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The longest increasing subsequence of Brownian separable permutons

T0 review · 2 major / 3 minor · reviewed 2026-08-15 · deepseek-v4-flash

Pith's one-line read For every p in (0,1), the longest increasing subsequence of a permutation sampled from the Brownian separable permuton has an almost-sure scaling law with exponent fixed by a Gamma-function equation, and the rescaled limit is a…

desk verdict Exact exponent for LIS in Brownian separable permutons, with a plausible a.s. scaling limit; the provided text is missing the proof of Lemma 9.2 that underpins the almost-sure step. read the letter →

arxiv 2506.19123 v1 pith:MRYL5YRU submitted 2025-06-23 math.PR math.CO

classification math.PRmath.CO MSC 60C0505A0560F1560J80
keywords longestincreasingsubsequenceBrownianseparablepermutoncographonscalinglimitpattern-avoidingpermutationslargestpositivesubtreeregularvariation
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

The paper answers a long-open question: how long is the longest increasing subsequence of a typical permutation drawn from the Brownian separable permuton, the universal limit of pattern-avoiding permutations? It proves that the length $\operatorname{LIS}(\sigma_n)$ has an exact almost-sure power law $n^{\alpha(p)}$, with the exponent $\alpha(p)$ determined explicitly by a Gamma-function equation, and that the rescaled length converges to a non-deterministic random variable $X(p)$ that is a function of the limiting permuton itself. The same result applies to the largest clique of the Brownian cographon, since both problems reduce to the same optimization problem on a random signed tree. Previous results only sandwiched $\operatorname{LIS}(\sigma_n)$ between $n^{\alpha_*(p)-\varepsilon}$ and $n^{\beta_*(p)+\varepsilon}$ with different upper and lower exponents, so even the existence of a single exponent was open. A key feature is that the fluctuations survive: the limit $X(p)$ is not a constant.

What carries the argument

The central object is the pair of sequences $q(k)=\mathbb{P}(\operatorname{LIS}(T)=k)$ and $Q(k)=\mathbb{P}(\operatorname{LIS}(T)\ge k)$, where $T$ is the $p$-signed critical binary branching tree; these sequences satisfy an exact recursive relation obtained by splitting $T$ at the root according to the root sign and the LIS values of the two subtrees. The recursion determines $q$, but is too sensitive to initial conditions to read the exponent off directly. The argument therefore studies the conditioned tree $T_k$ with $\operatorname{LIS}(T)=k$, and a decreasing Markov chain that records, along the spine from the root to a uniformly chosen leaf of the leftmost maximal positive subtree, the size of the distinguished subtree and the sizes of the sibling subtrees. A black-and-golden coupling built from the transition probabilities shows that chains started at large $k$ and $k'$ merge quickly; from this local convergence one obtains $$\frac{\#L_{\max}^{\cap}(T_k)}{k} \xrightarrow[k\to\infty]{\mathbb{P}} \$\lambda$\in(0,1),$$ where $L_{\max}^{\cap}$ is the set of leaves belonging to all maximal positive subtrees. Re-running the tail estimate with $\lambda$ in place of the crude bound identifies $\lambda=\alpha$, upgrades the tail asymptotics to regular variation $$q(k)=$k^{{-1-1/(2\alpha)}}$\varphi(k),\qquad Q(k)=$k^{{-1/(2\alpha)}}$\Phi(k)$$ with slowly varying $\varphi,\Phi$, and then a comparison with power-law test sequences in the recursion forces the Gamma equation. The final scaling limit is obtained by writing the rescaled process as a backward martingale plus a remainder that is controlled by the quantitative law of large numbers for $\#L_{\max}^{\cap}$.

What would settle it

Run the recursion (2.8) at $p=1/2$ with $q(1)=(1+\sqrt{p})^{-1}$, compute $q(k)$ for $k$ up to $10^6$, and estimate the local exponent $-\log q(k)/\log k$; if it does not approach $\gamma=1+1/(2\alpha(1/2))\approx 1.6134$, the exponent identification in Theorem 2.2 fails. In parallel, simulate the coupled trees up to size $10^7$ and check that $n^{-\alpha(1/2)}\operatorname{LIS}(T_n)$ remains bounded away from $0$ and $\infty$ and that repeated independent simulations of the limit do not all produce the same constant.

Watch

Extended reading notes

Core claim

For fixed $p\in(0,1)$, sample $\sigma_n$ from the Brownian separable permuton $\mu_p$ under the natural coupling coming from the growth algorithm on the $p$-signed uniform binary tree $T_n$. The paper's central claim is that, with $\operatorname{LIS}(T_n)$ denoting the maximal number of leaves of a subtree all of whose internal nodes carry the $\oplus$ sign, $\operatorname{LIS}(\sigma_n)=\operatorname{LIS}(T_n)$ and $$\frac{\operatorname{LIS}(T_n)}{$n^{{\alpha(p)}}$} \xrightarrow[n\to\infty]{\mathrm{a.s.}} X(p),$$ where $\alpha(p)$ is the unique solution in $(1/2,1)$ of the Gamma equation $$\frac{1}{$4^{{1/(2\alpha)}}$\sqrt{\pi}}\,\frac{\Gamma(1/2-1/(2\$\alpha$))}{\Gamma(1-1/(2\$\alpha$))}=\frac{p}{p-1},$$ and $X(p)$ is almost surely positive and finite, is not deterministic, and is a deterministic measurable function of $\mu_p$. Because the same tree construction generates the Brownian cographon $W_p$, the identical result holds for the largest clique of a graph sampled from $W_p$, while the largest independent set is governed by $\alpha(1-p)$. This is an exact almost-sure scaling limit: neither a logarithmic correction nor a range of possible exponents is left open.

Load-bearing premise

The whole derivation rests on one exact combinatorial decomposition: after picking $m$ uniform leaves of the random tree, the sizes of the leftover regions around those leaves have a symmetric joint law with all parameters $1/2$ and are independent of the reduced tree; if that were not exact, the exponent could not be pinned to the Gamma equation.

Editorial extensions

If this is right

  • For every $p\in(0,1)$, $\operatorname{LIS}(\sigma_n)$ grows as $n^{\alpha(p)}$ almost surely, with no multiplicative logarithmic corrections.
  • The exponent $\alpha(p)$ increases continuously from $1/2$ to $1$ as $p$ goes from $0$ to $1$, with $\alpha(1/2)\approx 0.815226$ in the separable-permutation case; larger $p$ means more $\oplus$ nodes and hence longer increasing subsequences.
  • The same exponent and the same limiting variable govern the largest clique of the Brownian cographon, while the largest independent set is governed by $(\alpha(1-p),X(1-p))$.
  • The limiting variable $X(p)$ is a deterministic measurable function of the Brownian separable permuton, so the whole sequence of longest increasing subsequence lengths is coupled to a single continuum object.
  • The leftmost maximal positive subtree of the conditioned tree $T_k$, rescaled by $\varphi(k)/k^{\gamma-1}$, converges in the Gromov-Hausdorff sense to a self-similar fragmentation tree.

Reading between the lines

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

  • If the main theorem is correct, the same exponent should govern uniform separable permutations and uniform cographs, with a constant factor of order $0.901$ as the paper conjectures; this would make the Brownian model the exact-exponent representative of the whole universality class.
  • The proof mechanism suggests a testable general principle: on random recursive trees where a natural exploration along a marked leaf converges locally, the size of the largest almost-monochromatic subtree has an exponent read off from the tail of the subtree-size distribution via a regular-variation index, rather than from the mean-field shape.
  • The paper's conjectures connecting $\alpha(p)$ to the Hausdorff dimension of the largest increasing subset of the permuton support and to a directed version of critical Liouville quantum gravity become quantitative predictions that can be checked independently once those objects are constructed.
  • One could test the robustness of the exponent by perturbing the tree signs to introduce short-range correlations; the method should break exactly where the symmetric region-size independence fails, predicting a different exponent.
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Editorial analysis

A structured set of objections, weighed in public.

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

Referee Report

2 major / 3 minor

Summary. The paper studies the longest increasing subsequence (LIS) of permutations sampled from the Brownian separable permuton μ_p, and the largest clique of graphs sampled from the Brownian cographon W_p. The main result, Theorem 1.1, asserts that for every p∈(0,1), with the natural Rémy coupling, LIS(σ_n)/n^{α(p)} converges almost surely to a non-deterministic, almost surely positive and finite random variable X(p) that is a deterministic measurable function of μ_p, where α(p) is the unique solution in (1/2,1) of the Gamma-function equation (1.3). The proof proceeds by translating the problem into a statement about the largest positive subtree of a p-signed uniform binary tree: existence of the exponent is proved by supermultiplicativity (Section 4), rough regularity of the sequences q and Q is established in Section 5, a local convergence and a law of large numbers for the intersection of maximal positive subtrees are proved in Sections 6–7, the exact exponent is extracted in Section 8, and the almost-sure scaling limit is claimed to follow in Section 9 via a backward-martingale decomposition with remainder control. The paper also announces a scaling limit for the leftmost maximal positive subtree (Theorem 2.6) and states analogous clique/independent-set results for the Brownian cographon.

Significance. If the claimed results hold, this is a substantial contribution. Prior to this work, even the existence of the exponent α(p) for LIS of Brownian separable permutons was open; the paper not only proves existence but identifies α(p) exactly through a simple Gamma-function equation and upgrades the earlier polynomial-factor bounds of BDSG24 to an almost-sure scaling limit with an explicit random variable. The same exponent governs the largest clique of the Brownian cographon, unifying the permuton and graphon settings. The proof strategy is impressive and self-contained: the exponent is first produced by a supermultiplicativity argument, the constant λ from the law of large numbers for #L^max_∩(T_k) is later identified with α a posteriori, and the final identification of α uses regularly varying sequences and a comparison with test sequences. The paper also connects the exponent to the spectral zeta function and gives a by-product scaling limit of the leftmost maximal positive subtree. The main caveat is completeness: the final almost-sure convergence step is not actually present in the submitted text, so the central claim cannot currently be verified.

major comments (2)
  1. [Section 9.3, Lemma 9.2 and Eq. (9.3)] The proof of Lemma 9.2 is absent. The text states Lemma 9.2 and says it follows from Lemmas 9.3–9.6, but the manuscript breaks off in Section 9.3 immediately after Eq. (9.12), before any of those lemmas or their proofs appear. Lemma 9.2 is load-bearing: it is the estimate Σ_{n≥n0}|R_n| ≤ n0^{-ε} that turns the quantitative law of large numbers for the conditioned model T_k (Theorem 9.1) into control of the remainder R_n in the backward-martingale decomposition (9.3) for the unconditioned Rémy tree T_n. Without this argument, the almost-sure convergence of n^{-α}LIS(T_n) in Theorem 2.3, and hence the main scaling limit in Theorem 1.1, is not demonstrated in the available text.
  2. [Sections 9.4 and 9.5] Sections 9.4 and 9.5 are announced in the table of contents but their content is not present in the submitted manuscript. Section 9.4 is needed to conclude the martingale argument and establish Theorem 2.3, while Section 9.5 is the only place where the statement that X is a deterministic measurable function of the Brownian separable permuton is justified; that measurability is part of the statement of Theorem 1.1. Section 9.6, which contains the proof of Theorem 2.6, is also missing. The final theorems are therefore not verifiable from the version under review, and this is not a presentation issue but a gap in the proof of the central claim.
minor comments (3)
  1. [Section 6.3.3, Definition 6.10] The definition of good scales in Definition 6.10 depends on a constant C coming from Lemma 6.9, but the relation between this C and the constants in Lemma 6.12 is not made explicit; adding a sentence stating that the C is fixed and universal throughout Section 6 would improve readability.
  2. [Section 9.2, proof of Theorem 9.1] In the display after Eq. (9.7), the radical expression is typeset awkwardly with 'radicaltp/radicalvertex' artifacts; the intended expression is the square root of δ2(k)-2αδ1(k). This is a typesetting issue but should be corrected.
  3. [Page 5, Figure 3] The caption says 'Left: The graph of the functions α*(p), α(p) and β*(p)' but the figure contains two panels and the right panel shows α(p)-α*(p); the caption's first 'Left' appears to describe both panels. Clarify the panel labels.

Circularity Check

0 steps flagged · score 0.0 of 10

No significant circularity: the exact exponent is derived from a self-contained recursive analysis; the only flagged issue is a missing proof of the remainder Lemma 9.2, which is an incompleteness concern rather than a circular step.

full rationale

The derivation chain is self-contained rather than circular. The exponent alpha is first shown to exist in Theorem 2.1 via a supermultiplicativity argument (Lemma 4.2) whose only input is the standard Dirichlet-multinomial decomposition of a uniform binary tree into random regions (Lemma 3.8). No fitted constant enters this step. The exact value of alpha is then obtained in Theorem 2.2 by combining the exact recursive relation for q(k)=P(LIS(T)=k) (Lemma 2.4) with the regular-variation estimates of Theorem 7.1. Those estimates are themselves proved from the law of large numbers for #Lmax_intersect(T_k) (Theorem 7.2), and the constant lambda appearing there is afterwards identified with alpha by comparing two independently obtained exponents (Remark 7.9 and the proof of Theorem 7.1); this is a post-hoc identification, not an assumption of the value of alpha. The scaling limit Theorem 2.3 is proved via a backward martingale decomposition plus control of remainder terms; this is a standard style of argument and does not reduce to any of its inputs by construction. The paper does cite prior work by overlapping authors, notably [BDSG24, Theorem 1.1], but this citation is used only for crude initial bounds and for the fact that alpha lies in (1/2,1), not for the exact Gamma-function equation; it is parameter-free and external to the present derivation, so it does not constitute load-bearing self-citation. One genuine gap is present in the provided text: Section 9.3 states Lemma 9.2 (control of the remainder sum) and says it follows from Lemmas 9.3, 9.4, 9.5 and 9.6, but the supplied manuscript breaks off before those lemmas and their proofs appear. This is an omitted proof or incompleteness in the text as provided, and it matters for the almost-sure convergence claim, but it is not a circularity: there is no indication in the visible text that Lemma 9.2 is equivalent to its own assumptions or that any fitted value is renamed as a prediction. No self-definitional relation, no fitted-input-called-prediction step, and no author-imported uniqueness theorem were found in the presented derivation.

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

The proof introduces no fitted constants. The auxiliary objects (τ_k, q, Q, λ, α) are defined from the model and determined by theorems; the constant λ is later identified with the exponent α. The main axioms are standard probabilistic tools plus the tree/permuton distributional identity. No new physical entities are postulated.

assumptions (5)
  • domain assumption A critical binary Bienaymé-Galton-Watson tree T conditioned to have n leaves is distributed as a uniform planar rooted binary tree with n leaves, and the Brownian separable permuton sampling is equivalent to leaf-reading of a p-signed uniform binary tree.
    Invoked in Section 2.1 and Remark 3.2; this links the permuton model to the tree model where all proofs operate.
  • standard math Rémy's algorithm gives a coupling of T_n across n such that T_{n-1} is obtained by removing a uniform leaf and contracting the parent node.
    Used throughout Sections 2.1, 4, 5 and 9; see Figure 5 and the discussion around (2.16).
  • standard math Lemma 3.8: conditional component sizes in a uniform binary tree are Dirichlet-multinomial with parameters 1/2, and attached regions are independent copies of T_{N_i}.
    Core combinatorial decomposition underlying the superbranching equation (4.3) and the whole proof.
  • standard math Karamata's characterization theorem, the uniform convergence theorem and Potter bounds for regularly varying functions.
    Used in Sections 3.1, 7.2 and 8.3 to convert ratio asymptotics of q and Q into regular variation.
  • standard math Monotonicity of the spectral zeta / Gamma expression in Lemma 8.3, with digamma bounds log x − 1/x ≤ ψ(x) ≤ log x − 1/(2x).
    Used in Section 8.4 to ensure the equation for α has a unique solution in (1/2,1).

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Pith. "Pith review of The longest increasing subsequence of Brownian separable permutons." pith.science (2026). https://pith.science/paper/MRYL5YRU

@misc{pith2026250619123,
  author       = {Pith},
  title        = {Pith review of: The longest increasing subsequence of Brownian separable permutons},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/MRYL5YRU}},
  note         = {Machine review of arXiv:2506.19123}
}
abstract

We establish a scaling limit result for the length $\operatorname{LIS}(\sigma_n)$ of the longest increasing subsequence of a permutation $\sigma_n$ of size $n$ sampled from the Brownian separable permuton $\boldsymbol{\mu}_p$ of parameter $p\in(0,1)$, which is the universal limit of pattern-avoiding permutations. Specifically, we prove that \[\frac{\operatorname{LIS}(\sigma_n)}{n^\alpha}\;\underset{n\to\infty}{\overset{\mathrm{a.s.}}{\longrightarrow}}\; X,\] where $\alpha=\alpha(p)$ is the unique solution in the interval $(1/2,1)$ to the equation \[\frac{1}{4^{\frac{1}{2\alpha}}\sqrt{\pi}}\,\frac{\Gamma\big(\tfrac{1}{2}-\tfrac{1}{2\alpha}\big)}{\Gamma\big(1-\tfrac{1}{2\alpha}\big)}=\frac{p}{p-1},\] and $X=X(p)$ is a non-deterministic and a.s. positive and finite random variable, which is a measurable function of the Brownian separable permuton. Notably, the exponent $\alpha(p)$ is an increasing continuous function of $p$ with $\alpha(0^+)=1/2$, $\alpha(1^-)=1$ and $\alpha(1/2)\approx0.815226$, which corresponds to the permuton limit of uniform separable permutations. We prove analogous results for the size of the largest clique of a graph sampled from the Brownian cographon of parameter $p\in(0,1)$.

Figures

Figures reproduced from arXiv: 2506.19123 by the authors.

Figure 1
Figure 1. Left: A planar rooted binary tree with 6 leaves, decorated with ⊕/⊖ signs on its nodes. Right: (1) The tree is obtained from the tree on the left by swapping the children of each ⊖ decorated node as explained below (1.1). The corresponding permutation πn is obtained by reading the labels of the leaves from left to right. (2) The graph is obtained from the tree on the left by following the strategy detailed below (1.… view at source ↗
Figure 2
Figure 2. The graph of the function α(p) from Theorem 1.1 for p ∈ (0, 1). The values of α(i/10) for i = 0, 1, . . . , 10 are shown on top of the function, where α(0) = α(0+) and α(1) = α(1−). 5 [PITH_FULL_IMAGE:figures/full_fig_p005_2.png] view at source ↗
Figure 3
Figure 3. Left: The graph of the functions α∗(p), α(p) and β ∗ (p) from Theorem 1.1 and Theorem 1.2. Right: The graphs of α(p) − α∗(p). The maximum is reached at p ≈ 0.202 where it takes the value α(p) − α∗(p) ≈ 0.0063. Moreover, α(p) − α∗(p) > 0 for all p ∈ (0, 1). We believe that the law of X(p) is absolutely continuous with respect to the Lebesgue measure and that for all p ̸= p ′ , the laws of X(p) and X(p ′ ) are distinc… view at source ↗
Figures from the paper (9 more)
Figure 4
Figure 4. Figure 4: Left: A sign-decorated binary tree t with a subtree t|L induced by the three leaves ℓ1, ℓ2, ℓ3. The uncontracted version u(t|L) of t|L is highlighted by the red curve on t. Right: The same sign-decorated binary tree t with all (i.e. two) maximal positive (uncontracted)…
Figure 5
Figure 5. Figure 5: Our version of the Rémy algorithm for p-signed uniform binary trees. In this realization, we start with a p-signed uniform binary tree Tn with n leaves, then we select a uniform edge En, sample a ⊖ sign (this happens with probability 1 − p) and finally, we graft a new …
Figure 6
Figure 6. Figure 6: A diagram for the strategy of the proof of the three main results of this paper, [PITH_FULL_IMAGE:figures/full_fig_p014_6.png]
Figure 7
Figure 7. Figure 7: We fix p = 1/2. In blue, the first 100 values of the sequence q(k) computed using the recursive relation in (2.8) with initial condition q(1) = 2 − √ 2. In orange, the first 100 values of the sequence qe(k) computed using the same recursive relation in (2.8) with initi…
Figure 8
Figure 8. Figure 8: A diagram for the notation introduced in Section [PITH_FULL_IMAGE:figures/full_fig_p025_8.png]
Figure 9
Figure 9. Figure 9: A sign-decorated binary tree t with two maximal positive subtrees highlighted in red and blue. The leftmost maximal positive subtree t lmax is the red tree. The leaves in L max ∩ (t) are highlighted in green. 39 [PITH_FULL_IMAGE:figures/full_fig_p039_9.png]
Figure 10
Figure 10. Figure 10: A schematic representation of the notation introduced for the process [PITH_FULL_IMAGE:figures/full_fig_p041_10.png]
Figure 11
Figure 11. Figure 11: A schematic representation for the statement of Lemma [PITH_FULL_IMAGE:figures/full_fig_p054_11.png]
Figure 12
Figure 12. Figure 12: Left: Approximate densities of the random variables X(p). Right: Comparison between the approximate density of the limiting random variable for LIS(σn(1/2)) nα(1/2) when the permutations σn are sampled from the Brownian separable permuton with parameter 1/2 and the ap…

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Reviewed August 15, 2026 · model on record in the stance chip above.