Almost all permutations and involutions are Kostant negative

Samuel Creedon; Volodymyr Mazorchuk

REVIEW 1 major objections 5 minor 58 references

Almost all permutations and involutions are Kostant negative

T0 review · 1 major / 5 minor · reviewed 2026-08-12 · deepseek-v4-flash

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Pith's one-line read 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.

desk verdict 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. read the letter →

arxiv 2411.13043 v2 pith:45K5RLJH submitted 2024-11-20 math.RT

Samuel Creedon , Volodymyr Mazorchuk This is my paper · ORCID

classification math.RT MSC 17B1005A0505A16

keywords Kostant'sproblemBGGcategoryOsimplehighestweightmodulesconsecutive2143-avoidinginvolutionssymmetricgroupasymptoticdensitytranslationfunctors

verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

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.

What carries the argument

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.

What would settle it

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.

Watch

Extended reading notes

Core claim

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.

Load-bearing premise

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.

Editorial extensions

If this is right

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.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

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 $Q_n$.

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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Desk Editor's Note

read the 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.

The 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.

I 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.

Bottom 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.

Referee Report

1 major / 5 minor

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.

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 (1)

[Section 2.3, Lemma 9] 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.

minor comments (5)

[Section 2.3, Lemma 8] 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 2.2, proof of Theorem 3] 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 2.3, Lemma 8, Cases 2–4] 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.
[References] Reference [KMM23] gives the page range '3329–373'; this appears to be a typo and should be corrected.
[Section 2.3] 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.

Circularity Check

0 steps flagged · score 0.0 of 10

No circular reduction: the core counting claims are derived from a new necessary condition and published external theorems, not from the conjectures they prove.

full rationale

The derivation chain behind Theorems 3 and 4 is: Proposition 5 supplies a combinatorial necessary condition for Kostant positivity; Theorem 3 counts permutations avoiding 2143 at disjoint blocks; Theorem 4 counts involutions in Q_n using Lemmas 7-9. None of these steps fits a parameter to the claimed answer or renames a known result as a new one. Proposition 5 is proved by applying published results ([KMM23, Theorem 8.16], [CMZ19, Propositions 2 and 46]) and adapting arguments from [MS08a, Theorem 12] and [MMM24, Section 5.5]. These works overlap with the second author, but they are external theorems whose assumptions do not include the conjectures being proved; under the stated review rules this is independent support and does not constitute circularity. The paper's own earlier conjectures [MMM24] are cited only as motivation, not as premises. The one mathematically serious weakness is not circular: Lemma 9 asserts independence of the events X_i from the fact that w(A_i) ∩ A_j = ∅, but the events are coupled through the common tail of the involution, so the product bound in Theorem 4 is not justified as written. That is a correctness gap in the probabilistic estimate, not a reduction of the conclusion to an input, and therefore it does not affect the circularity score.

Assumptions & free parameters 0 free parameters · 5 assumptions · 0 invented entities

No free parameters and no invented entities are introduced. The proof leans on a substantial body of prior representation theory and on Lemma 9's independence assertion, which is not justified and is in fact false as stated.

assumptions (5)

domain assumption KMM23, Theorem 8.16: if theta_{s_i} theta_{s_{i+1}} theta_{s_{i+2}} L_w is isomorphic to theta_{s_i} L_w, then L_w is not Kostant positive.
Used in the proof of Proposition 5 to convert a module isomorphism into Kostant negativity.
domain assumption Translation functor and Loewy structure facts from CMZ19 and MS08b, including the top, socle, and Jantzen middle of theta_s L_x and the conditions under which theta_s kills a simple module.
These structural facts drive the proof of Proposition 5.
standard math Knuth's asymptotic formula (1): i_n is asymptotic to a constant times n^{n/2} exp(-n/2 + sqrt(n)).
Used in Lemma 7 to show the exceptional set of involutions has density zero.
standard math For a uniformly random permutation, relative-order events on disjoint blocks of positions are independent.
Used in the proof of Theorem 3.
ad hoc to paper Lemma 9: for uniformly random w in Q_n, the events X_i are mutually independent because w(A_i) is disjoint from A_j.
This is the load-bearing independence claim in Theorem 4. It is not a standard theorem and is generally false because the events depend on the number of external partners of each block, and those counts are coupled through the shared tail set.

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Pith. "Pith review of Almost all permutations and involutions are Kostant negative." pith.science (2026). https://pith.science/paper/45K5RLJH

@misc{pith2026241113043,
  author       = {Pith},
  title        = {Pith review of: Almost all permutations and involutions are Kostant negative},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/45K5RLJH}},
  note         = {Machine review of arXiv:2411.13043}
}

abstract

We prove that, when $n$ goes to infinity, Kostant's problem has negative answer for almost all simple highest weight modules in the principal block of the BGG category $\mathcal{O}$ for the Lie algebra $\mathfrak{sl}_n(\mathbb{C})$.

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Reference graph

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Pith tools

Reviewed August 12, 2026 · model on record in the stance chip above.

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Primitive ideals and

Barbasch, D.; Vogan, D. Primitive ideals and

work page

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Barbasch, D.; Vogan, D

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