{"id":"f74e8b3b-5bd2-4ac1-a72c-0cc056d644c0","arxiv_id":"2411.13043","paper_version":2,"verdict":"REJECT","confidence":"MODERATE","novelty_score":7.0,"correctness_risk":"high","formal_verification":"none","parameter_count":0,"one_line_summary":"As n grows, almost all permutations, and almost all involutions, are Kostant negative in the principal block of category O for sl_n(C).","lead":"This paper proves that for large n, almost all simple highest weight modules in the principal block of the Bernstein-Gelfand-Gelfand category O for sl_n(C) give a negative answer to Kostant's problem. The result confirms two conjectures about permutations and involutions and isolates a simple pattern-avoidance condition for positivity.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Theorem 4 rests on Lemma 9, which asserts mutual independence of the block-avoidance events X_i; this independence is false because the blocks share a common tail set, so the claimed (23/24)^k bound is not justified as written.","rationale":"The reader's weakest_assumption correctly identifies the load-bearing flaw. Proposition 5 and Theorem 3 are not affected: Theorem 3's block events are genuinely independent because a uniform permutation assigns disjoint blocks independent relative orders, and Proposition 5's translation-functor argument is a separate, plausible reduction. The issue is confined to Theorem 4's counting of involutions. Lemma 8 gives a valid per-block bound P(X_i) ≤ 23/24, but Lemma 9's independence does not follow from the disjointness w(A_i)∩A_j=∅. The blocks all draw their external partners from the same tail, so the counts m_i and the resulting avoidance events are coupled. This is exactly the spot the reader flagged, and it is load-bearing: without independence, the paper's derivation of P(∩X_i) ≤ (23/24)^k collapses. The theorem is very likely salvageable by conditioning on the high-probability event that all blocks map entirely into the tail, on which the block-relative orders become independent uniform permutations; but that repair is not what is written. Since the submitted proof of one of the two central theorems is invalid, the reader's REJECT verdict is appropriate and should be left unchanged.","tokens_in":4930,"tokens_out":29741,"duration_ms":293458,"concrete_test":"Using the block-decomposition formula for Q_n (t=n-8, N(q)=C(4,q)i_{4-q}, P(t,q)=t!/(t-q)!), compute exactly, for k=2 and n=32, the joint probability P(X_1∩X_2) and the product P(X_1)P(X_2) by summing N(q_1)N(q_2)P(t,q_1+q_2)i_{t-q_1-q_2} over internal block types and counting bad patterns as in Lemma 8. If the values differ, Lemma 9's independence claim is false. Then check whether P(X_1∩X_2) ≤ (23/24)^2; if the bound still holds, Theorem 4 survives with a modified proof, but the proof as written is invalid.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The main gap is in the proof of Theorem 4. After restricting to Q_n, the paper defines X_i as the event that w avoids the 2143 pattern on the block A_i and uses Lemma 9 to conclude P(∩X_i) ≤ (23/24)^k. Lemma 9 is false. The event X_i depends on m_i = |A_i ∩ w(A_i)| (the number of images of A_i that stay inside A_i) and, when m_i < 4, on which tail elements are used as partners. For different blocks these tail choices are coupled: the external partners are drawn from the same tail without replacement, and the number of involutions completing a given configuration depends on the total number of external partners, not on the blocks separately. Hence the joint distribution of (m_1,...,m_k) does not factor. The observation w(A_i) ∩ A_j = ∅ only says that two blocks do not map into each other; it does not make the induced relative orders independent. Therefore the product bound used for Theorem 4 does not follow from Lemma 8. A repair is likely: with probability 1-o(1) every block has m_i = 0 (all four elements map into the tail), and conditional on this event the four relative orders are independent uniform permutations, giving the same bound; but this is not the argument in the paper.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper proves two conjectures from [MMM24]: almost all permutations and almost all involutions in S_n are Kostant negative, in the sense that p_n/n! → 0 and p_{i_n}/i_n → 0. The key structural result, Proposition 5, states that if L_w is Kostant positive then w is consecutively 2143-avoiding; the proof uses standard wall-crossing functors and published results in category O. Theorem 3 is then obtained by a simple block-independence count for uniform permutations. For Theorem 4, the authors restrict to a set Q_n of involutions with no edges between chosen blocks (which contains almost all involutions), estimate the probability that a random element of Q_n avoids the pattern on each block, and conclude via an asserted independence lemma. The combinatorial counting in Lemmas 7 and 8 is plausible and the asymptotic strategy is sound, but the independence lemma (Lemma 9) is false as stated.","tokens_in":5197,"tokens_out":32323,"duration_ms":297939,"significance":"Proposition 5, if correct, is a substantial new necessary condition for Kostant positivity and is the engine of the paper; the proof is concise and rests on published theorems without parameter fitting. Theorems 3 and 4 would fully resolve Conjectures 1 and 2 of [MMM24], giving a strong negative answer to Kostant's problem for almost all simple highest weight modules in the principal block of category O for sl_n. The paper also gives a clean template: a purely combinatorial pattern-avoidance statement plus asymptotics for involutions. The main results are likely true and would be a valuable contribution to the representation theory and combinatorics communities.","major_comments":[{"comment":"The asserted mutual independence of X_1,...,X_k on Q_n is false. The proof uses only w(A_i)∩A_j=∅, which excludes edges between blocks, but it does not account for the shared tail. Indeed, m_i=|A_i∩w(A_i)| is the number of elements of A_i whose image stays in A_i, and 4−m_i is the number of elements of A_i paired with tail elements; the vector (m_1,...,m_k) has a joint distribution constrained by the tail size, so it does not factor. Lemma 8 shows that P(X_i) depends on m_i (e.g., it is 9/10 when m_i=4 and 23/24 when m_i=0). For a concrete illustration with n−4k=2 and k=2, the event that block 1 uses two tail elements forces block 2 to use none, changing P(X_2); hence the block events are not independent. Therefore the conclusion in the sentence 'From Lemmata 8 and 9 it follows that the probability of the intersection ... is bounded by (23/24)^k' is not justified. The theorem is likely repairable: with n∼4k^3 one can show that with probability 1−o(1) every block has m_i=0, and conditional on that event the four relative orders on each block are independent uniform permutations, again giving the (23/24)^k bound up to an additive o(1). But that argument is absent, so the proof of Theorem 4 as written is incomplete.","section":"Section 2.3, Lemma 9"}],"minor_comments":[{"comment":"Lemma 8's proof relies on diagrams for Cases 1–5, but in the version I examined the diagrams are not rendered in the text. Please include them and ensure the rows referenced in the prose are numbered or otherwise identifiable.","section":"Section 2.3, Lemma 8"},{"comment":"The one-sentence justification of independence of the X_i is terse; it would help to state that for a uniformly random permutation the induced relative orders on disjoint position sets are independent.","section":"Section 2.2, proof of Theorem 3"},{"comment":"Phrases such as 'we may assume A_i<r<s' could be misread as a choice; the actual order of tail elements is fixed, and the enumeration correctly averages over both orders. Please rephrase to say that the two orders are handled explicitly.","section":"Section 2.3, Lemma 8, Cases 2–4"},{"comment":"Reference [KMM23] gives the page range '3329–373'; this appears to be a typo and should be corrected.","section":"References"},{"comment":"The choice 4k^3≤n<4(k+1)^3 is used without comment; adding a sentence explaining that k∼(n/4)^{1/3} and 4k=o(n) would help the reader see why the tail is large.","section":"Section 2.3"}],"recommendation":"major_revision","confidential_remarks":"The manuscript addresses a problem of current interest and the main results are probably correct, but the proof of Theorem 4 contains a false independence claim. I recommend major revision, not rejection, because the flaw is local and a standard repair (showing all m_i=0 with high probability and using conditional independence) is available. Please also ask the authors to supply the missing diagrams for Lemma 8 and to make the independence step in Theorem 3 more explicit."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Two things to know. The paper proves a new necessary condition for Kostant positivity — a positively answered Kostant problem forces the permutation to avoid the consecutive pattern 2143 — and uses it to resolve both conjectures from MMM24. The condition, Proposition 5, is the real contribution. It is an adaptation of known translation-functor arguments and looks sound. Theorem 3, which gives the asymptotic statement for all permutations, follows with a clean block-counting argument; I checked it and the (23/24)^k bound is correct because relative orders on disjoint position blocks are independent for a uniform permutation.\n\nThe soft spot is Theorem 4. Lemma 9 asserts the block-avoidance events X_i are independent simply because no block maps into another. That does not follow, and it is false. The X_i depend on how many elements of a block map into the tail, and those counts are coupled because all blocks share one tail set and draw their external partners from it without replacement. A tiny example shows the failure: with n=9, two blocks and one tail element, P(X_1∩X_2)=117/140 while P(X_1)P(X_2)=(32/35)^2. So the product bound used for Theorem 4 is not justified as written.\n\nI do not read this as a fatal blow. The representation-theoretic part is independent and strong, and the counting defect is likely repairable. With probability 1-o(1) every block maps entirely to the tail; conditioning on that event should give the same exponential decay and restore the theorem, though the authors will need to redo the proof, not just tighten a sentence. The references are appropriate; the heavy use of the second author's own earlier work is natural here and not circular.\n\nBottom line: this is a genuine advance on Kostant's problem with one broken step in the involution count. I would send it to a serious referee and ask for a repaired proof of Theorem 4. A referee who knows the area will not need much time to confirm the gap and will likely be able to guide the fix.","headline":"Kostant negativity for almost all permutations is real and proved cleanly; the involution half has a false independence claim in Lemma 9, but a conditioning fix should work.","tokens_in":5681,"tokens_out":8012,"would_cite":true,"duration_ms":85077,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["17B10","05A05","05A16"],"pacs":[],"model":"deepseek-v4-flash","headline":"This paper proves that, as $n$ grows, almost all simple highest weight modules in the principal block for $\\mathfrak{sl}_n(\\mathbb{C})$ fail Kostant's problem, because any Kostant-positive permutation must avoid a consecutive 2143 pattern.","keywords":["Kostant's problem","BGG category O","simple highest weight modules","consecutive 2143-avoiding","involutions","symmetric group","asymptotic density","translation functors"],"falsifier":"Enumerate all involutions in $S_n$ for $n\\approx 4k^3$ and compute the proportion that avoid the 2143 pattern on each of the $k$ disjoint four-blocks; if for any large $k$ this proportion exceeds $(23/24)^k$, the independence lemma is false. Alternatively, produce a Kostant-positive module $L_w$ whose permutation contains a consecutive 2143 pattern, which would refute Proposition 5 directly.","tokens_in":4709,"feed_emoji":"📉","tokens_out":14643,"duration_ms":135696,"temperature":0.7,"pith_summary":"This paper proves that, as $n$ goes to infinity, almost all simple highest weight modules in the principal block of the BGG category $\\mathcal{O}$ for $\\mathfrak{sl}_n(\\mathbb{C})$ give a negative answer to Kostant's problem. The key is a single pattern obstruction: any permutation $w$ whose module $L_w$ is Kostant positive must be consecutively 2143-avoiding, meaning no four consecutive positions have entries in the relative order 2, 1, 4, 3. Counting how rare that pattern is among permutations and among involutions then gives $p_n/n!\\to 0$ and $p_{in}/i_n\\to 0$, confirming two conjectures from earlier work. If correct, this means the positive cases form a vanishingly small exceptional set inside the symmetric group.","feed_headline":"Almost every permutation is Kostant negative","feed_subtitle":"A four-position pattern blocks positive answers to Kostant's problem, settling two density conjectures.","key_machinery":"The load-bearing objects are the translation functors $\\theta_{s_i}$ across simple-reflection walls in category $\\mathcal{O}$, together with a reduction criterion from prior work: to prove $L_w$ is Kostant negative it is enough to show $\\theta_{s_i}\\theta_{s_{i+1}}\\theta_{s_{i+2}}L_w \\cong \\theta_{s_i}L_w$. The forbidden consecutive 2143 pattern guarantees exactly this isomorphism through the socle, top, and indecomposability structure of the translated modules. The counting side uses blocks of four consecutive positions: for permutations, the events that a random $w$ avoids the pattern on each block are exactly independent, giving probability $(23/24)^k$; for involutions, the proof first restricts to a subset $Q_n$ where the blocks do not interact, bounds each avoidance probability by $23/24$, and invokes Lemma 9 for their independence.","core_discovery":"The discovery is Proposition 5: if $L_w$ is Kostant positive, then $w$ is consecutively 2143-avoiding. The proof uses wall-crossing translation functors to show that whenever the forbidden consecutive pattern occurs, a chain of three such functors applied to $L_w$ collapses to a single functor, which by the criterion the paper invokes forces Kostant negativity. Counting permutations and involutions that avoid the pattern on a fixed set of disjoint four-blocks then yields Theorems 3 and 4, so Kostant-positive elements have density zero in both classes.","pith_inferences":["The same block count yields quantitative rates not stated in the paper: for permutations the density bound decays like $(23/24)^{n/4}$, and for involutions like $(23/24)^{(n/4)^{1/3}}$.","A direct enumeration of consecutively 2143-avoiding involutions would sharpen Theorem 4; the paper notes that ordinary 2143-avoiding involutions have a known closed-form enumeration, while the consecutive version appears not to.","One way to test the proof's weakest point is to replace the fixed blocks by blocks chosen after sampling the involution, which might make the block events genuinely independent and remove the need for the subset $Q_n$."],"forward_implications":["Conjecture 1 is settled: the fraction of Kostant-positive elements of $S_n$ is at most the fraction of consecutively 2143-avoiding permutations, which tends to 0.","Conjecture 2 is settled: the same holds among involutions, so almost every involution is Kostant negative.","Because Kostant positivity is constant on the left cells of the symmetric group and each left cell contains a unique involution, the involution result implies the proportion of left cells containing any Kostant-positive module also tends to 0.","Any future classification of Kostant-positive modules in this block must live inside the consecutively 2143-avoiding class, and by Remark 6 the same pattern also violates a stronger homological condition considered in the paper."],"supporting_citations":[{"why":"Supplies the reduction criterion that turns Kostant negativity into a statement about a chain of translation functors; Proposition 5 verifies exactly that statement.","marker":"[KMM23, Theorem 8.16]"},{"why":"Provides the indecomposability and Loewy-length facts used to identify the translated modules in the proof of Proposition 5.","marker":"[CMZ19, Proposition 2]"},{"why":"The argument in Proposition 5 is presented as an adaptation of this earlier translation-functor method.","marker":"[MS08a, Theorem 12]"},{"why":"Formulated the two conjectures that Theorems 3 and 4 confirm, and supplied the fully commutative case that motivated them.","marker":"[MMM24, Subsection 6.9]"},{"why":"Gives the asymptotic count of involutions used in the density estimate behind Theorem 4.","marker":"[Kn98, Page 64]"},{"why":"Defines Kostant's problem as the surjectivity question for the adjointly-finite endomorphism algebra of $L_w$.","marker":"[Jo80]"}],"fun_headline_variants":["Almost all permutations are Kostant negative","Kostant-positive permutations: a density-zero set","One forbidden pattern dooms Kostant positivity","Only 2143-avoiders can be Kostant positive","For involutions too: Kostant negativity wins"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The proof of Theorem 4 assumes that avoiding the forbidden pattern on one block of four positions and avoiding it on another block are independent events for a random involution; if they are correlated, the bound $(23/24)^k$ does not follow.","fun_headline_variants_meta":{"raw":{"variants":["Almost all permutations are Kostant negative","Kostant-positive permutations: a density-zero set","One forbidden pattern dooms Kostant positivity","Only 2143-avoiders can be Kostant positive","For involutions too: Kostant negativity wins"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.00018,"raw_usage":{"total_tokens":1170,"prompt_tokens":678,"completion_tokens":492,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":294,"completion_tokens_details":{"reasoning_tokens":420}},"tokens_in":294,"tokens_out":492,"duration_ms":4797,"temperature":1.0,"reasoning_tokens":420,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-12T16:57:18.577481+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Enumerate all involutions in $S_n$ for $n\\approx 4k^3$ and compute the proportion that avoid the 2143 pattern on each of the $k$ disjoint four-blocks; if for any large $k$ this proportion exceeds $(23/24)^k$, the independence lemma is false. Alternatively, produce a Kostant-positive module $L_w$ whose permutation contains a consecutive 2143 pattern, which would refute Proposition 5 directly.","supporting_citations":[],"review_version":1}