An efficient sum of squares nonnegativity certificate for quaternary quartic

1. Problem Statement

The paper addresses the classical problem of certifying the global nonnegativity of polynomials. Specifically, it focuses on quaternary quartics (homogeneous polynomials of degree 4 in 4 variables, denoted $f \in \mathbb{R}[x_1, x_2, x_3, x_4]$).

• Context: Hilbert (1888) proved that not all nonnegative polynomials are sums of squares (s.o.s.) of polynomials, except in specific low-dimensional or low-degree cases.
• The Gap: For nonnegative polynomials that are not s.o.s., a theorem by Hilbert (1893) guarantees that they can be written as a sum of squares of rational functions. This implies the existence of a "multiplier" polynomial $q$ such that $qf$ is an s.o.s.
• The Specific Question: For a nonnegative quaternary quartic ($n=4, d=4$), does there always exist a single quadratic multiplier ($q \in \Sigma_{4,2}$) such that $qf$ is a sum of squares?
  • Previous work suggested that for ternary sextics ($n=3, d=3$), a quadratic multiplier suffices.
  • The author investigates whether this extends to quaternary quartics and whether the bound on the degree of the multiplier is sharp (i.e., can we always find one quadratic multiplier, or do we sometimes need two?).

2. Methodology

The paper employs a combination of algebraic geometry, polynomial optimization, and computational algebra (using Macaulay2).

A. Reduction to a Real Zero Case

The author first establishes that without loss of generality, one can assume the polynomial $f$ has a real zero.

• If $f > 0$, a coordinate transformation allows $f$ to be written as $u^4 + g(u, x)$ where $g \geq 0$.
• If $f$ has a zero, an orthogonal transformation moves the zero to $(1:0:0:0)$, allowing $f$ to be expressed as a quadratic in one variable $u$:
  $$f = au^2 + 2bu + c$$
  where $a, b, c$ are ternary forms (polynomials in 3 variables) of degrees 2, 3, and 4, respectively.

B. The Discriminant Approach

The nonnegativity of the quadratic form $f = au^2 + 2bu + c$ implies that its discriminant is nonnegative:
$$\Delta = ac - b^2 \geq 0$$
Here, $\Delta$ is a ternary sextic (degree 6 in 3 variables).

• The author invokes Hilbert's 1893 result for ternary sextics: For any nonnegative ternary sextic $\Delta$, there exists a quadratic form $q' \in \Sigma_{3,2}$ such that $q'\Delta$ is a sum of squares.
• By "completing the square," the author derives the identity:
  $$a f = (au + b)^2 + \Delta$$
  Multiplying by a suitable $q'$, we get $q' a f = q'(au+b)^2 + q'\Delta$. Since both terms on the RHS are s.o.s., the product $q' a f$ is an s.o.s.
• Conclusion of Method: This proves that two quadratic multipliers ($q=a$ and $q'$) always suffice to make $f$ an s.o.s.

C. Sharpness and Counter-Example Construction

To determine if one multiplier suffices (specifically, if $af$ is always s.o.s.), the author investigates whether the discriminant $\Delta = ac - b^2$ can be nonnegative but not a sum of squares ($\Delta \in P_{3,6} \setminus \Sigma_{3,6}$).

• If $\Delta$ is always an s.o.s., then $af$ would be an s.o.s. directly.
• The author constructs explicit examples where $\Delta$ is nonnegative but not an s.o.s., using the boundary of the cone of sums of squares ($\partial \Sigma_{3,6}$).
• Construction Strategy:
  1. Start with a known non-s.o.s. ternary sextic (Robinson's polynomial $R_1$).
  2. Perturb it to find an element $p$ on the boundary of the s.o.s. cone that is strictly positive but not s.o.s.
  3. Show that this $p$ can be represented in the form $ac - b^2$ (i.e., $p \in D_-$, the variety of discriminants).
  4. This yields a specific $f = au^2 + 2bu + c$ where $\Delta$ is non-s.o.s.

3. Key Contributions and Results

Theorem 2 (Main Existence Result)

For any nonnegative quaternary quartic $f \in P_{4,4}$, there exist two quadratic forms $q, q' \in \Sigma_{4,2}$ such that:
$$q q' f \in \Sigma_{4,8}$$
This confirms that two quadratic multipliers are always sufficient.

Theorem 3 (Existence of Non-S.O.S. Discriminants)

The author explicitly constructs a ternary sextic $\Delta = ac - b^2$ that is nonnegative but not a sum of squares.

• Parameters: $a = x^2 + y^2 + z^2$, $b = x(y^2 - z^2)$, and a specific $c$ with a parameter $\tau$.
• Result: For a specific range of $\tau$ (specifically $4.6875 < \tau \leq \tau^* \approx 4.922$), $\Delta_\tau$ is nonnegative but not an s.o.s.
• Significance: This proves that the discriminant of a nonnegative quaternary quartic is not necessarily an s.o.s., meaning the "straightforward" decomposition via $af$ is not guaranteed to work immediately without further multipliers.

Theorem 4 (Relationship between $f$ and $\Delta$)

The paper establishes a precise algebraic link between the s.o.s. decomposability of $f$ and its discriminant $\Delta$:

• If there exists a quadratic multiplier $q$ such that $qf$ is s.o.s., then $aq\Delta$ must be an s.o.s. (specifically, a sum of squares of forms in the ideal $(a, b)$).
• Conversely, if $a\Delta$ is an s.o.s., then $f$ is an s.o.s. (for the case $j=0$).
• Conjecture: The author conjectures that one multiplier ($q=a$) is always sufficient (i.e., $af$ is always s.o.s.), even though $\Delta$ itself might not be. The examples constructed show $\Delta$ is non-s.o.s., but in all tested cases, $af$ was found to be an s.o.s. (verified numerically).

4. Significance and Implications

1. Refinement of Hilbert's Theorem: The paper provides an "almost sharp" bound for the degree of multipliers needed for quaternary quartics. While Hilbert's general bound for rational functions is a tower of exponents, this work shows that for this specific case, the degree is very low (quadratic).
2. Clarification of the "One vs. Two" Multiplier Problem:
   • The paper proves that two multipliers are definitely sufficient.
   • It leaves open the question of whether one is sufficient.
   • Crucially, it demonstrates that the failure of $\Delta$ to be an s.o.s. does not imply the failure of $af$ to be an s.o.s. The "obstruction" in the discriminant can be resolved by the factor $a$.
3. Computational Algebra: The paper provides explicit code (Macaulay2) and numerical examples to verify these theoretical bounds, bridging the gap between abstract algebraic geometry and computational verification.
4. Geometric Insight: The construction of the non-s.o.s. discriminant relies on the geometry of the variety $D_-$ (discriminants of nonnegative forms) intersecting the boundary of the s.o.s. cone, offering new insights into the structure of these polynomial cones.

Summary Conclusion

Dmitrii Pasechnik proves that any nonnegative quaternary quartic can be made a sum of squares by multiplying it by two quadratic forms. While it remains an open conjecture whether a single quadratic multiplier ($q=a$) is always sufficient, the paper constructs explicit examples where the discriminant is not a sum of squares, yet the product $af$ appears to be. This suggests that the standard decomposition via the discriminant is not the optimal path, and the relationship between the polynomial and its discriminant is more subtle than previously assumed.