terms in the below presentation of the decomposition by this local symmetry.
We were unable to determine any global symmetry in this presentation of this
decomposition.

$\displaystyle T_{\mathfrak{sl}_{3}}=$
$\displaystyle(\frac{1}{3^{4}2})\begin{bmatrix}0&\zeta_{6}&0\\\ 0&0&0\\\
0&1&0\end{bmatrix}\otimes\begin{bmatrix}-6&-4\zeta_{6}^{2}&-6\zeta_{6}\\\
9\zeta_{6}&18&0\\\
6\zeta_{6}^{2}&0&-12\end{bmatrix}\otimes\begin{bmatrix}6\zeta_{6}&-4&0\\\
9\zeta_{6}^{2}&0&-9\\\ 0&4\zeta_{6}^{2}&-6\zeta_{6}\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{4}2^{3}})\begin{bmatrix}-6&-4\zeta_{6}^{2}&-6\zeta_{6}\\\
9\zeta_{6}&18&0\\\
6\zeta_{6}^{2}&0&-12\end{bmatrix}\otimes\begin{bmatrix}0&\zeta_{6}&0\\\
0&0&0\\\
0&1&0\end{bmatrix}\otimes\begin{bmatrix}-6\zeta_{6}&4&-6\zeta_{6}^{2}\\\
0&6\zeta_{6}&9\\\ -6&0&0\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{4}2})\begin{bmatrix}6&4\zeta_{6}^{2}&0\\\
9\zeta_{6}&0&9\zeta_{6}^{2}\\\
0&4\zeta_{6}&-6\end{bmatrix}\otimes\begin{bmatrix}12\zeta_{6}&-4&6\zeta_{6}^{2}\\\
0&-18\zeta_{6}&-9\\\
6&0&6\zeta_{6}\end{bmatrix}\otimes\begin{bmatrix}0&\zeta_{6}&0\\\ 0&0&0\\\
0&1&0\end{bmatrix}+$
$\displaystyle(\frac{\zeta_{6}}{3^{5}2})\begin{bmatrix}0&0&6\zeta_{6}\\\
-18\zeta_{6}&-18\zeta_{6}^{2}&-9\\\
-6&-4\zeta_{6}&18\zeta_{6}^{2}\end{bmatrix}\otimes\begin{bmatrix}18\zeta_{6}^{2}&0&6\zeta_{6}\\\
9\zeta_{6}&-18\zeta_{6}^{2}&18\\\
-6&-4\zeta_{6}&0\end{bmatrix}\otimes\begin{bmatrix}0&\zeta_{6}^{2}&0\\\
0&0&0\\\ 0&-1&0\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{4}2^{2}})\begin{bmatrix}-6\zeta_{6}&4\zeta_{6}^{2}&-6\\\
9&18\zeta_{6}&0\\\
-6\zeta_{6}^{2}&0&-12\zeta_{6}\end{bmatrix}\otimes\begin{bmatrix}0&\zeta_{6}^{2}&0\\\
0&0&0\\\ 0&-1&0\end{bmatrix}\otimes\begin{bmatrix}-12\zeta_{6}&0&-6\\\
-9&6\zeta_{6}&18\zeta_{6}^{2}\\\ -6\zeta_{6}^{2}&4&6\zeta_{6}\end{bmatrix}+$
$\displaystyle(\frac{\zeta_{6}}{3^{4}})\begin{bmatrix}0&1&0\\\ 0&0&0\\\
0&\zeta_{6}&0\end{bmatrix}\otimes\begin{bmatrix}-3\zeta_{6}^{2}&-2&-3\zeta_{6}\\\
0&9\zeta_{6}^{2}&0\\\
1&0&-6\zeta_{6}^{2}\end{bmatrix}\otimes\begin{bmatrix}6\zeta_{6}&0&-6\\\
1&6\zeta_{6}&-9\zeta_{6}^{2}\\\ -6\zeta_{6}^{2}&4&-12\zeta_{6}\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{4}2^{3}})\begin{bmatrix}-12&-4\zeta_{6}^{2}&-6\zeta_{6}\\\
0&18&-9\zeta_{6}^{2}\\\
6\zeta_{6}^{2}&0&-6\end{bmatrix}\otimes\begin{bmatrix}-6\zeta_{6}^{2}&4\zeta_{6}&0\\\
9&0&9\zeta_{6}\\\
0&4&6\zeta_{6}^{2}\end{bmatrix}\otimes\begin{bmatrix}0&-4&6\zeta_{6}^{2}\\\
-9\zeta_{6}^{2}&-6\zeta_{6}&0\\\ 6&0&6\zeta_{6}\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{5}2^{3}})\begin{bmatrix}-18&0&6\zeta_{6}^{2}\\\
9\zeta_{6}^{2}&18&18\zeta_{6}\\\
-6\zeta_{6}&-4\zeta_{6}^{2}&0\end{bmatrix}\otimes\begin{bmatrix}0&0&-6\zeta_{6}^{2}\\\
18\zeta_{6}^{2}&-18&9\zeta_{6}\\\
6\zeta_{6}&4\zeta_{6}^{2}&18\end{bmatrix}\otimes\begin{bmatrix}0&4\zeta_{6}&6\zeta_{6}^{2}\\\
-9\zeta_{6}^{2}&6&0\\\ -6\zeta_{6}&0&-6\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{3}2^{3}})\begin{bmatrix}-12\zeta_{6}&8&-6\zeta_{6}^{2}\\\
0&18\zeta_{6}&9\\\
-6&-4\zeta_{6}^{2}&-6\zeta_{6}\end{bmatrix}\otimes\begin{bmatrix}-2\zeta_{6}^{2}&0&0\\\
3&0&3\zeta_{6}\\\
0&0&2\zeta_{6}^{2}\end{bmatrix}\otimes\begin{bmatrix}0&0&-6\zeta_{6}\\\
9\zeta_{6}&6&0\\\ 6\zeta_{6}^{2}&4\zeta_{6}&-6\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{5}2^{3}})\begin{bmatrix}18\zeta_{6}^{2}&12&6\zeta_{6}\\\
9\zeta_{6}&-18\zeta_{6}^{2}&18\\\
-6&8\zeta_{6}&0\end{bmatrix}\otimes\begin{bmatrix}0&-6&-6\zeta_{6}\\\
18\zeta_{6}&18\zeta_{6}^{2}&9\\\
6&-2\zeta_{6}&-18\zeta_{6}^{2}\end{bmatrix}\otimes\begin{bmatrix}0&0&6\zeta_{6}\\\
-9\zeta_{6}&-6\zeta_{6}^{2}&0\\\ -6&-4\zeta_{6}&6\zeta_{6}^{2}\end{bmatrix}+$

$\displaystyle(\frac{1}{3^{4}2^{3}})\begin{bmatrix}12&4&-6\\\ 0&-18&9\\\
-6&0&6\end{bmatrix}\otimes\begin{bmatrix}-6&4&0\\\ 9&0&-9\\\
0&-4&6\end{bmatrix}\otimes\begin{bmatrix}0&0&6\\\ -9&6&0\\\
6&-4&-6\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{3}2^{3}})\begin{bmatrix}-12&0&6\\\ 0&18&-9\\\
6&-4&-6\end{bmatrix}\otimes\begin{bmatrix}2&0&0\\\ -3&0&3\\\
0&0&-2\end{bmatrix}\otimes\begin{bmatrix}0&4&-6\\\ 9&-6&0\\\
-6&0&6\end{bmatrix}+$ $\displaystyle(\frac{1}{3^{3}2})\begin{bmatrix}-2&0&0\\\
3&0&-3\\\ 0&0&2\end{bmatrix}\otimes\begin{bmatrix}12&0&-6\\\ 0&-18&9\\\
-6&4&6\end{bmatrix}\otimes\begin{bmatrix}0&-1&0\\\ 0&0&0\\\
0&1&0\end{bmatrix}+$ $\displaystyle(\frac{1}{3^{4}2})\begin{bmatrix}6&4&-6\\\
9&-18&0\\\ -6&0&12\end{bmatrix}\otimes\begin{bmatrix}0&1&0\\\ 0&0&0\\\
0&-1&0\end{bmatrix}\otimes\begin{bmatrix}6&0&-6\\\ 0&-6&9\\\
-6&4&0\end{bmatrix}+$ $\displaystyle(\frac{1}{3^{3}2})\begin{bmatrix}0&1&0\\\
0&0&0\\\ 0&-1&0\end{bmatrix}\otimes\begin{bmatrix}-6&-4&6\\\ -9&18&0\\\
6&0&-12\end{bmatrix}\otimes\begin{bmatrix}2&0&0\\\ -3&0&3\\\
0&0&-2\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{3}2^{5}})\begin{bmatrix}-4&-6&-6\\\ 0&0&0\\\
9&4&4\end{bmatrix}\otimes\begin{bmatrix}-4&6&-6\\\ 0&0&0\\\
-9&4&4\end{bmatrix}\otimes\begin{bmatrix}-4&-6&6\\\ 0&0&0\\\
9&-4&4\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{3}2^{5}})\begin{bmatrix}4&-6&-6\\\ 0&0&0\\\
9&4&-4\end{bmatrix}\otimes\begin{bmatrix}-4&-6&6\\\ 0&0&0\\\
9&-4&4\end{bmatrix}\otimes\begin{bmatrix}4&-6&6\\\ 0&0&0\\\
9&-4&-4\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{3}2^{5}})\begin{bmatrix}-4&-6&6\\\ 0&0&0\\\
9&-4&4\end{bmatrix}\otimes\begin{bmatrix}-4&6&6\\\ 0&0&0\\\
-9&-4&4\end{bmatrix}\otimes\begin{bmatrix}4&6&6\\\ 0&0&0\\\
-9&-4&-4\end{bmatrix}+$
$\displaystyle(\frac{1}{3^{3}2^{5}})\begin{bmatrix}-4&6&-6\\\ 0&0&0\\\
-9&4&4\end{bmatrix}\otimes\begin{bmatrix}-4&-6&-6\\\ 0&0&0\\\
9&4&4\end{bmatrix}\otimes\begin{bmatrix}4&-6&-6\\\ 0&0&0\\\
9&4&-4\end{bmatrix}+$ $\displaystyle(\frac{2}{3^{2}})\begin{bmatrix}-1&0&0\\\
0&0&0\\\ 0&0&1\end{bmatrix}\otimes\begin{bmatrix}0&0&3\\\ 0&0&0\\\
0&-2&0\end{bmatrix}\otimes\begin{bmatrix}0&-2&0\\\ 0&0&0\\\
3&0&0\end{bmatrix}$

In an attempt to find a smaller rank decomposition, we found numerical
evidence suggesting that
$\textbf{\text@underline{R}}(T_{\mathfrak{sl}_{3}})\leq 18$. The above method
was unable to determine exact algebraic numbers for it to be an honest border
rank decomposition. We include the approximate border rank decomposition,
which was obtained as a numerical solution to machine precision using Newton’s
method, in Appendix A. This decomposition is satisfies the equation
$T_{\mathfrak{sl}_{3}}=\sum_{k=1}^{18}a_{k}(t)\otimes b_{k}(t)\otimes
c_{k}(t)+O(t)$ to a maximum error in each entry of
$3.88578058618805\,10^{-16}$ ($\ell_{0}$ error). It also is satisfied with a
sum of squares error of $1.85900227125328\,10^{-15}$ ($\ell_{2}$ error), which
is the square root of the sum of the squares of all errors in each entry.

##  Appendix A Appendix

⬇

## Approximate Border Rank 18 Decomposition for sl3. B is list of length 18,

## Each element of B is 3 lists, one for each tensor factor

t = var(’t’)

decomp_sl3_br18 = [

[[0.8582131228193816*t^4, 1.0*t^3, -0.656183917735616, 0,
-0.6991867952664118*t^4, 0.48158077142326267*t, 0, 0],

[0.4276000554886944, 0, -0.18145759099190728*t^-4, -0.615151864463497*t,
0.3663065358105463, 0.41017982352375143*t^-3, 0.32919233217347255*t^4,
-0.37760897344619454*t^3],

[-0.1349207590097993*t^-4, -0.6002859548136603*t^-3, 1.0,
0.5115159306456755*t^-5, -0.1595527810915205*t^-4, -0.6048099252034903*t^-1,
-0.7277947872836206*t^-8, -0.40421722543545996*t^-7]],

[[-1.0*t^4, 0.35415463745033937*t^3, 0.0064263290832379735, 0,
-0.36545526641405424*t^4, -0.07853056373003943*t, -0.032960816562400096*t^8,
0],

[0, 0, 1.0*t^-4, -0.2566348901375672*t, -0.22472886423613508, 0,
-1.5211311526049585*t^4, 0],

[0.8999601061346961*t^-4, 0, -0.9891997951816988, 0.7757989654047442*t^-5,
0.6720573676204581*t^-4, 0, -0.7680275258871294*t^-8,
0.23098457888352464*t^-7]],

[[1.4652283519838352*t^4, 0, -0.49981037975335263, -0.1095789691297215*t^5, 0,
-0.39151017186719006*t, -0.0035109361876165053*t^8, 0],

[-0.0370198277154002, -0.4868421872551675*t^-1, -0.6112939642144736*t^-4,
1.0*t, 0, -0.2400052666531245*t^-3, 0.3994050510000636*t^4,
0.38578320751551654*t^3],

[-0.976481792017985*t^-4, -0.6067214925499242*t^-3, 0,
-0.723941360626237*t^-5, -1.0*t^-4, 0.5548674236521388*t^-1,
0.43144009347718587*t^-8, 0]],

[[0, 0, 0, 0.7384182957633163*t^5, 0, 0.644445145257651*t, 1.0*t^8,
0.14298165951600136*t^7],

[0.48917042278155703, -1.1676021066814524*t^-1, 0.0742778280974771*t^-4, 0, 0,
0.43490564244204044*t^-3, 0, -0.35107756870823037*t^3],

[0.35864016315209*t^-4, 0.08753789324397686*t^-3, 0.09745671407220928,
0.8747382868849779*t^-5, -0.08844676171127677*t^-4, 0, 1.0*t^-8,
0.26767660269283405*t^-7]],

[[0, 0.7154062494455652*t^3, 0.753065093225573, 0.25916815694138623*t^5,
1.0*t^4, -0.13328651389721624*t, -0.43754268152319187*t^8,
0.33712746822644685*t^7],

[0.31175249346291917, 0.4893435772426664*t^-1, -0.6750030230291015*t^-4, 0, 0,
0, -1.0*t^4, 0.2730200083141182*t^3],

[-0.48768995751368266*t^-4, 0.3666405285647919*t^-3, -0.509119354246841,
-0.4700930934822484*t^-5, -0.22797281366964972*t^-4,
-0.35924834770877284*t^-1, 0.21574925698409564*t^-8,
-0.4457498667959218*t^-7]],

[[-0.3726117655275953*t^4, -0.6850149555795361*t^3, -0.47485190359116686,
-0.3845493471094187*t^5, 1.3294054010695648*t^4, 0, 0,
-0.360876333169775*t^7],

[-0.20585599800494148, 0.14163424899135532*t^-1, -0.07154541239160381*t^-4, 0,
0, 0.6161959625923827*t^-3, -1.0*t^4, 1.1951737645411018*t^3],

[0.32862886361063015*t^-4, -1.0*t^-3, 0.02421044619636145,
-0.10037637090787854*t^-5, 0.8394716140987267*t^-4, 0.6969300342438073*t^-1,
0.33915176005821984*t^-8, -0.8710663402014699*t^-7]],

[[-0.26018060869681375*t^4, -0.5986510147484245*t^3, 0, 0,
0.34975430620458303*t^4, 0.6295753676920407*t, -0.008909285185907506*t^8,
-1.0*t^7],

[0.1869896012120452, 0.21074450964709926*t^-1, 0.15931993306826575*t^-4,
-0.5916688936748106*t, -1.0, 0.4704225477242227*t^-3, 0.4764298640490715*t^4,
0],

[-0.11381180152788646*t^-4, 0, -0.6321964884450085, 1.1046891281785145*t^-5,
0, 1.0*t^-1, -0.39603240282608254*t^-8, 0.9263232781144852*t^-7]],

[[0.5249458233325552*t^4, 0.48647993414509433*t^3, 0.288317913855889,
-0.6006289011129895*t^5, -0.43339360309318065*t^4, -0.9021189701254521*t,
-0.3580507322918174*t^8, 0],

[0, -0.6140778368802516*t^-1, 0, 0.3157042904634412*t, 1.0,
0.3354809468306432*t^-3, -0.4344836650252679*t^4, 0.3640820338419208*t^3],

[0, 0.021642017184031522*t^-3, 0, 1.0*t^-5, -0.16272749957295446*t^-4, 0,
0.3302270185683934*t^-8, -0.5277803864906128*t^-7]],

[[0, 0.5078943260216121*t^3, -0.8437068467257393, -0.16779079767696245*t^5, 0,
-0.22773097773605452*t, 1.0*t^8, 0],

[-0.6816689949383756, 0, -0.5858467479048473*t^-4, 0.4936500799168931*t, -1.0,
-0.06688550512654293*t^-3, 0, 0],

[0, 0.6059010244748405*t^-3, 0, 1.3353428165296193*t^-5, 0, 0,
-0.6434299512786738*t^-8, -0.4609346348556645*t^-7]],

[[0.5342376006675218*t^4, -0.6086422770257539*t^3, 0, 0.4884368142723877*t^5,
-0.718333221563019*t^4, 0, 0, -1.0*t^7],

[0.4628574417338272, 0.8309327397168714*t^-1, 0.21142383552160535*t^-4,
-1.0560448029801677*t, 0, -1.0*t^-3, 0, 0.5907349063234486*t^3],

[-0.19068598535726494*t^-4, 1.251113151532601*t^-3, 0, 0.801440153916977*t^-5,
-0.4621883155369931*t^-4, 0.649922008892636*t^-1, 0.4517624031930135*t^-8,
0.6177441885560955*t^-7]],

[[-0.46119950535589427*t^4, 0.7406376078020875*t^3, 0.29057187128303724, 0,
0.6143929403505343*t^4, 0.4568692437943906*t, 1.0*t^8,
0.36703376745263416*t^7],

[0, 1.080760392781858*t^-1, -0.14934226596262576*t^-4, 0, 0.463851046688496,
-0.5286724400096557*t^-3, 0, -0.5122850384466106*t^3],

[0, -0.17675493191050737*t^-3, -0.7935243857498888, 1.0525678466492765*t^-5,
0, 0, 0.8232503249794846*t^-8, 1.0*t^-7]],

[[-0.1128613231857281*t^4, -1.004601481243395*t^3, -0.18790258087897918,
0.40149916700646043*t^5, -0.14415337265479422*t^4, 0, 0, 1.0*t^7],

[-0.6862576195287183, 0.5664746389899703*t^-1, 0.2908675685084361*t^-4,
1.3714715312276369*t, 1.0, -0.5713223321771269*t^-3, 0.5522456890804313*t^4,
0],

[-0.3006833454545763*t^-4, -1.5673937490149197*t^-3, 0, 0,
0.38713909105007405*t^-4, 0.9955939462610987*t^-1, 0.5677563588587456*t^-8,
0.5353436694572539*t^-7]],

[[-0.6245100623794343*t^4, 0, -1.0, 0.5065225347182659*t^5,
0.5317253204886139*t^4, 0, 0, -0.3719669835035259*t^7],

[0.5699237870531472, -0.2500784148082814*t^-1, 0.442945128286494*t^-4,
0.45638933734578824*t, 1.3973251849132609, 0, 0, 0],

[0, 0.5427787886626944*t^-3, 0.6987561651998788, 1.0*t^-5, 0,
0.20324504499293194*t^-1, -0.21496324065341224*t^-8,
-0.35477523806474964*t^-7]],

[[0, -0.8732161724035096*t^3, -0.2040186481697669, -1.2010646422303124*t^5,
-0.23508531984800027*t^4, 0.5654807700289866*t, 0, 1.0788666817176849*t^7],

[-1.0, 0.23352368487408065*t^-1, 0.2848223349469312*t^-4,
0.1768073101741446*t, 1.1645543351076657, 0, 0, 0],

[0.4548846200775119*t^-4, 0, -1.0, 0.7165423092004359*t^-5,
-0.2616313451578387*t^-4, -1.2932734528263643*t^-1, -0.30361754210884495*t^-8,
0.12055753515645255*t^-7]],

[[-0.36174656010597567*t^4, 0, 0, 1.0*t^5, 0, -0.9529930944532633*t, 0, 0],

[-0.249684891380413, -0.48049270284391143*t^-1, 0.27222663948587644*t^-4,
0.06402554598955673*t, 0.32897469345165686, -0.29230508678584777*t^-3,
0.3854984554913255*t^4, 1.084386049943073*t^3],

[0.5943543096084136*t^-4, 0, -0.4586866892423329, -1.0*t^-5,
-0.2869864460438895*t^-4, 0.7692509696955846*t^-1, -0.5902387210012353*t^-8,
-0.10600670730550378*t^-7]],

[[0.2878925845174677*t^4, 0.5281046432775963*t^3, -0.2234313203594912, 0,
-0.14791845243016916*t^4, -0.33262458187432487*t, 1.108538091915835*t^8,
0.2489733522919065*t^7],

[-0.19980163401634468, -0.4519313364099573*t^-1, -0.9400278789365702*t^-4,
1.7023099438662026*t, 0.47743291480661737, 0.40637104558436565*t^-3,
-0.9678003132293527*t^4, -1.0*t^3],

[1.0*t^-4, 0.391031885184312*t^-3, 0, -1.0*t^-5, 0.8654063566747124*t^-4, 0,
0, 0]],

[[1.0110775513705723*t^4, -0.3307689454870473*t^3, -0.45870287159609613, 0,
-0.2918463909613691*t^4, 0.30167345561087666*t, 0.582399773106054*t^8, 0],

[0, -1.0*t^-1, 0.7191039431350641*t^-4, 0.5392536432576923*t,
-0.39197591630294476, -0.7143544436146056*t^-3, 0, 0.8086746481540814*t^3],

[0.4561800893644289*t^-4, -0.7356641704491395*t^-3, 0.4710082951189216, 0, 0,
0.7109669616748934*t^-1, -1.0*t^-8, 0]],

[[-0.41687446856120663*t^4, 0.3197337020122604*t^3, 0.3753986532686001, 0,
-0.40495125990774045*t^4, 0.2979428965247865*t, -0.6344817450528121*t^8,
-0.212014762523078*t^7],

[-0.2396929297153146, 0.7894128907681753*t^-1, 0, 0, 0.2995112979592937,
1.0*t^-3, 0.09278089610904368*t^4, 0.5160805291028968*t^3],

[0.4011063039077983*t^-4, 0, 0, 0.2503035377295029*t^-5, 1.0*t^-4,
-0.9730216219558445*t^-1, 0.6637049194937615*t^-8, -0.5073657590422873*t^-7]]

]

## References

  * [AFT11] Boris Alexeev, Michael A. Forbes, and Jacob Tsimerman.  Tensor rank: some lower and upper bounds.  In 26th Annual IEEE Conference on Computational Complexity, pages 283–291. IEEE Computer Soc., Los Alamitos, CA, 2011.
  * [BB20] Weronika Buczyńska and Jarosław Buczyński.  Apolarity, border rank and multigraded hilbert scheme, 2020, arXiv: 1910.01944 [math.AG].
  * [Bin80] D. Bini.  Relations between exact and approximate bilinear algorithms. Applications.  Calcolo, 17(1):87–97, 1980.
  * [CHI+18] Luca Chiantini, Jonathan D. Hauenstein, Christian Ikenmeyer, Joseph M. Landsberg, and Giorgio Ottaviani.  Polynomials and the exponent of matrix multiplication.  Bull. Lond. Math. Soc., 50(3):369–389, 2018.
  * [CHL19] Austin Conner, Alicia Harper, and J. M. Landsberg.  New lower bounds for matrix multiplication and the 3x3 determinant, 2019, arXiv: 1911.07981 [math.AG].
  * [CLGV20] Austin Conner, Joseph M. Landsberg, Fulvio Gesmundo, and Emanuele Ventura.  Kronecker Powers of Tensors and Strassen’s Laser Method.  In Thomas Vidick, editor, 11th Innovations in Theoretical Computer Science Conference (ITCS 2020), volume 151 of Leibniz International Proceedings in Informatics (LIPIcs), pages 10:1–10:28, Dagstuhl, Germany, 2020. Schloss Dagstuhl–Leibniz-Zentrum fuer Informatik.
  * [dGH86] Hans F. de Groote and Joos Heintz.  A lower bound for the bilinear complexity of some semisimple Lie algebras.  In Algebraic algorithms and error correcting codes (Grenoble, 1985), volume 229 of Lecture Notes in Comput. Sci., pages 211–222. Springer, Berlin, 1986.
  * [FH91] William Fulton and Joe Harris.  Representation theory, volume 129 of Graduate Texts in Mathematics.  Springer-Verlag, New York, 1991.  A first course, Readings in Mathematics.
  * [Lan17] J. M. Landsberg.  Geometry and complexity theory, volume 169 of Cambridge Studies in Advanced Mathematics.  Cambridge University Press, Cambridge, 2017.
  * [Lic85] Thomas Lickteig.  Typical tensorial rank.  Linear Algebra Appl., 69:95–100, 1985.
  * [LM17] J. M. Landsberg and Mateusz Michałek.  On the geometry of border rank decompositions for matrix multiplication and other tensors with symmetry.  SIAM J. Appl. Algebra Geom., 1(1):2–19, 2017.
  * [LM19] J. M. Landsberg and Mateusz Michałek.  Towards finding hay in a haystack: explicit tensors of border rank greater than $2.02m$ in $c^{m}\otimes c^{m}\otimes c^{m}$, 2019, arXiv: 1912.11927 [cs.CC].
  * [LO15] Joseph M. Landsberg and Giorgio Ottaviani.  New lower bounds for the border rank of matrix multiplication.  Theory Comput., 11:285–298, 2015.
  * [LY20] Lek-Heng Lim and Ke Ye.  Ubiquity of the exponent of matrix multiplication.  In Proceedings of the 45th International Symposium on Symbolic and Algebraic Computation, ISSAC ’20, page 8–11, New York, NY, USA, 2020. Association for Computing Machinery.
  * [Mir85] Roland Mirwald.  The algorithmic structure of sl(2, k).  In Proceedings of the 3rd International Conference on Algebraic Algorithms and Error-Correcting Codes, AAECC-3, page 274–287, Berlin, Heidelberg, 1985. Springer-Verlag.
  * [Mum95] David Mumford.  Algebraic geometry. I.  Classics in Mathematics. Springer-Verlag, Berlin, 1995.  Complex projective varieties, Reprint of the 1976 edition.
  * [Sha13] Igor R. Shafarevich.  Basic algebraic geometry. 1.  Springer, Heidelberg, third edition, 2013.  Varieties in projective space.
  * [Str69] Volker Strassen.  Gaussian elimination is not optimal.  Numer. Math., 13:354–356, 1969.
  * [Str83] V. Strassen.  Rank and optimal computation of generic tensors.  Linear Algebra Appl., 52/53:645–685, 1983.