# Decompose polynomial of degree 3 in a sum of cubed orthogonal linear forms

Given is a homogenous polynomial $$p$$ of degree $$3$$ with $$n$$ terms where the coefficients can only be $$\pm1$$. The polynomial has $$m$$ variables and in each term the variables have degree $$1$$, i.e.,

$$p(x_1,x_2,\dots,x_m)=\sum\limits_{l=1}^n s_l\; x_{i}\; x_{j}\; x_{k},$$

where $$s_l = \pm 1$$, and $$i, j, k \in \{1, \ldots, m\}$$ and $$i \ne j \ne k$$ for a given $$l$$. A proper example would be the polynomial $$x_1 x_2 x_5-x_2x_3x_4-x_1x_3x_5$$ with $$n=3$$, $$m=5$$, $$s_1=1$$, $$s_2=-1$$, $$s_3=-1$$.

Let's assume there is a decomposition of $$p$$ by a sum of cubes of orthogonal linear forms with coefficient vectors $${\bf a}_w=(a_{w1},a_{w2},\ldots,a_{wm})$$, i.e.

$$p(x_1,x_2,\dots,x_m)=\sum\limits_{w=1}^r \langle{\bf a}_w{\bf, x}\rangle^3,$$

where

$$\langle{\bf a}_u, {\bf a}_v\rangle\begin{cases}=0\text{ if } u\ne v\\\ne0\text{ if }u= v,\end{cases}$$

and due to requested orthogonality $$r\le m$$. Here, $${\bf x}=(x_1,x_2,\dots,x_m)$$ indicates the vector of the variables of the polynomial, and $$\langle\cdot,\cdot\rangle$$ is the dot product.

In case such a special polynomial is decomposable, I would know if there is a method to find the vectors $${\bf a}_w$$? I am only aware how to solve this if the degree of the polynomial would be $$2$$ but not if the degree is $$3$$.

Simple non-orthogonal example

For the polynomial $$x_1x_2x_3$$ we have $$n=1$$, $$m=3$$, $$s_1=1$$ and concerning this post and this paper$$\color{red}{^\star}$$ it can be decomposed in a sum of four cubes of linear forms, however the coefficient vectors are not mutually orthogonal.

$$x_1 x_2 x_3=\left(\frac{x_1}{2 \sqrt[3]{3}}+\frac{x_2}{2 \sqrt[3]{3}}+\frac{x_3}{2 \sqrt[3]{3}}\right)^3+\left(\frac{x_1}{2 \sqrt[3]{3}}-\frac{x_2}{2 \sqrt[3]{3}}-\frac{x_3}{2 \sqrt[3]{3}}\right)^3+\left(-\frac{x_1}{2 \sqrt[3]{3}}+\frac{x_2}{2 \sqrt[3]{3}}-\frac{x_3}{2 \sqrt[3]{3}}\right)^3+\left(-\frac{x_1}{2 \sqrt[3]{3}}-\frac{x_2}{2 \sqrt[3]{3}}+\frac{x_3}{2 \sqrt[3]{3}}\right)^3,$$

with coefficient vectors

$${\bf a}_1=\frac{1}{2\sqrt[3]{3}}\begin{pmatrix}1\\1\\1\end{pmatrix}, \quad {\bf a}_2=\frac{1}{2\sqrt[3]{3}}\begin{pmatrix}1\\-1\\-1\end{pmatrix}, \quad {\bf a}_3=\frac{1}{2\sqrt[3]{3}}\begin{pmatrix}-1\\1\\-1\end{pmatrix}, \quad {\bf a}_4=\frac{1}{2\sqrt[3] {3}}\begin{pmatrix}-1\\-1\\1\end{pmatrix}$$

I am not aware if for this polynomial exists a decomposition of orthogonal coefficients. Also I am not aware of another example where the coefficients are orthogonal.

Reference

$$\color{red}{\star}$$ Boris Reichstein, On expressing a cubic form as a sum of cubes of linear forms, Linear Algebra and its Applications, Volume 86, February 1987.

For your example $$x_1 x_2 x_3$$, it seems there is no decomposition into cubes of three linear forms. I used Maple to look at the system of $$10$$ equations in $$9$$ variables $$a_{ij}$$ obtained by taking the coefficients of $$x_1 x_2 x_3 - \sum_{i=1}^3 (a_{i1} x_1 + a_{i2} x_2 + a_{i3} x_3)^3$$; a Groebner basis was $$\{1\}$$, indicating that the system is inconsistent.
eqs:= convert(collect(x[1]*x[2]*x[3]-add((a[i,1]*x[1]+a[i,2]*x[2]+a[i,3]*x[3])^3,i=1..3),

• The example was only a demonstration and not all polynomials might be applicable. I am looking for a method for decomposable polynomials. Could the system of equations by generalized to larger polynomials, lets say for $m=n=10$? Commented Oct 7, 2022 at 14:02