{"id":"e925af09-dad3-489c-823a-bb071080f957","arxiv_id":"2506.17260","paper_version":4,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":6.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":1,"one_line_summary":"Any 2x2 PSD biquadratic polynomial is a sum of three squares of quadratic polynomials, via a variable-reduction technique and constructive decompositions.","lead":"The paper shows that every 2-by-2 biquadratic polynomial that is never negative can be written as the sum of three squares of quadratic polynomials. This drops the previous bound of nine squares and connects biquadratic forms to Hilbert's 1888 theorem on ternary quartics.","discovery_kind":"new_method","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Theorem 2.1's SOS equivalence is asserted, not proved; lifting h's SOS to a biquadratic SOS needs a Newton-polytope support argument, without which the 2x2 conclusion does not follow.","rationale":"The reader's weakest assumption identifies the same soft spot: the paper does not justify that an SOS of the reduced tripartite polynomial lifts to an SOS of the original biquadratic form. I checked whether the claim is actually false; it is not, because Reznick's Newton-polytope support theorem closes the gap and forces every square factor to be affine-bilinear, hence liftable. This means the concern is a proof gap rather than a counterexample, but it is genuinely load-bearing because Hilbert's ternary quartic theorem only applies to h, not directly to f. The Section 4 numerical discrepancies, e.g. the threshold in Theorem 4.8 versus Example 4.9, are real but they concern the constructive part, not the existence theorem. Since the main theorem is credible but not fully proved as written, CONDITIONAL remains the appropriate verdict: accept after the omitted Newton-polytope argument is supplied and the Section 4 formulas are corrected.","tokens_in":16408,"tokens_out":20501,"duration_ms":251515,"concrete_test":"Supply the missing step analytically: take an arbitrary SOS g = Σ p_t^2 with supp(g) ⊆ {(α,β) : |α| ≤ 2, |β| ≤ 2}; invoke Reznick's theorem to conclude supp(p_t) ⊆ ½ New(g), so no p_t contains a monomial with |α| = 2 or |β| = 2. Define q_t(x_hat, y_hat) by replacing each monomial x^α y^β with x^α x_m^{2−|α|} y^β y_n^{2−|β|}; verify q_t is bilinear and Σ q_t^2 dehomogenizes to g, hence equals the original f by multihomogeneity. If this derivation fails for any monomial with α_i = 2 or β_j = 2, the claimed equivalence is false; if it goes through, Theorem 2.1 holds.","verdict_should_be":"CONDITIONAL","load_bearing_attack":"The central claim Theorem 2.1 is built on the chain f PSD/SOS iff g PSD/SOS iff h PSD/SOS, stated in the proof after setting x_m = y_n = 1 and replacing x_m, y_n by z. PSD and the easy directions are fine. The load-bearing unproved step is: g SOS implies f SOS (equivalently h SOS implies f SOS). If h = Σ r_i(x,y,z)^2 with r_i a quadratic form, one cannot automatically write f = Σ q_i(x_hat,y_hat)^2 with q_i bilinear, because r_i may contain monomials x_i x_j or y_k y_l that are not in the image of the substitution x_m = y_n = z. The paper simply asserts 'g is SOS if and only if f is SOS'. The missing fact is Reznick's Newton-polytope lemma, cited later as [11]: every SOS factor of g or h has support in half the Newton polytope. Since New(g) has x-degree ≤ 2 and y-degree ≤ 2, each factor has x-degree ≤ 1 and y-degree ≤ 1, hence is an affine-bilinear polynomial and lifts uniquely to a bilinear form q_i; equality of dehomogenizations then forces Σ q_i^2 = f. This is repairable, but as written Theorem 2.1 and the 2x2 claim do not follow from Hilbert's theorem without it.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper studies the relationship between positive semidefinite (PSD) biquadratic polynomials and sum-of-squares (SOS) decompositions. The main theorem (Thm 2.1) claims that an m×n biquadratic f is PSD (resp. SOS) if and only if an associated tripartite homogeneous quartic h in m+n−1 variables is PSD (resp. SOS); for m=n=2 this gives, via Hilbert's theorem on ternary quartics, that every PSD 2×2 biquadratic is an SOS of three quadratic polynomials, improving Calderón's nine-square certificate. Theorem 3.1 reformulates SOS of a PSD biquadratic as existence of a PSD matrix M(Γ) in a gauge family, with SOS rank at most mn; Theorem 4.8 and the surrounding propositions provide explicit SOS constructions for three restricted families of 2×2 PSD biquadratics.","tokens_in":16701,"tokens_out":25601,"duration_ms":258286,"significance":"If the main theorem is fully established, the paper gives a clean and elegant route to a known but nontrivial result: every 2×2 PSD biquadratic is SOS, with the sharp number three of squares. The gauge-matrix characterization in Theorem 3.1 is a useful reformulation with a finite-dimensional certificate, and the constructive cases in Section 4 provide concrete decompositions. The use of Reznick's Newton-polytope lemma, though only cited, is appropriate. However, the central SOS lifting step is not proved in the manuscript, and there are local errors in Theorem 3.1 and Theorem 4.8, so the present version is not yet publication-ready.","major_comments":[{"comment":"The proof asserts 'f is SOS if and only if g is SOS' after setting x_m=y_n=1 and then 'g is SOS if and only if h is SOS' after homogenizing. The only nontrivial direction is lifting an SOS representation of g (or h) to an SOS representation of f as squares of bilinear forms. This requires the Newton-polytope support lemma of Reznick [11]: since every variable has exponent at most two in New(f) (resp. New(g)), half the Newton polytope forces each square factor to have exponent at most one in each x- and y-variable. Without this argument, the chain h SOS ⇒ g SOS ⇒ f SOS does not follow, and Hilbert's theorem cannot be applied. The fix is straightforward, but as written Theorem 2.1 and the paper's main claim are not proved.","section":"Theorem 2.1 and 2.2 (proof of SOS equivalence)"},{"comment":"The proof states that Cholesky decomposition M(Γ)=CC^T gives f=Σ(c_t^T z)^2 with orthogonal coefficients. For C with columns c_t, M=CC^T=Σ c_t c_t^T, and the c_t are not orthogonal in general. Orthogonality is instead obtained from an eigendecomposition M=QΛQ^T, taking coefficient vectors √λ_t q_t. The statement of (i) is true, but the supplied proof is incorrect.","section":"Theorem 3.1, proof of (ii)⇒(i)"},{"comment":"In the subcase cx=cy and c_y^2 ≤ a11 ≤ 5/4 c_y^2, the proof of Case (i2) derives the threshold a22 ≥ 1/(a11−c_y^2)=1/(a11−c_x^2), but the displayed formula in condition (c) for this subcase contains an additional term −4/(3c_y−√(9c_y^2−4a11))^2. This extra term is not justified and contradicts the proof; Example 4.9 uses the erroneous numerical value 4.8431 instead of 5. Please correct the theorem and the example.","section":"Theorem 4.8, condition (c), second case"}],"minor_comments":[{"comment":"The title misspells 'Positive' as 'Postive'.","section":"Title"},{"comment":"The abstract credits the first explicit PNS quartic in four variables to Choi (1975) [3], while the introduction credits Choi and Lam (1977) [4]; these statements should be harmonized.","section":"Abstract and Section 1"},{"comment":"The tensor symmetry is stated as a_{ijkl}=a_{klij}, while the polynomial representation (2) also requires symmetry under i↔j and k↔l; please state the full symmetry or explain the convention.","section":"Section 3, definition of B"},{"comment":"Expressions such as √a11a22 and √a12a21 are ambiguous; use √(a11a22) and √(a12a21).","section":"Lemmas 4.4 and 4.5, Theorem 4.5"},{"comment":"The displayed γ3* is written as c_y/(α*)^4 −1/(α*)^2; the preceding equations give c_y/(α*)^3 −1/(α*)^2, which matches the denominator (3c_y−√(9c_y^2−4a11))^3 in the theorem statement.","section":"Theorem 4.8, proof of Case (i1)"},{"comment":"The phrase 'the 2 × 2 bipartite polynomial' should be 'the 2 × 2 biquadratic polynomial'; also, the sentence 'Some final remarks and open questions are made and raised' is awkward and should be rewritten.","section":"Section 4, after Proposition 4.2"},{"comment":"The PSD equivalence after setting x_m=y_n=1 should be justified explicitly using the two-variable homogeneity f(λx,μy)=λ^2μ^2f(x,y), including continuity at points where the normalization variables vanish.","section":"Theorem 2.1 proof"}],"recommendation":"major_revision","confidential_remarks":"The paper is likely correct in its main mathematical direction, but the current manuscript has a significant proof gap in the central theorem and several internal inconsistencies. I would encourage revision rather than rejection, as the missing Newton-polytope argument is already cited later in the paper and is easily supplied. The novel contribution is moderate: the reduction to tripartite quartics is elegant, but much of Section 4 is a case-by-case construction covering only part of the 2×2 landscape."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Dear Colleague,\n\nThe headline is that this paper has a real new idea—the tripartite reduction of an m×n biquadratic form to a homogeneous quartic in m+n−1 variables—and it uses that reduction with Hilbert's theorem to claim that every 2×2 PSD biquadratic form is a sum of three squares of quadratics, improving Calderon's nine-square certificate. I believe the result is likely true, but the written proof leaves a load-bearing gap, and Section 4 has inconsistencies.\n\nWhat's good: Theorem 3.1 gives a clean characterization: a PSD biquadratic is SOS iff some matrix M(Γ) in the affine Gram-matrix set is PSD, and then the SOS rank is at most mn. The proof of that theorem is standard but correct. The main idea—reducing to the ternary quartic case where Hilbert's theorem applies—is elegant, and the 3-square conclusion is a genuine improvement over Calderon.\n\nThe soft spot is Theorem 2.1. The equivalence 'f is PSD iff g is PSD iff h is PSD' is fine. The SOS direction is not. When you dehomogenize and then homogenize with a single z, an SOS of h may have square factors that are not in the image of the substitution x_m = y_n = z. A factor containing x_i x_j or y_k y_l would produce monomials in its square that don't occur in h, so such factors can't appear; but this must be argued. The missing tool is Reznick's Newton-polytope lemma, which the paper actually cites later. With it, every square factor's support lies in half the Newton polytope, so each factor is affine-bilinear in (x,y,z) and lifts uniquely to a bilinear form. As written, the proof simply asserts 'f is SOS iff h is SOS' and the 2×2 result does not follow from Hilbert's theorem without this step. It's repairable, but it needs to be written down.\n\nSection 4 also needs care. The threshold formulas in Theorem 4.8 don't match the expressions derived in the proof; the case with c_x = c_y and a11 between c_y^2 and 5/4 c_y^2 derives γ*_3 = 1/(a11 − c_y^2) but the statement and Example 4.9 use a different, more elaborate expression. The numerical value 4.8431 in the example is inconsistent with the displayed formula. These are fixable mistakes, but they make the constructive part hard to trust as it stands.\n\nWho should read this: anyone working on nonnegative biquadratic forms or low-dimensional SOS theory. The main claim is new and likely correct, and the gap is fillable. I would not cite it in its current form, but I would send it to a serious referee and ask the authors to supply the lifting proof and correct Section 4. If those are done, the paper will be a solid contribution.","headline":"New 3-square certificate for 2x2 PSD biquadratic forms via a tripartite reduction; the proof has a repairable gap and Section 4 needs corrections.","tokens_in":17252,"tokens_out":8718,"would_cite":false,"duration_ms":89057,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11E25","12D15","14P10","15A69","90C23"],"pacs":[],"model":"deepseek-v4-flash","headline":"This paper claims that every 2×2 PSD biquadratic polynomial—a homogeneous quartic in four variables that is nonnegative everywhere—is a sum of squares of three quadratic polynomials, with the proof running through a reduction to a…","keywords":["biquadratic polynomials","sum of squares","positive semidefinite","tripartite quartic polynomials","SOS rank","ternary quartics","Newton polytope","biquadratic tensors"],"falsifier":"For a dense set of $2\\times2$ PSD biquadratic polynomials with coefficients satisfying the paper's inequalities (10), solve the semidefinite program that minimizes the rank of $M(\\Gamma)$ subject to $M(\\Gamma)\\succeq0$; if any such polynomial has minimum rank 4, Theorem 2.1's three-square conclusion is false, because a three-square certificate would give rank at most 3.","tokens_in":16210,"feed_emoji":"🧮","tokens_out":12170,"duration_ms":121273,"temperature":0.7,"pith_summary":"This paper's central claim is that every $2\\times 2$ biquadratic polynomial—a homogeneous quartic in four variables, two $x$-variables and two $y$-variables, of the form $f(x,y)=\\sum a_{ijkl}x_ix_jy_ky_l$—that is nonnegative on $\\mathbb{R}^4$ can be written as a sum of squares of three quadratic polynomials. The route is a reduction: any $m\\times n$ biquadratic is shown to be equivalent, for both nonnegativity and sum-of-squares status, to a tripartite quartic in $m+n-1$ variables, and for $m=n=2$ that reduced object is exactly a ternary quartic, where a classical 1888 theorem guarantees three squares. If this is right, the known certificate size for $2\\times2$ biquadratics drops from nine squares to three. The paper also gives a necessary and sufficient condition for a general $m\\times n$ PSD biquadratic to be SOS, with SOS rank at most $mn$, and constructs explicit three-square decompositions in three $2\\times2$ coefficient regimes.","feed_headline":"Every 2×2 nonnegative biquadratic has a 3-square certificate","feed_subtitle":"A variable-counting reduction turns these quartics into ternary quartics, where three squares always suffice.","key_machinery":"The central object is the tripartite quartic polynomial $h(x,y,z)=h_0z^4+h_1(x,y)z^3+h_2(x,y)z^2+h_3(x,y)z+h_4(x,y)$, obtained by setting $x_m=y_n=1$ and homogenizing with a single flexible variable $z$. It carries the argument because the paper proves that $f$ is PSD if and only if $h$ is PSD, and $f$ is SOS if and only if $h$ is SOS; the 1888 ternary-quartic theorem then applies exactly when $m=n=2$. The second piece of machinery is the parameterized flattening $M(\\Gamma)=B+P(\\Gamma)$, with $z=x\\otimes y$, where $\\Gamma$ encodes the non-unique split of full-cross terms; existence of $\\Gamma$ with $M(\\Gamma)\\succeq0$ characterizes SOS biquadratics.","core_discovery":"On the paper's own terms, the discovery is that biquadratic structure compensates for having four variables: although a general PSD quartic in four variables need not be SOS, a $2\\times2$ biquadratic always is. The proof places the object in a tripartite quartic $h(x,y,z)=h_0z^4+h_1(x,y)z^3+h_2(x,y)z^2+h_3(x,y)z+h_4(x,y)$ in three variables, preserving PSD and SOS equivalence. The 1888 theorem for ternary quartics then gives three quadratic summands for $h$, and the paper asserts these convert back to three quadratic summands for the original biquadratic. In the general $m\\times n$ setting, the paper characterizes SOS by the existence of a parameter matrix $\\Gamma$ such that the flattened matrix $M(\\Gamma)=B+P(\\Gamma)$ is positive semidefinite, which forces SOS rank at most $mn$ whenever $f$ is SOS.","pith_inferences":["The rank bound $mn$ from the general theorem is likely not tight: for $m=n=2$ it gives 4 while the paper's main theorem gives 3, so a natural extension is to determine the maximal SOS rank for $m\\times2$ and $3\\times3$ biquadratics, which the paper leaves open.","Because the reduction identifies biquadratics with tripartite quartics, a numerical SDP that solves the ternary-quartic SOS problem could serve as a practical SOS test for $2\\times2$ biquadratics without expanding to the full four-variable certificate.","The three explicit cases cover no half-cross terms, one half-cross term, and two neighbor half-cross terms without a full-cross term; the omitted case with two half-cross terms plus a full-cross term appears to be the next testbed for a fully constructive classification of all $2\\times2$ SOS biquadratics."],"forward_implications":["For $2\\times2$ PSD biquadratics, the SOS certificate uses three squares, improving the previous nine-square bound from 1973.","If an $m\\times n$ PSD biquadratic is SOS, it admits an SOS decomposition with at most $mn$ squares whose coefficient vectors can be taken orthogonal; by the paper's equivalence, this amounts to choosing $\\Gamma$ with $M(\\Gamma)\\succeq0$.","The PSD-SOS problem for biquadratics is equivalently a problem about tripartite quartics in one fewer variable, so any algorithmic or theoretical progress on tripartite quartics transfers directly.","When the tripartite quartic is degenerated, meaning $h_0=0$, PSD is equivalent to $h_1\\equiv0$ and three explicit PSD conditions involving $h_2$, $h_4$, and $4h_2h_4-h_3^2$, giving a checkable certificate in that regime.","For the three $2\\times2$ cases treated constructively, the coefficient inequalities given are necessary and sufficient for PSD and simultaneously yield an explicit SOS decomposition."],"supporting_citations":[{"why":"Supplies the 1888 result that every PSD ternary quartic is a sum of three squares, which is applied to the reduced tripartite quartic in Theorem 2.1.","marker":"[6]"},{"why":"Gives the nine-square SOS bound for 2×2 biquadratics that this paper improves, and states the m×2 SOS result that motivates the rank questions.","marker":"[1]"},{"why":"Provides the Newton-polytope theorem used in Theorem 3.1 to force each SOS summand to be bilinear, yielding the SOS rank bound at most mn.","marker":"[11]"},{"why":"Supplies the flattened biquadratic tensor/M-eigenvalue setup and the 2×2 numerical example used in Example 3.2 for the M(Γ) construction.","marker":"[10]"}],"fun_headline_variants":["2×2 PSD biquadratics always need only 3 squares","Three squares suffice for every 2×2 PSD biquadratic","Hilbert's 1888 theorem yields 3-square certificate for 2×2 biquadratics","Calderon's 9 squares reduced to 3 for 2×2 PSD biquadratics","Every 2×2 nonnegative biquadratic is sum of 3 squares"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The argument's weakest point is the unproved assertion that after setting one $x$ and one $y$ variable to 1, a sum-of-squares decomposition of the smaller polynomial can always be lifted back to a sum of squares of bilinear forms in the original variables.","fun_headline_variants_meta":{"raw":{"variants":["2×2 PSD biquadratics always need only 3 squares","Three squares suffice for every 2×2 PSD biquadratic","Hilbert's 1888 theorem yields 3-square certificate for 2×2 biquadratics","Calderon's 9 squares reduced to 3 for 2×2 PSD biquadratics","Every 2×2 nonnegative biquadratic is sum of 3 squares"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000937,"raw_usage":{"total_tokens":4061,"prompt_tokens":1053,"completion_tokens":3008,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":669,"completion_tokens_details":{"reasoning_tokens":2891}},"tokens_in":669,"tokens_out":3008,"duration_ms":25037,"temperature":1.0,"reasoning_tokens":2891,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-07T05:33:02.387360+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"For a dense set of $2\\times2$ PSD biquadratic polynomials with coefficients satisfying the paper's inequalities (10), solve the semidefinite program that minimizes the rank of $M(\\Gamma)$ subject to $M(\\Gamma)\\succeq0$; if any such polynomial has minimum rank 4, Theorem 2.1's three-square conclusion is false, because a three-square certificate would give rank at most 3.","supporting_citations":[{"cited_title":"Fidalgo and A","cited_arxiv_id":null,"evidence_quote":"Supplies the 1888 result that every PSD ternary quartic is a sum of three squares, which is applied to the reduced tripartite quartic in Theorem 2.1."},{"cited_title":"Calder\\' o n, ``A note on biquadratic forms'', Linear Algebra and Its Applications 7 (1973) 175-177","cited_arxiv_id":null,"evidence_quote":"Gives the nine-square SOS bound for 2×2 biquadratics that this paper improves, and states the m×2 SOS result that motivates the rank questions."},{"cited_title":"An elementary proof of Hilbert’s theorem on ternary quartics","cited_arxiv_id":null,"evidence_quote":"Provides the Newton-polytope theorem used in Theorem 3.1 to force each SOS summand to be bilinear, yielding the SOS rank bound at most mn."},{"cited_title":"Motzkin, ``The arithmetic geometry inequality'', in: Proceedings of the Symposium , Wright-Patterson Air Force Base, Ohio, 1965, pp.205-224","cited_arxiv_id":null,"evidence_quote":"Supplies the flattened biquadratic tensor/M-eigenvalue setup and the 2×2 numerical example used in Example 3.2 for the M(Γ) construction."}],"review_version":1}