{"id":"495af241-cf4f-489d-93d7-19501914a8e1","arxiv_id":"2608.11182","paper_version":1,"verdict":"REJECT","confidence":"MODERATE","novelty_score":7.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"The Kronecker product of any two symmetric persistent tensors is again a symmetric persistent tensor.","lead":"This paper claims that Kronecker products preserve the persistence property for symmetric tensors, which would yield new tensor rank lower bounds. The claim is plausible, but the proof as written jumps from decomposable inputs to all inputs, so the central theorem is not established.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Proposition 10 proves (22) only on decomposable tuples; since the Segre cone is not Zariski dense in (V⊗W)^r, the extension to all tuples — on which Theorem 11 rests — is unjustified.","rationale":"The reader's weakest_assumption identifies exactly the same load-bearing concern: the polarized Hessian determinant identity is established only on decomposable tuples, and the extension to all tuples is invalid because the Segre variety is not Zariski dense. My independent reading of Proposition 10 confirms this. The universal-property construction of P_{f⊠g} on all of (V⊗W)^r is legitimate, but the proof never shows that the left-hand side of (22) equals P_{f⊠g}^{d1d2} off the Segre cone. Since Remark 9 itself notes that determinant identities of this type can fail away from the Segre variety, the gap is not merely cosmetic. The example in the paper is unhelpful for resolving the gap because it falls into the rank-one case where Remark 12 provides a separate argument. The theorem may be true, and the counterexample of Shitov does not bear on the symmetric case, but the submitted proof is incomplete. Therefore the reader's REJECT verdict is appropriate, and my stress-test does not change it. I set verdict_should_be to UNCHANGED because the reader already reached the correct verdict. A single computational test on a non-rank-one persistent pair would settle whether the gap is repairable or fatal to Proposition 10.","tokens_in":12095,"tokens_out":10280,"duration_ms":88533,"concrete_test":"Use a computer algebra system to search for persistent tensors f∈Sym^4 C^3 and g∈Sym^4 C^3 satisfying Theorem 3 with P_f and P_g non-rank-one bilinear forms (if no such pair exists, the theorem reduces to the rank-one case covered by Remark 12, but that must be proved). For any such pair, evaluate both sides of (22) at the non-decomposable tuple U^(1)=U^(2)=e_0⊗e_0+e_1⊗e_1 in C^3⊗C^3, keeping z symbolic. If the difference D(z)=Hess((f⊠g)_{U^(1),U^(2)})(z) - P_{f⊠g}(U^(1),U^(2))^9 is nonzero, Proposition 10 is false and the proof of Theorem 11 collapses; if D is identically zero for all such pairs, the extension step needs a new proof, and the current manuscript must be revised.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The proof of Proposition 10 derives (22) for U^(i)=v^(i)⊗w^(i), i.e., on the product of Segre cones. The definition of P_{f⊠g} on all of (V⊗W)^r via the universal property is valid, but it only defines the right-hand side; it does not extend the equality. Two polynomials that agree on the Segre cone need not agree on the ambient space, because the affine Segre cone is a proper closed subvariety of V⊗W; its ideal is generated by the 2×2 minors, as Remark 9 explicitly notes. The left-hand side of (22) is a polynomial on (V⊗W)^r, and the right-hand side is P_{f⊠g}^{d1d2}; both are multihomogeneous of degree d1d2 in each slot. Agreement on decomposable tuples implies only that their difference lies in the Segre ideal, not that the difference is zero. No further argument — such as showing the difference vanishes modulo that ideal, or that persistence forces the difference to be zero — is supplied. The proof of Theorem 11 is a direct application of Theorem 3 and inherits this gap. The worked Example 13 does not test the gap: there P_f and P_g are rank-one, so Remark 12 gives a separate proof of (22) for all tuples; the general case with non-rank-one P_f or P_g is precisely where the unproved extension is needed. The central claim is therefore not established by the submitted argument.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper defines Kronecker products of symmetric tensors and claims that if f∈Sym^n C^{d1} and g∈Sym^n C^{d2} are persistent, then f⊠g is persistent (Theorem 11). The proof route is: Proposition 7/Corollary 8 establish differentiation identities for Hessian matrices; Proposition 10 claims a polarized perfect-power identity for Hessian determinants; Theorem 11 applies the Hessian characterization of persistence from [GO]. An appendix discusses normalized Hessian spaces for cubic forms.","tokens_in":12357,"tokens_out":13938,"duration_ms":129713,"significance":"The claimed closure result would be significant: it would yield iterated Kronecker products and powers of persistent symmetric tensors, hence new families of tensors with certified lower bounds on tensor rank. The differentiation identities (Prop. 7, Cor. 8) are clean, and the determinant computation on decomposable tuples in Prop. 10 is correct as far as it goes. However, the key extension from decomposable to all direction vectors is not justified, so the central claim is not established by the submitted argument.","major_comments":[{"comment":"The proof verifies Eq. (22) only for tuples U^{(i)}=v^{(i)}⊗w^{(i)}. The sentence \"Performing the Kronecker substitution therefore gives (22)\" assumes that agreement on the product of Segre cones implies equality as polynomials on (V⊗W)^{×r}. This is false: the affine Segre cone is a proper closed subvariety whenever d1,d2≥2, and its ideal is generated by the 2×2 minors, as Remark 9 explicitly notes. Both sides of (22) are multihomogeneous of degree d1d2 in each slot, so the difference lies in the Segre ideal but need not vanish. No argument shows that the difference vanishes modulo that ideal, nor that persistence of f and g forces it to vanish. Thus Proposition 10 is unproved.","section":"§3, Proposition 10, Eq. (22)"},{"comment":"The proofs of Theorem 11 and its corollaries are direct applications of Proposition 10 and therefore inherit the gap. The worked Example 13 does not test the gap: it verifies the identity only for decomposable U^{(1)},U^{(2)}, and in that example P_f and P_g are rank-one, so the special case of Remark 12 applies. The general case with non-rank-one P_f or P_g is precisely where the missing extension is needed.","section":"§3, Theorem 11 and Corollaries 14–15"},{"comment":"The diagonal identity (23) is presented as a consequence of Proposition 10 and is then used to derive (29). Since (22) is not established for non-decomposable U, Eq. (23) is conditional, and so is the derivation of (29) even in the rank-one case. The direct computation in Example 13 verifies a single instance but does not replace a proof of the general statement.","section":"§3, Remark 12, Eq. (23)"}],"minor_comments":[{"comment":"There is a typographical error in the sentence defining the polynomials: \"P_f (v(1), . . . , v(r) and P_g(w(1), . . . , w(r))\" is missing a closing parenthesis after v^{(r)}.","section":"§3, Proposition 10, proof"},{"comment":"The paper relies on the Hessian characterization of [GO], an arXiv preprint by the same research group. Since this is a load-bearing external result, the authors should state explicitly that it is a preprint and provide the latest version or a proof sketch.","section":"§1 and References"},{"comment":"The symbol ⊠ is used for both the polynomial Kronecker product and, after evaluation, the ordinary matrix Kronecker product. The distinction is explained, but the multiple uses make Remark 12 harder to follow; a dedicated notation for the entrywise polynomial Kronecker product would improve clarity.","section":"§3, Corollary 8 and Remark 12"}],"recommendation":"reject","confidential_remarks":"The main theorem is unsupported because of the unjustified extension in Proposition 10. The differentiation identities are sound and might be publishable independently, but the closure result requires a new proof. If the authors can supply a valid proof of Eq. (22) on all tuples, a resubmission would be appropriate. The dependence on [GO], a same-group preprint, should also be disclosed more prominently."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Short version: the paper is aiming at a good theorem, and the first half is solid, but the proof of the main closure result has a real gap. I'd send it to a referee, but the referee should be told to focus on Proposition 10.\n\nWhat's actually new: the Hessian-matrix identity (Corollary 8) is a clean and correct global statement, and the differentiation rules (Proposition 7) are useful. The strategy—use the [GO] Hessian criterion and prove a polarized perfect-power identity for the Kronecker product—is sensible. For decomposable direction tensors U^(i)=v^(i)⊗w^(i), the computation goes through: the determinant reduces to (1/n! P_f P_g)^{d1d2}, which is exactly what you want. The binary quartic example is worked carefully and checks out, though it falls in the rank-one case.\n\nThe soft spot is the last step of Proposition 10. The paper proves (22) on the product of Segre cones, then defines P_{f⊠g} on all of (V⊗W)^r via the universal property, and then says 'performing the Kronecker substitution therefore gives (22).' That 'therefore' doesn't follow. The two sides of (22) are polynomials on (V⊗W)^r; agreeing on the Segre cone only tells you their difference lies in the Segre ideal. The Segre cone is not Zariski dense when both factors have dimension >1, so equality there doesn't force global equality. Remark 9 explicitly acknowledges this issue for the ordinary Hessian determinant, and the same obstruction applies here. The proof supplies no argument that the difference vanishes modulo the Segre ideal. Theorem 11 is a direct application and inherits the gap. The main result may be true—the examples are consistent, and the structure is plausible—but the submitted argument doesn't prove it.\n\nOne more caveat: the theorem depends essentially on the Hessian characterization from [GO], which is a preprint from the same group. That is not circular, but it does mean the paper is conditional on an unverified result. That's a normal state of affairs but worth noting.\n\nBottom line: a serious referee should see this. The result is significant if true, and the gap is possibly repairable. If I were the editor, I'd send it out with a clear instruction to scrutinize Proposition 10.","headline":"A nice result is likely true, but the proof of the key step has a real gap: equality on decomposable tensors does not extend to all tensors without an argument about the Segre ideal.","tokens_in":12886,"tokens_out":3072,"would_cite":false,"duration_ms":27910,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["15A69","15A03","15A72","81P40"],"pacs":[],"model":"deepseek-v4-flash","headline":"This paper proves that the Kronecker product of any two symmetric persistent tensors is again persistent, closing the symmetric case of a conjecture that fails in general.","keywords":["symmetric persistent tensors","Kronecker products","Hessian determinants","tensor rank lower bounds","substitution method","perfect-power identity","symmetric tensors"],"falsifier":"Compute the $(n-2)$-fold polarized Hessian determinant of $f\\boxtimes g$ at a tuple $U^{(1)},\\dots,U^{(n-2)}\\in\\mathbb{C}^{d_1d_2}$ in which at least one $U^{(i)}$ is not of the form $v\\otimes w$ — for the quartics of Example 13, take $U^{(1)}=e_0\\otimes e_0+e_1\\otimes e_1$. If the determinant is not the $(d_1d_2)$-th power of a multihomogeneous expression in the $U^{(i)}$, Proposition 10 and Theorem 11 are false.","tokens_in":11855,"feed_emoji":"🔗","tokens_out":10015,"duration_ms":78356,"temperature":0.7,"pith_summary":"Persistent tensors are a recursively defined class that yields certified lower bounds on tensor rank through repeated substitution. Earlier work conjectured that the class is closed under Kronecker products, but a nonsymmetric counterexample showed the unrestricted conjecture is false. This paper establishes the symmetric case: if $f$ and $g$ are symmetric persistent tensors of the same order $n$, then their Kronecker product $f\\boxtimes g$ is persistent. The proof first shows that the Hessian matrix of a Kronecker product is the Kronecker product of the factor Hessians up to a normalization, then demonstrates that for persistent factors the polarized Hessian determinant is a perfect power. It follows that symmetric persistence is closed under iterated Kronecker products and Kronecker powers, generating new infinite families of tensors with certified tensor-rank lower bounds.","feed_headline":"Symmetric tensor persistence closes under Kronecker products","feed_subtitle":"The recursive persistence test now applies to Kronecker powers, yielding certified tensor-rank lower bounds.","key_machinery":"The load-bearing machinery is the Hessian characterization of symmetric persistence, combined with two identities for Kronecker products of homogeneous polynomials. The Hessian characterization (Theorem 3, from [GO]) says $f\\in\\operatorname{Sym}^n\\mathbb{C}^d$ is persistent iff there is a nonzero multihomogeneous polynomial $P_f$ of multidegree $(1,\\ldots,1)$ such that $\\operatorname{Hess}\\big(f_{v^{(1)},\\ldots,v^{(n-2)}}(x)\\big) = P_f(v^{(1)},\\ldots,v^{(n-2)})^d$ for every $(n-2)$-tuple of vectors. Proposition 7 and Corollary 8 show that partial differentiation and Hessians are compatible with Kronecker products: $H_{f\\boxtimes g} = \\frac{1}{n(n-1)} H_f \\boxtimes H_g$. Proposition 10 then combines these: for persistent factors, the determinant of the polarized Hessian of $f\\boxtimes g$ reduces, on decomposable tuples $v^{(i)}\\otimes w^{(i)}$, to the $(d_1d_2)$-th power of the product $\\frac{1}{n!}P_fP_g$; multilinearity in each slot lets the author define $P_{f\\boxtimes g}$ on the full space $(\\mathbb{C}^{d_1}\\otimes\\mathbb{C}^{d_2})^{\\times(n-2)}$, which is exactly the data the Hessian criterion requires.","core_discovery":"The paper's central claim is Theorem 11: for $f\\in \\operatorname{Sym}^n\\mathbb{C}^{d_1}$ and $g\\in\\operatorname{Sym}^n\\mathbb{C}^{d_2}$ that are persistent, $f\\boxtimes g\\in\\operatorname{Sym}^n(\\mathbb{C}^{d_1}\\otimes\\mathbb{C}^{d_2})$ is persistent. The proof rests on the Hessian characterization of symmetric persistence from [GO]: a symmetric tensor is persistent exactly when the Hessian determinant of every $(n-2)$-fold partial polarization is a perfect power. Using the differentiation identity $\\partial(f\\boxtimes g)/\\partial z_{ij} = \\frac{1}{n}(\\partial f/\\partial x_i)\\boxtimes(\\partial g/\\partial y_j)$, and hence $H_{f\\boxtimes g} = \\frac{1}{n(n-1)}H_f\\boxtimes H_g$, the author derives a polarized perfect-power identity: the Hessian determinant of any partial polarization of $f\\boxtimes g$ is the $(d_1d_2)$-th power of a single multihomogeneous polynomial built from the corresponding polynomials of $f$ and $g$. This exactly matches the Hessian criterion, so persistence is inherited.","pith_inferences":["Editorial inference: if the extension from decomposable to arbitrary tuples is supplied, the same Hessian-perfect-power route could prove persistence for other symmetry classes closed under Kronecker products, since the obstruction is purely about the embedding of decomposable tensors being nondegenerate.","Editorial inference: Corollary 18's one-sided triangularizability criterion suggests that for cubic forms the product may be persistent under conditions weaker than persistence of both factors, such as triangularizability of one normalized Hessian space; this is not explored in the paper.","Editorial inference: the Kronecker powers of the W-state give explicit persistent tensors in dimensions $2^k$, which may be useful as test cases for numerical tensor-rank algorithms because persistence provides a lower bound that can be compared against constructive upper bounds."],"forward_implications":["If $f_1,\\dots,f_k$ are persistent symmetric tensors of the same order $n$, the iterated Kronecker product $f_1\\boxtimes\\cdots\\boxtimes f_k$ is persistent.","Every Kronecker power $f^{\\boxtimes k}$ of a persistent symmetric tensor is persistent, so repeated self-products create persistent tensors in exponentially growing dimensions.","The binary W-state tensor $W_n = x_0^{n-1}x_1$ yields persistent Kronecker powers in $\\operatorname{Sym}^n(\\mathbb{C}^{2^k})$, giving explicit infinite families of tensors with certified lower bounds on tensor rank.","The Hessian-matrix identity $H_{f\\boxtimes g} = \\frac{1}{n(n-1)}H_f\\boxtimes H_g$ holds for arbitrary symmetric tensors and gives a new tool for studying differential invariants of Kronecker products.","The closure converts the recursive persistence test into a practical certificate: to certify a large Kronecker product, it suffices to certify its smaller factors."],"supporting_citations":[{"why":"Supplies the Hessian characterization of symmetric persistence (perfect-power condition on polarized Hessian determinants) that the proof invokes as Theorem 3.","marker":"[GO]"},{"why":"Introduces persistent tensors, the recursive substitution method, and the conjecture that persistence might be preserved under Kronecker products.","marker":"[GL]"},{"why":"Develops persistent tensors and shows the binary W-state $x_0^{n-1}x_1$ is persistent, used in Example 16.","marker":"[Gh]"},{"why":"Gives the nonsymmetric counterexample to the unrestricted conjecture, motivating the symmetric case treated here.","marker":"[Sh]"},{"why":"Provides the polarization/spanning principle used in Lemma 6 to pass from decomposable tuples to all tuples in the differentiation identities.","marker":"[IK, Lan]"}],"fun_headline_variants":["Symmetric persistence survives Kronecker multiplication","Kronecker products preserve tensor persistence in symmetric case","Symmetric tensors: Kronecker product keeps persistence property","Closure: symmetric persistence closed under Kronecker powers","Persistence test works for Kronecker products of symmetric tensors"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The proof depends on a step in which an identity verified only for tuples of the form one vector in the first space times one vector in the second is asserted to hold for every vector in the larger tensor space, even though Remark 9 notes that a similar determinant identity can fail for vectors not of that product form.","fun_headline_variants_meta":{"raw":{"variants":["Symmetric persistence survives Kronecker multiplication","Kronecker products preserve tensor persistence in symmetric case","Symmetric tensors: Kronecker product keeps persistence property","Closure: symmetric persistence closed under Kronecker powers","Persistence test works for Kronecker products of symmetric tensors"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000374,"raw_usage":{"total_tokens":2001,"prompt_tokens":956,"completion_tokens":1045,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":572,"completion_tokens_details":{"reasoning_tokens":965}},"tokens_in":572,"tokens_out":1045,"duration_ms":7302,"temperature":1.0,"reasoning_tokens":965,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-12T04:32:38.072001+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Compute the $(n-2)$-fold polarized Hessian determinant of $f\\boxtimes g$ at a tuple $U^{(1)},\\dots,U^{(n-2)}\\in\\mathbb{C}^{d_1d_2}$ in which at least one $U^{(i)}$ is not of the form $v\\otimes w$ — for the quartics of Example 13, take $U^{(1)}=e_0\\otimes e_0+e_1\\otimes e_1$. If the determinant is not the $(d_1d_2)$-th power of a multihomogeneous expression in the $U^{(i)}$, Proposition 10 and Theorem 11 are false.","supporting_citations":[],"review_version":1}