{"id":"d58b9a89-a518-48c4-a016-fa0ded25b09f","arxiv_id":"2608.02170","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":6.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"For weighted oriented graphs, the homological shift ideals have linear quotients exactly when the underlying tree is a star or broom and the weighted oriented graph avoids D1,D2,D5,D6,D8 as induced subgraphs.","lead":"This paper studies algebraic objects called homological shift ideals attached to edge ideals of weighted oriented graphs, and proves when these ideals have a property called linear quotients. It gives a complete answer for trees: linear quotients hold for all levels exactly when the graph is a star or a broom and five small forbidden patterns are absent.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Theorem 4.15's tree classification is not established: the proof reduces to 'co-chordal ⇒ star or broom', but co-chordal trees include every double-star; the necessary H_6^c-free condition from Corollary 4.9 is never used in that step.","rationale":"The reader's verdict was CONDITIONAL, and the reader's weakest assumption already flagged the star/broom classification in Theorem 4.15 as stated rather than proved. My stress-test agrees but sharpens the concern: the published proof's only stated reason for the classification is co-chordality, which is insufficient because co-chordal trees include all double-stars. The proof derives H_n^c-freeness in Corollary 4.9 but never uses it in Theorem 4.15. Since this is the step that reduces arbitrary trees to the two shapes in the main characterization, it is load-bearing. The issue is a proof gap, not an evident counterexample, and it is addressable by inserting the missing H_6^c-free argument. I therefore do not move the reader's conditional verdict. I also note the abstract's omission of the star/broom condition, which makes the headline claim literally false as stated, though the full theorem restores the condition.","tokens_in":26410,"tokens_out":25293,"duration_ms":278856,"concrete_test":"Enumerate all unlabeled trees on n=5,...,8 vertices. For each tree, check (a) co-chordality, (b) H_n^c-freeness for all n≥6, and (c) whether the tree is a star or broom. If any tree passes (a) and (b) but fails (c), Theorem 4.15 is false. Independently, re-derive the necessity proof of Theorem 4.15 without using the one-line 'co-chordal ⇒ star/broom' claim, replacing it with the explicit 2K2/H_6^c analysis; if that re-derivation requires adding H_6^c-freeness at a step where the published proof does not, the gap is confirmed.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The necessity direction of Theorem 4.15 relies on the assertion: 'Since G is co-chordal, then G cannot have a path of length ≥4. This leads to G is same as Figure 3.' This inference is not valid as written. Co-chordality of a tree is equivalent to 2K2-freeness, and 2K2-free trees include all double-stars: for example, the graph H_6^c from Theorem 2.9 is exactly the double-star with two leaves on each of the two central vertices, and it is co-chordal. To conclude that the underlying tree is a star or a broom, one must additionally use the H_n^c-free condition obtained in Corollary 4.9. That condition is not invoked anywhere in the proof of Theorem 4.15. Thus the structural reduction underpinning the tree characterization has a missing argument: the proof, as written, would also allow co-chordal double-stars that are not stars or brooms. The theorem may still be true once H_6^c-freeness is added, but the one-line classification is not sufficient and needs to be replaced by an explicit argument showing co-chordal + H_6^c-free forces the star/broom shapes. A related presentation issue is that the abstract states the tree characterization as 'D is D_i-free for i=1,2,5,6,8' without mentioning the star/broom condition, which is false as stated (e.g., an unweighted P5 is D_i-free).","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper studies the k-th homological shift ideals HS_k(I(D)) of edge ideals of weighted oriented graphs. The main results are: (1) Theorem 3.15, asserting that for vertex-splittable I(D), HS_1(I(D)) has linear quotients if and only if the induced subgraphs D_6, D_7, D_8 are absent; (2) Theorem 4.6, asserting that if I(D) has linear quotients then the radical of HS_k(I(D)) equals HS_k(I(G)) for all k, with descent consequences for homological linear quotients and Cohen-Macaulay-type properties; and (3) Theorem 4.15, a characterization for weighted oriented trees: HS_k(I(D)) has linear quotients for all k iff the underlying simple graph is a star or broom graph and the induced subgraphs D_1, D_2, D_5, D_6, D_8 are absent. The abstract, however, states the tree characterization without the star/broom condition, which is false as written.","tokens_in":26795,"tokens_out":20474,"duration_ms":490873,"significance":"If the results hold, the paper makes a useful contribution to the active study of homological shift ideals: it provides a radical reduction from weighted oriented graphs to their underlying simple graphs (Theorem 4.6), gives necessary conditions for homological linear quotients (Corollaries 4.8 and 4.9), and offers a clean forbidden-subgraph characterization for trees. The proofs are mostly built from standard tools (vertex splittings, linear quotient orders, Lemma 2.7, Lemma 2.8), and the paper is carefully structured. However, the proof of the tree characterization contains a significant gap in the structural reduction, and the abstract misstates the main tree theorem. Both issues are localized and appear fixable, but they currently prevent the paper from being accepted as is.","major_comments":[{"comment":"The abstract's tree characterization omits the star/broom condition. It states that HS_k(I(D)) has linear quotients for all k≥0 iff D is D_i-free for i=1,2,5,6,8, whereas Theorem 4.15 requires in addition that the underlying simple graph G is a star or broom graph. The abstract statement is false: take a double-star with two leaves on each central vertex and set all weights equal to 1. None of D_1, D_2, D_5, D_6, D_8 can occur as induced weighted oriented subgraphs because all weights are 1, but this graph is neither a star nor a broom. The abstract must be corrected to match Theorem 4.15.","section":"Abstract"},{"comment":"The structural reduction is incomplete. After obtaining from Corollary 4.9 that G is co-chordal and H_6^c-free, the proof asserts: \"Since G is co-chordal, then G can not have a path of length≥4. This leads to that G is same as Figure 3.\" Co-chordality of a tree only gives diameter at most 3, i.e., G is a double-star (possibly a star). It does not exclude a double-star with two or more leaves on each of the two central vertices; H_6^c is exactly such a double-star. The H_6^c-free condition is not used in this step. To justify \"star or broom\", one must argue that H_6^c-freeness rules out double-stars with at least two leaves on both sides. This missing argument is load-bearing for the necessity direction.","section":"Theorem 4.15, proof of necessity"}],"minor_comments":[{"comment":"In the converse direction the hypothesis is written as \"D_i are not induced subgraphs of D for all i∈{1,2,5,6}\", but the proof later invokes D_8-freeness (e.g., in Case 2). The list should include 8 to match the theorem statement.","section":"Theorem 4.15, converse"},{"comment":"The expressions \"I(D)=x_1(x_1,...,x_i^{w_i},...,x_n)\" and \"HS_k(I(D))=x_1 HS_k((x_1,...,x_i^{w_i},...,x_n))\" appear to repeat x_1; the ideal should be generated by the remaining variables. Also, \"N_D[{...}]\" should be \"D[{...}]\" throughout the proof.","section":"Proposition 4.14"},{"comment":"In the proof, \"Assume i≥1\" should read \"Assume k≥1\"; the index i is not defined at that point.","section":"Lemma 4.12"},{"comment":"The displayed computation ends with a stray \"=\", and the claim that HS_1(I(G)) is Cohen-Macaulay while HS_1(I(D)) is not would benefit from a brief justification or reference.","section":"Example 4.11"}],"recommendation":"major_revision","confidential_remarks":"The main gap is localized to the necessity proof of Theorem 4.15 and the abstract's misstatement of the tree characterization; both are fixable within the manuscript's scope. The rest of the paper appears technically sound, though the multi-case proofs in Section 3 and 4 are dense and would benefit from expansion. I see no indication of misconduct or citation problems."},"author_rebuttal":null,"desk_editor":null,"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["13D02","16E05","05E40"],"pacs":[],"model":"deepseek-v4-flash","headline":"For weighted oriented trees, all homological shift ideals have linear quotients exactly when the tree is a star or a broom and five small subgraphs are absent.","keywords":["homological shift ideals","weighted oriented graphs","linear quotients","edge ideals","vertex splittable ideals","forbidden induced subgraphs","radical of monomial ideals","tree classification"],"falsifier":"Take the weighted oriented double-star tree with two adjacent centers, at least two leaves on each center, all weights 1 and any orientation that avoids D1, D2, D5, D6, D8; compute its homological shift ideals $HS_k$ for all $k$. If every $HS_k$ has linear quotients, Theorem 4.15's 'only if' direction fails; if some $HS_k$ fails, the missing $H_n^c$-free argument is needed to explain why.","tokens_in":26302,"feed_emoji":"🌳","tokens_out":7456,"duration_ms":60899,"temperature":0.7,"texified_at":"2026-08-05T21:59:18.965914+00:00","pith_summary":"This paper studies homological shift ideals of edge ideals of weighted oriented graphs—ideals generated by the multigraded shifts appearing in the minimal free resolution. It proves that when $I(D)$ is vertex-splittable, $HS_1(I(D))$ has linear quotients exactly when three small weighted oriented graphs D6, D7, D8 are absent. It also proves a radical-descent theorem: if $I(D)$ has linear quotients, then the radical of every $HS_k(I(D))$ equals $HS_k(I(G))$ of the underlying simple graph. The payoff is a complete tree classification: all homological shift ideals of a weighted oriented tree have linear quotients precisely when the underlying tree is a star or a broom and five forbidden induced subgraphs D1, D2, D5, D6, D8 do not occur.","texify_model":"deepseek-v4-flash","texify_usage":{"total_tokens":5083,"prompt_tokens":785,"completion_tokens":4298,"prompt_tokens_details":{"cached_tokens":0},"prompt_cache_hit_tokens":0,"prompt_cache_miss_tokens":785,"completion_tokens_details":{"reasoning_tokens":3574}},"feed_headline":"Star or broom: the only trees with linear shift ideals","feed_subtitle":"Five forbidden subgraphs decide when every homological shift ideal of a weighted oriented tree has linear quotients.","key_machinery":"The paper's engine is the homological shift ideal $HS_k(I)$, defined as the monomial ideal generated by all multigraded shifts $x^a$ appearing in the $k$th module of the minimal free resolution of $I$. It combines this with the linear-quotient property (an ordering of generators whose colon ideals are variable-generated), the vertex-splitting decomposition $I = x I_1 + I_2$ of vertex-splittable ideals, and a family D1,...,D8 of small weighted oriented graphs used as forbidden induced subgraphs. A technical equality, $\\mathrm{set}_I(u_i) = \\mathrm{set}_{\\sqrt I}(\\sqrt u_i)$, transfers the linear quotient order from the weighted ideal to its radical.","core_discovery":"The central discovery is that, for edge ideals of weighted oriented graphs, good homological behaviour of the weighted object is controlled by the unweighted skeleton after taking radicals, together with a short list of forbidden induced weighted subgraphs. More precisely, under the hypothesis that $I(D)$ has linear quotients, the paper proves that $\\sqrt{HS_k(I(D))} = HS_k(I(G))$ for all $k$; consequently, if the weighted ideal has homological linear quotients, so does the edge ideal of the underlying graph. For trees this becomes a sharp if-and-only-if: the underlying graph must be a star or a broom, and the weighted oriented graph must avoid the five induced subgraphs D1, D2, D5, D6, D8.","pith_inferences":["If the set_I(u) = set_{\\sqrt I}(\\sqrt u) equality is robust beyond the vertex-splittable setting, the radical-descent theorem may extend to larger classes of monomial ideals that admit degree-increasing linear quotient orders.","The star/broom classification suggests an inductive generation strategy: all weighted oriented graphs with homological linear quotients might be built from stars and brooms by gluing operations that never create the forbidden D_i, giving a recursive classification beyond trees.","Because known results on simple graphs can now be imported as necessary conditions for weighted graphs, one can narrow the search for weighted counterexamples by first checking co-chordality and H_n^c-freeness of the underlying graph.","A testable extension: compute the radicals of HS_k for small weighted oriented cycles and compare with HS_k of the underlying cycle; the paper's Example 4.7 shows what failure looks like when linear quotients are absent, so a broader equivalence could be probed numerically."],"forward_implications":["If I(D) has homological linear quotients, then I(G) does as well, so any orientation or weighting of a graph whose edge ideal lacks homological linear quotients also lacks them.","The radical of every homological shift ideal of a weighted oriented graph with linear quotients coincides with that of the underlying simple graph; hence Cohen–Macaulay, Gorenstein, Buchsbaum, and related properties descend from HS_k(I(D)) to HS_k(I(G)).","For trees, the property 'all HS_k have linear quotients' is a finite forbidden-subgraph condition: the underlying graph must be a star or a broom, and D must avoid D1, D2, D5, D6, D8.","No tree shape other than star or broom can have all homological shift ideals linear, regardless of orientation or vertex weights—a complete structural obstruction.","The same forbidden-subgraph list gives a practical test: to check whether a weighted oriented tree has homological linear quotients, one only needs to inspect its induced subgraphs of size at most four and the underlying tree shape."],"fun_headline_variants":["Star or broom: the only trees with linear homological shift ideals","Weighted tree shift ideals: linear quotients iff star or broom","Five forbidden subgraphs decide tree shift ideal linearity","For weighted trees, linear quotients come only from stars and brooms"],"cache_read_input_tokens":2304,"weakest_assumption_plain":"The tree classification assumes that every co-chordal tree whose edge ideal could have homological linear quotients must be a star or a broom; the proof asserts this classification from 'no path of length at least four' without deriving it, so a tree of diameter three that is not a broom (a double-star with leaves on both centers) is the load-bearing gap.","fun_headline_variants_meta":{"raw":{"variants":["Star or broom: the only trees with linear homological shift ideals","Weighted tree shift ideals: linear quotients iff star or broom","Five forbidden subgraphs decide tree shift ideal linearity","For weighted trees, linear quotients come only from stars and brooms"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000901,"raw_usage":{"total_tokens":3716,"prompt_tokens":744,"completion_tokens":2972,"prompt_tokens_details":{"cached_tokens":256},"prompt_cache_hit_tokens":256,"prompt_cache_miss_tokens":488,"completion_tokens_details":{"reasoning_tokens":2899}},"tokens_in":488,"tokens_out":2972,"duration_ms":23679,"temperature":1.0,"reasoning_tokens":2899,"cache_read_input_tokens":256,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-04T12:56:21.675693+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Take the weighted oriented double-star tree with two adjacent centers, at least two leaves on each center, all weights 1 and any orientation that avoids D1, D2, D5, D6, D8; compute its homological shift ideals $HS_k$ for all $k$. If every $HS_k$ has linear quotients, Theorem 4.15's 'only if' direction fails; if some $HS_k$ fails, the missing $H_n^c$-free argument is needed to explain why.","supporting_citations":[],"review_version":2}