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I came across two polynomials that Choi and Lam gave in 1976, that are not sum of squares of polynomials, despite being evidently non-negative by AM-GM $$ S(x,y,z) = x^4 y^2 + y^4 z^2 + z^4 x^2 - 3 x^2 y^2 z^2 $$ $$Q(w,x,y,z) = w^4 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 4wxyz $$ The sextic was asked about in 2015 Representing as sum of squares of polynomials From what I can see, proving a polynomial is not sos can be done by hand.

From a 2003 article by Reznick, I know that $$ S_8(x,y,z) = (x^2 + y^2 + z^2) S(x,y,z) $$ really is the sum of squares of (quartic homogeneous) polynomials.

That is question 1, can anybody supply an sos version of $S_8 \; ? \;$

From context, it seems likely that $$ Q_6(w,x,y,z) = (w^2 +x^2 + y^2 + z^2) Q(w,x,y,z) $$ is also the sum of polynomial squares. The context is that Delzell's examples are rather later, I would think some author would point out that the ``bad points'' property holds for this Choi-Lam example, if it were true.

That would be question 2, can anybody supply an sos version of $Q_6 \; ? \;$

I have requested a book edited by Delzell from the library, and ordered a used copy of his (2003?) book Positive Polynomials (with Prestel)

confirming Willie's answer for $Q_8$ by CAS:

? g= 2* x^2* y^2 * (x^2-z^2)^2 +  y^4 * (x^2-z^2)^2 + 2* y^2* z^2 * (y^2-x^2)^2 +  z^4 * (y^2-x^2)^2 + 2* z^2* x^2 * (z^2-y^2)^2 +  x^4 * (z^2-y^2)^2  
%1 = 2*y^2*x^6 + (2*y^4 - 4*z^2*y^2 + 2*z^4)*x^4 + (-4*z^2*y^4 - 4*z^4*y^2 + 2*z^6)*x^2 + (2*z^2*y^6 + 2*z^4*y^4)
? 
? h = ( x^2 + y^2 + z^2) * (x^4*y^2 + y^4*z^2 + z^4 * x^2 - 3 * x^2 * y^2 * z^2 )
%2 = y^2*x^6 + (y^4 - 2*z^2*y^2 + z^4)*x^4 + (-2*z^2*y^4 - 2*z^4*y^2 + z^6)*x^2 + (z^2*y^6 + z^4*y^4)
? 
? 2*h
%3 = 2*y^2*x^6 + (2*y^4 - 4*z^2*y^2 + 2*z^4)*x^4 + (-4*z^2*y^4 - 4*z^4*y^2 + 2*z^6)*x^2 + (2*z^2*y^6 + 2*z^4*y^4)
? g
%4 = 2*y^2*x^6 + (2*y^4 - 4*z^2*y^2 + 2*z^4)*x^4 + (-4*z^2*y^4 - 4*z^4*y^2 + 2*z^6)*x^2 + (2*z^2*y^6 + 2*z^4*y^4)
? 2*h - g
%5 = 0
?

Might as well: Reznick proved that a positive definite polynomial can changed into an sos by multiplying by $(\sum x_i^2)^N.$ Eventually an example of Delzell showed up, for which that failed; in this case the source gave the altered product that does become an SOS.

$$ D = z^8 + w^2 x^4 y^2 + w^2 y^4 z^2 + w^2 z^4 x^2 - 3 w^2 x^2 y^2 z^2 $$

However, $D_{10}=(x^2 + y^2 + z^2)D$ really is an sos of polynomials, the article says eight squares, integer or rational coefficients with small denominatros.

Rodrigo de Azevedo
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Will Jagy
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  • My thought is to attack Question 1 by evaluating the polynomial at integer arguments for some (perhaps all) of the variables. The result must also be a sum of squares, and interpolation can be attempted to reconstruct the multivariate squares. – hardmath Mar 16 '21 at 21:31
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    $Q_6$ would be a bit harder for me, if true. $S$ is a symmetric polynomial, as is $S_8$, so one may expect that the factoring in to sums of squares is also symmetric under cyclic permutations. The same is not true for $Q$. – Willie Wong Mar 16 '21 at 21:55

1 Answers1

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Question 1 can be done by hand pretty simply.

$$ S_8 = (x^2 + y^2 + z^2) (x^4 y^2 + y^4 z^2 + z^4 x^2 - 3 x^2 y^3 z^2) $$

So

$$ = x^6 y^2 + x^4 y^4 - 2 x^4 y^2 z^2 + \text{cyclic cycle indices} $$

Observe that

$$ x^6 y^2 - 2 x^4 y^2 z^2 + x^2 y^2 z^4 = x^2 y^2 (x^2 - z^2)^2 $$

And that

$$ \frac12 z^4 y^4 + \frac12 x^4 z^4 - x^2 y^2 z^4 = \frac12 z^4 (x^2 - y^2)^2 $$

we find

$$ S_8 = x^2 y^2 (x^2 - z^2)^2 + \text{ cyclic } + \frac12 x^4 (y^2 - z^2)^2 + \text{ cyclic } $$

Willie Wong
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  • Wow, you're quick. – Will Jagy Mar 16 '21 at 21:41
  • Sorry, a typo. Needs to be fixed. – Willie Wong Mar 16 '21 at 21:41
  • Okay, this should be fine. You should check it using a CAS to make sure I didn't screw something up. – Willie Wong Mar 16 '21 at 21:46
  • Thanks again. I have requested a Contemporary Mathematics, number 253, edited by Charles N. Delzell ( I keep thinking Denzel Washington) and purchased (ordered) a used book, Positive Polynomials by Prestel and Pretzel . see https://www.math.lsu.edu/~delzell/positive_updates.html Second author is actually Delzell again – Will Jagy Mar 19 '21 at 17:16