A positive polynomial of Schur as a sum of squares of polynomials

Question

This is a follow-up this question on MathOverflow. The previous time I asked about SOS was Choi Lam homogeneous polynomials as sums of squares.

With positive integer $k,$ and real variables $x,y,z$ that are allowed to take on positive or negative or zero values. Schur defines a positive semi-definite polynomial $ F = x^{2k}(x-y)(x-z) + y^{2k}(y-x)(y-z) +z^{2k}(z-x)(z-y)$

Quick proof by Peter Mueller: If there are two or three negative values, negate all three, so that now there is at most one negative value. Name them in decreasing order, we have $$ x \geq y \geq z \; , \; \; \; y \geq 0 $$ so that $|x| \geq |y| $ and $x^{2k} \geq y^{2k}.$ Define the Schur expression
$$ F = x^{2k}(x-y)(x-z) + y^{2k}(y-x)(y-z) +z^{2k}(z-x)(z-y)$$ We see that $F \geq 0$ from the rearranged $$ \color{teal}{ F = (x-y) \left( x^{2k}(x-y) + (y-z)(x^{2k} - y^{2k}) \right) + z^{2k}(x-z)(y-z) } $$

The best thing for me would be information about how I can download software that can handle SOS problems. But for the first specific pair of problems: the conjecture is that $$ ( x^4 (x-y)(x-z) + y^4 (y-z)(y-x) + z^4(z-y)(z-x) )$$ is NOT a sum of squares of (real coefficient) polynomials. The second conjecture is that $$(x^2 + y^2 + z^2) ( x^4 (x-y)(x-z) + y^4(y-z)(y-x) + z^4(z-y)(z-x) )$$ really is a sum of squares. That is the specific question: somehow verify both sides of the conjecture. I have been trying with the second half, no luck so far.

It is a result of Claus Scheiderer that there is a non-negative integer $N$ such that $$(x^2 + y^2 + z^2)^N ( x^4 (x-y)(x-z) + y^4(y-z)(y-x) + z^4(z-y)(z-x) )$$ is a sum of squares of real polynomials. Actually the result applies to any positive semidefinite polynomial in three real variables. He has a recent book. ADDED: The eBook is on sale, I actually managed to buy it!

So that is the QUESTION: show the original form is not SOS (or quote relevant software), give an expression for the multiplied expression as SOS, and tell me what software I can download.

Answer 1

As indicated in a comment, I have used this Python wrapper of Chenyang Yuan.

The setup is as follows:

import sympy as sp
import numpy as np
from SumOfSquares import SOSProblem, poly_opt_prob
x,y,z = sp.symbols('x y z')
poly = x4(x-y)(x-z) + y4(y-z)(y-x) + z*4  (z-y)*(z-x)
prob1 = SOSProblem()
const1 = prob1.add_sos_constraint(poly, [x,y,z])
prob2 = SOSProblem()

const2 = prob.add_sos_constraint(poly(x2+y2+z*2), [x,y,z])

When I run prob1.solve(), I receive a picos.modeling.problem.SolutionFailure, which indicates that this the polynomial $$P(x,y,z):=x^4(x-y)(x-z)+y^4(y-x)(y-z)+z^4(z-x)(z-y)$$ has no sum-of-squares representation. I do not know how to turn this into a human-checkable proof of this fact.

When I run prob2.solve(), no such error occurs, and I can use const.get_sos_decomp() to extract a sum-of-squares decomposition for $(x^2+y^2+z^2)P$. It's not pretty. However, I was able to turn it into a nicer form as follows:

1. I spotted the term 0.619*(1.0*x**4 - 0.999*x**3*y - x**3*z + 1.0*x**2*y*z)**2. Up to rounding error, this is a constant times $x^4(x-y)^2(x-z)^2$. So, I figured out the largest constant $c$ for which $$(x^2+y^2+z^2)P-c\sum_{cyc}x^4(x-y)^2(x-z)^2$$ is a sum of squares. (The packages will do this for you if you set up the problem right, but I found it easier to just binary search.) This turns out to be $2/3$. Let $Q$ be the remainder.

2. When I computed a sum-of-squares decomposition of $Q$, the only expression I found which looked like a rational polynomial with small denominators was (0.5*x**4 + 1.0*x**3*y - 0.5*x**3*z - 1.0*x**2*y*z - x*y**3 + 1.0*x*y**2*z - 0.5*y**4 + 0.5*y**3*z)**2. So, I took the symmetric sum $S$ of this and found the largest constant $c'$ for which $Q-c'S$ is a sum of squares. Let $R$ be the remainder.

3. Finally, the sum of squares decomposition of $R$ was quite nice.

So, in the end, I got the following human-readable sum-of-squares decomposition: \begin{align*} (x^2+y^2+z^2)P &=\frac23\sum_{\mathrm{cyc}}\left(x^2(x-y)(x-z)\right)^2\\ &\ \ +\frac19\sum_{\mathrm{cyc}}(y-z)^2\big((y+z)^3-x(y^2+3yz+z^2)\big)^2\\ &\ \ +\frac1{18}\sum_{\mathrm{cyc}}(y-z)^2\big(x^3-3x^2(y+z)+x(y+z)^2+(y+z)(y^2+z^2)\big)^2. \end{align*}

Answer 2

The Macaulay2 package SumsOfSquares$^\color{magenta}{\dagger}$ can handle sums of squares (SOS). Using the web interface:

Macaulay2, version 1.24.11-1695-gf35df1017f (vanilla)
Macaulay2, version 1.24.11-1695-gf35df1017f (vanilla)
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, Isomorphism, LLLBases, MinimalPrimes, OnlineLookup, PackageCitations, Polyhedra, PrimaryDecomposition, ReesAlgebra, Saturation, TangentCone, Truncations, Varieties
i1 : needsPackage( "SumsOfSquares" );
-- storing configuration for package NumericalAlgebraicGeometry in /home/m2user/.Macaulay2/init-NumericalAlgebraicGeometry.m2
-- storing configuration for package Bertini in /home/m2user/.Macaulay2/init-Bertini.m2
-- storing configuration for package SemidefiniteProgramming in /home/m2user/.Macaulay2/init-SemidefiniteProgramming.m2
i2 : R = QQ[x,y,z];
i3 : p = x^4 * (x-y)(x-z) + y^4  (y-z) * (y-x) + z^4 * (z-y) * (z-x);
i4 : sosPoly solveSOS p

Since nothing was output after input i4, I assume that no SOS decomposition was found. Multiplying $p$ by $x^2 + y^2 + z^2$ and trying again,

i5 : f = p * (x^2 + y^2 + z^2);
i6 : sosPoly solveSOS f

Since the output o6 is hard to read on the web interface, one can use tex o6 to generate the $\TeX$ for output o6:

$$\texttt{SOSPoly}\left\{\texttt{coefficients}\ \Rightarrow \ \left\{{2.27005}\cdot 10^{-9},\,{4.98058}\cdot 10^{-9},\,{4.98064}\cdot 10^{-9},\,{2.81877}\cdot 10^{-8},\,{.000202688},\,{.000202688},\,{.000699075},\,{.966712},\,{.966738},\,{1.40704},\,{2.32435},\,{2.32438},\,{4.33843},\,{12.5545},\,{12.5545}\right\},\,\texttt{generators}\ \Rightarrow \ \left\{-{.45095}\,x^{4}-{.20703}\,x^{3}y-{.207093}\,x^{2}y^{2}-{.20703}\,x\,y^{3}-{.450957}\,y^{4}-{.207031}\,x^{3}z+{.0368588}\,x^{2}y\,z+{.0368588}\,x\,y^{2}z-{.207039}\,y^{3}z-{.207094}\,x^{2}z^{2}+{.0368588}\,x\,y\,z^{2}-{.207101}\,y^{2}z^{2}-{.207031}\,x\,z^{3}-{.207039}\,y\,z^{3}-{.450959}\,z^{4},\,{.018992}\,x^{4}+{.362657}\,x^{3}y+{.362636}\,x^{2}y^{2}+{.362685}\,x\,y^{3}+{.343697}\,y^{4}-{.343663}\,x^{3}z-{.00000102535}\,x^{2}y\,z-{.0000186032}\,x\,y^{2}z-{.0189588}\,y^{3}z-{.343646}\,x^{2}z^{2}+{.0000196385}\,x\,y\,z^{2}-{.0189874}\,y^{2}z^{2}-{.343696}\,x\,z^{3}-{.0190197}\,y\,z^{3}-{.362685}\,z^{4},\,-{.407835}\,x^{4}-{.18749}\,x^{3}y-{.187445}\,x^{2}y^{2}-{.187436}\,x\,y^{3}+{.220355}\,y^{4}-{.22038}\,x^{3}z+{.000022099}\,x^{2}y\,z-{.0000119085}\,x\,y^{2}z+{.40781}\,y^{3}z-{.220333}\,x^{2}z^{2}-{.0000101284}\,x\,y\,z^{2}+{.407769}\,y^{2}z^{2}-{.220329}\,x\,z^{3}+{.407808}\,y\,z^{3}+{.187463}\,z^{4},\,-{.17309}\,x^{4}+{.154303}\,x^{3}y+{.154118}\,x^{2}y^{2}+{.154303}\,x\,y^{3}-{.173089}\,y^{4}+{.154303}\,x^{3}z+{.481666}\,x^{2}y\,z+{.481666}\,x\,y^{2}z+{.154303}\,y^{3}z+{.154119}\,x^{2}z^{2}+{.481666}\,x\,y\,z^{2}+{.154119}\,y^{2}z^{2}+{.154303}\,x\,z^{3}+{.154303}\,y\,z^{3}-{.173089}\,z^{4},\,{.017071}\,x^{4}-{.274117}\,x^{3}y+{.114089}\,x^{2}y^{2}+{.0458084}\,x\,y^{3}+{.0971204}\,y^{4}+{.35939}\,x^{3}z+{.0682285}\,x^{2}y\,z+{.388166}\,x\,y^{2}z+{.439329}\,y^{3}z-{.0970331}\,x^{2}z^{2}-{.456394}\,x\,y\,z^{2}-{.0170557}\,y^{2}z^{2}-{.165212}\,x\,z^{3}-{.405198}\,y\,z^{3}-{.114191}\,z^{4},\,{.122001}\,x^{4}+{.349032}\,x^{3}y+{.0461748}\,x^{2}y^{2}-{.441435}\,x\,y^{3}-{.0757843}\,y^{4}+{.260389}\,x^{3}z+{.487607}\,x^{2}y\,z-{.302891}\,x\,y^{2}z+{.0628758}\,y^{3}z+{.075716}\,x^{2}z^{2}-{.184716}\,x\,y\,z^{2}-{.121891}\,y^{2}z^{2}-{.411908}\,x\,z^{3}+{.181046}\,y\,z^{3}-{.0462165}\,z^{4},\,{.129086}\,x^{4}+{.129174}\,x^{3}y-{.516422}\,x^{2}y^{2}+{.129175}\,x\,y^{3}+{.129086}\,y^{4}+{.129174}\,x^{3}z+{.128864}\,x^{2}y\,z+{.128865}\,x\,y^{2}z+{.129174}\,y^{3}z-{.516422}\,x^{2}z^{2}+{.128865}\,x\,y\,z^{2}-{.516422}\,y^{2}z^{2}+{.129175}\,x\,z^{3}+{.129174}\,y\,z^{3}+{.129086}\,z^{4},\,-{.455461}\,x^{4}+{.307095}\,x^{3}y+{.0581603}\,x^{2}y^{2}+{.277735}\,x\,y^{3}-{.187495}\,y^{4}-{.0250904}\,x^{3}z+{.132422}\,x^{2}y\,z+{.0545113}\,x\,y^{2}z-{.161645}\,y^{3}z-{.016961}\,x^{2}z^{2}-{.186932}\,x\,y\,z^{2}-{.0411993}\,y^{2}z^{2}-{.145455}\,x\,z^{3}-{.252643}\,y\,z^{3}+{.642958}\,z^{4},\,{.479473}\,x^{4}+{.0093414}\,x^{3}y+{.0139964}\,x^{2}y^{2}+{.131362}\,x\,y^{3}-{.634165}\,y^{4}-{.306219}\,x^{3}z-{.139391}\,x^{2}y\,z+{.18439}\,x\,y^{2}z+{.261269}\,y^{3}z-{.0573736}\,x^{2}z^{2}-{.0449692}\,x\,y\,z^{2}+{.0433772}\,y^{2}z^{2}-{.270635}\,x\,z^{3}+{.174823}\,y\,z^{3}+{.154721}\,z^{4},\,{.288685}\,x^{4}-{.288646}\,x^{3}y-{.000168526}\,x^{2}y^{2}-{.288634}\,x\,y^{3}+{.288727}\,y^{4}-{.288634}\,x^{3}z+{.288748}\,x^{2}y\,z+{.288725}\,x\,y^{2}z-{.288637}\,y^{3}z-{.000171322}\,x^{2}z^{2}+{.288725}\,x\,y\,z^{2}-{.000198927}\,y^{2}z^{2}-{.288615}\,x\,z^{3}-{.288626}\,y\,z^{3}+{.288699}\,z^{4},\,-{.0463119}\,x^{4}-{.180577}\,x^{3}y+{.54128}\,x^{2}y^{2}-{.245933}\,x\,y^{3}-{.0683413}\,y^{4}-{.111555}\,x^{3}z+{.119838}\,x^{2}y\,z+{.176867}\,x\,y^{2}z-{.185216}\,y^{3}z-{.322653}\,x^{2}z^{2}-{.296723}\,x\,y\,z^{2}-{.218627}\,y^{2}z^{2}+{.365816}\,x\,z^{3}+{.357501}\,y\,z^{3}+{.114635}\,z^{4},\,{.105646}\,x^{4}+{.318124}\,x^{3}y-{.0600633}\,x^{2}y^{2}-{.270814}\,x\,y^{3}-{.0929063}\,y^{4}+{.34839}\,x^{3}z-{.273411}\,x^{2}y\,z+{.240505}\,x\,y^{2}z-{.315477}\,y^{3}z-{.438733}\,x^{2}z^{2}+{.0329303}\,x\,y\,z^{2}+{.498797}\,y^{2}z^{2}-{.00268313}\,x\,z^{3}-{.0775897}\,y\,z^{3}-{.0127151}\,z^{4},\,-{.00000235896}\,x^{4}+{.408254}\,x^{3}y-{.00000263972}\,x^{2}y^{2}-{.408251}\,x\,y^{3}+{.00000221554}\,y^{4}-{.408244}\,x^{3}z-{.00000393591}\,x^{2}y\,z+{.00000332025}\,x\,y^{2}z+{.408244}\,y^{3}z-{.00000105854}\,x^{2}z^{2}+{.00000155861}\,x\,y\,z^{2}+{.00000369752}\,y^{2}z^{2}+{.408246}\,x\,z^{3}-{.408251}\,y\,z^{3}+{.0000010862}\,z^{4},\,{.165248}\,x^{4}-{.0322529}\,x^{3}y-{.358917}\,x^{2}y^{2}+{.236807}\,x\,y^{3}-{.0108709}\,y^{4}-{.236949}\,x^{3}z+{.488121}\,x^{2}y\,z-{.0321114}\,x\,y^{2}z-{.219092}\,y^{3}z-{.0252692}\,x^{2}z^{2}-{.456012}\,x\,y\,z^{2}+{.384187}\,y^{2}z^{2}+{.251351}\,x\,z^{3}+{.000141831}\,y\,z^{3}-{.15438}\,z^{4},\,-{.0828541}\,x^{4}+{.271614}\,x^{3}y-{.236405}\,x^{2}y^{2}-{.136879}\,x\,y^{3}+{.184534}\,y^{4}-{.136636}\,x^{3}z-{.244742}\,x^{2}y\,z+{.545095}\,x\,y^{2}z-{.163738}\,y^{3}z+{.429033}\,x^{2}z^{2}-{.300354}\,x\,y\,z^{2}-{.192628}\,y^{2}z^{2}-{.107878}\,x\,z^{3}+{.273521}\,y\,z^{3}-{.101681}\,z^{4}\right\},\,\texttt{ring}\ \Rightarrow \ {\mathbb R}_{53}\mathopen{}\left[x\,{.}{.}\,z\right]\right\}$$

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A few questions on Mathematics SE that I have answered using SumsOfSquares:

• Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms

• How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?

• Turkevicius inequality

It seems that there is an "equivalent" package for Julia. There is also Parrilo et al.'s SOSTOOLS for Matlab, which is over two decades old.

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_{$\color{magenta}{\dagger}$ Diego Cifuentes, Thomas Kahle, Pablo Parrilo, Sums of squares in Macaulay2, Journal of Software for Algebra and Geometry, Volume 10, 2020.}