Polynomials and the exponent of matrix multiplication

TL;DR

This paper introduces polynomial-based tensors, including the symmetrized matrix multiplication tensor, to leverage algebraic geometry techniques for studying the matrix multiplication exponent , aiming to deepen understanding of computational complexity.

Contribution

It defines polynomial tensors with the same exponent as matrix multiplication and explores algebraic geometry methods to analyze , especially through the symmetrized tensor sM_n.

Findings

Polynomials can represent matrix multiplication tensors with the same exponent .

Algebraic geometry techniques can be applied to study  using these polynomial tensors.

The symmetrized tensor sM_n provides a new perspective on matrix multiplication complexity.

Abstract

We define tensors, corresponding to cubic polynomials, which have the same exponent $\omega$ as the matrix multiplication tensor. In particular, we study the symmetrized matrix multiplication tensor $sM_n$ defined on an $n\times n$ matrix $A$ by $sM_n(A)=trace(A^3)$. The use of polynomials enables the introduction of additional techniques from algebraic geometry in the study of the matrix multiplication exponent $\omega$.

Equations

$$2\le \omega <2.374.$$

$$M_{⟨n⟩}(A,B,C)=\mathrm{trace}(ABC),$$

$$[({\mathbb{C}}^N{)}^{\otimes 3}{]}^{{\mathbb{Z}}_3}={\mathrm{Sym}}^3{\mathbb{C}}^N\oplus {\Lambda }^3{\mathbb{C}}^N.$$

$$s{M}_{⟨n⟩}(A,B,C)$$

$$\Lambda M_{⟨n⟩}(A,B,C)$$

$$M_{⟨n⟩}=s{M}_{⟨n⟩}+\Lambda M_{⟨n⟩}.$$

$$s{M}_{⟨n⟩}(A)=\mathrm{trace}(A^3).$$

$$\mathbb{R}(P)\le {\mathbb{R}}_s(P)\text{leavevmode}\text{nobreak}\text{leavevmode}\text{nobreak}\text{leavevmode}\text{nobreak}and\text{leavevmode}\text{nobreak}\text{leavevmode}\text{nobreak}\text{leavevmode}\text{nobreak}\underline{\mathbb{R}}(P)\le {\underline{\mathbb{R}}}_s(P).$$

$$\omega$$

$$=\underset{n}{\mathrm{liminf}}\left[{\log }_n{\mathbb{R}}_s(s{M}_{⟨n⟩})\right]=\underset{n}{\mathrm{liminf}}\left[{\log }_n{\underline{\mathbb{R}}}_s(s{M}_{⟨n⟩})\right]$$

$$=\underset{n}{\mathrm{liminf}}\left[{\log }_n\mathbb{R}(s{M}_{⟨n⟩}^S)\right]=\underset{n}{\mathrm{liminf}}\left[{\log }_n\underline{\mathbb{R}}(s{M}_{⟨n⟩}^S)\right]$$

$$=\underset{n}{\mathrm{liminf}}\left[{\log }_n{\mathbb{R}}_s(s{M}_{⟨n⟩}^S)\right]=\underset{n}{\mathrm{liminf}}\left[{\log }_n{\underline{\mathbb{R}}}_s(s{M}_{⟨n⟩}^S)\right]$$

$$=\underset{n}{\mathrm{liminf}}\left[{\log }_n\mathbb{R}(s{M}_{⟨n⟩}^{S,0})\right]=\underset{n}{\mathrm{liminf}}\left[{\log }_n\underline{\mathbb{R}}(s{M}_{⟨n⟩}^{S,0})\right]$$

$$=\underset{n}{\mathrm{liminf}}\left[{\log }_n{\mathbb{R}}_s(s{M}_{⟨n⟩}^{S,0})\right]=\underset{n}{\mathrm{liminf}}\left[{\log }_n{\underline{\mathbb{R}}}_s(s{M}_{⟨n⟩}^{S,0})\right]$$

$$=\underset{n}{\mathrm{liminf}}\left[{\log }_n\mathbb{R}(s{M}_{⟨n⟩}^Z)\right]=\underset{n}{\mathrm{liminf}}\left[{\log }_n\underline{\mathbb{R}}(s{M}_{⟨n⟩}^Z)\right]$$

$$=\underset{n}{\mathrm{liminf}}\left[{\log }_n{\mathbb{R}}_s(s{M}_{⟨n⟩}^Z)\right]=\underset{n}{\mathrm{liminf}}\left[{\log }_n{\underline{\mathbb{R}}}_s(s{M}_{⟨n⟩}^Z)\right].$$

$$\mathrm{trace}(A^3)=\sum _{i=1}^k{\left[\mathrm{trace}(L_i^{T}A)\right]}^3.$$

$$4\mathbb{R}(M_{⟨n⟩})\ge {\mathbb{R}}_s(s{M}_{⟨n⟩})\ge \mathbb{R}(s{M}_{⟨n⟩})$$

$$\omega \ge \underset{n}{\mathrm{liminf}}\left[{\log }_n{\mathbb{R}}_s(s{M}_{⟨n⟩})\right]\ge \underset{n}{\mathrm{liminf}}\left[{\log }_n\mathbb{R}(s{M}_{⟨n⟩})\right].$$

$${\mathrm{Sym}}^3(\mathrm{End}(V))$$

$$=\mathrm{End}({\mathrm{Sym}}^{3}V)\oplus \mathrm{End}(S_{21}V)\oplus \mathrm{End}({\Lambda }^{3}V).$$

$$\mathrm{End}({\mathrm{Sym}}^{3}V)\oplus \left(\mathrm{End}(S_{21}V{)}^2\right)\oplus \mathrm{End}({\Lambda }^{3}V).$$

$$\{[A]\in \mathbb{P}{\mathrm{Mat}}_n\mid A^2=0\}.$$

$$s{M}_{⟨2⟩}=$$

$$=$$

$$s{M}_{⟨2⟩}=\sum _{i=1}^6{L}_i^3$$

$$A_{\varphi }:V^{\vee }\otimes {\Lambda }^{4}V\to {\Lambda }^{5}V\otimes V$$

$$S^{2}V\otimes {\Lambda }^{4}V\to {\Lambda }^{5}V\otimes V$$

$${\mathbb{R}}_s(s{M}_{⟨n⟩}^S)\le {\mathbb{R}}_s(s{M}_{⟨n⟩})\text{leavevmode}\text{nobreak}\text{leavevmode}\text{nobreak}\text{leavevmode}\text{nobreak}and\text{leavevmode}\text{nobreak}\text{leavevmode}\text{nobreak}\text{leavevmode}\text{nobreak}\mathbb{R}(s{M}_{⟨n⟩}^S)\le \mathbb{R}(s{M}_{⟨n⟩}).$$

$$X=\left(\begin{array}{ccc}0& C^T& A\\ C& 0& B^T\\ A^T& B& 0\end{array}\right).$$

$$X^3=\left(\begin{array}{ccc}ABC+C^T{B}^T{A}^T& \ast & \ast \\ \ast & CAB+B^T{A}^T{C}^T& \ast \\ \ast & \ast & BCA+A^T{C}^T{B}^T\end{array}\right).$$

$$s{M}_{⟨2⟩}^S\left(\begin{array}{cc}{a}_0& a_1\\ a_1& a_2\end{array}\right)=(a_0+a_2)(a_0^2+3a_1^2-a_0{a}_2+a_2^2),$$

$$6\cdot s{M}_{⟨2⟩}^S=L_1^3+L_2^3-2L_3^3-2L_4^3$$

$$L_1=\left(\begin{array}{cc}2& \frac{\sqrt{-1}}{2}\\ \frac{\sqrt{-1}}{2}& 0\end{array}\right),L_2=\left(\begin{array}{cc}0& -\frac{\sqrt{-1}}{2}\\ -\frac{\sqrt{-1}}{2}& 2\end{array}\right),L_3=\left(\begin{array}{cc}1& \sqrt{-1}\\ \sqrt{-1}& 0\end{array}\right),L_4=\left(\begin{array}{cc}0& -\sqrt{-1}\\ -\sqrt{-1}& 1\end{array}\right).$$

$$T_{10}=\begin{array}{c}\text{young}(11122,23334,44555,66767,78898,91091010).\end{array}$$

Full text

1. Introduction

The exponent of matrix multiplication is the smallest constant $\omega$ such that two $n\times n$ matrices may be multiplied by performing $O(n^{\omega+\epsilon})$ arithmetic operations for every $\epsilon>0$. It is a central open problem to estimate $\omega$ since it governs the complexity of many basic algorithms in linear algebra. The current state of the art [14, 37, 27, 34] is

[TABLE]

A tensor $t\in\mathbb{C}^{N}\otimes\mathbb{C}^{N}\otimes\mathbb{C}^{N}$ has (tensor) rank $r$ if $r$ is the minimum such that there exists $u_{i},v_{i},w_{i}\in\mathbb{C}^{N}$ with $t=\sum_{i=1}^{r}u_{i}\otimes v_{i}\otimes w_{i}$. In this case, we write ${\mathbb{R}}(t)=r$. Let $V=\mathbb{C}^{n}$ and $\operatorname{End}(V)={\rm{Mat}}_{n}$ be the vector space of $n\times n$ matrices over $\mathbb{C}$. The matrix multiplication tensor $M_{\langle n\rangle}\in{\rm{Mat}}_{n}^{\vee}\otimes{\rm{Mat}}_{n}^{\vee}\otimes{\rm{Mat}}_{n}^{\vee}$ is

[TABLE]

where ${\rm{Mat}}_{n}^{\vee}$ is the vector space dual to ${\rm{Mat}}_{n}$.

Strassen [35] showed that $\omega=\liminf[\operatorname{log}_{n}(\mathbb{R}(M_{\langle n\rangle}))]$. If the tensor $t$ can be expressed as a limit of tensors of rank $s$ (but not a limit of tensors of rank at most $s-1$), then $t$ has border rank $s$, denoted $\underline{\mathbb{R}}(t)=s$. This is equivalent to $t$ being in the Zariski closure of the set of tensors of rank $s$ but not in the Zariski closure of the set of tensors of rank at most $s-1$, see, e.g., [28, Thm. 2.33]. This was rediscovered in complexity theory in [3]. Bini [6] showed that $\omega=\liminf[\operatorname{log}_{n}(\underline{\mathbb{R}}(M_{\langle n\rangle}))]$.

The determination of the fundamental constant $\omega$ is a central question in algebraic complexity theory. In 1981, Schönhage [32] showed the exponent $\omega$ could be bounded using disjoint sums of matrix multiplication tensors. Then, in 1987, Strassen [36] proposed using tensors other than $M_{\langle n\rangle}$ which are easier to analyze due to their combinatorial properties to prove upper bounds on $\omega$. These other tensors are then degenerated to disjoint matrix multiplication tensors. The main goal of this paper is to open a different path to bounding $\omega$ by introducing polynomials that are closely related to matrix multiplication.

We expect these polynomials are easier to work with in two ways. First, we want to take advantage of the vast literature in algebraic geometry regarding the geometry of cubic hypersurfaces. Second, we want to exploit recent numerical computational techniques. The difficulty of the usual matrix multiplication tensor is the sheer size of the problem, even for relatively small $n$. Despite considerable effort, no $4\times 4$ decompositions, other than the standard rank 64 decomposition and the rank 49 decomposition obtained by squaring Strassen’s $2\times 2$ decomposition, have appeared in the literature. With our approach, the polynomials are defined on much smaller spaces thereby allowing one to perform more computational experiments and produce additional data for forming conjectures.

Let $\mathrm{Sym}^{3}\mathbb{C}^{N}\subset(\mathbb{C}^{N})^{\otimes 3}$ and $\Lambda^{3}\mathbb{C}^{N}\subset(\mathbb{C}^{N})^{\otimes 3}$ respectively denote the space of symmetric and skew-symmetric tensors. Tensors in $\mathrm{Sym}^{3}\mathbb{C}^{N}$ may be viewed as homogeneous cubic polynomials in $N$ variables. While the matrix multiplication tensor $M_{\langle n\rangle}$ is neither symmetric nor skew-symmetric, it is $\mathbb{Z}_{3}$-invariant where $\mathbb{Z}_{3}$ denotes the cyclic group on three elements permuting the factors since $\operatorname{trace}(ABC)=\operatorname{trace}(BCA)$. The space of $\mathbb{Z}_{3}$-invariant tensors in $(\mathbb{C}^{N})^{\otimes 3}$ is

[TABLE]

Thus, respectively define the symmetrized and skew-symmetrized part of the matrix multiplication tensor, namely

[TABLE]

so that

[TABLE]

The $\mathbb{Z}_{3}$-invariance implies $sM_{\langle n\rangle}\in\mathrm{Sym}^{3}\mathbb{C}^{N}$ and $\Lambda M_{\langle n\rangle}\in\Lambda^{3}\mathbb{C}^{N}$.

The tensor $M_{\langle n\rangle}$ is the structure tensor for the algebra ${\rm{Mat}}_{n}$. Similarly, the skew-symmetrized matrix multiplication tensor $\Lambda M_{\langle n\rangle}$ is (if one ignores the $\frac{1}{2}$) the structure tensor for the Lie algebra $\mathfrak{g}\mathfrak{l}(V)$. The symmetrized matrix multiplication tensor $sM_{\langle n\rangle}$ is the structure tensor for ${\rm{Mat}}_{n}$ considered as a Jordan algebra, i.e., with the multiplication $A\circ B=\frac{1}{2}(AB+BA)$. In particular, considered as a cubic polynomial on ${\rm{Mat}}_{n}$,

[TABLE]

We further define the following cubic polynomials (symmetric tensors):

• •

$sM_{\langle n\rangle}^{S}$: restriction of $sM_{\langle n\rangle}$ to symmetric matrices $\mathrm{Sym}^{2}V$,

• •

$sM_{\langle n\rangle}^{S,0}$: restriction of $sM_{\langle n\rangle}^{S}$ to traceless symmetric matrices, and

• •

$sM_{\langle n\rangle}^{Z}$: restriction of $sM_{\langle n\rangle}^{S}$ to symmetric matrices with zeros on diagonal.

In order to have an invariant definition of $sM_{\langle n\rangle}^{S}$ and $sM_{\langle n\rangle}^{S,0}$, one needs an identification of $V$ with $V^{*}$. Two natural ways of obtaining this identification are via a nondegenerate symmetric quadratic form or, when $\operatorname{dim}V$ is even, a skew-symmetric form. We will often use the former, which reduces the symmetry group from the general linear group to the orthogonal group. We do not know of a nice invariant definition for the polynomial $sM_{\langle n\rangle}^{Z}$.

For a homogeneous degree $d$ polynomial $P$, the symmetric or Waring rank $\mathbb{R}_{s}(P)$ is the smallest $r$ such that $P=\sum_{j=1}^{r}\ell_{j}^{d}$, where $\ell_{j}$ are linear forms. The symmetric border rank $\underline{\mathbb{R}}_{s}(P)$ is the smallest $r$ such that $P$ is a limit of polynomials of symmetric rank at most $r$. Note that

[TABLE]

We notice that there are several general cases where equality holds in both of these relations. We refer to [13, 7] for a discussion.

Our main result is that one can compute the exponent $\omega$ of matrix multiplication using these polynomials even when considering symmetric rank and border rank.

Theorem 1.1**.**

Let $\omega$ denote the exponent of matrix multiplication. Then

[TABLE]

Proofs are given in §2 for (1.6), §3 for (1.7) and (1.8), and §4 for (1.9).

1.1. Explicit ranks and border ranks

For any $t\in\mathbb{C}^{N}\otimes\mathbb{C}^{N}\otimes\mathbb{C}^{N}$, the symmetrization of $t$ is $\mathcal{S}(t):=\frac{1}{6}\sum_{\pi\in\mathfrak{S}_{3}}\pi(t)\in\mathrm{Sym}^{3}\mathbb{C}^{N}$. In particular, $\mathcal{S}(t)=t$ if and only if $t\in\mathrm{Sym}^{3}\mathbb{C}^{N}$. The following provides bounds relating $t$ and $\mathcal{S}(t)$.

Lemma 1.2**.**

For $t\in\mathbb{C}^{N}\otimes\mathbb{C}^{N}\otimes\mathbb{C}^{N}$, ${\mathbb{R}}_{s}(\mathcal{S}(t))\leq 4{{\mathbb{R}}}(t)$ and $\underline{\mathbb{R}}_{s}(\mathcal{S}(t))\leq 4{\underline{\mathbb{R}}}(t)$.

Proof.

If $t=\sum_{i=1}^{r}u_{i}\otimes v_{i}\otimes w_{i}$ with $u_{i},v_{i},w_{i},\in V$, then $\mathcal{S}(t)=\sum_{i=1}^{r}(u_{i}v_{i}w_{i})$. Since ${\mathbb{R}}_{s}(xyz)=4$ (see, e.g., [22, §10.4]), this immediately yields that ${\mathbb{R}}_{s}(\mathcal{S}(t))\leq 4{{\mathbb{R}}}(t)$. In the same way, if $t$ is a limit of tensors of the form $\sum_{i=1}^{r}u_{i}\otimes v_{i}\otimes w_{i}$, this yields $\underline{\mathbb{R}}_{s}(\mathcal{S}(t))\leq 4{\underline{\mathbb{R}}}(t)$. ∎

In particular, $\mathbb{R}(sM_{\langle n\rangle})\leq 2\mathbb{R}(M_{\langle n\rangle})<2n^{3}$ (as $sM_{\langle n\rangle}$ is the sum of two matrix multiplications, by (1.2)) so that $\mathbb{R}_{s}(sM_{\langle n\rangle})\leq 8\mathbb{R}(M_{\langle n\rangle})<8n^{3}$ and similarly for all its degenerations.

The following summarizes some results about small cases.

Theorem 1.3**.**

---

1. (1)

${\mathbb{R}}_{s}(sM_{\langle 2\rangle})=6$* and $\underline{\mathbb{R}}_{s}(sM_{\langle 2\rangle})=5$ ([33, IV, §97], [25, Prop. 7.2]). * 2. (2)

$\underline{\mathbb{R}}_{s}(sM_{\langle 3\rangle})\geq 14$. 3. (3)

${\mathbb{R}}_{s}(sM_{\langle 2\rangle}^{S})=\underline{\mathbb{R}}_{s}(sM_{\langle 2\rangle}^{S})=4$* ([33, IV §96] or [25, §8]).* 4. (4)

$\underline{\mathbb{R}}_{s}(sM^{S}_{\langle 3\rangle})=10$. 5. (5)

$sM^{S,0}_{\langle 2\rangle}=0$* while $\underline{\mathbb{R}}_{s}(sM^{S,0}_{\langle 3\rangle})=\mathbb{R}_{s}(sM^{S,0}_{\langle 3\rangle})=8$.* 6. (6)

$\underline{\mathbb{R}}_{s}(sM^{S,0}_{\langle 4\rangle})\geq 14$. 7. (7)

$sM^{Z}_{\langle 2\rangle}=0$* while $\underline{\mathbb{R}}_{s}(sM_{\langle n\rangle}^{Z})=\mathbb{R}_{s}(sM_{\langle n\rangle}^{Z})=2^{n-1}$ for $n=3,4,5$, with $\mathbb{R}_{s}(sM^{Z}_{\langle 6\rangle})\leq 30$, $\mathbb{R}_{s}(sM^{Z}_{\langle 7\rangle})\leq 48$, and $\mathbb{R}_{s}(sM^{Z}_{\langle 8\rangle})\leq 64$.*

The cases (1) and (3) are discussed respectively in §2.1 and §3.1. The case (4) is proved in §3.2 with a tableau evaluation. The cases (2), (5), (6) are proved with the technique of Young flattenings introduced in [24] which has already been used in the unsymmetric case in [23]. In particular, Proposition 2.6 below considers (2) with the other cases following analogously. The case (7) is proved by exhibiting explicit decompositions in Theorems 4.2, 4.3, and 4.4.

Since one of our goals is to simplify the problem in order to further exploit numerical computations, we experiment with numerical tools and probabilistic methods via Bertini [5]. We believe the computations could likely be converted to rigorous proofs, e.g., by showing that an overdetermined system has a solution nearby the given numerical approximation [2]. We write Theorem* when we mean the result of a numerical computation.

Theorem 1.4***.**

${\mathbb{R}}_{s}(sM_{\langle 3\rangle})\leq 18$.

We show this in Theorem* 2.7 with data regarding this and other computations available at http://dx.doi.org/10.7274/R0VT1Q1J.

Notation and conventions

The group of invertible linear maps $\mathbb{C}^{N}\rightarrow\mathbb{C}^{N}$ is denoted $GL_{N}$ and the permutation group on $d$ elements by $\mathfrak{S}_{d}$. For $u,v,w\in\mathbb{C}^{N}$, we have $u\otimes v\otimes w\in(\mathbb{C}^{N})^{\otimes 3}$ and $uvw\in\mathrm{Sym}^{3}\mathbb{C}^{N}$. The space ${\rm{Mat}}_{n}$ is canonically self-dual. Given a matrix $L$, when we consider $L\in{\rm{Mat}}_{n}^{\vee}$, we write $L^{3}\in\mathrm{Sym}^{3}({\rm{Mat}}_{n}^{\vee})$ for the cubic polynomial function which sends the matrix $A$ to $\left[\mathrm{trace}(L^{T}A)\right]^{3}$, where $L^{T}$ is the transpose of $L$. Note that $L^{3}$ is a function and not the cube of the matrix $L$. In particular, $sM_{\langle n\rangle}=\sum_{i=1}^{k}L_{i}^{3}$ means that

[TABLE]

For a partition $\pi$ of $d$, $S_{\pi}\mathbb{C}^{N}$ denotes the corresponding $GL_{N}$-module and $[\pi]$ the corresponding $\mathfrak{S}_{d}$-module. In particular $S_{(d)}\mathbb{C}^{N}=\mathrm{Sym}^{d}\mathbb{C}^{N}$ and $S_{(1^{d})}\mathbb{C}^{N}=\Lambda^{d}\mathbb{C}^{N}$.

Acknowledgement

This project began during the Fall 2014 program Algorithms and Complexity in Algebraic Geometry. The authors thank the Simons Institute for providing a wonderful research environment.

2. The polynomial $sM_{\langle n\rangle}$

We start with the first statement from Theorem 1.1.

Proof of (1.6).

Lemma 1.2 and (1.5) imply

[TABLE]

so that

[TABLE]

For $n\times n$ matrices $A$, $B$, $C$ consider the $3n\times 3n$ matrix $X=\begin{pmatrix}0&0&A\\ C&0&0\\ 0&B&0\end{pmatrix}$. Then, $X^{3}=\begin{pmatrix}ABC&0&0\\ 0&CAB&0\\ 0&0&BCA\end{pmatrix}$ and $\operatorname{trace}(X^{3})=3\operatorname{trace}(ABC)$. This shows that ${\mathbb{R}}(M_{\langle n\rangle})\leq{\mathbb{R}}(sM_{\langle 3n\rangle})$ yielding the inequality $\omega\leq\liminf_{n}\left[\log_{n}{\mathbb{R}}(sM_{\langle n\rangle})\right]$. The border rank statement follows similarly by taking limits. ∎

As a $GL_{N}$-module via the Cauchy formula, $\mathrm{Sym}^{3}(\operatorname{End}(V))=\mathrm{Sym}^{3}(V\otimes V^{*})$ decomposes as

[TABLE]

The tensor $M_{\langle n\rangle}\in{\rm{Mat}}_{n}^{\vee}\otimes{\rm{Mat}}_{n}^{\vee}\otimes{\rm{Mat}}_{n}^{\vee}=\mathrm{End}(V^{\otimes 3})$ corresponds to the identity endomorphism. Since $V^{\otimes 3}=\mathrm{Sym}^{3}V{\mathord{\,\oplus}\,}(S_{21}V)^{{{\mathord{\,\oplus}\,}}2}{\mathord{\,\oplus}\,}\Lambda^{3}V$, it follows that $\mathrm{End}(V^{\otimes 3})$, as a $GL(V)$-module, contains the submodule

[TABLE]

The projection of $sM_{\langle n\rangle}$ onto each of the three summands in (2.2) is the identity endomorphism (the last summand requires $n\geq 3$ to be nonzero). In particular, all three projections are nonzero when $n\geq 3$.

For $n\geq 2$, the following shows that in any symmetric rank decomposition of $sM_{\langle n\rangle}$, it is impossible to have all summands corresponding to matrices $L_{i}$ of rank one. Moreover, for $n\geq 3$, at least one summand corresponds to a matrix having rank at least $3$. We note that this statement is in contrast to tensor decompositions of $sM_{\langle n\rangle}$ where there do exist decompositions constructed from rank one tensors. In fact, the $n=3$ case is in stark contrast to usual matrix multiplication where there exist decompositions for which no matrix appearing has rank greater than one, e.g., the standard decomposition.

Theorem 2.1**.**

Suppose that $sM_{\langle n\rangle}=\sum_{i=1}^{k}L_{i}^{3}$ is a symmetric rank decomposition. If $n=2$, there exists $i$ with $\operatorname{rank}(L_{i})=2$. Moreover, if $n\geq 3$, $\max_{i}\operatorname{rank}(L_{i})\geq 3$.

Proof.

Any summand $L_{i}^{3}$ with $\operatorname{rank}L_{i}=1$ is of the form $L_{i}=v_{i}\otimes\omega_{i}\in V\otimes V^{\vee}$ and induces an element of rank one that takes $a\otimes b\otimes c$ to $\omega_{i}(a)\omega_{i}(b)\omega_{i}(c)v_{i}^{3}$ which vanishes outside $\mathrm{Sym}^{3}V$. This element lies in $\operatorname{End}(\mathrm{Sym}^{3}V)$ in the decomposition (2.2). Hence, any sum of these elements lies in this subspace and thus projects to zero in the second and third factors in (2.2).

Similarly, any summand of rank two only gives rise to a term appearing in $\mathrm{Sym}^{3}V\otimes\mathrm{Sym}^{3}V^{*}{\mathord{\,\oplus}\,}S_{21}V\otimes S_{21}V^{*}$ because one needs three independent vectors for a term in $\Lambda^{3}V\otimes\Lambda^{3}V^{*}$. ∎

This following provides a slight improvement over the naïve bound of $8n^{3}$.

Proposition 2.2** (A modest upper bound).**

${\mathbb{R}_{s}}(sM_{\langle n\rangle})\leq 8{n\choose 3}+4{n\choose 2}+n$.

Proof.

Every monomial appearing in $sM_{\langle n\rangle}$ has the form $a_{ij}a_{jk}a_{ki}$. This bound arises from considering the symmetric ranks of each of these monomials. There are $2{n\choose 3}$ monomials corresponding to distinct cardinality 3 sets $\{i,j,k\}\subset\{1,\ldots,n\}$ and each monomial has symmetric rank $4$. There are $2{n\choose 2}$ monomials corresponding to distinct cardinality 2 sets $\{i,j\}\subset\{1,\ldots,n\}$ and they group together in ${n\choose 2}$ pairs as $a_{ij}a_{ji}(a_{ii}+a_{jj})$ with each such term having symmetric rank two. Finally, there are $n$ monomials of the form $a_{ii}^{3}$ for $i=1,\ldots,n$. ∎

The following considers algebraic geometric aspects of $sM_{\langle n\rangle}$.

Proposition 2.3**.**

(i) The singular locus of $\{sM_{\langle n\rangle}=0\}\subset\mathbb{P}{\rm{Mat}}_{n}$ is

[TABLE]

(ii) The polynomial $sM_{\langle 2\rangle}$ is reducible, while $sM_{\langle n\rangle}$ is irreducible for $n\geq 3$.

Proof.

The map $(A,B)\mapsto\mathrm{tr}(AB^{t})$ is a nondegenerate pairing. Since we can write $\mathrm{tr}(A^{3})=\mathrm{tr}(A\cdot A^{2})$, this proves $(i)$.

Note that the $(i,j)$ entry of $A^{2}$ coincides, up to scalar multiple, with the partial derivative $\dfrac{\partial\left(sM_{\langle n\rangle}\right)}{\partial a_{j,i}}$. In order to prove $(ii)$, we estimate the dimension of the singular locus computed in $(i)$. If $A$ belongs to the singular locus of $\{sM_{\langle n\rangle}=0\}$, we know $\ker(A)\subseteq\mathrm{im}(A)$ so that $\operatorname{rank}(A)\leq n/2$. It follows that the singular locus of $\{sM_{\langle n\rangle}=0\}$ has codimension $\geq 3$ for $n\geq 3$ showing that $sM_{\langle n\rangle}$ must be irreducible. If not, the singular locus contains the intersection of any two irreducible components, having codimension $\leq 2$. The $n=2$ case follows from (2.3) below. ∎

2.1. Decomposition of $sM_{\langle 2\rangle}$

The reducibility of $sM_{\langle 2\rangle}$ is as follows:

[TABLE]

In particular, for this classically studied polynomial, its zero set is the union of a smooth quadric and a non-tangent hyperplane. A general cubic surface has a unique Waring decomposition as a sum of $5$ summands by the Sylvester Pentahedral Theorem [29, Theor. 3.9]. Hence, every $f\in\mathrm{Sym}^{3}\mathbb{C}^{4}$ has $\underline{\mathbb{R}}_{s}(f)\leq 5$. However, $\mathbb{R}_{s}(sM_{\langle 2\rangle})=6$ (see [33, IV, §97]) with a minimal Waring decomposition given by

[TABLE]

where $L_{1}=\dfrac{1}{2}\begin{pmatrix}-1&1\\ -1&-1\end{pmatrix}$, $L_{2}=\dfrac{1}{2}\begin{pmatrix}-1&-1\\ 1&-1\end{pmatrix}$, $L_{3}=\dfrac{1}{2}\begin{pmatrix}1&1\\ 1&1\end{pmatrix}$, $L_{4}=\dfrac{1}{2}\begin{pmatrix}1&-1\\ -1&1\end{pmatrix}$, $L_{5}=\begin{pmatrix}1&0\\ 0&0\end{pmatrix}$, and $L_{6}=\begin{pmatrix}0&0\\ 0&1\end{pmatrix}$.

Remark 2.4** (A remark on 5 summands).**

For the decomposition presented in (2.4), $\operatorname{rank}(L_{i})=2$ for $i=1,2$ while $\operatorname{rank}(L_{i})=1$ for $i=3,4,5,6$ in agreement with Theorem 2.1. Since $sM_{\langle 2\rangle}$ is $GL_{2}$-invariant for the conjugate action which takes $A$ to $G^{-1}AG$ for every $G\in GL_{2}$, the matrices $L_{i}$ can be replaced in (2.4) with $G^{-1}L_{i}G$ for any $G\in GL_{2}$.

Consider a family $f_{{2},\epsilon}$ which has a Waring decomposition given by five matrices $L_{i,\epsilon}$ for $\epsilon\neq 0$ and $f_{{2},0}=sM_{\langle{2}\rangle}$. In all the examples we have found, the five matrices $L_{i,\epsilon}$ converge as $\epsilon\to 0$ to the identity matrix that is indeed a fixed point for the conjugate action.

The following Remark provides a geometric description for decompositions of $sM_{\langle 2\rangle}$ using six terms.

Remark 2.5**.**

Identify the projective space of $2\times 2$ matrices with $\mathbb{P}^{3}$. Let $\mathcal{Q}$ be the quadric of matrices of rank $1$ and let $\ell$ denote the line spanned by the identity $I$ and the skew-symmetric point $\Lambda$.

For a choice of $3$ points $Q_{1},Q_{2},Q_{3}$ in the intersection of $\mathcal{Q}$ with the plane of traceless matrices, let $A_{1},A_{2},B_{1},B_{2},C_{1},C_{2}$ denote the 6 points of intersection of the two rulings of $\mathcal{Q}$ passing through each $Q_{i}$. These points, together with $I$, determine a minimal decomposition of the general tensor $M_{\langle 2\rangle}$, as explained in [11].

A decomposition of $sM_{\langle 2\rangle}$ is determined as follows: let $Q_{3}$ be the intersection of the lines $(B_{1}C_{1})$ and $(B_{2}C_{2})$. Then the six points $L_{1}\ldots L_{6}$ are obtained by taking $L_{6}=A_{2}$, $L_{5}=A_{1}$, $L_{4}=$ the intersection of $(B_{1},C_{1})$ with the plane $\pi$ of symmetric matrices, $L_{3}=$ the intersection of $(B_{2},C_{2})$ with $\pi$, $L_{2}=$ the intersection of the line $(Q_{3}A_{2})$ with $\ell$ (they meet), and $L_{1}=$ the intersection of the line $(Q_{3},A_{1})$ with $\ell$.

For instance, starting with $Q_{1}=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$, $Q_{2}=\begin{pmatrix}0&0\\ 1&0\end{pmatrix}$, and $Q_{3}=\begin{pmatrix}1&1\\ -1&-1\end{pmatrix}$, we obtain the six points $L_{1},\dots,L_{6}$ of the decomposition (2.4) described above.

We ask if an analogous geometric description could provide small decompositions of $sM_{\langle n\rangle}$ for $n\geq 3$.

2.2. Case of $sM_{\langle 3\rangle}$

The polynomial $sM_{\langle 3\rangle}$ is irreducible by Proposition 2.3 with the following lower bound on border rank.

Proposition 2.6**.**

$\underline{\mathbb{R}}_{s}(sM_{\langle 3\rangle})\geq 14.$**

Proof.

Let $V=\mathbb{C}^{9}$. For any $\phi\in\mathrm{Sym}^{3}V$ we have the linear map

[TABLE]

which is defined by contracting the elements of the source with $\phi$ and then projecting to the target. This projection is well-defined because the map

[TABLE]

is a $GL(V)$-module map and the image of the projection is the unique copy of $S_{411}V\subset\Lambda^{5}V\otimes V$. This map was denoted as $\mathrm{YF}_{3,8}(\phi)$ in [24, Eq. (2)]. Using [17], direct computation shows that $\operatorname{rank}A_{v^{3}}=70$ for a nonzero $v\in V$ and $\operatorname{rank}A_{sM_{\langle 3\rangle}}=950$. By linearity $\underline{\mathbb{R}}_{s}(sM_{\langle 3\rangle})\geq\lceil\frac{950}{70}\rceil=14.$ ∎

The following provides information on the rank.

Theorem 2.7***.**

${\mathbb{R}}_{s}(sM_{\langle 3\rangle})\leq 18$* with a Waring decomposition of $sM_{\langle 3\rangle}$ with $18$ summands found numerically by Bertini [5] with all $18$ summations having rank $3$.*

Proof.

After numerically approximating a decomposition with Bertini [5], applying the isosingular local dimension test [21] suggested that there is at least one $9$-dimensional family of decompositions. We used the extra $9$ degrees of freedom to set $9$ entries to [math], $1$, or $-1$ producing a polynomial system which has an isolated nonsingular root with an approximation given in Appendix A and electronically available at http://dx.doi.org/10.7274/R0VT1Q1J. ∎

Decompositions with $18$ summands were highly structured leading to the following.

Conjecture 2.8**.**

${\mathbb{R}}_{s}(sM_{\langle 3\rangle})=18$.

In our experiments, we were unable to compute a decomposition of $sM_{\langle 3\rangle}$ using $18$ summands with real matrices.

3. The polynomials $sM_{\langle n\rangle}^{S}$ and $sM_{\langle n\rangle}^{S,0}$

We start with statements from Theorem 1.1.

Proof of (1.7) and (1.8).

The following two inequalities are trivial since $sM_{\langle n\rangle}^{S}$ is a specialization of $sM_{\langle n\rangle}$:

[TABLE]

For $n\times n$ matrices $A$, $B$, $C$ consider the $3n\times 3n$ symmetric matrix

[TABLE]

We have $\operatorname{trace}(X^{3})=6\operatorname{trace}(ABC)$ since

[TABLE]

It immediately follows ${\mathbb{R}}(M_{\langle n\rangle})\leq{\mathbb{R}}(sM^{S}_{\langle 3n\rangle})$. Hence, (1.7) follows by a similar argument as in the proof of (1.6).

Since $X$ is traceless, the same argument also proves (1.8). ∎

3.1. Decomposition of $sM_{\langle 2\rangle}^{S}$

As in the general case (2.3), $sM_{\langle 2\rangle}^{S}$ is a reducible polynomial while $sM_{\langle n\rangle}^{S}$ is irreducible for $n\geq 3$ (the same argument as in Proposition 2.3 works). In fact,

[TABLE]

which corresponds to the union of a smooth conic with a secant (not tangent) line. Moreover, it was known classically that $\underline{\mathbb{R}}_{s}(sM_{\langle 2\rangle}^{S})=\mathbb{R}_{s}(sM_{\langle 2\rangle}^{S})=4$, which is the generic rank in $\mathbb{P}(\mathrm{Sym}^{3}\mathbb{C}^{3})$ with a minimal Waring decomposition given by

[TABLE]

where

[TABLE]

We note that $L_{1}$ and $L_{2}$ are similar as well as $L_{3}$ and $L_{4}$ all have rank $2$.

3.2. Case of $sM^{S}_{\langle 3\rangle}$

We consider $\underline{\mathbb{R}}_{s}(sM^{S}_{\langle 3\rangle})$ as a cubic polynomial on $\mathbb{C}^{6}$. Since the generic rank in $\mathbb{P}(\mathrm{Sym}^{3}\mathbb{C}^{6})$ is $10$ (see [4]), we have $\underline{\mathbb{R}}_{s}(sM^{S}_{\langle 3\rangle})\leq 10$. To show that equality holds, consider the degree $10$ invariant in $\mathrm{Sym}^{10}(\mathrm{Sym}^{3}\mathbb{C}^{6})$ corresponding to the following Young diagram (see, e.g., [31, §3.9], for the symbolic notation of invariants):

[TABLE]

This invariant is generalized in [8, Prop. 3.25].

Lemma 3.1**.**

The polynomial $T_{10}$ defined by (3.2) forms a basis of the $SL_{6}$-invariant space $(\mathrm{Sym}^{10}(\mathrm{Sym}^{3}\mathbb{C}^{6}))^{SL_{6}}$ and is in the ideal of $\sigma_{9}(\nu_{3}(\mathbb{P}^{5}))$. Moreover, $T_{10}(sM^{S}_{\langle 3\rangle})\neq 0$ showing that $\underline{\mathbb{R}}_{s}(sM^{S}_{\langle 3\rangle})>9$.

Proof.

A plethysm calculation, e.g., using Schur [9], shows that

[TABLE]

We explicitly evaluated $T_{10}(sM^{S}_{\langle 3\rangle})$ using the same algorithm as in [1] and [10] which phrases the evaluation as a tensor contraction and ignores summands that contribute zero to the result. The result was that $T_{10}(sM^{S}_{\langle 3\rangle})\neq 0$.

We now consider evaluating $T_{10}$ on all cubics of the form $f=\sum_{i=1}^{9}\ell_{i}^{3}$. The expression $T_{10}(f)$ splits as the sum of several terms of the form $T_{10}(\ell_{i_{1}}^{3},\ldots,\ell_{i_{10}}^{3})$ where, in each of these summands, there is a repetition of some $\ell_{i}$. We claim that every $T_{10}(\ell_{i_{1}}^{3},\ldots,\ell_{i_{10}}^{3})$ vanishes due to this repetition. Indeed, each pair $(i,j)$ with $1\leq i<j\leq 10$ appears in at least one column of (3.2). In other words, for any $g\colon\{1,\ldots,10\}\to\{1,\ldots,9\}$, the tableau evaluation $g(T_{10})$ has a repetition in at least one column and thus vanishes. This approach is the main tool used in [1]. ∎

Since the polynomial $T_{10}$ vanishes on $\sigma_{9}(\nu_{3}(\mathbb{P}^{5}))$, Lemma 3.1 immediately yields that $\underline{\mathbb{R}}_{s}(sM^{S}_{\langle 3\rangle})=10$, because $\sigma_{10}(v_{3}(\mathbb{P}^{5}))$ equals $\mathbb{P}\mathrm{Sym}^{3}\mathbb{C}^{6}$. The following considers decompositions with $10$ summands.

Proposition 3.2***.**

${\mathbb{R}}_{s}(sM^{S}_{\langle 3\rangle})=10$.

Proof.

Consider decompositions consisting of 3 symmetric matrices each of the form

[TABLE]

each of which is clearly rank deficient, and one symmetric matrix of the form

[TABLE]

which is clearly traceless.

Upon substituting these forms, which have a total of $3\cdot 10=30$ unknowns, into the $\binom{5+3}{3}=56$ equations which describe the decompositions of $sM^{S}_{\langle 3\rangle}$, there are $28$ equations which vanish identically leaving $28$ polynomial equations in $30$ affine variables. The isosingular local dimension test [21] in Bertini [5] suggests that this system has at least one $3$-dimensional solution component which we utilize the $3$ extra degrees of freedom to make one entry either $\pm 1$ in one of each of the three types of matrices in (3.3). The resulting system has an isolated solution which we present one here to $4$ significant digits:

[TABLE]

The eigenvalues $\lambda_{1},\lambda_{2},\lambda_{3}$ of the first $9$ summands satisfy

[TABLE]

while the eigenvalues of the traceless matrix satisfy

[TABLE]

∎

The variety $\sigma_{9}(\nu_{3}(\mathbb{P}^{5}))$ has codimension $2$ as expected. The following describes generators of its ideal.

Theorem 3.3***.**

The variety $\sigma_{9}(\nu_{3}(\mathbb{P}^{5}))$ has codimension $2$ and degree $280$. It is the complete intersection of the solution set of $T_{10}$ and a hypersurface of degree $28$.

Proof.

It is easy to computationally verify that the variety $X:=\sigma_{9}(\nu_{3}(\mathbb{P}^{5}))\subset\mathbb{P}^{55}$ has the expected dimension of $53$, e.g., via [19, Lemma 3]. This also follows from the Alexander-Hirschowitz Theorem [4]. We used the approach in [18, §2] with Bertini [5] to compute a so-called pseudowitness set [19] for $X$ yielding $\deg X=280$. With this pseudowitness set, [15, 16] shows that $X$ is arithmetically Cohen-Macaulay and arithmetically Gorenstein. In particular, the Hilbert function of the finite set $X\cap\mathcal{L}$ where $\mathcal{L}\subset\mathbb{P}^{55}$ is a general linear space of dimension $2$ is

[TABLE]

Thus, the ideal of $X\cap\mathcal{L}$ is minimally generated by a degree $10$ polynomial (corresponding to $T_{10}$) and a polynomial of degree $28$. The same holds for $X$, i.e., $X$ is a complete intersection defined by the vanishing of $T_{10}$ and a polynomial of degree $28$, since $X$ is arithmetically Cohen-Macaulay. The Hilbert series of $X$ is

[TABLE]

∎

Remark 3.4**.**

The generic polynomial in $\sigma_{9}(\nu_{3}(\mathbb{P}^{5}))$ has exactly two Waring decompositions, this is the last subgeneric $\sigma_{k}(\nu_{d}(\mathbb{P}^{n}))$ whose generic member has a non-unique Waring decomposition [12, Thm. 1.1].

We close with the traceless $3\times 3$ case $sM^{S,0}_{\langle 3\rangle}$ where we take $a_{5}=-(a_{0}+a_{3})$.

Proposition 3.5**.**

$\underline{\mathbb{R}}_{s}(sM^{S,0}_{\langle 3\rangle})=\mathbb{R}_{s}(sM^{S,0}_{\langle 3\rangle})=8$**

Proof.

Although $\sigma_{7}(\nu_{3}(\mathbb{P}^{4}))\subset\mathbb{P}^{34}$ is expected to fill the ambient space, it is defective: it is a hypersurface of degree $15$ defined by the cubic root of the determinant of a 45$\times$45 matrix, e.g., see [30, 1]. This $45\times 45$ matrix evaluated at $sM^{S,0}_{\langle 3\rangle}$ has full rank showing that $\underline{\mathbb{R}}_{s}(sM^{S,0}_{\langle 3\rangle})>7$. Since $8$ is the generic rank, $\underline{\mathbb{R}}_{s}(sM^{S,0}_{\langle 3\rangle})=8$.

To show the existence of a decomposition using $8$ summands, we need to solve a system of $\binom{4+3}{3}=35$ polynomials in $40$ affine variables. By including the determinant of the matrices corresponding to the first $5$ summands, we produce a square system with $40$ polynomials in $40$ variables. We prove the existence of a solution via $\alpha$-theory using alphaCertified [20] starting with the following approximation:

[TABLE]

where $i=\sqrt{-1}$. ∎

4. The polynomial $sM_{\langle n\rangle}^{Z}$

Let $Z_{n}$ be the space of symmetric matrices with zeros on the diagonal. The cubic $sM_{\langle n\rangle}^{Z}(A)$ is a polynomial in ${{n}\choose 2}$ indeterminates, its naïve expression has ${{n}\choose 3}$ terms:

[TABLE]

The proof of (1.9) is similar to the others and thus omitted.

Since $sM_{\langle 2\rangle}^{Z}$ is identically zero, we take $n\geq 3$. Let $P_{n}$ denote the finite set of $2^{n-1}$ vectors of the form $v=\left(1,\pm 1,\ldots,\pm 1\right)^{T}\in\mathbb{Z}^{n}$. In Theorem 4.1, we use $P_{n}$ to construct a decomposition of $sM_{\langle n\rangle}^{Z}$. Although such a decomposition is not minimal for $n\geq 6$ (see Proposition 4.3), a modification of it constructs the decomposition (4.3) for $n=8$ which we believe is minimal (see Remark 4.5).

For each $v\in P_{n}$, $vv^{T}-I_{n}\in Z_{n}$ with eigenvalues $\{-1,\ldots,-1,n-1\}$ and off-diagonal elements $\pm 1$.

Theorem 4.1**.**

For $n\geq 3$, we have the decomposition of $2^{n-1}$ summands:

[TABLE]

Proof.

If $v=\left(v_{1},\ldots,v_{n}\right)^{T}\in P_{n}$, then the $(i,j)$-entry of $(vv^{T}-I_{n})$ is $v_{i}v_{j}-\delta_{ij}$. The monomials appearing in $\left(\operatorname{trace}[(vv^{T}-I_{n})\cdot A]\right)^{3}=\left(\sum_{i<j}a_{ij}v_{i}v_{j}\right)^{3}$ divide into three groups

1. (1)

$a_{ij}^{3}v_{i}v_{j}$ for $i<j$, 2. (2)

$a_{ij}^{2}a_{pq}v_{p}v_{q}$ for $i<j$ and $p<q$, 3. (3)

$a_{ij}a_{pq}a_{rs}v_{i}v_{j}v_{p}v_{q}v_{r}v_{s}$.

Summing over $P_{n}$, the monomials of the first group cancel each other because, for any fixed value $v_{i}\in\{-1,+1\}$, the vectors $v\in P_{n}$ having this fixed value divide into two subsets of equal size, having respectively $v_{j}=-1$ or $v_{j}=1$. This argument includes the case $i=0$, when $v_{0}=1$.

For the same reason the monomials of the second group cancel each other.

In the third group, all monomials when $\#\{i,j,p,q,r,s\}\geq 4$ cancel each other because there is an index which appear only once, and the above argument shows that the sum over this index makes zero. If $\#\{i,j,p,q,r,s\}=3$ and the monomial is not in the first or second group, then each index appears exactly twice and we get exactly all the summands which appear in (4.1).

Since these cover all cases, the right-hand side of (4.2) sums up to a scalar multiple of the left-hand side. ∎

Proposition 4.2**.**

The decompositions in (4.2) are minimal for $n=3,4,5$. In particular, $\underline{\mathbb{R}}_{s}(sM_{\langle n\rangle}^{Z})=\mathbb{R}_{s}(sM_{\langle n\rangle}^{Z})=2^{n-1}$ for $n=3,4,5$.

Proof.

We compute the Koszul flattening $\mathrm{YF}_{3,n^{2}-1}(sM_{\langle n\rangle}^{Z})$ as in [24, (2)] where $r_{n}$ is its rank. Let $q_{n}=\operatorname{rank}\mathrm{YF}_{3,n^{2}-1}(\ell^{3})={{m_{n}}\choose{\lfloor m_{n}/2\rfloor}}$ for any linear form $\ell$, where $m_{n}=n(n-1)/2-1$. With this setup, [24, Prop. 4.1.1] and Theorem 4.1 provide

[TABLE]

The result follows immediately from the following table:

[TABLE]

∎

For comparison, the known lower bounds on the border rank of $M_{\langle n\rangle}$ when $n=3,4,5$ are $15,29,47$, respectively, with the general lower bound from [26] of $\underline{\mathbb{R}}(M_{\langle n\rangle})\geq 2n^{2}-\operatorname{log}_{2}(n)-1$.

For $n=6,7$, the decomposition (4.2) has $32$ and $64$ summands, respectively. The following shows that such decompositions are not minimal (as well for any $n\geq 6$).

Proposition 4.3**.**

$\mathbb{R}_{s}(sM_{\langle 6\rangle}^{Z})\leq 30$* and $\mathbb{R}_{s}(sM_{\langle 7\rangle}^{Z})\leq 48$.*

Proof.

See Appendix A for a decomposition of $32sM_{\langle 6\rangle}^{Z}$ with $30$ summands having integer coefficients and a decomposition of $160sM_{\langle 7\rangle}^{Z}$ with $48$ summands having coefficients in $\mathbb{Q}[\sqrt{5}]$. ∎

When $n=8$, the following provides a decomposition using $64$ summands.

Proposition 4.4**.**

Let $P_{8}^{+}$ be the subset of $P_{8}$ consisting of $v$ such that $+1$ appears an even number of times, so $\#P_{8}^{+}=64$. Then,

[TABLE]

Proof.

This is easy to verify by direct computation. ∎

Remark 4.5**.**

*We expect the decomposition (4.3) with $64$ summands is minimal. In the $\binom{8}{2}=28$ indeterminants of the matrix $A$, consider the polynomial $f(A;\ell_{1},\ldots,\ell_{64})=\sum_{i=1}^{64}\left(\operatorname{trace}[(\ell_{i}^{T}A]\right)^{3}$. Evaluated at $\ell_{i}=(v_{i}v_{i}^{T}-I_{n})$ where $v_{i}\in P_{8}^{+}$, the polynomial $f$ has maximal rank of $1792=28\cdot 64$. *

Proposition 4.4 suggests one should look for strategic subsets $P_{n}^{?}\subset P_{n}$ analogous to $P_{8}^{+}\subset P_{8}$ to produce minimal decompositions. For $n=9$ and $10$, we can use $P_{9}^{+}$ and $P_{10}^{+}$ to obtain (again by direct computation) $\mathbb{R}_{s}(sM_{\langle 9\rangle}^{Z})\leq 128$ and $\mathbb{R}_{s}(sM_{\langle 10\rangle}^{Z})\leq 256$, but both seem to not be sharp.

Appendix A Decompositions

The following $18$ matrices of rank $3$ form a numerical approximation of a decomposition of $sM_{\langle 3\rangle}$ where $i=\sqrt{-1}$:

[TABLE]

[TABLE]

A decomposition of $32sM^{Z}_{\langle 6\rangle}$ using $30$ summands is:

[TABLE]

With $\beta=\sqrt{5}/2$, a decomposition of $160sM^{Z}_{\langle 7\rangle}$ using $48$ summands is:

[TABLE]