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Characters of symmetric groups: sharp bounds on virtual degrees and the Witten zeta function

Lucas Teyssier University of British Columbia, [email protected] Paul Thévenin Université d’Angers, [email protected]
Abstract

We prove sharp bounds on the virtual degrees introduced by Larsen and Shalev. This leads to improved bounds on characters of symmetric groups. We then sharpen bounds of Liebeck and Shalev concerning the Witten zeta function. Our main application is a characterization of the fixed-point free conjugacy classes whose associated random walk mixes in 2 steps.

Résumé

Nous prouvons des bornes précises sur les degrés virtuels introduits par Larsen et Shalev. Cela induit de meilleures bornes sur les caractères du groupe symétrique. Dans un second temps, nous améliorons certaines bornes de Liebeck et Shalev sur la fonction zeta de Witten. Notre application principale est une caractérisation des classes de conjugaison sans point fixe dont la marche aléatoire associée est mélangée au temps 2.

Resumo

Ni pruvas precizajn barojn je la virtualaj gradoj enkondukitaj de Larsen kaj Shalev. Tio induktas plibonigitajn barojn je karakteroj de la simetria grupo. Ni sekve pliakrigas barojn de Liebeck kaj Shalev pri la zeto-funkcio de Witten. Nia ĉefa aplikaĵo estas karakterizo de la senfikspunktaj konjugklasoj kies asociata hazarda promenado miksiĝas post 2 paŝoj.

1 Introduction

1.1 Context

Representation theory has been used to solve various problems in different areas of mathematics. A striking example is the pioneering work of Diaconis and Shahshahani on a random process called the random transposition shuffle [DS81]. They used character estimates to prove a sharp phase transition – now known as the cutoff phenomenon – concerning the minimum number of random transpositions whose product yields an almost uniform random permutation. This was the starting point of the field of mixing times, whose goal is to understand how long it takes for a random process to approach stationarity.

Other techniques to study mixing times were developed in the following years, notably by Aldous and Diaconis [Ald83, AD86], and mixing properties of emblematic card shufflings were precisely understood, also for small decks of cards [BD92]. We refer to [LP17] for an introduction to mixing times and to [DF23] for the mathematics of card shuffling.

A natural way to generalize Diaconis and Shahshahani’s random transposition shuffle on the symmetric group 𝔖n{\mathfrak{S}}_{n} is to replace the conjugacy class of transpositions – from which we pick uniform elements – by another conjugacy class. In specific cases, a similar cutoff phenomenon was proved ; for kk-cycles (where k>n/2k>n/2) in [LP02] ; for kk-cycles (where kk is fixed) in [BSZ11] ; for kk-cycles (where k=o(n)k=o(n)) in [Hou16] ; and for all conjugacy classes whose support has size o(n)o(n) in [BŞ19] (the support of a permutation is the set of its non-fixed points). Most of these results rely on representation theory. However, interestingly, the proof of [BSZ11] follows the cycle structure of the permutation obtained after several multiplications via probabilistic arguments, and that of [BŞ19], which is currently the best result for conjugacy classes with small support, relies on a curvature argument.

For conjugacy classes with cycle type [rn/r][r^{n/r}], Lulov [Lul96] proved that the mixing time is 33 when r=2r=2 and 22 when r3r\geq 3. Lulov and Pak later conjectured in [LP02, Conjecture 4.1] that the mixing time is at most 3 for all fixed-point free conjugacy classes (i.e., without fixed points), and Larsen and Shalev [LS08] proved that conjecture.

In recent years, results were obtained concerning the cutoff profile of such processes, that is, what happens within the cutoff window (when we zoom in, around the phase transition). The cutoff profile of the random transposition shuffle was determined by the first author in [Tey20], and this was later generalized to kk-cycles (for k=o(n)k=o(n)) by Nestoridi and Olesker-Taylor [NOT21]. Recently, another proof of the cutoff profile for transpositions was obtained by Jain and Sawhney [JS24], using a hybrid technique that combines probabilistic and representation theoretic arguments.

Motivated by various problems, including the study of mixing times of Markov chains, uniform bounds on characters have been established and sharpened over the past few decades. Roichman [Roi96] proved bounds that led to the correct order of the mixing time of conjugacy class walks on 𝔖n\mathfrak{S}_{n} that have a number of fixed points of order nn. This result was later strengthened by Müller and Schlage-Puchta [MSP07], who obtained results that are uniform over all conjugacy classes.

Sharp bounds for characters of permutations with few fixed points were proved in the landmark paper [LS08] of Larsen and Shalev. They proved several conjectures, among which the Lulov–Pak conjecture as discussed earlier. The bounds that we obtain here build on this paper ; see Section 1.2 for a more detailed description of their character bounds.

An important tool to estimate characters of 𝔖n{\mathfrak{S}}_{n} is the Murnaghan–Nakayama rule (see e.g. Theorem 3.10 in [Mé17]), which is a key element in the proofs of [Roi96, MSP07, LS08]. On the other hand, other techniques have been proved to be more powerful for certain diagrams. Féray and Śniady [FS11] found uniform bounds on characters that are especially good for diagrams that do not have a long first line or column, using a reformulation of Stanley’s conjectured character formula. More precisely, [Fé10] proves Conjecture 3 from [Sta06], while the reformulation of Stanley’s formula appears as Theorem 2 in [FS11].

In a different direction, Lifschitz and Marmor [LM23] recently used a new technique called hypercontractivity to establish character bounds. They obtain uniform bounds on characters for permutations with at most a given number of cycles (instead of considering the full cycle structure, as done in the previously mentioned works). This allows them to consider permutations having up to n/(lnn)O(1)n/(\ln n)^{O(1)} cycles, thus covering new cases.

Bounds on characters play a crucial role in the broader context of finite groups, and were notably used to prove a conjecture due to Ore [LOST10]. General bounds on characters of finite classical groups were recently obtained by Guralnick, Larsen and Tiep [GLT20, GLT24], and uniform bounds on characters for all finite quasisimple groups of Lie type were established by Larsen and Tiep [LT24].

A canonical tool used to convert character bounds for a group GG into practical estimates for applications is the Witten zeta function. It is defined as ζ(s)=α(dα)s\zeta(s)=\sum_{\alpha}(d_{\alpha})^{-s} for s>0s>0, where the sum is taken over all irreducible representations α\alpha of GG, and dαd_{\alpha} is the dimension of α\alpha. Such a zeta function initially appeared in the computation of volumes of moduli spaces in Witten’s work [Wit91] (see Equation (4.72) there). Liebeck and Shalev proved bounds for the symmetric group [LS04] and for Fuchsian groups [LS05], which were useful for a variety of applications, including mixing times and covering numbers [LS08, LST24, KLS24]. We refer to [LS08], [LT24], and the survey article [Lie17] for further applications of characters bounds.

Throughout the paper, 𝔖n\mathfrak{S}_{n} denotes the symmetric group of order nn, and 𝔖n^\widehat{\mathfrak{S}_{n}} the set of its irreducible representations. For a permutation σ\sigma and i{1,,n}i\in\{1,\ldots,n\}, fi(σ)f_{i}(\sigma) denotes the number of cycles of length ii in σ\sigma. By extension, for a conjugacy class 𝒞{\mathcal{C}}, we denote by fi(𝒞)f_{i}({\mathcal{C}}) the number of cycles of length ii in any permutation σ𝒞\sigma\in{\mathcal{C}}. In addition, we will use the convenient notation f=max(f1,1)f=\max(f_{1},1). For convenience, we will use λ\lambda to simultaneously denote a representation, the associated integer partition, and the associated Young diagram (see Section 2.1 for definitions). If EE is a finite set, we denote by UnifE\operatorname{\mathrm{Unif}}_{E} the uniform probability measure on EE.

1.2 The Larsen–Shalev character bounds

We describe here the results of Larsen and Shalev [LS08], whose improvement is the main purpose of the paper. We assume familiarity with the representation theory of symmetric groups and will follow the notation from the textbook [Mé17]: we denote by dλd_{\lambda} the dimension of a representation λ\lambda, chλ(σ)\operatorname{\mathrm{ch}}^{\lambda}(\sigma) the character of λ\lambda evaluated at a permutation σ𝔖n\sigma\in\mathfrak{S}_{n}, and χλ(σ)=chλ(σ)dλ\chi^{\lambda}(\sigma)=\frac{\operatorname{\mathrm{ch}}^{\lambda}(\sigma)}{d_{\lambda}} the associated normalized character.

Larsen and Shalev obtained character bounds by introducing and studying new objects. One of them is called the virtual degree D(λ)D(\lambda) of a representation λ\lambda. It is defined as

D(λ):=(n1)!iai!bi!,D(\lambda):=\frac{(n-1)!}{\prod_{i}a_{i}!b_{i}!}, (1.1)

where ai=λiia_{i}=\lambda_{i}-i and bi=λiib_{i}=\lambda^{\prime}_{i}-i. Here λi\lambda_{i} (resp. λi\lambda^{\prime}_{i}) denotes the size of the ii-th row (resp. ii-th column) of the Young diagram associated to λ\lambda (see Section 2.1 for precise definitions). The virtual degree is a convenient substitute for the dimension dλd_{\lambda}, which allows for easier computations. See Section 2.3 for more details.

Another object introduced in [LS08] is the orbit growth exponent E(σ)E(\sigma). Let n1n\geq 1 and σ𝔖n\sigma\in\mathfrak{S}_{n}. Denote by fi(σ)f_{i}(\sigma) the number of cycles of length i1i\geq 1 of σ\sigma, and for k1k\geq 1, set Σk=Σk(σ):=1ikifi(σ)\Sigma_{k}=\Sigma_{k}(\sigma):=\sum_{1\leq i\leq k}if_{i}(\sigma). We define the orbit growth sequence (ei)i1(e_{i})_{i\geq 1} by ne1++ei=max(Σi,1)n^{e_{1}+\ldots+e_{i}}=\max\left(\Sigma_{i},1\right) for all i1i\geq 1, and set

E(σ)=i1eii.E(\sigma)=\sum_{i\geq 1}\frac{e_{i}}{i}. (1.2)

A key intermediate result that Larsen and Shalev proved is the following, which we will refer to as a theorem.

Theorem 1.1 ([LS08]).

For any n1n\geq 1, σ𝔖n\sigma\in\mathfrak{S}_{n} and λ𝔖n^\lambda\in\widehat{\mathfrak{S}_{n}}, we have the following character bound:

|chλ(σ)|D(λ)E(σ).\left|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)\right|\leq D(\lambda)^{E(\sigma)}. (1.3)

This is proved via an elegant induction in the proof of their main theorem (for reference, the induction hypothesis is [LS08, Eq. (17)]), where the orbit growth exponent E(σ)E(\sigma) naturally appears. They also showed that D(λ)D(\lambda) cannot be much larger than dλd_{\lambda}.

Theorem 1.2 ([LS08], Theorem 2.2).

As |λ||\lambda|\to\infty, we have

D(λ)dλ1+o(1).D(\lambda)\leq d_{\lambda}^{1+o(1)}. (1.4)

Combining Theorem 1.1, Theorem 1.2 and the fact that E(σ)1E(\sigma)\leq 1 for all σ\sigma, they obtain the following character bound.

Theorem 1.3 ([LS08], Theorem 1.1 (a)).

As nn\to\infty, we have uniformly over all σ𝔖n\sigma\in\mathfrak{S}_{n} and λ𝔖n^\lambda\in\widehat{\mathfrak{S}_{n}},

|chλ(σ)|dλE(σ)+o(1).\left|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)\right|\leq d_{\lambda}^{E(\sigma)+o(1)}. (1.5)

The orbit growth exponent E(σ)E(\sigma) defined in (1.2) looks complicated at first glance, but it is actually simple to compute and appears to lead to the best possible character bounds in different regimes such as in (1.6) below.

Let us give a few important examples of values of E(σ)E(\sigma). We always denote by f1f_{1} the number of fixed points of a permutation σ\sigma and set f=max(f1,1)f=\max(f_{1},1).

Example 1.4.
  1. (a)

    Assume that σ[2n/2]\sigma\sim[2^{n/2}], then e2=1e_{2}=1 and ei=0e_{i}=0 for i2i\neq 2, so E(σ)=1/2E(\sigma)=1/2.

  2. (b)

    Assume that σ\sigma has no fixed point, then e2+e3+=1e_{2}+e_{3}+...=1 so E(σ)1/2E(\sigma)\leq 1/2.

  3. (c)

    Assume that σ\sigma has at least one fixed point, then e1=lnflnne_{1}=\frac{\ln f}{\ln n} so E(σ)e1+12(1e1)=1+e12=12+lnf2lnnE(\sigma)\leq e_{1}+\frac{1}{2}(1-e_{1})=\frac{1+e_{1}}{2}=\frac{1}{2}+\frac{\ln f}{2\ln n}, that is, E(σ)112ln(n/f)lnnE(\sigma)-1\leq-\frac{1}{2}\frac{\ln(n/f)}{\ln n}

  4. (d)

    Assume that σ[k,1nk]\sigma\sim[k,1^{n-k}], then E(σ)=1/nE(\sigma)=1/n if k=nk=n (i.e. if σ\sigma is an nn-cycle) and if f11f_{1}\geq 1, E(σ)=e1+ekk=e1+1e1k=lnflnn+ln(n/f)klnnE(\sigma)=e_{1}+\frac{e_{k}}{k}=e_{1}+\frac{1-e_{1}}{k}=\frac{\ln f}{\ln n}+\frac{\ln(n/f)}{k\ln n}.

Combining Theorem 1.3 with Example 1.4 (b), Larsen and Shalev obtained the following bound as nn\to\infty, uniform over fixed-point free permutations σ𝔖n\sigma\in\mathfrak{S}_{n} and irreducible representations λ𝔖n^\lambda\in\widehat{\mathfrak{S}_{n}},

|chλ(σ)|dλ1/2+o(1).\left|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)\right|\leq d_{\lambda}^{1/2+o(1)}. (1.6)

This allowed them to prove the abovementioned conjecture of Lulov and Pak, and even extend it (in Theorem 1.8) to the mixing time of conjugacy classes that have a small number of fixed points.

1.3 Main results

We present here our main contributions: the main result, Theorem 1.5, which improves on Theorem 1.2, provides uniform bounds on virtual degrees and allows us to obtain refined character bounds (Theorem 1.6). Proposition 1.8 provides bounds on the Witten zeta function, which we apply together with Theorem 1.6 to characterize fixed-point free conjugacy classes that mix in 2 steps (Theorem 1.9). We also deduce from our bounds that conjugacy classes with o(n)o(\sqrt{n}) cycles mix fast (Theorem 1.10), and finally that there is a cutoff whenever the conjugacy class has a large support and no short cycles (Theorem 1.12).

Theorem 1.5.

There exists a universal constant CC such that, for every n2n\geq 2 and any integer partition λn\lambda\vdash n, we have

D(λ)dλ1+Clnn.D(\lambda)\leq d_{\lambda}^{1+\frac{C}{\ln n}}. (1.7)

We will show in Section 4.4 that Theorem 1.5 is sharp for different shapes of diagrams.

Plugging this into Theorem 1.1, we get the following character bound.

Theorem 1.6.

For every n2n\geq 2, any permutation σ𝔖n\sigma\in\mathfrak{S}_{n} and any integer partition λn\lambda\vdash n, we have

|chλ(σ)|dλ(1+Clnn)E(σ),\left|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)\right|\leq d_{\lambda}^{\left(1+\frac{C}{\ln n}\right)E(\sigma)}, (1.8)

where CC is the universal constant from Theorem 1.5.

In Proposition 5.1 we prove that E(σ)ln(max(cyc(σ),2))lnnE(\sigma)\leq\frac{\ln\left(\max(\operatorname{\mathrm{cyc}}(\sigma),2)\right)}{\ln n}, where cyc(σ)\operatorname{\mathrm{cyc}}(\sigma) is the number of cycles of σ\sigma. This allows us to deduce the following character bound, which improves on [LS08, Theorem 1.4] and [LM23, Theorem 1.7].

Theorem 1.7.

As nn\to\infty, uniformly over all σ𝔖n\sigma\in{\mathfrak{S}}_{n} and λ𝔖n^\lambda\in\widehat{{\mathfrak{S}}_{n}}, we have

|chλ(σ)|dλlncyc(σ)lnn+O(1lnn).\left|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)\right|\leq d_{\lambda}^{\frac{\ln\operatorname{\mathrm{cyc}}(\sigma)}{\ln n}+O\left(\frac{1}{\ln n}\right)}. (1.9)

We now define the Witten zeta function. For n1n\geq 1, A𝔖n^A\subset\widehat{\mathfrak{S}_{n}} and s0s\geq 0,

ζn(A,s):=λA1(dλ)s.\zeta_{n}(A,s):=\sum_{\lambda\in A}\frac{1}{\left(d_{\lambda}\right)^{s}}. (1.10)

Note that compared to the usual definition where the sum is over all irreducible representations, we add the subset of representations AA on which we sum as a parameter of the function ζn\zeta_{n}.

In [LS04], Liebeck and Shalev proved that ζn(𝔖n^,s)=O(ns)\zeta_{n}(\widehat{\mathfrak{S}_{n}}^{**},s)=O(n^{-s}), if s>0s>0 is fixed and nn\to\infty, where 𝔖n^=𝔖n^\{[n],[1n]}\widehat{\mathfrak{S}_{n}}^{**}=\widehat{\mathfrak{S}_{n}}\backslash\left\{[n],[1^{n}]\right\}. We improve their result to allow the argument ss to tend to 0.

Proposition 1.8.

Let (sn)(s_{n}) be a sequence of positive real numbers. We have

ζn(𝔖n^,sn)n0 if and only if snlnnn.\zeta_{n}(\widehat{\mathfrak{S}_{n}}^{**},s_{n})\xrightarrow[n\to\infty]{}0\quad\text{ if and only if }\quad s_{n}\ln n\xrightarrow[n\to\infty]{}\infty. (1.11)

In Section 6, we will also prove variants of Proposition 1.8, for other subsets A𝔖n^A\subset\widehat{{\mathfrak{S}}_{n}}, and when sns_{n} is of order 1lnn\frac{1}{\ln n}.

As an application of our bounds on characters and on the Witten zeta function, we are able to characterize which fixed-point free conjugacy classes mix in 2 steps. We recall that the convolution product of two measures μ,ν\mu,\nu on a group GG is defined by μν(g)=xGμ(x)ν(gx1)\mu\ast\nu(g)=\sum_{x\in G}\mu(x)\nu(gx^{-1}) for gGg\in G. As such, if SGS\subset G, then UnifSt\operatorname{\mathrm{Unif}}_{S}^{*t} is the distribution of the simple random walk (started at the neutral element) on the Cayley graph Cay(G,S)\operatorname{\mathrm{Cay}}(G,S) after tt steps.

Theorem 1.9.

For each n2n\geq 2 let 𝒞(n){\mathcal{C}}^{(n)} be a conjugacy class of 𝔖n{\mathfrak{S}}_{n}, which is fixed-point free (i.e. f1(𝒞(n))=0f_{1}({\mathcal{C}}^{(n)})=0). Recall that f2(𝒞(n))f_{2}({\mathcal{C}}^{(n)}) denotes the number of transpositions of a permutation σ𝒞(n)\sigma\in{\mathcal{C}}^{(n)}. Then we have

dTV(Unif𝒞(n)2,Unif𝔄n)n0 if and only if f2(𝒞(n))=o(n).\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right)\xrightarrow[n\to\infty]{}0\quad\quad\text{ if and only if }\quad\quad f_{2}({\mathcal{C}}^{(n)})=o(n). (1.12)

We also prove that conjugacy classes with few cycles mix fast.

Theorem 1.10.

For each n2n\geq 2 let 𝒞(n){\mathcal{C}}^{(n)} be a conjugacy class of 𝔖n{\mathfrak{S}}_{n}. Assume that cyc(σ)=o(n)\operatorname{\mathrm{cyc}}(\sigma)=o\left(\sqrt{n}\right). Then

dTV(Unif𝒞(n)2,Unif𝔄n)n0.\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right)\xrightarrow[n\to\infty]{}0. (1.13)

We believe that the only parts of the cycle structure that affect mixing times are the number of fixed points and the number of transpositions. We make the following conjecture.

Conjecture 1.11.

For each n2n\geq 2 let 𝒞(n){\mathcal{C}}^{(n)} be a conjugacy class of 𝔖n{\mathfrak{S}}_{n}. Then

dTV(Unif𝒞(n)2,Unif𝔄n)n0 if and only if [f1=o(n) and f2=o(n)],\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right)\xrightarrow[n\to\infty]{}0\quad\text{ if and only if }\quad[f_{1}=o(\sqrt{n})\text{ and }f_{2}=o(n)], (1.14)

where f1=f1(𝒞(n))f_{1}=f_{1}({\mathcal{C}}^{(n)}) and f2=f2(𝒞(n))f_{2}=f_{2}({\mathcal{C}}^{(n)}).

As a third application, we finally show that conjugacy classes with a large support and no short cycles exhibit cutoff. If 𝒞{\mathcal{C}} is a conjugacy class on 𝔖n{\mathfrak{S}}_{n} and t0t\geq 0, we denote 𝔈(𝒞,t)\mathfrak{E}({\mathcal{C}},t) the set 𝔖n\𝔄n\mathfrak{S}_{n}\backslash\mathfrak{A}_{n} if 𝒞𝔖n\𝔄n{\mathcal{C}}\subset\mathfrak{S}_{n}\backslash\mathfrak{A}_{n} and tt is odd, and 𝔄n\mathfrak{A}_{n} otherwise.

Theorem 1.12.

For each n2n\geq 2 let 𝒞(n){Id}{\mathcal{C}}^{(n)}\neq\left\{\operatorname{\mathrm{Id}}\right\} be a conjugacy class of 𝔖n{\mathfrak{S}}_{n}. Write f=max(f1(𝒞(n)),1)f=\max(f_{1}({\mathcal{C}}^{(n)}),1). Assume that f=o(n)f=o(n) and min{i2:fi(𝒞(n))1}n\min\left\{i\geq 2{\;:\;}f_{i}({\mathcal{C}}^{(n)})\geq 1\right\}\xrightarrow[n\to\infty]{}\infty. Then for every ε>0\varepsilon>0, we have

d(n)(lnnln(n/f)(1ε))n1 and d(n)(lnnln(n/f)(1+ε))n0,\mathrm{d}^{(n)}\left(\left\lfloor\frac{\ln n}{\ln(n/f)}(1-\varepsilon)\right\rfloor\right)\xrightarrow[n\to\infty]{}1\quad\text{ and }\quad\mathrm{d}^{(n)}\left(\left\lceil\frac{\ln n}{\ln(n/f)}(1+\varepsilon)\right\rceil\right)\xrightarrow[n\to\infty]{}0, (1.15)

where d(n)(t):=dTV(Unif𝒞(n)t,Unif𝔈n(𝒞(n),t))\mathrm{d}^{(n)}(t):=\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*t},\operatorname{\mathrm{Unif}}_{{\mathfrak{E}}_{n}({\mathcal{C}}^{(n)},t)}\right) for t0t\geq 0.

Remark 1.1.

Our improved bounds are useful in several ways. They lead to sharp results for permutations with few fixed points or few cycles, as we saw in Theorem 1.9 and Theorem 1.10. They also extend the applicability of the Larsen–Shalev character bounds (recalled as Theorem 1.3) from f=n1ε(n)f=n^{1-\varepsilon(n)} (where ε(n)\varepsilon(n) is the non-explicit o(1)o(1) from Theorem 1.3) to f=o(n)f=o(n). In the regime f=o(n)f=o(n), recalling Example 1.4 (c), this leads to the bound

|chλ(σ)|dλdλ(12+o(1))ln(n/f)lnn.\frac{|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|}{d_{\lambda}}\leq d_{\lambda}^{-\left(\frac{1}{2}+o(1)\right)\frac{\ln(n/f)}{\ln n}}. (1.16)

The constant 1/21/2 above cannot be improved, due to fixed point free involutions.

We can also derive bounds in the case where ff is of order nn. More precisely, when fe6Cnf\leq e^{-6C}n (where CC is the constant appearing in Theorems 1.5 and 1.6), we obtain from Example 1.4 (c) and Theorem 1.6:

|chλ(σ)|dλdλ13ln(n/f)lnn.\frac{|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|}{d_{\lambda}}\leq d_{\lambda}^{-\frac{1}{3}\frac{\ln(n/f)}{\ln n}}. (1.17)

Combining (1.17) with Roichman’s character bound [Roi96] (applied with δ=e6C\delta=e^{-6C}), it can be shown that there exists a constant q>0q>0 such that uniformly over all σ,λ\sigma,\lambda we have

|chλ(σ)|dλdλqln(n/f)lnn.\frac{|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|}{d_{\lambda}}\leq d_{\lambda}^{-q\frac{\ln(n/f)}{\ln n}}. (1.18)

Up to the value of the constant qq, this is the character bound of Müller and Schlage-Puchta [MSP07, Theorem B (i)]. Note that another combination, namely combining [LS08, Theorem 1.3] and [MSP07, Theorem B (i)], shows that one can take q=112+o(1)q=\frac{1}{12}+o(1) in (1.18).

1.4 Structure of the paper

Section 2 is devoted to preliminaries on Young diagrams and some elementary results. We recall there in particular the celebrated hook length formula.

In Section 3.1, we introduce the sliced hook products. This generalizes the idea of replacing the whole hook product of a diagram by a simpler product to any set partition of a diagram. In Section 3.2, we study sliced hook products and obtain precise approximations of (classical) hook products. A key result is the bound dλ(n|c|)dsdce6|c|d_{\lambda}\geq\binom{n}{|c|}d_{s}d_{c}e^{6\sqrt{|c|}} presented in Proposition 3.8 (b), which improves on [LS08, Lemma 2.1]. Here, ss denotes the external hook of λ\lambda and c=λ\sc=\lambda\backslash s is the center of λ\lambda (as drawn in Section 2.3). In Section 3.3, we introduce the notion of augmented dimension dλ+d_{\lambda}^{+} and consider some of its useful properties.

In Section 4 we use the results of Section 3.2 to prove Theorem 1.5 in two steps. We first show in Section 4.2 that the virtual degree D(λ)D(\lambda) and the augmented dimension dλ+d_{\lambda}^{+} are close to each other. We then prove in Section 4.3, by induction, slicing the external hooks of diagrams λ\lambda, that dλ+=dλ1+O(1/ln|λ|)d_{\lambda}^{+}=d_{\lambda}^{1+O(1/\ln|\lambda|)}. Finally, in Section 4.4 we provide examples that show that Theorem 1.5 is sharp up to the value of the constant.

Section 5 is devoted to the proof of Theorem 1.7, which shows character bounds in terms of the number of cycles.

In Section 6, we improve the bounds from Liebeck and Shalev [LS04] on the Witten zeta function and prove Proposition 1.8.

In Section 7, we apply the previous results and characterize which fixed-point free conjugacy classes mix in 2 steps, proving Theorem 1.9.

Finally, we prove in Section 8 Theorems 1.10 and 1.12 about mixing time estimates for specific conjugacy classes.

2 Preliminaries

2.1 Young diagrams, representations and integer partitions

We say that λ:=[λ1,,λk]\lambda:=[\lambda_{1},\ldots,\lambda_{k}] is a partition of the integer nn if λi1:={1,2,}\lambda_{i}\in\mathbb{Z}_{\geq 1}:=\{1,2,\ldots\} for all 1ik1\leq i\leq k, λiλi+1\lambda_{i}\geq\lambda_{i+1} for all ik1i\leq k-1, and i=1kλi=n\sum_{i=1}^{k}\lambda_{i}=n. We write λn\lambda\vdash n if λ\lambda is a partition of nn. It turns out that integers partitions of nn are in bijection with irreducible representations of 𝔖n{\mathfrak{S}}_{n}, which makes it a good tool to study characters.

A common and useful way to code an integer partition (and thus a representation of 𝔖n{\mathfrak{S}}_{n}) is through Young diagrams. The Young diagram of shape λ:=[λ1,,λk]\lambda:=\left[\lambda_{1},\ldots,\lambda_{k}\right] (where λ\lambda is a partition of an integer nn) is a table of boxes, whose ii-th line is made of λi\lambda_{i} boxes, see Fig. 1 for an example.

{ytableau}\none

&

\none
\none
\none
\none
\none
\none
Figure 1: The Young diagram coding the partition [5,2,1][5,2,1] of the integer 88. It has 5 boxes on the first line, 2 on the second, and 1 on the third.

We also denote by λ\lambda this Young diagram, and say that nn is its size.

2.2 Hook lengths, hook products and the hook-length formula

Let λ\lambda be a Young diagram. If uλu\in\lambda, the hook of uu in λ\lambda is the set of boxes on the right or above uu, including uu. We call hook length the size of this set and denote it by H(λ,u)H(\lambda,u). See Fig. 2 for an example.

{ytableau}\none

&

\none
\none

*(orange)

\none
\none
\none
\none{ytableau}
\none

&

\none

*(pink!60)

\none

*(pink!60) *(pink!60) *(pink!60) *(pink!60)

\none
Figure 2: Left: the Young diagram associated to λ=[7,5,4,1]\lambda=[7,5,4,1], with a box uu colored in orange. Right: the hook associated to uu is in pink. Its length is H(λ,u)=5H(\lambda,u)=5.

Let us now consider a subset of boxes EλE\subset\lambda. We define

H(λ,E):=uEH(λ,u)H(\lambda,E):=\prod_{u\in E}H(\lambda,u) (2.1)

the hook-product of EE in λ\lambda. See Fig. 3 for an example.

{ytableau}\none

&

\none

*(orange)

\none

*(orange) *(orange)

\none

*(orange) *(orange) *(orange) \none\none\none

Figure 3: The Young diagram coding λ=[7,5,4,1]\lambda=[7,5,4,1], with the set E={(1,1),(3,1),(6,1),(1,2),(2,2),(3,3)}E=\left\{(1,1),(3,1),(6,1),(1,2),(2,2),(3,3)\right\} in orange. Here, H(λ,E)=1072752=9800H(\lambda,E)=10\cdot 7\cdot 2\cdot 7\cdot 5\cdot 2=9800.

We can now state the hook-length formula. Let λ\lambda be a Young diagram and n:=|λ|n:=|\lambda| its number of boxes. A standard tableau of λ\lambda is a filling of λ\lambda with the numbers from 1 to nn such that the numbers are increasing on each line and column. We denote by ST(λ)\operatorname{ST}(\lambda) the set of standard tableaux of λ\lambda, and dλ:=|ST(λ)|d_{\lambda}:=|\operatorname{ST}(\lambda)|, see Fig. 4. It is well-known that dλd_{\lambda} is the dimension of the irreducible representation of the symmetric group associated to λ\lambda.

{ytableau}\none

& *(green!60)4 *(cyan!60)5

\none

*(red!60)1 *(orange!60)2 *(yellow!60)3 \none {ytableau} \none& *(yellow!60)3 *(cyan!60)5

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*(red!60)1 *(orange!60)2 *(green!60)4 \none{ytableau} \none& *(yellow!60)3 *(green!60)4

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*(red!60)1 *(orange!60)2 *(cyan!60)5 \none{ytableau} \none& *(orange!60)2 *(cyan!60)5

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*(red!60)1 *(yellow!60)3 *(green!60)4 \none{ytableau} \none& *(orange!60)2 *(green!60)4

\none

*(red!60)1 *(yellow!60)3 *(cyan!60)5 \none[.]

Figure 4: The diagram λ=[3,2]\lambda=[3,2] and its dλ=|ST(λ)|=5d_{\lambda}=|\operatorname{ST}(\lambda)|=5 standard tableaux.

The hook-length formula was discovered in 1953 by Frame, Robinson, and Thrall [FRT54], and allows to compute the number of standard tableaux of a diagram looking only at its hook lengths: for any diagram λ\lambda of size nn, we have

dλ=n!H(λ,λ).d_{\lambda}=\frac{n!}{H(\lambda,\lambda)}. (2.2)
Example 2.1.

Let us consider again the λ=[3,2]\lambda=[3,2], which has size 5. We have n!H(λ,λ)=5!43121=12024=5\frac{n!}{H(\lambda,\lambda)}=\frac{5!}{4\cdot 3\cdot 1\cdot 2\cdot 1}=\frac{120}{24}=5. Therefore, by the hook length formula, we recover that dλ=5d_{\lambda}=5.

2.3 Diagram notation

Let us define some notation that we will use for diagrams. To keep the notation light, we will usually use the same pieces of notation for both the diagrams and their lengths. For example we will say the first line of λ\lambda is λ1\lambda_{1} and has length λ1\lambda_{1}. If we want to emphasize that we are considering a diagram or a size, we will add brackets or absolute values. For example we can say that the first line [λ1]\left[\lambda_{1}\right] has length |λ1|\left|\lambda_{1}\right|.

Definition 2.2 (Definition-notation).

Let λ\lambda be a Young diagram.

  • We denote by λi\lambda_{i} the ii-th line of λ\lambda and by λi\lambda_{i}^{\prime} its ii-th column.

  • We denote by δi\delta_{i} the ii-th diagonal box of λ\lambda, that is, δi=λiλi\delta_{i}=\lambda_{i}\cap\lambda^{\prime}_{i}.

  • We denote by δ=δ(λ)\delta=\delta(\lambda) the diagonal of λ\lambda, that is, δ=iδi\delta=\cup_{i}\delta_{i}.

  • We denote by rr the truncated diagram (removing the first line of λ\lambda, s=s(λ):=λ1λ1s=s(\lambda):=\lambda_{1}\cup\lambda_{1}^{\prime} the external hook of λ\lambda, and by c=λ\sc=\lambda\backslash s its center (i.e. the diagram, except the first line and the first column).

  • We denote λi:=jiλj\lambda_{\geq i}:=\cup_{j\geq i}\lambda_{j} and λi:=jiλj\lambda_{\leq i}:=\cup_{j\leq i}\lambda_{j}.

  • Similarly, we define λi:=jiλj\lambda^{\prime}_{\geq i}:=\cup_{j\geq i}\lambda^{\prime}_{j}, λi:=jiλj\lambda^{\prime}_{\leq i}:=\cup_{j\leq i}\lambda^{\prime}_{j}, and λi1i2:=i1jiλj\lambda^{\prime}_{i_{1}\to i_{2}}:=\cup_{i_{1}\leq j\leq i}\lambda^{\prime}_{j}.

  • We denote by λai=ai\lambda_{a}^{i}=a_{i} and λbi=bi\lambda_{b}^{i}=b_{i} the truncated ii-th lines and columns of λ\lambda.

  • We denote by λi\lambda^{i} the ithi-th hook of λ\lambda, that is the boxes that are on the right of δi\delta_{i} on λi\lambda_{i} or above δi\delta_{i} on λi\lambda^{\prime}_{i}. Formally we can write λi=(λiλi)(λiλi)\lambda^{i}=(\lambda_{i}\cap\lambda^{\prime}_{\geq i})\cup(\lambda_{\geq i}\cap\lambda^{\prime}_{i}).

  • We define λi:=jiλj\lambda^{\geq i}:=\cup_{j\geq i}\lambda^{j}, λi:=jiλj\lambda^{\leq i}:=\cup_{j\leq i}\lambda^{j}.

  • We denote by uiu_{i} the part of the ii-th column which is above the first line. That is ui=λiλ2u_{i}=\lambda_{i}^{\prime}\cap\lambda_{\geq 2}.

  • We define ui:=jiuju_{\geq i}:=\cup_{j\geq i}u_{j}, ui:=jiuju_{\leq i}:=\cup_{j\leq i}u_{j}.

  • We denote by λi,j\lambda_{\geq i,\geq j} the subdiagram of λ\lambda whose boxes are on a line after the ii-th and on a column after the jj-th.

Fig. 5 illustrates these definitions on diagrams.

{ytableau}\none
\none

&*(green!60)\none[λ_4]

\none

*(yellow!60)*(yellow!60)*(yellow!60)\none[λ_3]

\none

*(orange!60)*(orange!60)*(orange!60)*(orange!60) \none[λ_2]

\none

*(red!60)*(red!60)*(red!60)*(red!60)*(red!60) \none[λ_1] {ytableau} \none&\none[λ_1’]

\none

*(red!60)\none[λ_2’]\none[λ_3’]

\none

*(red!60)*(orange!60) *(yellow!60)\none[λ_4’]

\none

*(red!60)*(orange!60)*(yellow!60)*(green!60)\none[λ_5’]

\none

*(red!60)*(orange!60)*(yellow!60) *(green!60) *(cyan!60) \none {ytableau} \none

\none

& \none\none\none[δ]

\none

*(red!60)\none

\none

*(red!60)

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*(red!60) \none

{ytableau}\none
\none

&*(red!60) \none\none\none[r = λ_≥2]

\none

*(red!60)*(red!60)*(red!60)

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*(red!60)*(red!60)*(red!60)*(red!60)

\none
\none{ytableau}
\none
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&*(red!60) \none\none[s = λ^1]\none

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*(red!60)\none

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*(red!60)

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*(red!60)*(red!60)*(red!60) *(red!60) *(red!60) \none {ytableau} \none

\none

&\none\none\none[c = λ^≥2]

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*(red!60)*(red!60)\none

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*(red!60)*(red!60)*(red!60)

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\none{ytableau}
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&\none

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[u_2]

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*(red!60)\none

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*(red!60)

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\none{ytableau}
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&*(red!60) \none\none[u_≤3]

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*(red!60)*(red!60)*(red!60)\none

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*(red!60)*(red!60)*(red!60)

\none
\none{ytableau}
\none

&\none[b_1 = λ_b^1]

\none

*(cyan!60) \none\none[b_2 = λ_b^2] \none\none

\none

*(cyan!60)*(green!60)*(yellow!60) δ_3\none

\none

*(cyan!60)*(yellow!60)δ_2*(orange!60)*(orange!60)\none\none[a_2 = λ_a^1]

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*(yellow!60)δ_1*(red!60)*(red!60)*(red!60)*(red!60) \none\none[a_1 = λ_a^1]

{ytableau}\none
\none

&

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*(orange!90)*(orange!90)*(orange!90) \none\none[λ’_≥2]

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*(orange!90)*(orange!90)*(orange!90) *(orange!90)*(orange!90)

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*(orange!90)*(orange!90)*(orange!90) *(orange!90) *(orange!90) *(orange!90)

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*(orange!90) *(orange!90) *(orange!90)*(orange!90)*(orange!90)*(orange!90)*(orange!90) \none\none {ytableau} \none

\none

& *(orange!90)

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*(orange!90)*(orange!90) \none\none[λ_≥2,≥3]

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*(orange!90)*(orange!90)*(orange!90)

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*(orange!90)*(orange!90)*(orange!90)*(orange!90)*(orange!90)

\none
Figure 5: Examples of Definition 2.2.

2.4 Some elementary results

We collect here standard or elementary results that we will us throughout the paper. We give proofs for completeness.

Lemma 2.3.
  1. (a)

    For any n1n\geq 1 and λn\lambda\vdash n we have dλn!d_{\lambda}\leq\sqrt{n!}.

  2. (b)

    For n6n\geq 6, we have n!(n/2)nn!\leq(n/2)^{n}.

  3. (c)

    Let n1n\geq 1 and λn\lambda\vdash n. Denote by s=s(λ)s=s(\lambda) the external hook of λ\lambda. Then ds=(s1λa1)d_{s}=\binom{s-1}{\lambda_{a}^{1}}.

Proof.
  1. (a)

    A classical formula from representation theory is the following: for any finite group GG we have |G|=ρG^dρ2|G|=\sum_{\rho\in\widehat{G}}d_{\rho}^{2}, where G^\widehat{G} is the set of all irreducible representations of GG. This identity represents for example the dimensions on both sides of the Fourier isomorphism (see [Mé17, Section 1.3]). Hence, for any λG^\lambda\in\widehat{G} of a finite group GG we have

    dλ2ρG^dρ2=|G|.d_{\lambda}^{2}\leq\sum_{\rho\in\widehat{G}}d_{\rho}^{2}=|G|. (2.3)

    The result follows taking square roots on both sides, since for n1n\geq 1 we have |𝔖n|=n!|\mathfrak{S}_{n}|=n!.

  2. (b)

    We proceed by induction. The result holds for n=6n=6 since 6!=720729=36=(6/2)66!=720\leq 729=3^{6}=(6/2)^{6}. Let now n6n\geq 6 and assume that n!(n/2)nn!\leq(n/2)^{n}. Then

    (n+1)!((n+1)/2)n+1=n!(n/2)n(n+1)2n+1(nn+1)n2(nn+1)n.\frac{(n+1)!}{((n+1)/2)^{n+1}}=\frac{n!}{(n/2)^{n}}(n+1)\frac{2}{n+1}\left(\frac{n}{n+1}\right)^{n}\leq 2\left(\frac{n}{n+1}\right)^{n}. (2.4)

    Since ln\ln is concave, we have ln(1+x)xln2\ln(1+x)\geq x\ln 2 for 0x10\leq x\leq 1, and therefore, we have for n1n\geq 1

    2(nn+1)n=2enln(1+1/n)2enln2n=1,2\left(\frac{n}{n+1}\right)^{n}=2e^{-n\ln(1+1/n)}\leq 2e^{-n\frac{\ln 2}{n}}=1, (2.5)

    concluding the proof.

  3. (c)

    Recall that dsd_{s} is the number of standard tableaux of ss. Observe that a standard tableau necessarily has 1 placed in the bottom-left corner. Furthermore, since ss is a hook, the rest of the standard diagram is determined by which numbers we chose to place on the first line, for which there are (s1λa1)\binom{s-1}{\lambda_{a}^{1}} possibilities.

3 Sliced hook products

We introduce here a new tool in the study of Young diagrams, which we call sliced hook products. This notion of hook product, which consists in cutting a diagram along its lines and columns, turns out to be suited for proofs by induction.

3.1 Definitions

Despite being elegant, the hook-length formula may be tricky to use, because of how hard it is to estimate hook products. The idea is to take into account only part of the hook lengths, making the resulting product easier to compute and to use.

We first extend the definition of hook lengths to any set of boxes. We represent boxes by the coordinates of their top right corner in the plane 2{\mathbb{Z}}^{2}. As such a box can be seen as an element of 2{\mathbb{Z}}^{2} and a set of boxes is a subset S2S\subset{\mathbb{Z}}^{2}, see Fig. 6.

Definition 3.1.

Let S2S\subset{\mathbb{Z}}^{2} be a finite set of boxes, and let u=(x,y)u=(x,y)\in{\mathbb{Z}} (not necessarily in SS). We define the hook length of uu with respect to SS by

H(S,u)=|{(x,y)S:[x=x and yy] or [y=y and xx]}|.H(S,u)=\left|\left\{(x^{\prime},y^{\prime})\in S{\;:\;}[x=x^{\prime}\text{ and }y\leq y^{\prime}]\text{ or }[y=y^{\prime}\text{ and }x\leq x^{\prime}]\right\}\right|. (3.1)
{ytableau}\none

& \none

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\none{ytableau}
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& \none*(pink!60)

\none
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*(pink!60)

\none
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*(pink!60)*(pink!60)\none *(pink!60)

Figure 6: The hook of u=(3,1)u=(3,1) in the set of boxes S={(1,1),(4,1),(5,1),(7,1),(3,2),(4,2),(1,3),(1,4),(3,4)}S=\left\{(1,1),(4,1),(5,1),(7,1),(3,2),(4,2),(1,3),(1,4),(3,4)\right\}. In pink are the boxes in the hook of uu, and we have H(S,u)=5H(S,u)=5.
Definition 3.2.

Let λ\lambda be an integer partition and P={ν1,ν2,,νr}P=\left\{\nu_{1},\nu_{2},...,\nu_{r}\right\} be a set partition of λ\lambda (i.e. such that no νi\nu_{i} is empty and iνi=λ\sqcup_{i}\nu_{i}=\lambda). The hook product sliced along PP is the product

H(λ,λ)=HP(λ,λ):=i=1rH(νi,νi).H^{*}(\lambda,\lambda)=H^{*P}(\lambda,\lambda):=\prod_{i=1}^{r}H(\nu_{i},\nu_{i}). (3.2)

If uνiu\in\nu_{i}, we call the quantity H(νi,u)H(\nu_{i},u) the hook length of uu sliced along PP, or sliced hook length.

More generally, if EE is a subset of λ\lambda, we define the following (partial) sliced hook product:

HP(λ,E):=i=1rH(νi,νiE).H^{*P}(\lambda,E):=\prod_{i=1}^{r}H(\nu_{i},\nu_{i}\cap E). (3.3)

We will refer to “slicing” for the procedure consisting of replacing H(λ,λ)H(\lambda,\lambda) by H(λ,λ)H^{*}(\lambda,\lambda), as well as for the associated partition PP.

Let us give some examples of slicings. We will each time provide an example where the νi\nu_{i} are represented in different colors, and the boxes of the diagrams λ\lambda are filled with their sliced hook lengths.

Definition 3.3.

Let λ\lambda be a Young diagram.

  • We call λ1\lambda_{1}-slicing (of λ\lambda) a slicing along {λ1,λ2}\left\{\lambda_{1},\lambda_{\geq 2}\right\}, see Fig. 7.

  • We call λ1\lambda^{1}-slicing (of λ\lambda) a slicing along {λ1,λ2}\left\{\lambda^{1},\lambda^{\geq 2}\right\}, see Fig. 8.

  • We call abδab\delta-slicing (of λ\lambda) a slicing along P:={λai}i{λbi}i{δi}iP:=\left\{\lambda^{i}_{a}\right\}_{i}\sqcup\left\{\lambda^{i}_{b}\right\}_{i}\sqcup\left\{\delta_{i}\right\}_{i}, see Fig. 9. We denote the (partial) abδab\delta-sliced hook products by Habδ(λ,)H^{*ab\delta}(\lambda,\cdot).

{ytableau}\none

&*(red!60)1

\none

*(red!60)3*(red!60)1

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*(red!60)4*(red!60)2

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*(red!60)6*(red!60)4*(red!60)1 \none\none[λ_≥2]

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*(red!60)8*(red!60)6*(red!60)3*(red!60)1

\none

*(blue!60)14*(blue!60)13*(blue!60)12*(blue!60)11*(blue!60)10*(blue!60)9*(blue!60)8*(blue!60)7*(blue!60)6*(blue!60)5*(blue!60)4*(blue!60)3*(blue!60)2*(blue!60)1 \none[λ_1]

Figure 7: The λ1\lambda_{1}-slicing for λ=[14,4,3,2,2,1]\lambda=[14,4,3,2,2,1]. Numbers in the boxes corespond to the hook lengths after slicing.
{ytableau}\none

&*(blue!60)1

\none

*(blue!60)2*(red!60)1

\none

*(blue!60)3*(red!60)2

\none

*(blue!60)4*(red!60)4*(red!60)1 \none\none[λ^≥2]

\none

*(blue!60)5*(red!60)6*(red!60)3*(red!60)1

\none

*(blue!60)19*(blue!60)13*(blue!60)12*(blue!60)11*(blue!60)10*(blue!60)9*(blue!60)8*(blue!60)7*(blue!60)6*(blue!60)5*(blue!60)4*(blue!60)3*(blue!60)2*(blue!60)1 \none[λ^1]

Figure 8: The λ1\lambda^{1}-slicing for λ=[14,4,3,2,2,1]\lambda=[14,4,3,2,2,1]. Numbers in the boxes corespond to the hook lengths after slicing.

The abδab\delta-slicing of the diagram λ=[14,4,3,2,2,1]\lambda=[14,4,3,2,2,1] can be represented as follows.

{ytableau}\none

&\none[λ_b^1]

\none

*(orange!90)1 \none[λ_b^2]

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*(orange!90)2*(orange!60)1

\none

*(orange!90)3*(orange!60)2

\none

*(orange!90)4*(orange!60)3*(green!30)1 \none[δ_3]

\none

*(orange!90)5*(green!60)1*(cyan!60)2*(cyan!60)1 \none[λ_a^2]

\none

*(green!90)1*(cyan!90)13*(cyan!90)12*(cyan!90)11*(cyan!90)10*(cyan!90)9*(cyan!90)8*(cyan!90)7*(cyan!90)6*(cyan!90)5*(cyan!90)4*(cyan!90)3*(cyan!90)2*(cyan!90)1 \none[λ_a^1]

\none
\none

[δ_1]

Figure 9: The abδab\delta-slicing for λ=[14,4,3,2,2,1]\lambda=[14,4,3,2,2,1]. Numbers in the boxes corespond to the hook lengths after slicing. If EE is the subdiagram [3,2][3,2] of λ\lambda (i.e. E:={(1,1),(2,1),(3,1),(1,2),(2,2)}E:=\{(1,1),(2,1),(3,1),(1,2),(2,2)\}), we have Habδ(λ,E)=1131251=780H^{*ab\delta}(\lambda,E)=1\cdot 13\cdot 12\cdot 5\cdot 1=780.

3.2 Bounds on sliced hook products

Let us define, for any Young diagrams μλ\mu\subset\lambda and any set partition PP of λ\lambda, the ratio

RP(λ,μ):=H(λ,μ)HP(λ,μ).R_{P}(\lambda,\mu):=\frac{H(\lambda,\mu)}{H^{*P}(\lambda,\mu)}. (3.4)

If PP is the abδab\delta-(resp. λ1\lambda_{1}-, λ1\lambda^{1}-)slicing, we denote the corresponding ratio by RabδR_{ab\delta} (resp. Rλ1R_{\lambda_{1}}, Rλ1R_{\lambda^{1}}).

We start with rewriting the ratio, in the case of a slicing with respect to the first line.

Lemma 3.4.

If μ,λ\mu,\lambda are two Young diagrams such that μλ\mu\subset\lambda, then we have:

  • (i)

    Rλ1(λ,μ)=H(λ,μ1)H(λ1,μ1)R_{\lambda_{1}}(\lambda,\mu)=\frac{H(\lambda,\mu_{1})}{H(\lambda_{1},\mu_{1})};

  • (ii)

    Rλ1(λ,μ)Rλ1(λ,λ)R_{\lambda_{1}}(\lambda,\mu)\leq R_{\lambda_{1}}(\lambda,\lambda).

Proof.

Let us first prove (i). Since the first line of λ\lambda does not appear in the hook products of boxes on the second line and above, we have H(λ,μ2)=H(λ2,μ2)H(\lambda,\mu_{\geq 2})=H(\lambda_{\geq 2},\mu_{\geq 2}). Therefore,

Rλ1(λ,μ)=H(λ,μ1)H(λ,μ2)H(λ1,μ1)H(λ2,μ2)=H(λ,μ1)H(λ1,μ1).R_{\lambda_{1}}(\lambda,\mu)=\frac{H(\lambda,\mu_{1})H(\lambda,\mu_{\geq 2})}{H(\lambda_{1},\mu_{1})H(\lambda_{\geq 2},\mu_{\geq 2})}=\frac{H(\lambda,\mu_{1})}{H(\lambda_{1},\mu_{1})}. (3.5)

Now we prove (ii). For each uλu\in\lambda, we have Hλ1(λ,u)H(λ,u)H_{\lambda_{1}}(\lambda,u)\leq H(\lambda,u), i.e. Hλ1(λ,u)H(λ,u)1\frac{H_{\lambda_{1}}(\lambda,u)}{H(\lambda,u)}\leq 1. Using (i), we get that

Rλ1(λ,μ)Rλ1(λ,λ)=H(λ,μ1)/H(λ1,μ1)H(λ,λ1)/H(λ1,λ1)=H(λ1,λ1)/H(λ1,μ1)H(λ,λ1)/H(λ,μ1)=H(λ1,λ1\μ1)H(λ,λ1\μ1)=uλ1\μ1H(λ1,u)H(λ,u)1,\begin{split}\frac{R_{\lambda_{1}}(\lambda,\mu)}{R_{\lambda_{1}}(\lambda,\lambda)}&=\frac{H(\lambda,\mu_{1})/H(\lambda_{1},\mu_{1})}{H(\lambda,\lambda_{1})/H(\lambda_{1},\lambda_{1})}=\frac{H(\lambda_{1},\lambda_{1})/H(\lambda_{1},\mu_{1})}{H(\lambda,\lambda_{1})/H(\lambda,\mu_{1})}=\frac{H(\lambda_{1},\lambda_{1}\backslash\mu_{1})}{H(\lambda,\lambda_{1}\backslash\mu_{1})}\\ &=\prod_{u\in\lambda_{1}\backslash\mu_{1}}\frac{H(\lambda_{1},u)}{H(\lambda,u)}\leq 1,\end{split} (3.6)

where λ1\μ1\lambda_{1}\backslash\mu_{1} denotes the set of boxes that are in λ1\lambda_{1} but not in μ1\mu_{1}, that is the set of boxes whose top right angle is in {(i,1):μ1+1iλ1}\left\{(i,1){\;:\;}\mu_{1}+1\leq i\leq\lambda_{1}\right\}. ∎

Let us now prove various bounds on λ1\lambda_{1}-slicings. To ease notations, we set as before r:=λ2r:=\lambda_{\geq 2}. We start with a lemma which will be useful at several places.

Definition 3.5.

Let p1p\geq 1 and (bi)1ip,(bi)1ip(b_{i})_{1\leq i\leq p},(b^{\prime}_{i})_{1\leq i\leq p} be two (weakly) decreasing pp-tuples of positive real numbers. We say that (b1,,bp)(b1,,bp)(b_{1},\ldots,b_{p})\succeq(b^{\prime}_{1},\ldots,b^{\prime}_{p}) if, for all 1p1\leq\ell\leq p:

i=1bii=1bi,\displaystyle\sum_{i=1}^{\ell}b_{i}\geq\sum_{i=1}^{\ell}b^{\prime}_{i}, (3.7)

and

i=1pbi=i=1pbi.\sum_{i=1}^{p}b_{i}=\sum_{i=1}^{p}b^{\prime}_{i}. (3.8)

We say that (b1,,bp)(b1,,bp)(b_{1},\ldots,b_{p})\succ(b^{\prime}_{1},\ldots,b^{\prime}_{p}) if, in addition, they are not equal.

Lemma 3.6.

Let p1p\geq 1 and (ai)1ip,(bi)1ip,(bi)1ip(a_{i})_{1\leq i\leq p},(b_{i})_{1\leq i\leq p},(b^{\prime}_{i})_{1\leq i\leq p} be three (weakly) decreasing pp-tuples of positive real numbers such that (b1,,bp)(b1,,bp)(b_{1},\ldots,b_{p})\succeq(b^{\prime}_{1},\ldots,b^{\prime}_{p}). Then,

i=1p(ai+bi)i=1p(ai+bi).\prod_{i=1}^{p}(a_{i}+b_{i})\leq\prod_{i=1}^{p}(a_{i}+b^{\prime}_{i}). (3.9)
Proof.

Define f:(z1,,zp)i=1p(ai+zi)f:(z_{1},\ldots,z_{p})\mapsto\prod_{i=1}^{p}(a_{i}+z_{i}), on the simplex

Δ:={(x1,,xp),x1xp>0,i=1pxi=1}.\Delta:=\left\{(x_{1},\ldots,x_{p}),x_{1}\geq\ldots\geq x_{p}>0,\sum_{i=1}^{p}x_{i}=1\right\}. (3.10)

We will prove that, for all (b1,,bp)(b1,,bp)(b_{1},\ldots,b_{p})\succ(b^{\prime}_{1},\ldots,b^{\prime}_{p}), there exists (c1,,cp)Δ(c_{1},\ldots,c_{p})\in\Delta such that (b1,,bp)(c1,,cp)(b1,,bp)(b_{1},\ldots,b_{p})\succ(c_{1},\ldots,c_{p})\succeq(b^{\prime}_{1},\ldots,b^{\prime}_{p}) and f(b1,,bp)f(c1,,cp)f(b_{1},\ldots,b_{p})\leq f(c_{1},\ldots,c_{p}). By continuity of ff (which is polynomial), this will allow us to conclude that f(b1,,bp)f(b1,,bp)f(b_{1},\ldots,b_{p})\leq f(b^{\prime}_{1},\ldots,b^{\prime}_{p}). Extending the result on simplices of the form Δs:={(x1,,xp),x1xp>0,i=1pxi=s}\Delta_{s}:=\left\{(x_{1},\ldots,x_{p}),x_{1}\geq\ldots\geq x_{p}>0,\sum_{i=1}^{p}x_{i}=s\right\} for s1s\neq 1 is immediate.

Fix (b1,,bp)(b1,,bp)(b_{1},\ldots,b_{p})\succ(b^{\prime}_{1},\ldots,b^{\prime}_{p}), and let j:=min{i1,p,bi>bi}j:=\min\{i\in\llbracket 1,p\rrbracket,b_{i}>b^{\prime}_{i}\}. Necessarily jp1j\leq p-1. Let also k:=max{ij,bi=bj}k:=\max\{i\geq j,b_{i}=b_{j}\} (in particular k=jk=j if bj+1<bjb_{j+1}<b_{j}). By assumption, jkp1j\leq k\leq p-1. Let ε(0,min{bkbk2,bkbk+12})\varepsilon\in(0,\min\{\frac{b_{k}-b^{\prime}_{k}}{2},\frac{b_{k}-b_{k+1}}{2}\}) to be fixed later, and define (c1,,cp)(c_{1},\ldots,c_{p}) as: ci=bic_{i}=b_{i} if i{k,k+1},ck=bkε,ck+1=bk+1+εi\notin\{k,k+1\},c_{k}=b_{k}-\varepsilon,c_{k+1}=b_{k+1}+\varepsilon. By assumption, we have (c1,,cp)Δ(c_{1},\ldots,c_{p})\in\Delta and (b1,,bp)(c1,,cp)(b1,,bp)(b_{1},\ldots,b_{p})\succ(c_{1},\ldots,c_{p})\succeq(b^{\prime}_{1},\ldots,b^{\prime}_{p}). In addition, we have

f(c1,,cp)f(b1,,bp)=(ak+ck)(ak+1+ck+1)(ak+bk)(ak+1+bk+1)1.\displaystyle\frac{f(c_{1},\ldots,c_{p})}{f(b_{1},\ldots,b_{p})}=\frac{(a_{k}+c_{k})(a_{k+1}+c_{k+1})}{(a_{k}+b_{k})(a_{k+1}+b_{k+1})}\geq 1. (3.11)

Indeed,

(ak+ck)(ak+1+ck+1)(ak+bk)(ak+1+bk+1)=ak(ck+1bk+1)+ak+1(ckbk)+ckck+1bkbk+1=ε(akak+1+bkbk+1)ε2>0\begin{split}&(a_{k}+c_{k})(a_{k+1}+c_{k+1})-(a_{k}+b_{k})(a_{k+1}+b_{k+1})\\ =&a_{k}(c_{k+1}-b_{k+1})+a_{k+1}(c_{k}-b_{k})+c_{k}c_{k+1}-b_{k}b_{k+1}\\ =&\varepsilon(a_{k}-a_{k+1}+b_{k}-b_{k+1})-\varepsilon^{2}>0\end{split} (3.12)

for ε>0\varepsilon>0 small enough, since bk>bk+1b_{k}>b_{k+1} by assumption. The result follows. ∎

For knk\leq n integers, we write

nk:=n(n1)(nk+1)=0ik1(ni)=n!(nk)!=k!(nk).n^{\downarrow k}:=n(n-1)...(n-k+1)=\prod_{0\leq i\leq k-1}(n-i)=\frac{n!}{(n-k)!}=k!\binom{n}{k}. (3.13)
Proposition 3.7.

Let μλ\mu\subset\lambda be (non-empty) Young diagrams. Then we have

  1. (a)

    We have

    Rλ1(λ,μ)(λ1+uμ1(λ)μ1)μ1λ1μ1.R_{\lambda_{1}}(\lambda,\mu)\leq\frac{\left(\lambda_{1}+\left\lceil\frac{u_{\leq\mu_{1}}(\lambda)}{\mu_{1}}\right\rceil\right)^{\downarrow\mu_{1}}}{\lambda_{1}^{\downarrow\mu_{1}}}. (3.14)
  2. (b)

    Recalling the notation r=λ2r=\lambda_{\geq 2}, we have:

    Rλ1(λ,λ)(λ1+rλ1λ1).R_{\lambda_{1}}(\lambda,\lambda)\leq\binom{\lambda_{1}+\left\lceil\frac{r}{\lambda_{1}}\right\rceil}{\lambda_{1}}. (3.15)
  3. (c)

    If rλ1r\leq\lambda_{1}, then

    Rλ1(λ,λ)1+rλ1r+1.R_{\lambda_{1}}(\lambda,\lambda)\leq 1+\frac{r}{\lambda_{1}-r+1}. (3.16)
  4. (d)

    Finally, we have for all λ\lambda:

    Rλ1(λ,λ)e3r.R_{\lambda_{1}}(\lambda,\lambda)\leq e^{3\sqrt{r}}. (3.17)
Proof.
  1. (a)

    First observe that, as a consequence of Lemma 3.6, we have

    i=1μ1(λ1(i1)+ui)i=1μ1(λ1(i1)+uμ1(λ)μ1).\prod_{i=1}^{\mu_{1}}(\lambda_{1}-(i-1)+u_{i})\leq\prod_{i=1}^{\mu_{1}}\left(\lambda_{1}-(i-1)+\frac{u_{\leq\mu_{1}}(\lambda)}{\mu_{1}}\right). (3.18)

    This implies that, setting m:=uμ1(λ)μ1m:=\left\lceil\frac{u_{\leq\mu_{1}}(\lambda)}{\mu_{1}}\right\rceil, we have by Lemma 3.4 (i):

    Rλ1(λ,μ)=H(λ,μ1)H(λ1,μ1)=i=1μ1λ1(i1)+uiλ1(i1)i=1μ1λ1(i1)+mλ1(i1)=(λ1+m)μ1λ1μ1.\begin{split}R_{\lambda_{1}}(\lambda,\mu)&=\frac{H(\lambda,\mu_{1})}{H(\lambda_{1},\mu_{1})}=\prod_{i=1}^{\mu_{1}}\frac{\lambda_{1}-(i-1)+u_{i}}{\lambda_{1}-(i-1)}\\ &\leq\prod_{i=1}^{\mu_{1}}\frac{\lambda_{1}-(i-1)+m}{\lambda_{1}-(i-1)}=\frac{(\lambda_{1}+m)^{\downarrow\mu_{1}}}{\lambda_{1}^{\downarrow\mu_{1}}}.\end{split} (3.19)

    This concludes the proof of (a).

  2. (b)

    In the case μ=λ\mu=\lambda, we have uμ1(λ)μ1=rλ1\frac{u_{\leq\mu_{1}}(\lambda)}{\mu_{1}}=\frac{r}{\lambda_{1}}. We therefore have directly, by (a):

    Rλ1(λ,λ)(λ1+m)λ1λ1λ1=(λ1+mλ1).R_{\lambda_{1}}(\lambda,\lambda)\leq\frac{(\lambda_{1}+m)^{\downarrow\lambda_{1}}}{\lambda_{1}^{\downarrow\lambda_{1}}}=\binom{\lambda_{1}+m}{\lambda_{1}}. (3.20)
  3. (c)

    If r=λ1r=\lambda_{1}, then (b) yields

    Rλ1(λ,λ)(λ1+rλ1λ1)=λ1+1,\displaystyle R_{\lambda_{1}}(\lambda,\lambda)\leq\binom{\lambda_{1}+\left\lceil\frac{r}{\lambda_{1}}\right\rceil}{\lambda_{1}}=\lambda_{1}+1,

    which is what we want. If r<λ1r<\lambda_{1}, then, by Lemma 3.6, λ1\lambda_{1} and rr being fixed, the smallest value of Rλ1(λ,λ)R_{\lambda_{1}}(\lambda,\lambda) is reached when rr is flat, that is, λ=[λ1,nλ1]\lambda=[\lambda_{1},n-\lambda_{1}]. We therefore get

    Rλ1(λ,λ)Rλ1([λ1,nλ1],[λ1,nλ1])=λ1+1λ1r+1.R_{\lambda_{1}}(\lambda,\lambda)\leq R_{\lambda_{1}}([\lambda_{1},n-\lambda_{1}],[\lambda_{1},n-\lambda_{1}])=\frac{\lambda_{1}+1}{\lambda_{1}-r+1}. (3.21)
  4. (d)

    We split the proof according to the value of rr.

    • If r<λ1r<\lambda_{1}, then starting from (c) we have

      Rλ1(λ,λ)λ1+1λ1r+1r+1rr+1=r+1er,R_{\lambda_{1}}(\lambda,\lambda)\leq\frac{\lambda_{1}+1}{\lambda_{1}-r+1}\leq\frac{r+1}{r-r+1}=r+1\leq e^{\sqrt{r}}, (3.22)

      and thus Rλ1(λ,λ)e3rR_{\lambda_{1}}(\lambda,\lambda)\leq e^{3\sqrt{r}}.

    • Assume now that rλ1r\geq\lambda_{1}. Let n=|λ|n=|\lambda|. Note that here again μ=λ\mu=\lambda, so that μ1=λ1\mu_{1}=\lambda_{1} and uμ1=ru_{\leq\mu_{1}}=r. Hence, m=rλ1m=\left\lceil\frac{r}{\lambda_{1}}\right\rceil, and since rλ1r\geq\lambda_{1} we have m1m\geq 1. Hence, m=rλ1rλ1+12rλ1m=\left\lceil\frac{r}{\lambda_{1}}\right\rceil\leq\frac{r}{\lambda_{1}}+1\leq 2\frac{r}{\lambda_{1}}, so that λ12rm\lambda_{1}\leq\frac{2r}{m}. First assume that λ1m\lambda_{1}\leq m. Using that pp=p!(p/e)pp^{\downarrow p}=p!\geq(p/e)^{p} for all p0p\geq 0, we therefore have, using m2r/λ1m\leq 2r/\lambda_{1} in the last inequality,

      (λ1+mλ1)=(λ1+m)λ1λ1λ1(2m)λ1(λ1/e)λ1=(2emλ1)λ1(4erλ12)λ1.\binom{\lambda_{1}+m}{\lambda_{1}}=\frac{(\lambda_{1}+m)^{\downarrow\lambda_{1}}}{\lambda_{1}^{\downarrow\lambda_{1}}}\leq\frac{(2m)^{\lambda_{1}}}{(\lambda_{1}/e)^{\lambda_{1}}}=\left(\frac{2em}{\lambda_{1}}\right)^{\lambda_{1}}\leq\left(\frac{4er}{\lambda_{1}^{2}}\right)^{\lambda_{1}}. (3.23)

      Now observe that we can rewrite (4erλ12)λ1=((2erλ1)λ1)2\left(\frac{4er}{\lambda_{1}^{2}}\right)^{\lambda_{1}}=\left(\left(\frac{2\sqrt{er}}{\lambda_{1}}\right)^{\lambda_{1}}\right)^{2}. Furthermore, for T>0T>0 the function x+(T/x)xx\in{\mathbb{R}}^{+}\mapsto(T/x)^{x} is maximal at x=T/ex=T/e, we get, with T=2erT=2\sqrt{er},

      (λ1+mλ1)(eT/e)2=e4ree3r.\binom{\lambda_{1}+m}{\lambda_{1}}\leq\left(e^{T/e}\right)^{2}=e^{\frac{4\sqrt{r}}{\sqrt{e}}}\leq e^{3\sqrt{r}}. (3.24)

      The case mλ1m\leq\lambda_{1} is proved the same way by symmetry, since (λ1+mλ1)=(λ1+mm)\binom{\lambda_{1}+m}{\lambda_{1}}=\binom{\lambda_{1}+m}{m}. The result follows by (b). This concludes the proof of (d).

We can now use these results to bound classical hook products and obtain a crucial inequality involving dλ,dsd_{\lambda},d_{s} and dcd_{c}.

Proposition 3.8.

Let λ\lambda be a (non-empty) integer partition. Then we have

  1. (a)

    We have

    H(λ,λ)=s!(s1λb1)H(λ,λa1)H(λa1,λa1)H(λ,λb1)H(λb1,λb1)H(c,c),H(\lambda,\lambda)=\frac{s!}{\binom{s-1}{\lambda_{b}^{1}}}\frac{H(\lambda,\lambda_{a}^{1})}{H(\lambda_{a}^{1},\lambda_{a}^{1})}\frac{H(\lambda,\lambda_{b}^{1})}{H(\lambda_{b}^{1},\lambda_{b}^{1})}H(c,c), (3.25)

    and in particular

    dλ(nc)dsdc=H(λa1,λa1)H(λ,λa1)H(λb1,λb1)H(λ,λb1).\frac{d_{\lambda}}{\binom{n}{c}d_{s}d_{c}}=\frac{H(\lambda_{a}^{1},\lambda_{a}^{1})}{H(\lambda,\lambda_{a}^{1})}\frac{H(\lambda_{b}^{1},\lambda_{b}^{1})}{H(\lambda,\lambda_{b}^{1})}. (3.26)
  2. (b)
    dλ(nc)dsdce6c,\frac{d_{\lambda}}{\binom{n}{c}d_{s}d_{c}}\geq e^{-6\sqrt{c}}, (3.27)

    where we recall that n=|λ|n=|\lambda|, c=λ2c=\lambda^{\geq 2}, and s=nc=λ1s=n-c=\lambda^{1}.

Proof.
  1. (a)

    First, we have

    H(λ,λ)=sH(λ,λa1)H(λ,λb1)H(λ,c)=sλa1!λb1!H(λ,λa1)H(λa1,λa1)H(λ,λb1)H(λb1,λb1)H(c,c)=s!(s1λb1)H(λ,λa1)H(λa1,λa1)H(λ,λb1)H(λb1,λb1)H(c,c).\begin{split}H(\lambda,\lambda)&=sH(\lambda,\lambda_{a}^{1})H(\lambda,\lambda_{b}^{1})H(\lambda,c)\\ &=s\cdot\lambda_{a}^{1}!\cdot\lambda_{b}^{1}!\frac{H(\lambda,\lambda_{a}^{1})}{H(\lambda_{a}^{1},\lambda_{a}^{1})}\frac{H(\lambda,\lambda_{b}^{1})}{H(\lambda_{b}^{1},\lambda_{b}^{1})}H(c,c)\\ &=\frac{s!}{\binom{s-1}{\lambda_{b}^{1}}}\frac{H(\lambda,\lambda_{a}^{1})}{H(\lambda_{a}^{1},\lambda_{a}^{1})}\frac{H(\lambda,\lambda_{b}^{1})}{H(\lambda_{b}^{1},\lambda_{b}^{1})}H(c,c).\end{split} (3.28)

    We therefore deduce from the hook length formula and Lemma 2.3 that

    dλ(nc)=c!s!n!n!H(λ,λ)=(s1λb1)H(λa1,λa1)H(λ,λa1)H(λb1,λb1)H(λ,λb1)c!H(c,c)=(s1λb1)H(λa1,λa1)H(λ,λa1)H(λb1,λb1)H(λ,λb1)dc=H(λa1,λa1)H(λ,λa1)H(λb1,λb1)H(λ,λb1)dsdc,\begin{split}\frac{d_{\lambda}}{\binom{n}{c}}=\frac{c!s!}{n!}\frac{n!}{H(\lambda,\lambda)}&=\binom{s-1}{\lambda_{b}^{1}}\frac{H(\lambda_{a}^{1},\lambda_{a}^{1})}{H(\lambda,\lambda_{a}^{1})}\frac{H(\lambda_{b}^{1},\lambda_{b}^{1})}{H(\lambda,\lambda_{b}^{1})}\frac{c!}{H(c,c)}\\ &=\binom{s-1}{\lambda_{b}^{1}}\frac{H(\lambda_{a}^{1},\lambda_{a}^{1})}{H(\lambda,\lambda_{a}^{1})}\frac{H(\lambda_{b}^{1},\lambda_{b}^{1})}{H(\lambda,\lambda_{b}^{1})}d_{c}\\ &=\frac{H(\lambda_{a}^{1},\lambda_{a}^{1})}{H(\lambda,\lambda_{a}^{1})}\frac{H(\lambda_{b}^{1},\lambda_{b}^{1})}{H(\lambda,\lambda_{b}^{1})}d_{s}d_{c},\end{split} (3.29)

    as desired.

  2. (b)

    By Proposition 3.7 (d) applied twice respectively to the first line and the first column of the diagrams λ2\lambda^{\prime}_{\geq 2} and λ2\lambda_{\geq 2}, we get

    H(λa1,λa1)H(λ,λa1)H(λb1,λb1)H(λ,λb1)e3ce3c=e6c.\frac{H(\lambda_{a}^{1},\lambda_{a}^{1})}{H(\lambda,\lambda_{a}^{1})}\frac{H(\lambda_{b}^{1},\lambda_{b}^{1})}{H(\lambda,\lambda_{b}^{1})}\geq e^{-3\sqrt{c}}e^{-3\sqrt{c}}=e^{-6\sqrt{c}}. (3.30)

    Plugging this into (b) concludes the proof of (c).

3.3 Virtual degrees and augmented dimensions

If PP is a set partition of a diagram λn\lambda\vdash n, we can associate to it a notion of PP-dimension via the analog of the hook length formula:

dλP:=n!HP(λ,λ).d_{\lambda}^{*P}:=\frac{n!}{H^{*P}(\lambda,\lambda)}. (3.31)

The virtual degree and abδab\delta-dimension are closely related: we have

D(λ)=dλabδn,D(\lambda)=\frac{d_{\lambda}^{*ab\delta}}{n}, (3.32)

where we recall that

D(λ):=(n1)!iai!bi!.D(\lambda):=\frac{(n-1)!}{\prod_{i}a_{i}!b_{i}!}. (3.33)

Let us now define a last notion of dimension, which will prove to be very convenient in the proof of Theorem 1.5.

If λ\lambda is a Young diagram, we define the augmented dimension dλ+d_{\lambda}^{+} of λ\lambda by

dλ+:=n!H+(λ,λ),d^{+}_{\lambda}:=\frac{n!}{H^{+}(\lambda,\lambda)}, (3.34)

where

H+(λ,λ)=(isi)(iai!bi!).H^{+}(\lambda,\lambda)=\left(\prod_{i}s_{i}\right)\left(\prod_{i}a_{i}!b_{i}!\right). (3.35)

(Here sis_{i} denotes the hook started at δi\delta_{i} and we recall that ai=λaia_{i}=\lambda_{a}^{i} and bi=λbib_{i}=\lambda_{b}^{i}.)

The augmented dimension dλ+d_{\lambda}^{+} is hybrid between dλd_{\lambda} and D(λ)D(\lambda): the on-diagonal augmented hook lengths are the usual hook lengths, and the off-diagonal augmented hook-lengths are the virtual hook lengths.

{ytableau}\none

&\none[λ_b^1]

\none

*(orange!90)1 \none[λ_b^2]

\none

*(orange!90)2*(orange!60)1

\none

*(orange!90)3*(orange!60)2

\none

*(orange!90)4*(orange!60)3*(red!30)1 \none[δ_3]

\none

*(orange!90)5*(red!60)6*(cyan!60)2*(cyan!60)1 \none[λ_a^2]

\none

*(red!90)19*(cyan!90)13*(cyan!90)12*(cyan!90)11*(cyan!90)10*(cyan!90)9*(cyan!90)8*(cyan!90)7*(cyan!90)6*(cyan!90)5*(cyan!90)4*(cyan!90)3*(cyan!90)2*(cyan!90)1 \none[λ_a^1]

\none
\none

[δ_1]

Figure 10: The augmented hook lengths for λ=[14,4,3,2,2,1]\lambda=[14,4,3,2,2,1]. Numbers in the boxes corespond to the augmented hook lengths.

Note that H+(λ,λ)H^{+}(\lambda,\lambda) is not a sliced hook product, since we keep the full hook lengths of the boxes on the diagonal.

By definition we have

D(λ)dλ+=(n1)!iai!bi!n!(iai!bi!)(isi)=isin.\frac{D(\lambda)}{d_{\lambda}^{+}}=\frac{\frac{(n-1)!}{\prod_{i}a_{i}!b_{i}!}}{\frac{n!}{\left(\prod_{i}a_{i}!b_{i}!\right)\left(\prod_{i}s_{i}\right)}}=\frac{\prod_{i}s_{i}}{n}. (3.36)

One advantage of using dλ+d_{\lambda}^{+} is the following identity (which we prove later on, in Lemma 4.7):

dλ+=(nc)ds+dc+.d_{\lambda}^{+}=\binom{n}{c}d_{s}^{+}d_{c}^{+}. (3.37)

This will be convenient when bounding dλ+d_{\lambda}^{+} by induction on the size of the center cc of λ\lambda.

4 Sharp bounds on virtual degrees

The aim of this section is to use the results of the previous sections to prove Theorem 1.5.

4.1 Strategy

In regards of (3.36), which rewrites as

D(λ)=isindλ+,D(\lambda)=\frac{\prod_{i}s_{i}}{n}d_{\lambda}^{+}, (4.1)

it is enough to prove the two following statements.

Proposition 4.1.

There exists a constant Cdiag>0C_{diag}>0 such that, for every n2n\geq 2 and every diagram λn\lambda\vdash n, we have

isindλCdiag/lnn.\frac{\prod_{i}s_{i}}{n}\leq d_{\lambda}^{C_{diag}/\ln n}. (4.2)
Proposition 4.2.

There exists a constant Caug>0C_{aug}>0 such that, for every n2n\geq 2 and every diagram λn\lambda\vdash n, we have

dλ+dλ1+Caug/lnn.d_{\lambda}^{+}\leq d_{\lambda}^{1+C_{aug}/\ln n}. (4.3)

We will prove Proposition 4.1 in Section 4.2 and Proposition 4.2 in Section 4.3.

4.2 Proof of Proposition 4.1

Let us first give two general upper bounds on isin\frac{\prod_{i}s_{i}}{n}.

Lemma 4.3.

Let n1n\geq 1 and λn\lambda\vdash n such that c=c(λ)1c=c(\lambda)\geq 1. Then we have

  1. (a)
    isin(c/δ(c))δ(c),\frac{\prod_{i}s_{i}}{n}\leq(c/\delta(c))^{\delta(c)}, (4.4)

    where c=λ2c=\lambda^{\geq 2} is the center of λ\lambda and δ(c):=δ(λ)1\delta(c):=\delta(\lambda)-1 is the diagonal length of cc.

  2. (b)

    Furthermore,

    isineclnc.\frac{\prod_{i}s_{i}}{n}\leq e^{\sqrt{c}\ln c}. (4.5)
Proof.
  1. (a)

    First, since s1ns_{1}\leq n, we have

    isini2si.\frac{\prod_{i}s_{i}}{n}\leq\prod_{i\geq 2}s_{i}. (4.6)

    Moreover, by concavity of the logarithm, we have (using that δ(c)=δ(λ)1\delta(c)=\delta(\lambda)-1)

    i2si=exp(2iδ(λ)ln(si))exp(2iδ(λ)ln(cδ(c)))=(c/δ(c))δ(c),\prod_{i\geq 2}s_{i}=\exp\left(\sum_{2\leq i\leq\delta(\lambda)}\ln(s_{i})\right)\leq\exp\left(\sum_{2\leq i\leq\delta(\lambda)}\ln\left(\frac{c}{\delta(c)}\right)\right)=(c/\delta(c))^{\delta(c)}, (4.7)

    concluding the proof of (a).

  2. (b)

    We have δ(c)c\delta(c)\leq\sqrt{c} (since the square of side length δ(c)\delta(c) is included in cc), and sics_{i}\leq c for each i2i\geq 2. It therefore follows that

    isini2sicδ(c)cc=eclnc,\frac{\prod_{i}s_{i}}{n}\leq\prod_{i\geq 2}s_{i}\leq c^{\delta(c)}\leq c^{\sqrt{c}}=e^{\sqrt{c}\ln c}, (4.8)

    as desired.

In Proposition 3.8 (b), we showed that dλ(s1λb1)e6c(nc)dcd_{\lambda}\geq\binom{s-1}{\lambda_{b}^{1}}e^{-6\sqrt{c}}\binom{n}{c}d_{c}. Depending on the size of cc only some of the terms in the lower bound will be useful. We give here simple lower bounds on some of these terms.

Lemma 4.4.

Let n1n\geq 1 and λn\lambda\vdash n such that c1c\geq 1. Then

  1. (a)
    (s1λb1)s1.\binom{s-1}{\lambda_{b}^{1}}\geq s-1. (4.9)
  2. (b)
    (nc)max((n/c)c,(n/s)s).\binom{n}{c}\geq\max\left((n/c)^{c},(n/s)^{s}\right). (4.10)
Proof.
  1. (a)

    Since c1c\geq 1, we have 1λb1s21\leq\lambda_{b}^{1}\leq s-2. We therefore have

    (s1λb1)(s11)=s1.\binom{s-1}{\lambda_{b}^{1}}\geq\binom{s-1}{1}=s-1. (4.11)
  2. (b)

    This bound is a classical bound on binomial coefficients. We have

    (nc)=ncc!(n/c)c,\binom{n}{c}=\frac{n^{\downarrow c}}{c!}\geq(n/c)^{c}, (4.12)

    and since (nc)=(ns)\binom{n}{c}=\binom{n}{s}, we also have (nc)(n/s)s\binom{n}{c}\geq(n/s)^{s} by symmetry, concluding the proof.

We finally give a lower bound on dλd_{\lambda} depending only on its diagonal length.

Lemma 4.5.

Let n1n\geq 1 and λn\lambda\vdash n. Then

dλ(δ/2)δ(δ1),d_{\lambda}\geq(\delta/2)^{\delta(\delta-1)}, (4.13)

where δ\delta is the diagonal length of λ\lambda. In particular, if n(2e)3n\geq(2e)^{3} and δn1/3\delta\geq n^{1/3}, we have

dλen2/3/2.d_{\lambda}\geq e^{n^{2/3}/2}. (4.14)
Proof.

Since dλd_{\lambda} is the number of standard tableaux of λ\lambda, and λ\lambda contains the square μ:=[δδ]\mu:=\left[\delta^{\delta}\right], we have

dλdμ.d_{\lambda}\geq d_{\mu}. (4.15)

Moreover, we claim that

H(μ,μ)δ!((2δ)δ)δ1.H(\mu,\mu)\leq\delta!\left((2\delta)^{\downarrow\delta}\right)^{\delta-1}. (4.16)

Indeed, the product of hook lengths on the top row of μ\mu is δ!\delta!. Furthermore, the hook length of a box in column ii is at most 2δi+12\delta-i+1. Taking the product proves (4.16).

Using (4.16) and the hook length formula, we have

dμ=δ2!H(μ,μ)δ2!δ!((2δ)δ)δ1=(δ2)(δ2δ)((2δ)δ)δ1=k=2δ(kδ)δ(2δ)δk=2δ(kδ)δ(2δ)δ=(δ2)δ2δ((2δ)δ)δ1=(δ/2)δ(δ1).\begin{split}d_{\mu}&=\frac{\delta^{2}!}{H(\mu,\mu)}\geq\frac{\delta^{2}!}{\delta!\left((2\delta)^{\downarrow\delta}\right)^{\delta-1}}=\frac{\left(\delta^{2}\right)^{\downarrow(\delta^{2}-\delta)}}{\left((2\delta)^{\downarrow\delta}\right)^{\delta-1}}=\prod_{k=2}^{\delta}\frac{(k\delta)^{\downarrow\delta}}{(2\delta)^{\downarrow\delta}}\\ &\geq\prod_{k=2}^{\delta}\frac{(k\delta)^{\delta}}{(2\delta)^{\delta}}=\frac{\left(\delta^{2}\right)^{\delta^{2}-\delta}}{\left((2\delta)^{\delta}\right)^{\delta-1}}\\ &=(\delta/2)^{\delta(\delta-1)}.\end{split} (4.17)

Assume now that n(2e)3n\geq(2e)^{3} and δn1/3\delta\geq n^{1/3}. Then we have δ((2e)3)1/3=2e\delta\geq\left((2e)^{3}\right)^{1/3}=2e so δ/2e\delta/2\geq e, and we also have δ(δ1)δ2/2\delta(\delta-1)\geq\delta^{2}/2 since δ2\delta\geq 2, so (δ/2)δ(δ1)eδ2/2en2/3/2(\delta/2)^{\delta(\delta-1)}\geq e^{\delta^{2}/2}\geq e^{n^{2/3}/2}. ∎

We now prove Proposition 4.1 for small values of nn:

Lemma 4.6.

For any n02n_{0}\geq 2, for every 2nn02\leq n\leq n_{0} and every λn\lambda\vdash n, we have

1nisidλC/lnn,\frac{1}{n}\prod_{i}s_{i}\leq d_{\lambda}^{C/\ln n}, (4.18)

where C=n0(lnn0)2ln2C=\frac{n_{0}(\ln n_{0})^{2}}{\ln 2}.

Proof.

Let n02n_{0}\geq 2, 2nn02\leq n\leq n_{0} and λn\lambda\vdash n. If λ\lambda is flat (horizontal or vertical), then dλ=1d_{\lambda}=1 and 1nisi=1\frac{1}{n}\prod_{i}s_{i}=1 so the result holds. Assume now that λ\lambda is not flat. Then dλ2d_{\lambda}\geq 2 so

1nisi1iδ(λ)sin0n0=21ln2n0lnn0dλ1ln2n0lnn0dλ1ln2(n0lnn0)lnn0lnn=dλC/lnn.\frac{1}{n}\prod_{i}s_{i}\leq\prod_{1\leq i\leq\delta(\lambda)}s_{i}\leq n_{0}^{n_{0}}=2^{\frac{1}{\ln 2}n_{0}\ln n_{0}}\leq d_{\lambda}^{\frac{1}{\ln 2}n_{0}\ln n_{0}}\leq d_{\lambda}^{\frac{1}{\ln 2}(n_{0}\ln n_{0})\frac{\ln n_{0}}{\ln n}}=d_{\lambda}^{C/\ln n}. (4.19)

We now have all the tools to prove Proposition 4.1.

Proof of Proposition 4.1.

We prove that the result holds for C:=Cdiag=500C:=C_{diag}=500 and ne55n\geq e^{55}. By Lemma 4.6, the result then extends to all n2n\geq 2, up to taking a larger constant CdiagC_{diag}.

We will split the proof into several cases, depending on how large cc is.

  • If c=0c=0 then 1nisi=nn=1\frac{1}{n}\prod_{i}s_{i}=\frac{n}{n}=1 so the result holds.

  • Assume that 1cn8/91\leq c\leq n^{8/9}. First, by Proposition 3.8 (b) we have dλ(nc)e6cd_{\lambda}\geq\binom{n}{c}e^{-6\sqrt{c}}. We deduce from Lemma 4.4 (b) that

    dλ(n/c)ce6cnc/9e6cnc/500.d_{\lambda}\geq(n/c)^{c}e^{-6\sqrt{c}}\geq n^{c/9}e^{-6\sqrt{c}}\geq n^{c/500}. (4.20)

    On the other hand, we have 1nisieclnc\frac{1}{n}\prod_{i}s_{i}\leq e^{\sqrt{c}\ln c} by Lemma 4.3 (b). We deduce that (using that lncc\ln c\leq\sqrt{c} for all c1c\geq 1, so that clncc\sqrt{c}\ln c\leq c),

    1nisieclncec=(nc/500)C/lnndλC/lnn.\frac{1}{n}\prod_{i}s_{i}\leq e^{\sqrt{c}\ln c}\leq e^{c}=\left(n^{c/500}\right)^{C/\ln n}\leq d_{\lambda}^{C/\ln n}. (4.21)
  • Assume that n8/9cnn2/3n^{8/9}\leq c\leq n-n^{2/3}. Then

    ncnnn2/3=11n1/31en1/3=en1/3\frac{n}{c}\geq\frac{n}{n-n^{2/3}}=\frac{1}{1-n^{-1/3}}\geq\frac{1}{e^{-n^{-1/3}}}=e^{n^{-1/3}} (4.22)

    so (using again Proposition 3.8 (b) and Lemma 4.4 (b) for the first inequality)

    dλ(n/c)ce6c(en1/3)n8/9e6c=en5/9e6cen5/9/2.d_{\lambda}\geq(n/c)^{c}e^{-6\sqrt{c}}\geq\left(e^{n^{-1/3}}\right)^{n^{8/9}}e^{-6\sqrt{c}}=e^{n^{5/9}}e^{-6\sqrt{c}}\geq e^{n^{5/9}/2}. (4.23)

    We conclude using Lemma 4.3 (b) (and that cnc\leq n) that

    1nisieclncenlnne500n5/92lnndλ500/lnndλC/lnn.\frac{1}{n}\prod_{i}s_{i}\leq e^{\sqrt{c}\ln c}\leq e^{\sqrt{n}\ln n}\leq e^{500\frac{n^{5/9}}{2\ln n}}\leq d_{\lambda}^{500/\ln n}\leq d_{\lambda}^{C/\ln n}. (4.24)
  • Assume now that cnn2/3c\geq n-n^{2/3}, i.e. sn2/3s\leq n^{2/3}. Then we have δ(λ)n1/3\delta(\lambda)\geq n^{1/3} and therefore by Lemma 4.5 we have dλen2/3/2d_{\lambda}\geq e^{n^{2/3}/2}. We conclude, proceeding as in the previous case, that

    1nisieclncenlnne500n2/32lnndλ500/lnndλC/lnn.\frac{1}{n}\prod_{i}s_{i}\leq e^{\sqrt{c}\ln c}\leq e^{\sqrt{n}\ln n}\leq e^{500\frac{n^{2/3}}{2\ln n}}\leq d_{\lambda}^{500/\ln n}\leq d_{\lambda}^{C/\ln n}. (4.25)

This concludes the proof. ∎

Remark 4.1.

In the proof of Proposition 4.1 above, (4.20) holds since ne5400/91n\geq e^{5400/91} and (4.23) holds since n1218n\geq 12^{18}. Taking n0=n55n_{0}=n^{55} in Lemma 4.6, we deduce that Proposition 4.1 holds (for all n2n\geq 2 if Cdiagmax(500,n0(lnn0)2ln2)=e55(552)/ln2C_{diag}\geq\max(500,\frac{n_{0}(\ln n_{0})^{2}}{\ln 2})=e^{55}(55^{2})/\ln 2, and we can therefore take Cdiag=e64C_{diag}=e^{64}.

4.3 Proof of Proposition 4.2

Our aim is to show that dλ+dλ\frac{d_{\lambda}^{+}}{d_{\lambda}} is small. We will proceed by induction, slicing the external hook s=λ1s=\lambda^{1}.

First, we recall from Proposition 3.8 (b) that the dimension of a diagram dλd_{\lambda} and the dimension of its center dcd_{c} are related as follows: we have dλe6c(nc)dsdcd_{\lambda}\geq e^{-6\sqrt{c}}\binom{n}{c}d_{s}d_{c}, i.e. (recalling from Lemma 2.3 that ds=(s1λa1)d_{s}=\binom{s-1}{\lambda_{a}^{1}}):

dcdλe6c(s1λa1)(ns).\frac{d_{c}}{d_{\lambda}}\leq\frac{e^{6\sqrt{c}}}{\binom{s-1}{\lambda_{a}^{1}}\binom{n}{s}}. (4.26)
Remark 4.2.

We also always have dλ(nc)dsdcd_{\lambda}\leq\binom{n}{c}d_{s}d_{c}. Indeed, by first choosing which numbers from 1 to nn we place in cc, we obtain |ST(λ)|(nc)|ST(s)||ST(c)||\operatorname{ST}(\lambda)|\leq\binom{n}{c}|\operatorname{ST}(s)||\operatorname{ST}(c)|. This argument was written by Diaconis and Shahshahani in their proof of cutoff for random transpositions (with r=λ2r=\lambda_{\geq 2} in place of cc), see [Dia88, Fact 1 in Chapter 3D, p. 39-40] . One can therefore see Proposition 3.8 (b) as a quantified version of the intuitive approximation dλ(nc)dsdcd_{\lambda}\approx\binom{n}{c}d_{s}d_{c}, for large values of ss and cc.

Comparing the augmented dimensions dλ+d_{\lambda}^{+} and dc+d_{c}^{+} is much simpler by design, since the hook products involved there are already sliced by definition. Also, by definition of ds+d_{s}^{+} we have ds+=n!nλa1!λb1!=s1λa1!λb1!=(s1λa1)d_{s}^{+}=\frac{n!}{n\lambda_{a}^{1}!\lambda_{b}^{1}!}=\frac{s-1}{\lambda_{a}^{1}!\lambda_{b}^{1}!}=\binom{s-1}{\lambda_{a}^{1}}, and we recall from Lemma 2.3 that ds=(s1λa1)d_{s}=\binom{s-1}{\lambda_{a}^{1}}, so that

ds+=ds=(s1λa1).d_{s}^{+}=d_{s}=\binom{s-1}{\lambda_{a}^{1}}. (4.27)

In the next lemma we show that the aforementioned intuitive approximation for dimensions actually becomes an equality when it comes to augmented dimensions.

Lemma 4.7.

Let n1n\geq 1 and λn\lambda\vdash n. Then

dλ+=(ns)ds+dc+.d_{\lambda}^{+}=\binom{n}{s}d_{s}^{+}d_{c}^{+}. (4.28)

In other words, we have

dλ+dc+=(ns)(s1λa1).\frac{d_{\lambda}^{+}}{d_{c}^{+}}=\binom{n}{s}{\binom{s-1}{\lambda_{a}^{1}}}. (4.29)
Proof.

By definition we have

H+(λ,λ)=sλa1!λb1!H+(c,c).H^{+}(\lambda,\lambda)=s\lambda_{a}^{1}!\lambda_{b}^{1}!H^{+}(c,c). (4.30)

We therefore get

dλ+dc+=n!/H+(λ,λ)c!/H+(c,c)=nssλa1!λb1!=nss!(s1)!λa1!λb1!=(ns)(s1λa1).\frac{d_{\lambda}^{+}}{d_{c}^{+}}=\frac{n!/H^{+}(\lambda,\lambda)}{c!/H^{+}(c,c)}=\frac{n^{\downarrow s}}{s\lambda_{a}^{1}!\lambda_{b}^{1}!}=\frac{n^{\downarrow s}}{s!}\frac{(s-1)!}{\lambda_{a}^{1}!\lambda_{b}^{1}!}=\binom{n}{s}{\binom{s-1}{\lambda_{a}^{1}}}. (4.31)

Combining (4.26) and Lemma 4.7, we obtain the following bound.

Corollary 4.8.

Let n1n\geq 1 and λn\lambda\vdash n. Then

dλ+dλe6cdc+dc.\frac{d_{\lambda}^{+}}{d_{\lambda}}\leq e^{6\sqrt{c}}\frac{d_{c}^{+}}{d_{c}}. (4.32)

We now fix the constants of Proposition 4.2. The next lemma combines the bounds which we will use just after in the proof of our result.

Lemma 4.9.

There exist constants n03,c06n_{0}\geq 3,c_{0}\geq 6 and C103C\geq 10^{3} such that the following hold.

  1. (i)

    For all nn0n\geq n_{0}, we have

    • (i1i_{1})

      nn8\sqrt{n}\leq\frac{n}{8},

    • (i2i_{2})

      n4nln(n)n\geq 4\sqrt{n}\,\ln(n),

    • (i3i_{3})

      ln2ln(n)12\frac{\ln 2}{\ln(n)}\leq\frac{1}{2},

    • (i4i_{4})

      6+6103ln(n)ln(n)86+\frac{6}{10^{3}}\ln(n)\leq\frac{\ln(n)}{8};

  2. (ii)

    for all cc0c\geq c_{0}, 6+6ln(c)c4ln(8/7)6+6\ln(c)\leq\frac{\sqrt{c}}{4}\ln(8/7);

  3. (iii)

    we have

    • (iii1iii_{1})

      Cln(n0!)ln(n0)ln(2)C\geq\frac{\ln(n_{0}!)\ln(n_{0})}{\ln(2)},

    • (iii2iii_{2})

      C12c0ln(c0!)C\geq 12\sqrt{c_{0}}\ln(c_{0}!).

Proof.

Assertions (i) and (ii) are clearly true for c0,n0c_{0},n_{0} large enough. Fix c0,n0c_{0},n_{0} so that it holds. Observe finally that, having fixed c0c_{0} and n0n_{0}, (iii) clearly holds for CC large enough. ∎

For the rest of the section, we set c0c_{0}, n0n_{0} and CC satisfying the conditions of Lemma 4.9.

Remark 4.3.

One can check that, for example, the values c0=108,n0=e50,C=e60c_{0}=10^{8},n_{0}=e^{50},C=e^{60} satisfy the conditions of Lemma 4.9. In other words we can take Caug=e60C_{aug}=e^{60} in Proposition 4.2. Combining this with Remark 4.1, we deduce that in Theorem 1.5, we can take C=e65C=e^{65}.

We first check that our result holds for small values of nn.

Lemma 4.10.

Let nn such that 2nn02\leq n\leq n_{0} and let λn\lambda\vdash n. Then

dλ+dλ1+C/lnn.d_{\lambda}^{+}\leq d_{\lambda}^{1+C/\ln n}. (4.33)
Proof.

Let 2nn02\leq n\leq n_{0} and λn\lambda\vdash n. If λ\lambda is flat (horizontal or vertical), that is, λa1=0\lambda_{a}^{1}=0 or λb1=0\lambda_{b}^{1}=0, then dλ=dλ+=1d_{\lambda}=d_{\lambda}^{+}=1 and the result holds. Assume now that λ\lambda is not flat. Then dλ2d_{\lambda}\geq 2 and

dλ+n!n0!=2ln(n0!)ln2dλln(n0!)ln2dλln(n0!)ln2lnn0lnndλC/lnndλ1+C/lnn,d_{\lambda}^{+}\leq n!\leq n_{0}!=2^{\frac{\ln(n_{0}!)}{\ln 2}}\leq d_{\lambda}^{\frac{\ln(n_{0}!)}{\ln 2}}\leq d_{\lambda}^{\frac{\ln(n_{0}!)}{\ln 2}\frac{\ln n_{0}}{\ln n}}\leq d_{\lambda}^{C/\ln n}\leq d_{\lambda}^{1+C/\ln n}, (4.34)

by Lemma 4.9 (iii1iii_{1}). ∎

We now consider small values of cc, assuming that nn is large enough.

Lemma 4.11.

Let nn0n\geq n_{0} and λn\lambda\vdash n such that c(λ)c0c(\lambda)\leq c_{0}. Then

dλ+dλ1+C/lnn.d_{\lambda}^{+}\leq d_{\lambda}^{1+C/\ln n}. (4.35)
Proof.

If c=0c=0, we have dλ+=dλd_{\lambda}^{+}=d_{\lambda} so the result holds. Assume now that 1cc01\leq c\leq c_{0}. Then λb11\lambda_{b}^{1}\geq 1, so that dλds(s11)=s1nd_{\lambda}\geq d_{s}\geq\binom{s-1}{1}=s-1\geq\sqrt{n} (since nn03n\geq n_{0}\geq 3). Then by Corollary 4.8,

dλ+dλe6cdc+dcdλe6cdc+dλe6c0c0!=dλn12c0ln(c0!)lnndλ1+C/lnn,d_{\lambda}^{+}\leq d_{\lambda}e^{6\sqrt{c}}\frac{d_{c}^{+}}{d_{c}}\leq d_{\lambda}e^{6\sqrt{c}}d_{c}^{+}\leq d_{\lambda}e^{6\sqrt{c_{0}}}c_{0}!=d_{\lambda}\sqrt{n}^{\frac{12\sqrt{c_{0}}\ln(c_{0}!)}{\ln n}}\leq d_{\lambda}^{1+C/\ln n}, (4.36)

by Lemma 4.9 (iii2iii_{2}). ∎

The next result concerns large values of nn and specific values of ss.

Lemma 4.12.

Let nn0n\geq n_{0} and ss such that nsn/8\sqrt{n}\leq s\leq n/8. Then

(esn)s/2enlnn8.\left(\frac{es}{n}\right)^{s/2}\leq e^{-\frac{\sqrt{n}\ln n}{8}}. (4.37)
Proof.

First note that since nn0n\geq n_{0}, we have nn/8\sqrt{n}\leq n/8 by Lemma 4.9 (i1i_{1}). We define

f:[n,n/8]x(exn)x/2\begin{array}[]{lrcl}f:&\left[\sqrt{n},n/8\right]&\longrightarrow&{\mathbb{R}}\\ &x&\longmapsto&\left(\frac{ex}{n}\right)^{x/2}\end{array} (4.38)

and observe that g:=lnfg:=\ln\circ f is convex (since g′′(x)=12x>0g^{\prime\prime}(x)=\frac{1}{2x}>0 for x[n,n/8]x\in\left[\sqrt{n},n/8\right]). Hence, gg reaches its maximum value when xx is either minimal or maximal and therefore the same holds also for ff. Moreover, e28e^{2}\leq 8 and n4nlnnn\geq 4\sqrt{n}\ln n (by Lemma 4.9 (i2i_{2})) so we have

f(n/8)f(n)=(e/8)n/16(e/n)n/2nn/2en/16=nn/4en/161.\frac{f(n/8)}{f(\sqrt{n})}=\frac{(e/8)^{n/16}}{(e/\sqrt{n})^{{\sqrt{n}/2}}}\leq\frac{\sqrt{n}^{\sqrt{n}/2}}{e^{n/16}}=\frac{n^{\sqrt{n}/4}}{e^{n/16}}\leq 1. (4.39)

We deduce that f(n)f(\sqrt{n}) is a maximum of ff, and conclude that

maxnsn/8f(s)f(n)=(e/n)n/2(n1/4)n/2=enlnn8,\max_{\sqrt{n}\leq s\leq n/8}f(s)\leq f(\sqrt{n})=(e/\sqrt{n})^{\sqrt{n}/2}\leq(n^{-1/4})^{\sqrt{n}/2}=e^{-\frac{\sqrt{n}\ln n}{8}}, (4.40)

as desired. ∎

We now have all the tools to prove Proposition 4.2.

Proof of Proposition 4.2.

We proceed by (strong) induction on nn. Our recurrence hypothesis is

H(n):dλ+dλ1+C/lnn for every λn.H(n):\text{\textquotedblleft}d_{\lambda}^{+}\leq d_{\lambda}^{1+C/\ln n}\text{ for every }\lambda\vdash n\text{\textquotedblright}. (4.41)

By Lemma 4.10, H(n)H(n) holds for all 2nn02\leq n\leq n_{0}, which initializes the induction.

Let now nn0n\geq n_{0} and assume that H(m)H(m) holds for every mnm\leq n. Let λn+1\lambda\vdash n+1.

The case 0cc00\leq c\leq c_{0} follows from Lemma 4.11. Since, in addition, sns\geq\sqrt{n} for every diagram, we can assume from now on that c0cnnc_{0}\leq c\leq n-\sqrt{n}.

By Corollary 4.8,

dλ+dλ1+Clnn=dλ+dλdλClnne6cdc+dcdλClnn=e6cdc+dc1+Clnc(dcdλ)C/lncdλClncClnn.\frac{d_{\lambda}^{+}}{d_{\lambda}^{1+\frac{C}{\ln n}}}=\frac{d_{\lambda}^{+}}{d_{\lambda}}d_{\lambda}^{-\frac{C}{\ln n}}\leq e^{6\sqrt{c}}\frac{d_{c}^{+}}{d_{c}}d_{\lambda}^{-\frac{C}{\ln n}}=e^{6\sqrt{c}}\frac{d_{c}^{+}}{d_{c}^{1+\frac{C}{\ln c}}}\left(\frac{d_{c}}{d_{\lambda}}\right)^{C/\ln c}d_{\lambda}^{\frac{C}{\ln c}-\frac{C}{\ln n}}. (4.42)

Applying the induction hypothesis to cc, we deduce that

dλ+dλ1+Clnne6c(dcdλ)C/lncdλClncClnn.\frac{d_{\lambda}^{+}}{d_{\lambda}^{1+\frac{C}{\ln n}}}\leq e^{6\sqrt{c}}\left(\frac{d_{c}}{d_{\lambda}}\right)^{C/\ln c}d_{\lambda}^{\frac{C}{\ln c}-\frac{C}{\ln n}}. (4.43)

By (4.26) and using that dλ(nc)dcdsd_{\lambda}\leq\binom{n}{c}d_{c}d_{s} (as explained in Remark 4.2), we deduce that

dλ+dλ1+Clnn\displaystyle\frac{d_{\lambda}^{+}}{d_{\lambda}^{1+\frac{C}{\ln n}}} e6c(e6cds(ns))Clnc((nc)dcds)ClncClnn\displaystyle\leq e^{6\sqrt{c}}\left(\frac{e^{6\sqrt{c}}}{d_{s}\binom{n}{s}}\right)^{\frac{C}{\ln c}}\left(\binom{n}{c}d_{c}d_{s}\right)^{\frac{C}{\ln c}-\frac{C}{\ln n}}
=e6c(1+Clnc)(ds(nc))C/lnndcClncClnn\displaystyle=e^{6\sqrt{c}\left(1+\frac{C}{\ln c}\right)}\left(d_{s}\binom{n}{c}\right)^{-C/\ln n}d_{c}^{\frac{C}{\ln c}-\frac{C}{\ln n}}
=(e6c(1+lncC)(ds(nc))lnclnndc1lnclnn)C/lnc\displaystyle=\left(\frac{e^{6\sqrt{c}(1+\frac{\ln c}{C})}}{\left(d_{s}\binom{n}{c}\right)^{\frac{\ln c}{\ln n}}}d_{c}^{1-\frac{\ln c}{\ln n}}\right)^{C/\ln c} (4.44)
(e6c(1+lncC)(nc)lnclnndc1lnclnn)C/lnc\displaystyle\leq\left(e^{6\sqrt{c}(1+\frac{\ln c}{C})}\binom{n}{c}^{-\frac{\ln c}{\ln n}}d_{c}^{1-\frac{\ln c}{\ln n}}\right)^{C/\ln c} (4.45)

To show our result, it is therefore enough to prove that

e6c(1+lncC)(nc)lnclnndc1lnclnn1.e^{6\sqrt{c}(1+\frac{\ln c}{C})}\binom{n}{c}^{-\frac{\ln c}{\ln n}}d_{c}^{1-\frac{\ln c}{\ln n}}\leq 1. (4.46)

We split the proof of (4.46) depending on the value of cc.

nnc7n/8n-\sqrt{n}\geq c\geq 7n/8

In this case, we have nsn/8\sqrt{n}\leq s\leq n/8. Then

1lnclnn=lnnlnclnn=ln(c+sc)lnn=ln(1+sc)lnnsclnn.1-\frac{\ln c}{\ln n}=\frac{\ln n-\ln c}{\ln n}=\frac{\ln(\frac{c+s}{c})}{\ln n}=\frac{\ln(1+\frac{s}{c})}{\ln n}\leq\frac{s}{c\ln n}. (4.47)

In addition, by Lemma 2.3, dcc!cc/2nc/2d_{c}\leq\sqrt{c!}\leq c^{c/2}\leq n^{c/2}, so that

dc1lnclnnes/2.d_{c}^{1-\frac{\ln c}{\ln n}}\leq e^{s/2}. (4.48)

On the other hand, since 1s/n1/21-s/n\geq 1/2 by assumption, we have

lnclnn=1+ln(c/n)lnn=1+ln(1s/n)lnn1ln2lnn,\frac{\ln c}{\ln n}=1+\frac{\ln(c/n)}{\ln n}=1+\frac{\ln(1-s/n)}{\ln n}\geq 1-\frac{\ln 2}{\ln n}, (4.49)

and in particular lnclnn1/2\frac{\ln c}{\ln n}\geq 1/2 by Lemma 4.9 (i3i_{3}).

Recall also from Lemma 4.4 (b) that (nc)=(ns)(n/s)s\binom{n}{c}=\binom{n}{s}\geq(n/s)^{s}. We therefore have, by (4.48) and Lemma 4.12,

[(nc)]lnclnndc1lnclnn(sn)s/2es/2=(esn)s/2enlnn8.\left[\binom{n}{c}\right]^{-\frac{\ln c}{\ln n}}d_{c}^{1-\frac{\ln c}{\ln n}}\leq\left(\frac{s}{n}\right)^{s/2}e^{s/2}=\left(\frac{es}{n}\right)^{s/2}\leq e^{-\frac{\sqrt{n}\ln n}{8}}. (4.50)

We conclude, using that lnclnn\ln c\leq\ln n and cn\sqrt{c}\leq\sqrt{n}, that

e6c(1+lncC)(nc)lnclnndc1lnclnnen(6+6lnnClnn8)1,e^{6\sqrt{c}(1+\frac{\ln c}{C})}\binom{n}{c}^{-\frac{\ln c}{\ln n}}d_{c}^{1-\frac{\ln c}{\ln n}}\leq e^{\sqrt{n}\left(6+\frac{6\ln n}{C}-\frac{\ln n}{8}\right)}\leq 1, (4.51)

by Lemma 4.9 (i4i_{4}) (that we can apply since C103C\geq 10^{3}).


c0c7n/8c_{0}\leq c\leq 7n/8 Since cc06c\geq c_{0}\geq 6, we have by Lemma 2.3 that dcc!(c/2)c/2d_{c}\leq\sqrt{c!}\leq(c/2)^{c/2}. Therefore:

dc1lnclnn=dcln(n/c)lnn((c/2)c/2)ln(n/c)lnn=((c/n)c/2)ln(n/c)lnn(nc/2)ln(n/c)lnn((1/2)c/2)ln(n/c)lnn=((c/n)c/2)ln(n/c)lnn(n/c)c/2 2c2ln(n/c)lnn=(nc)c2(1ln(n/c)lnn)2c2ln(n/c)lnn.\begin{split}d_{c}^{1-\frac{\ln c}{\ln n}}=d_{c}^{\frac{\ln(n/c)}{\ln n}}\leq\left((c/2)^{c/2}\right)^{\frac{\ln(n/c)}{\ln n}}&=\left((c/n)^{c/2}\right)^{\frac{\ln(n/c)}{\ln n}}\left(n^{c/2}\right)^{\frac{\ln(n/c)}{\ln n}}\left((1/2)^{c/2}\right)^{\frac{\ln(n/c)}{\ln n}}\\ &=\left((c/n)^{c/2}\right)^{\frac{\ln(n/c)}{\ln n}}(n/c)^{c/2}\,\cdot\,2^{-\frac{c}{2}\frac{\ln(n/c)}{\ln n}}\\ &=\left(\frac{n}{c}\right)^{\frac{c}{2}\left(1-\frac{\ln(n/c)}{\ln n}\right)}2^{-\frac{c}{2}\frac{\ln(n/c)}{\ln n}}.\end{split} (4.52)

On the other hand we still have

(nc)(n/c)c\binom{n}{c}\geq(n/c)^{c} (4.53)

and

lnclnn=1ln(n/c)lnn,\frac{\ln c}{\ln n}=1-\frac{\ln(n/c)}{\ln n}, (4.54)

so

[(nc)]lnclnn(nc)c(1ln(n/c)lnn),\left[\binom{n}{c}\right]^{-\frac{\ln c}{\ln n}}\leq\left(\frac{n}{c}\right)^{-c\left(1-\frac{\ln(n/c)}{\ln n}\right)}, (4.55)

and overall we get

(nc)lnclnndc1lnclnn(nc)c(1ln(n/c)lnn)(nc)c2(1ln(n/c)lnn)2c2ln(n/c)lnn=(cn)c2(1ln(n/c)lnn)2c2ln(n/c)lnn.\begin{split}\binom{n}{c}^{-\frac{\ln c}{\ln n}}d_{c}^{1-\frac{\ln c}{\ln n}}&\leq\left(\frac{n}{c}\right)^{-c\left(1-\frac{\ln(n/c)}{\ln n}\right)}\left(\frac{n}{c}\right)^{\frac{c}{2}\left(1-\frac{\ln(n/c)}{\ln n}\right)}2^{-\frac{c}{2}\frac{\ln(n/c)}{\ln n}}\\ &=\left(\frac{c}{n}\right)^{\frac{c}{2}\left(1-\frac{\ln(n/c)}{\ln n}\right)}2^{-\frac{c}{2}\frac{\ln(n/c)}{\ln n}}.\end{split} (4.56)

Observe that both factors are clearly 1\leq 1. Now, if cnc\geq\sqrt{n} we have 1ln(n/c)lnn1/21-\frac{\ln(n/c)}{\ln n}\geq 1/2 so

(cn)c2(1ln(n/c)lnn)(cn)c4(7/8)c/4.\left(\frac{c}{n}\right)^{\frac{c}{2}\left(1-\frac{\ln(n/c)}{\ln n}\right)}\leq\left(\frac{c}{n}\right)^{\frac{c}{4}}\leq(7/8)^{c/4}. (4.57)

On the other hand, if cnc\leq\sqrt{n}, we have ln(n/c)lnn1/2\frac{\ln(n/c)}{\ln n}\geq 1/2 so

2c2ln(n/c)lnn2c4(7/8)c/4,2^{-\frac{c}{2}\frac{\ln(n/c)}{\ln n}}\leq 2^{-\frac{c}{4}}\leq(7/8)^{c/4}, (4.58)

so we always have

[(nc)]lnclnndc1lnclnn(7/8)c/4.\left[\binom{n}{c}\right]^{\frac{-\ln c}{\ln n}}d_{c}^{1-\frac{\ln c}{\ln n}}\leq(7/8)^{c/4}. (4.59)

Hence, we get (using Lemma 4.9 (ii) and the fact that C1C\geq 1) that

e6c(1+lncC)(nc)lnclnndc1lnclnne6c(1+lncC)c4ln(8/7)1.e^{6\sqrt{c}(1+\frac{\ln c}{C})}\binom{n}{c}^{-\frac{\ln c}{\ln n}}d_{c}^{1-\frac{\ln c}{\ln n}}\leq e^{6\sqrt{c}\left(1+\frac{\ln c}{C}\right)-\frac{c}{4}\ln(8/7)}\leq 1. (4.60)

4.4 Sharpness of the bound

We consider here two examples, namely square-shaped diagrams and almost-flat diagrams, which show sharpness of Theorem 1.5. In other words, Theorem 1.5 provides the best possible asymptotic bound that is uniform over all irreducible characters. For square-shaped diagrams λ=[pp]\lambda=[p^{p}], we show that D(λ)=dλ1+4ln2lnn(1+o(1))D(\lambda)=d_{\lambda}^{1+\frac{4\ln 2}{\ln n}(1+o(1))} where n=p2n=p^{2}, and for λ=[n2,2]\lambda=[n-2,2], we show that D(λ)=dλ1+ln22lnn(1+o(1))D(\lambda)=d_{\lambda}^{1+\frac{\ln 2}{2\ln n}(1+o(1))}. We did not optimize the value of the constant CC in Theorem 1.5, but we conjecture that as nn\to\infty we have D(λ)dλ1+4ln2+o(1)lnnD(\lambda)\leq d_{\lambda}^{1+\frac{4\ln 2+o(1)}{\ln n}} uniformly for any λn\lambda\vdash n, and that the constant 4ln24\ln 2 is optimal.

4.4.1 Square-shaped diagrams

For all n1n\geq 1 such that nn is a perfect square (i.e. p:=np:=\sqrt{n}\in\mathbb{N}), let λn:=[p,,p]\lambda_{n}:=[p,\ldots,p] be the Young diagram of squared shape (see Fig. 11).

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Figure 11: The Young diagram λ25\lambda_{25}

Let us compute dλnd_{\lambda_{n}}. We have

H(λn,λn)=i,j=1p(i+j1)=i=1p(i+p1)!(i1)!=j=02p1j!(i=0p1i!)2=G(2p+1)G(p+1)2,H(\lambda_{n},\lambda_{n})=\prod_{i,j=1}^{p}(i+j-1)=\prod_{i=1}^{p}\frac{(i+p-1)!}{(i-1)!}=\frac{\prod_{j=0}^{2p-1}j!}{(\prod_{i=0}^{p-1}i!)^{2}}=\frac{G(2p+1)}{G(p+1)^{2}}, (4.61)

where GG denotes Barnes’ GG-function.

In particular, the asymptotic expansion of its logarithm is known (see e.g. [Ada14, Lemma 5.15.1]): as zz\rightarrow\infty,

lnG(1+z)=z22lnz3z24+O(z).\ln G(1+z)=\frac{z^{2}}{2}\ln z-\frac{3z^{2}}{4}+O(z). (4.62)

This provides

lnH(λn,λn)=lnG(2p+1)2lnG(p+1)=(2p)22ln(2p)3(2p)242p22lnp+23p24+O(p)=2p2lnp+2p2ln23p2p2lnp+32p2+O(p)=p2lnp+(2ln232)p2+O(p).\begin{split}\ln H(\lambda_{n},\lambda_{n})&=\ln G(2p+1)-2\ln G(p+1)\\ &=\frac{(2p)^{2}}{2}\ln(2p)-\frac{3(2p)^{2}}{4}-2\frac{p^{2}}{2}\ln p+2\frac{3p^{2}}{4}+O(p)\\ &=2p^{2}\ln p+2p^{2}\ln 2-3p^{2}-p^{2}\ln p+\frac{3}{2}p^{2}+O(p)\\ &=p^{2}\ln p+(2\ln 2-\frac{3}{2})p^{2}+O(p).\end{split} (4.63)

Thus, we get using Stirling’s approximation:

lndλn=lnn!lnH(λn,λn)=nlnnn+O(lnn)p2lnp(2ln232)p2+O(p)=2p2lnpp2p2lnp(2ln232)p2+O(p)=p2lnp+(122ln2)p2+O(p).\begin{split}\ln d_{\lambda_{n}}&=\ln n!-\ln H(\lambda_{n},\lambda_{n})\\ &=n\ln n-n+O(\ln n)-p^{2}\ln p-(2\ln 2-\frac{3}{2})p^{2}+O(p)\\ &=2p^{2}\ln p-p^{2}-p^{2}\ln p-(2\ln 2-\frac{3}{2})p^{2}+O(p)\\ &=p^{2}\ln p+(\frac{1}{2}-2\ln 2)p^{2}+O(p).\end{split} (4.64)

On the other hand, we have

D(λn)=(n1)!ai!bi!=(n1)!(i=1p1i!)2=(n1)!G(1+p)2.\begin{split}D(\lambda_{n})=\frac{(n-1)!}{\prod a_{i}!\prod b_{i}!}=\frac{(n-1)!}{(\prod_{i=1}^{p-1}i!)^{2}}=\frac{(n-1)!}{G(1+p)^{2}}.\end{split} (4.65)

Using again Stirling’s formula along with (4.62), we get that

lnD(λn)=ln((n1)!)2lnG(p+1)=nlnnn+O(lnn)2p22lnp+23p24+O(p)=2p2lnpp2p2lnp+32p2+O(p)=p2lnp+12p2+O(p).\begin{split}\ln D(\lambda_{n})&=\ln((n-1)!)-2\ln G(p+1)\\ &=n\ln n-n+O(\ln n)-2\frac{p^{2}}{2}\ln p+2\frac{3p^{2}}{4}+O(p)\\ &=2p^{2}\ln p-p^{2}-p^{2}\ln p+\frac{3}{2}p^{2}+O(p)\\ &=p^{2}\ln p+\frac{1}{2}p^{2}+O(p).\end{split} (4.66)

We deduce that

lnD(λn)lndλn=(2ln2)p2+O(p)=4ln2lnnlndλn(1+o(1)).\begin{split}\ln D(\lambda_{n})-\ln d_{\lambda_{n}}=\left(2\ln 2\right)p^{2}+O(p)=\frac{4\ln 2}{\ln n}\ln d_{\lambda_{n}}(1+o(1)).\end{split} (4.67)

Hence, as nn\rightarrow\infty:

D(λn)=dλn1+4ln2lnn(1+o(1)).D(\lambda_{n})=d_{\lambda_{n}}^{1+\frac{4\ln 2}{\ln n}(1+o(1))}. (4.68)

4.4.2 Almost-flat diagrams

The second case that we consider is the case of a diagram μn\mu_{n} of size nn with two lines, one of length n2n-2 and the second of length 22 (see Fig. 12).

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Figure 12: The Young diagram μ10\mu_{10}

In this case, we can also compute

H(μn,μn)=(n4)!(n2)(n1)2=2n!n(n3),H(\mu_{n},\mu_{n})=(n-4)!*(n-2)*(n-1)*2=2\frac{n!}{n(n-3)}, (4.69)

so that

dμn=n(n3)2.d_{\mu_{n}}=\frac{n(n-3)}{2}. (4.70)

On the other hand, we have

D(μn)=(n1)!(n3)!=(n1)(n2).D(\mu_{n})=\frac{(n-1)!}{(n-3)!}=(n-1)(n-2). (4.71)

Hence, we have

ln(D(μn)dμn)=ln(2(n1)(n2)n(n3))=ln2+o(1)=ln22lnnlndμn(1+o(1)).\begin{split}\ln\left(\frac{D(\mu_{n})}{d_{\mu_{n}}}\right)&=\ln\left(\frac{2(n-1)(n-2)}{n(n-3)}\right)\\ &=\ln 2+o(1)\\ &=\frac{\ln 2}{2\ln n}\ln d_{\mu_{n}}(1+o(1)).\end{split} (4.72)

Hence, we get

D(μn)=dμn1+ln22lnn(1+o(1)).D(\mu_{n})=d_{\mu_{n}}^{1+\frac{\ln 2}{2\ln n}(1+o(1))}. (4.73)

5 Character bounds in function of the number of cycles

The goal of this section is to prove Theorem 1.7, which provide bounds on characters only in terms of the number of cycles of the permutation σ\sigma

To prove bounds on characters depending on the number of cycles in the permutation, Larsen and Shalev [LS08] introduced the notion of cycle growth sequence. The cycle growth sequence (bk)k1(b_{k})_{k\geq 1} is defined by nbk=max(Fk(σ),1)n^{b_{k}}=\max(F_{k}(\sigma),1), and Larsen and Shalev set

B(σ)=k1bkk(k+1).B(\sigma)=\sum_{k\geq 1}\frac{b_{k}}{k(k+1)}. (5.1)

They showed that asymptotically E(σ)B(σ)+o(1)E(\sigma)\leq B(\sigma)+o(1), and that if cyc(σ)nα\operatorname{\mathrm{cyc}}(\sigma)\leq n^{\alpha} then B(σ)αB(\sigma)\leq\alpha. Plugging these bounds into Theorem 1.3, they deduced in [LS08, Theorem 1.4] that if cyc(σ)α\operatorname{\mathrm{cyc}}(\sigma)\leq\alpha then

|chλ(σ)|dλdλα+o(1),\frac{|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|}{d_{\lambda}}\leq d_{\lambda}^{\alpha+o(1)}, (5.2)

improving on their bound from [LS09].

Remark 5.1.

The o(1)o(1) term above has two components: one coming from the bound D(λ)dλ1+o(1)D(\lambda)\leq d_{\lambda}^{1+o(1)} (Theorem 1.3) and one from the bound E(σ)B(σ)+o(1)E(\sigma)\leq B(\sigma)+o(1). The bounds at the end of the proof of Theorem 1.1 in [LS08] actually show that E(σ)B(σ)+O(lnlnnlnn)E(\sigma)\leq B(\sigma)+O\left(\frac{\ln\ln n}{\ln n}\right). Combining this remark with Theorem 1.6 would give the bound |chλ(σ)|dλdλα+O(lnlnnlnn)\frac{|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|}{d_{\lambda}}\leq d_{\lambda}^{\alpha+O\left(\frac{\ln\ln n}{\ln n}\right)} if cyc(σ)α\operatorname{\mathrm{cyc}}(\sigma)\leq\alpha, which is another formulation of the character bound from Lifschitz and Marmor [LM23, Theorem 1.7]. The first aim of this section is to prove Theorem 1.7, which provides a sharper bound.

An equivalent formulation (and a more direct proof) of the bound E(σ)B(σ)+o(1)E(\sigma)\leq B(\sigma)+o(1) is given in [KLS24, Lemma 2.8]: ε>0\varepsilon>0 and α>0\alpha>0 being fixed, then for nn large enough, if σ𝔖n\sigma\in{\mathfrak{S}}_{n} satisfies cyc(σ)nα\operatorname{\mathrm{cyc}}(\sigma)\leq n^{\alpha}, then E(σ)α+εE(\sigma)\leq\alpha+\varepsilon.

In Proposition 5.1 (and Corollary 5.2), we show the following stronger result: if 2cyc(σ)nα2\leq\operatorname{\mathrm{cyc}}(\sigma)\leq n^{\alpha} then E(σ)αE(\sigma)\leq\alpha. The proof relies on a convexity argument and a telescoping product. Combining it with Theorem 1.6, it allows us to extend the character bounds of Lifschitz and Marmor [LM23, Theorem 1.7] – which hold when cyc(σ)n(lnn)C\operatorname{\mathrm{cyc}}(\sigma)\leq\frac{n}{(\ln n)^{C}} for some constant C>0C>0 – to permutations with cyc(σ)cn\operatorname{\mathrm{cyc}}(\sigma)\leq cn for some constant c>0c>0. This is the content of Theorem 1.7.

5.1 A simple bound on orbit growth exponents

Let n2n\geq 2 and σ𝔖n\sigma\in{\mathfrak{S}}_{n}. We start by showing the following surprisingly simple bound on E(σ)E(\sigma) that depends only on the number of cycles cyc(σ)\operatorname{\mathrm{cyc}}(\sigma).

Proposition 5.1.

Let n2n\geq 2 and σ𝔖n\sigma\in\mathfrak{S}_{n}. Then

E(σ)ln(max(cyc(σ),2))lnn.E(\sigma)\leq\frac{\ln\left(\max(\operatorname{\mathrm{cyc}}(\sigma),2)\right)}{\ln n}. (5.3)

We postpone the proof to the end of this subsection. It implies the following corollary.

Corollary 5.2.

Let n2n\geq 2 and σ𝔖n\sigma\in\mathfrak{S}_{n}. Assume that cyc(σ)2\operatorname{\mathrm{cyc}}(\sigma)\geq 2, and let 0<α10<\alpha\leq 1 such that cyc(σ)=nα\operatorname{\mathrm{cyc}}(\sigma)=n^{\alpha}. Then

E(σ)α.E(\sigma)\leq\alpha. (5.4)
Proof.

Since cyc(σ)2\operatorname{\mathrm{cyc}}(\sigma)\geq 2, from Proposition 5.1 we have E(σ)lncyc(σ)lnn=lnnαlnn=αE(\sigma)\leq\frac{\ln\operatorname{\mathrm{cyc}}(\sigma)}{\ln n}=\frac{\ln n^{\alpha}}{\ln n}=\alpha. ∎

Let imin:=imin(σ)=min{i1:fi1}\operatorname{i_{\mathrm{min}}}:=\operatorname{i_{\mathrm{min}}}(\sigma)=\min\left\{i\geq 1{\;:\;}f_{i}\geq 1\right\} be the length of the smallest cycle in σ\sigma.

We can rewrite the definition of the orbit growth sequence in terms of imini_{min}.

Lemma 5.3.

Let n2n\geq 2 and σ𝔖n\sigma\in\mathfrak{S}_{n}. Then

ei={0 for i<iminln(iminfimin)lnn for i=iminln(Σi/Σi1)lnn=ln(1+ifi/Σi1)lnn for imin<in.e_{i}=\begin{cases}0&\text{ for }i<\operatorname{i_{\mathrm{min}}}\\ \frac{\ln\left(\operatorname{i_{\mathrm{min}}}f_{\operatorname{i_{\mathrm{min}}}}\right)}{\ln n}&\text{ for }i=\operatorname{i_{\mathrm{min}}}\\ \frac{\ln(\Sigma_{i}/\Sigma_{i-1})}{\ln n}=\frac{\ln(1+if_{i}/\Sigma_{i-1})}{\ln n}&\text{ for }\operatorname{i_{\mathrm{min}}}<i\leq n.\end{cases} (5.5)
Proof.

By definition of imin\operatorname{i_{\mathrm{min}}} we have fi=0f_{i}=0 for i<imini<\operatorname{i_{\mathrm{min}}}. Therefore for i<imini<\operatorname{i_{\mathrm{min}}} we have Σi=0\Sigma_{i}=0 and therefore ei=0e_{i}=0. Therefore, neimin=ne1++eimin=Σimin=iminfiminn^{e_{\operatorname{i_{\mathrm{min}}}}}=n^{e_{1}+...+e_{\operatorname{i_{\mathrm{min}}}}}=\Sigma_{\operatorname{i_{\mathrm{min}}}}=\operatorname{i_{\mathrm{min}}}f_{\operatorname{i_{\mathrm{min}}}}, and we deduce the formula for eimine_{\operatorname{i_{\mathrm{min}}}} taking logarithms.

Let now i>imini>\operatorname{i_{\mathrm{min}}}. Note that Σi1Σimin1\Sigma_{i-1}\geq\Sigma_{\operatorname{i_{\mathrm{min}}}}\geq 1. Therefore

Σi=ne1++ei=ne1++ei1nei=Σi1nei.\Sigma_{i}=n^{e_{1}+...+e_{i}}=n^{e_{1}+...+e_{i-1}}n^{e_{i}}=\Sigma_{i-1}n^{e_{i}}. (5.6)

We then obtain the first equality for eie_{i} by dividing on both sides by Σi1\Sigma_{i-1} and taking the logarithm, and the second inequality by rewriting Σi=Σi1+ifi\Sigma_{i}=\Sigma_{i-1}+if_{i}. ∎

We can now prove a general simple bound on E(σ)E(\sigma). For 1in1\leq i\leq n, we denote by Fi=f1++fiF_{i}=f_{1}+...+f_{i} the number of cycles of σ\sigma of length at most ii.

Proposition 5.4.

Let n2n\geq 2 and σ𝔖n\sigma\in\mathfrak{S}_{n}. Then

E(σ)1iminln(imincyc(σ))lnn.E(\sigma)\leq\frac{1}{\operatorname{i_{\mathrm{min}}}}\frac{\ln\left(\operatorname{i_{\mathrm{min}}}\operatorname{\mathrm{cyc}}(\sigma)\right)}{\ln n}. (5.7)
Proof.

Let i>imini>\operatorname{i_{\mathrm{min}}}. By Lemma 5.3, and using that (1+x)a1+ax(1+x)^{a}\leq 1+ax for x0x\geq 0 and 0a10\leq a\leq 1, we have

iminiei=ln((1+ifiΣi1)imin/i)lnnln(1+iminfiΣi1)lnn.\frac{\operatorname{i_{\mathrm{min}}}}{i}e_{i}=\frac{\ln\left(\left(1+\frac{if_{i}}{\Sigma_{i-1}}\right)^{\operatorname{i_{\mathrm{min}}}/i}\right)}{\ln n}\leq\frac{\ln\left(1+\frac{\operatorname{i_{\mathrm{min}}}f_{i}}{\Sigma_{i-1}}\right)}{\ln n}. (5.8)

Moreover since fj=0f_{j}=0 for 1j<imin1\leq j<\operatorname{i_{\mathrm{min}}}, we have Σi1iminFi1\Sigma_{i-1}\geq\operatorname{i_{\mathrm{min}}}F_{i-1}, and therefore

iminieiln(1+fiFi1)lnn=ln(Fi/Fi1)lnn.\frac{\operatorname{i_{\mathrm{min}}}}{i}e_{i}\leq\frac{\ln\left(1+\frac{f_{i}}{F_{i-1}}\right)}{\ln n}=\frac{\ln\left(F_{i}/F_{i-1}\right)}{\ln n}. (5.9)

We deduce, recalling from Lemma 5.3 that eimin=ln(iminfimin)lnn=ln(iminFimin)lnne_{\operatorname{i_{\mathrm{min}}}}=\frac{\ln\left(\operatorname{i_{\mathrm{min}}}f_{\operatorname{i_{\mathrm{min}}}}\right)}{\ln n}=\frac{\ln\left(\operatorname{i_{\mathrm{min}}}F_{\operatorname{i_{\mathrm{min}}}}\right)}{\ln n}, that

E(σ)=i1eii=eiminimin+1imini>iminiminiei1iminlnn(ln(iminFimin)+imin<inln(FiFi1))=1iminlnnln(iminFiminimin<inFiFi1)=ln(iminFn)iminlnn,\begin{split}E(\sigma)=\sum_{i\geq 1}\frac{e_{i}}{i}&=\frac{e_{\operatorname{i_{\mathrm{min}}}}}{\operatorname{i_{\mathrm{min}}}}+\frac{1}{\operatorname{i_{\mathrm{min}}}}\sum_{i>\operatorname{i_{\mathrm{min}}}}\frac{\operatorname{i_{\mathrm{min}}}}{i}e_{i}\\ &\leq\frac{1}{\operatorname{i_{\mathrm{min}}}\ln n}\left(\ln\left(\operatorname{i_{\mathrm{min}}}F_{\operatorname{i_{\mathrm{min}}}}\right)+\sum_{\operatorname{i_{\mathrm{min}}}<i\leq n}\ln\left(\frac{F_{i}}{F_{i-1}}\right)\right)\\ &=\frac{1}{\operatorname{i_{\mathrm{min}}}\ln n}\ln\left(\operatorname{i_{\mathrm{min}}}F_{\operatorname{i_{\mathrm{min}}}}\prod_{\operatorname{i_{\mathrm{min}}}<i\leq n}\frac{F_{i}}{F_{i-1}}\right)\\ &=\frac{\ln(\operatorname{i_{\mathrm{min}}}F_{n})}{\operatorname{i_{\mathrm{min}}}\ln n},\end{split} (5.10)

which concludes the proof since Fn=cyc(σ)F_{n}=\operatorname{\mathrm{cyc}}(\sigma). ∎

We can now prove Proposition 5.1.

Proof of Proposition 5.1.

First, if imin(σ)=1\operatorname{i_{\mathrm{min}}}(\sigma)=1, this follows immediately from Proposition 5.4.

Second, if cyc(σ)=1\operatorname{\mathrm{cyc}}(\sigma)=1 then σ\sigma is an nn-cycle. In this case we have E(σ)=1nln2lnnE(\sigma)=\frac{1}{n}\leq\frac{\ln 2}{\ln n} and the result holds.

Assume finally that imin2\operatorname{i_{\mathrm{min}}}\geq 2 and cyc(σ)2\operatorname{\mathrm{cyc}}(\sigma)\geq 2. Then we have

lncyc(σ)lniminimin1,\ln\operatorname{\mathrm{cyc}}(\sigma)\geq\frac{\ln\operatorname{i_{\mathrm{min}}}}{\operatorname{i_{\mathrm{min}}}-1}, (5.11)

which can be rewritten as

1iminln(imincyc(σ))lnnln(cyc(σ))lnn.\frac{1}{\operatorname{i_{\mathrm{min}}}}\frac{\ln\left(\operatorname{i_{\mathrm{min}}}\operatorname{\mathrm{cyc}}(\sigma)\right)}{\ln n}\leq\frac{\ln\left(\operatorname{\mathrm{cyc}}(\sigma)\right)}{\ln n}. (5.12)

Again, we conclude using Proposition 5.4. ∎

5.2 Character bound

Combining Theorem 1.6 and Proposition 5.1, we can prove Theorem 1.7.

Proof of Theorem 1.7.

. By Theorem 1.6 and Proposition 5.1, we have (since E(σ)1E(\sigma)\leq 1)

|chλ(σ)|dλE(σ)(1+Clnn)dλE(σ)+Clnndλln(max(cyc(σ),2))lnn+Clnn.\left|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)\right|\leq d_{\lambda}^{E(\sigma)\left(1+\frac{C}{\ln n}\right)}\leq d_{\lambda}^{E(\sigma)+\frac{C}{\ln n}}\leq d_{\lambda}^{\frac{\ln\left(\max(\operatorname{\mathrm{cyc}}(\sigma),2)\right)}{\ln n}+\frac{C}{\ln n}}. (5.13)

This allows to conclude since ln(max(cyc(σ),2))ln(2cyc(σ))ln(cyc(σ))+1\ln\left(\max(\operatorname{\mathrm{cyc}}(\sigma),2)\right)\leq\ln\left(2\operatorname{\mathrm{cyc}}(\sigma)\right)\leq\ln\left(\operatorname{\mathrm{cyc}}(\sigma)\right)+1. ∎

6 Bounds on the Witten zeta function

Recall our definition of the Witten zeta function (1.10). For n1n\geq 1, A𝔖n^A\subset\widehat{\mathfrak{S}_{n}}, we denote the associated Witten zeta function by

ζn(A,s):=λA1dλs\zeta_{n}(A,s):=\sum_{\lambda\in A}\frac{1}{d_{\lambda}^{s}} (6.1)

for s0s\geq 0.

Let Λn(k)={λn:λ1λ1nk}\Lambda^{n}(k)=\left\{\lambda\vdash n{\;:\;}\lambda_{1}^{\prime}\leq\lambda_{1}\leq n-k\right\} (recall that λ1\lambda_{1} denotes the size of the first row and λ1\lambda_{1}^{\prime} the size of the first column).

Liebeck and Shalev [LS04, Proposition 2.5] proved that if s>0s>0 and kk are fixed, then as nn\to\infty we have the bound

ζn(Λn(k),s)=O(nks).\zeta_{n}(\Lambda^{n}(k),s)=O\left(n^{-ks}\right). (6.2)

This bound is very convenient. It was used for example by Larsen and Shalev [LS08] in combination with the Diaconis–Shahshahani upper bound lemma ([Dia88, Lemma 1, page 24 in Chapter 3B]), to derive mixing time results from character bounds.

However, for some applications that we will consider, we want asymptotic bounds on ζn(Λkn,sn)\zeta_{n}(\Lambda_{k}^{n},s_{n}) where the argument sns_{n} is not fixed, but tends to 0 as nn\rightarrow\infty instead. The goal of this section is to adapt the estimates from [LS04] to such cases.

Proposition 6.1.

Let α>0\alpha>0. Let (sn)n3(s_{n})_{n\geq 3} such that snαlnns_{n}\geq\frac{\alpha}{\ln n} for all n3n\geq 3. There exist constants k0=k0(α)k_{0}=k_{0}(\alpha), n0=n0(α)n_{0}=n_{0}(\alpha), and C=C(α)C=C(\alpha) such that for every kk0k\geq k_{0}, for every nn0n\geq n_{0}, we have

ζn(Λn(k),sn)Cek12snlnn.\zeta_{n}\left(\Lambda^{n}(k),s_{n}\right)\leq Ce^{-\frac{k}{12}s_{n}\ln n}. (6.3)

Moreover, if α>12ln2\alpha>12\ln 2 we can take k0=1k_{0}=1 and C=2C=2.

Proof.

We follow the proof of [LS04, Proposition 2.52.5]. To ease notation we will write Λ\Lambda for Λ(n)(k)\Lambda^{(n)}(k). We also set Λ1=Λ1n(k):={λΛn(k):λ12n/3}\Lambda_{1}=\Lambda_{1}^{n}(k):=\left\{\lambda\in\Lambda^{n}(k){\;:\;}\lambda_{1}\geq 2n/3\right\}, and Λ2n(k)={λΛn(k):λ1<2n/3}\Lambda_{2}^{n}(k)=\left\{\lambda\in\Lambda^{n}(k){\;:\;}\lambda_{1}<2n/3\right\}. we will write in the proof Λ,Λ1,Λ2\Lambda,\Lambda_{1},\Lambda_{2} for Λn(k)\Lambda^{n}(k), Λ1n(k),Λ2n(k)\Lambda_{1}^{n}(k),\Lambda_{2}^{n}(k) respectively. We also set Σi:=ζn(Λi,sn)\Sigma_{i}:=\zeta_{n}(\Lambda_{i},s_{n}) for i{1,2}i\in\left\{1,2\right\}, so that ζn(Λ,sn)=Σ1+Σ2\zeta_{n}(\Lambda,s_{n})=\Sigma_{1}+\Sigma_{2}.

The bound obtained from [LS04] for Σ2:=ζn(Λ2,sn)\Sigma_{2}:=\zeta_{n}(\Lambda_{2},s_{n}) works ad verbum. The authors prove that there exists a constant c>1c>1 such that Σ2p(n)cnsn\Sigma_{2}\leq p(n)c^{-ns_{n}}, where p(n)p(n) is the number of partitions of the integer nn. Since p(n)=eO(n)p(n)=e^{O(\sqrt{n})} and sn1/lnns_{n}\gtrsim 1/\ln n, we have in particular

Σ2=O(en3/4sn).\Sigma_{2}=O\left(e^{-n^{3/4}s_{n}}\right). (6.4)

Let us now bound Σ1\Sigma_{1}. We start from an intermediate bound (see [LS04, Proof of Proposition 2.52.5]), namely

Σ1kn/3p()(n)sn.\Sigma_{1}\leq\sum_{k\leq\ell\leq n/3}\frac{p(\ell)}{\binom{n-\ell}{\ell}^{s_{n}}}. (6.5)

From this point our proof diverges from that of [LS04]. Observe immediately that we get from (6.5) and Lemma 4.4 (b):

Σ1kn/3p()(n)sn.\Sigma_{1}\leq\sum_{k\leq\ell\leq n/3}p(\ell)\left(\frac{\ell}{n-\ell}\right)^{\ell s_{n}}.

We slice this sum according to whether >n2/3\ell>n^{2/3} or n2/3\ell\leq n^{2/3}. We have

Σ1Σ1+Σ1,\begin{split}\Sigma_{1}\leq\Sigma_{1}^{*}+\Sigma_{1}^{**},\end{split} (6.6)

where Σ1=kn2/3p()(n)sn\Sigma_{1}^{*}=\sum_{k\leq\ell\leq n^{2/3}}p(\ell)\left(\frac{\ell}{n-\ell}\right)^{\ell s_{n}} and Σ1=n2/3<n/3p()(n)sn\Sigma_{1}^{**}=\sum_{n^{2/3}<\ell\leq n/3}p(\ell)\left(\frac{\ell}{n-\ell}\right)^{\ell s_{n}}.

Let us first consider Σ1\Sigma_{1}^{**}. Since n/3\ell\leq n/3, we have n12\frac{\ell}{n-\ell}\leq\frac{1}{2}. Since p(n)=eO(n)p(n)=e^{O(\sqrt{n})}, we have p()p(n)en3/5p(\ell)\leq p(n)\leq e^{n^{3/5}} for nn large enough. We therefore have for nn large enough, using that sn1/lnns_{n}\gtrsim 1/\ln n,

Σ1n2/3<n/3en3/5(12)n2/3snn3en3/5(12)n2/3snen0.65sn.\Sigma_{1}^{**}\leq\sum_{n^{2/3}<\ell\leq n/3}e^{n^{3/5}}\left(\frac{1}{2}\right)^{n^{2/3}s_{n}}\leq\frac{n}{3}e^{n^{3/5}}\left(\frac{1}{2}\right)^{n^{2/3}s_{n}}\leq e^{-n^{0.65}s_{n}}. (6.7)

We finally bound Σ1\Sigma_{1}^{*}. Lower bounding the denominator by n5/6n^{5/6} and upper bounding \ell by n2/3n^{2/3}, we get

Σ1kn2/3p()(n2/3n5/6)sn=kn2/3p()e6snlnn.\Sigma_{1}^{*}\leq\sum_{k\leq\ell\leq n^{2/3}}p(\ell)\left(\frac{n^{2/3}}{n^{5/6}}\right)^{\ell s_{n}}=\sum_{k\leq\ell\leq n^{2/3}}p(\ell)e^{-\frac{\ell}{6}s_{n}\ln n}. (6.8)

Moreover since p()=eO()p(\ell)=e^{O(\sqrt{\ell})} as \ell\to\infty, there exists k0=k0(α)k_{0}=k_{0}(\alpha) such that, for every k0\ell\geq k_{0}, we have p()eα12p(\ell)\leq e^{\frac{\alpha}{12}\ell}. Therefore for kk0k\geq k_{0} we have

Σ1\displaystyle\Sigma_{1}^{*} kn2/3e(α1216snlnn)=ke(α1216snlnn)\displaystyle\leq\sum_{k\leq\ell\leq n^{2/3}}e^{\ell\left(\frac{\alpha}{12}-\frac{1}{6}s_{n}\ln n\right)}\leq\sum_{\ell=k}^{\infty}e^{\ell\left(\frac{\alpha}{12}-\frac{1}{6}s_{n}\ln n\right)}
=ek(α1216snlnn)1e(α1216snlnn)\displaystyle=\frac{e^{k\left(\frac{\alpha}{12}-\frac{1}{6}s_{n}\ln n\right)}}{1-e^{\left(\frac{\alpha}{12}-\frac{1}{6}s_{n}\ln n\right)}}
ek12snlnn1eα12,\displaystyle\leq\frac{e^{-\frac{k}{12}s_{n}\ln n}}{1-e^{-\frac{\alpha}{12}}}, (6.9)

using the fact that snlnnαs_{n}\ln n\geq\alpha.

Together with (6.7) and (6.4), this ends the proof. Finally note that, if α>12ln28.32\alpha>12\ln 2\approx 8.32, we can simply use the bound p()2eα12p(\ell)\leq 2^{\ell}\leq e^{\frac{\alpha}{12}} so that (6) holds with k0=1k_{0}=1 and C=2C=2. This concludes the proof, as the bounds (6.7) and (6.4) are independent of the value of k0k_{0}. ∎

In what follows, we will write f(n)g(n)f(n)\lll g(n) if f(n)g(n)n0\frac{f(n)}{g(n)}\xrightarrow[n\to\infty]{}0, that is if f(n)=o(g(n))f(n)=o(g(n)).

Corollary 6.2.

Let (sn)n3(s_{n})_{n\geq 3} be a sequence of positive real numbers. Then for every k1k\geq 1 we have

ζn(Λn(k),sn)n0 if and only if sn1lnn.\zeta_{n}(\Lambda^{n}(k),s_{n})\xrightarrow[n\to\infty]{}0\quad\text{ if and only if }\quad s_{n}\ggg\frac{1}{\ln n}. (6.10)
Proof.

First, if sn1lnns_{n}\ggg\frac{1}{\ln n} then sn>9lnns_{n}>\frac{9}{\ln n} (for nn large enough), and, for any fixed k1k\geq 1, ζn(Λn(k),sn)n0\zeta_{n}(\Lambda^{n}(k),s_{n})\xrightarrow[n\to\infty]{}0 by Proposition 6.1. Conversely, assume that sn1lnns_{n}\ggg\frac{1}{\ln n} does not hold. Then there exists a constant BB and an increasing function φ:\varphi{\;:\;}{\mathbb{N}}\to{\mathbb{N}} such that, for n3n\geq 3, sφ(n)Blnφ(n)s_{\varphi(n)}\leq\frac{B}{\ln\varphi(n)}. Fix k1k\geq 1 and, for nn large enough so that nkkn-k\geq k, consider the diagram [nk,1k][n-k,1^{k}]. We have for all nn, by the hook-length formula, d[nk,1k]=(n1k)nkd_{[n-k,1^{k}]}=\binom{n-1}{k}\leq n^{k}, so

ζφ(n)(Λφ(n)(k),sφ(n))(d[φ(n)k,1k])B/lnφ(n)eBk>0,\zeta_{\varphi(n)}(\Lambda^{\varphi(n)}(k),s_{\varphi(n)})\geq(d_{[\varphi(n)-k,1^{k}]})^{-B/\ln\varphi(n)}\geq e^{-Bk}>0, (6.11)

and therefore ζn(Λn(k),sn)\zeta_{n}(\Lambda^{n}(k),s_{n}) does not converge to 0. ∎

For n1n\geq 1 and k0k\geq 0, we denote

Λsymn(k)={λ𝔖n^:max(λ1,λ1)nk}.\Lambda^{n}_{\mathrm{sym}}(k)=\left\{\lambda\in\widehat{\mathfrak{S}_{n}}{\;:\;}\max(\lambda_{1},\lambda^{\prime}_{1})\leq n-k\right\}. (6.12)
Proposition 6.3.

Let (sn)n3(s_{n})_{n\geq 3} be a sequence of positive real numbers, and let k1k\geq 1. Then

ζn(Λsymn(k),sn)n0 if and only if sn1lnn.\zeta_{n}(\Lambda_{\mathrm{sym}}^{n}(k),s_{n})\xrightarrow[n\to\infty]{}0\quad\text{ if and only if }\quad s_{n}\ggg\frac{1}{\ln n}. (6.13)

In particular,

λ𝔖n^1dλsnn0 if and only if sn1lnn.\sum_{\lambda\in\widehat{\mathfrak{S}_{n}}^{**}}\frac{1}{d_{\lambda}^{s_{n}}}\xrightarrow[n\to\infty]{}0\quad\text{ if and only if }\quad s_{n}\ggg\frac{1}{\ln n}. (6.14)

Observe that the second point is exactly Proposition 1.8.

Proof.

Let n1n\geq 1. By definition we have

Λsymn(k)={λ𝔖n^:λΛn(k) or λΛn(k)}.\Lambda_{\mathrm{sym}}^{n}(k)=\left\{\lambda\in\widehat{\mathfrak{S}_{n}}{\;:\;}\lambda\in\Lambda^{n}(k)\text{ or }\lambda^{\prime}\in\Lambda^{n}(k)\right\}. (6.15)

Since the dimension of a diagram λ\lambda is the same as the dimension of its transpose λ\lambda^{\prime}, we deduce that

ζn(Λn(k),sn)ζn(Λsymn(k),sn)2ζn(Λn(k),sn).\zeta_{n}\left(\Lambda^{n}(k),s_{n}\right)\leq\zeta_{n}\left(\Lambda^{n}_{\mathrm{sym}}(k),s_{n}\right)\leq 2\zeta_{n}\left(\Lambda^{n}(k),s_{n}\right). (6.16)

The first statement then follows from Corollary 6.2, and the second is an application of the first one with k=1k=1, since by definition Λn(1)=𝔖n^\Lambda^{n}(1)=\widehat{\mathfrak{S}_{n}}^{**}. ∎

7 Characterization of fixed point free conjugacy classes that mix in two steps

7.1 Random walks and fixed point free conjugacy classes

Let (𝒞(n))({\mathcal{C}}^{(n)}) be a sequence of fixed-point free conjugacy classes (that is, such that f1(𝒞(n))=0f_{1}({\mathcal{C}}^{(n)})=0). We consider the sequence of random walks on 𝔖n{\mathfrak{S}}_{n}, with respective increment measures Unif𝒞(n)\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}. Let

𝔈(𝒞,t):={𝔖n\𝔄n if 𝒞𝔖n\𝔄n and t is odd,𝔄n otherwise,\mathfrak{E}({\mathcal{C}},t):=\begin{cases}\mathfrak{S}_{n}\backslash\mathfrak{A}_{n}&\text{ if }{\mathcal{C}}\subset\mathfrak{S}_{n}\backslash\mathfrak{A}_{n}\text{ and }t\text{ is odd},\\ \mathfrak{A}_{n}&\text{ otherwise,}\end{cases} (7.1)

be the coset of 𝔄n{\mathfrak{A}}_{n} on which the walk is supported after tt steps, and set

d(n)(t):=dTV(Unif𝒞(n)t,Unif𝔈n)\mathrm{d}^{(n)}(t):=\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*t},\operatorname{\mathrm{Unif}}_{{\mathfrak{E}}_{n}}\right) (7.2)

for t0t\geq 0.

Note that after 1 step the walk (started at Id\operatorname{\mathrm{Id}}) is concentrated on 𝒞(n){\mathcal{C}}^{(n)} so d(n)(1)=1o(1)\mathrm{d}^{(n)}(1)=1-o(1) and the walk cannot have mixed yet. Larsen and Shalev proved in [LS08] that the mixing time of such a sequence of walks is 2 or 3, i.e. (in regard of what precedes) that d(n)(3)0\mathrm{d}^{(n)}(3)\to 0. This settled a conjecture due to Lulov and Pak ([LP02, Conjecture 4.1]).

A natural extension is then to understand for which conjugacy classes the mixing time is 2 and for which ones it is 3 ; in other words to understand when we have d(n)(2)0\mathrm{d}^{(n)}(2)\to 0. The goal of this section is to prove such a characterization, Theorem 1.9.

7.2 Lower bound via a suitable observable

The study of fixed points of random permutations has a long story that dates back at least to de Montmort. In the chapter on jeu de Treize of [dM13], he describes computes the proportion of permutations with at least one fixed point (which is the probability to win a simplified version of the game called ’jeu de Treize’), via a general recursive formula, and computes this proportion for decks of 1 to 13 cards in the Table on Pages 59-60. He then shows in Remark I on Pages 60-61 that it converges to 1e11-e^{-1} as the number of cards tends to infinity. (The arguments of De Montmort make reference to Leibniz’s work about logarithms, and the notation ee for the natural base of the logarithm had not yet been introduced by Euler.)

A way to prove a lower bound on the total variation distance between two measures μ\mu and ν\nu is to find a good splitting event AA, which is an event AA such that μ(A)\mu(A) is large and ν(A)\nu(A) is small. For random transpositions, Diaconis and Shashahani considered the number of fixed points, which corresponds to the number of untouched cards. Here we also consider the number of fixed points but in a different way. We show that if 𝒞(n){\mathcal{C}}^{(n)} has order nn transpositions and x,yUnif𝒞(n)x,y\sim\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}} are independent uniform elements of the conjugacy class, then in the product permutation xyxy many transpositions compensate, leading to a significantly higher probability to have many fixed points than for a uniform permutation.

For n1n\geq 1 and m0m\geq 0 we set

Em=Em(n)={σ𝔖n:f1(σ)m}.E_{m}=E_{m}^{(n)}=\left\{\sigma\in{\mathfrak{S}}_{n}{\;:\;}f_{1}(\sigma)\geq m\right\}. (7.3)

For n1n\geq 1 we set X=XnUnif𝔄nX=X_{n}\sim\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}, Y=YnUnif𝔖nY=Y_{n}\sim\operatorname{\mathrm{Unif}}_{{\mathfrak{S}}_{n}}, and Z=ZnPoiss(1)Z=Z_{n}\sim\operatorname{Poiss}(1).

We recall a classical fact on the distribution of the number of fixed points of random permutations, usually attributed to Goncharov.

Lemma 7.1 ([Gon42]).

We have

dTV(f1(Yn),Poiss(1))n0.\operatorname{d_{TV}}\left(f_{1}(Y_{n}),\operatorname{Poiss}(1)\right)\xrightarrow[n\to\infty]{}0. (7.4)

There are many proofs of this result and the convergence is very fast, see the discussions in [DM23] and [Ful24]. Here we will only need rough bounds on the tails for the uniform distribution on the alternating group 𝔄n{\mathfrak{A}}_{n}.

Lemma 7.2.

Let us fix m0m\geq 0. Then there exists n0=n0(m)n_{0}=n_{0}(m) such that for all nn0n\geq n_{0} we have

Unif𝔄n(Em)3m!.\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}(E_{m})\leq\frac{3}{m!}. (7.5)
Proof.

First, since |𝔖n|=2|𝔄n||{\mathfrak{S}}_{n}|=2|{\mathfrak{A}}_{n}|, we have (f1(Xn)m)2(f1(Yn)m){\mathbb{P}}(f_{1}(X_{n})\geq m)\leq 2{\mathbb{P}}(f_{1}(Y_{n})\geq m) for all nn. Moreover, by Lemma 7.1, for nn large enough we have

2(f1(Yn)m)e(Znm)=jm1j!3m!.2{\mathbb{P}}(f_{1}(Y_{n})\geq m)\leq e{\mathbb{P}}(Z_{n}\geq m)=\sum_{j\geq m}\frac{1}{j!}\leq\frac{3}{m!}. (7.6)

This concludes the proof since (f1(Xn)m)=Unif𝔄n(Em){\mathbb{P}}(f_{1}(X_{n})\geq m)=\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}(E_{m}). ∎

Let us also show that the number of fixed points in subsets of [n][n] also asymptotically behaves as a Poisson variable. We believe that this result is part of the folklore, but add a proof for completeness.

Lemma 7.3.

Let 0α10\leq\alpha\leq 1. For n1n\geq 1, fix An[n]A_{n}\subset[n], let YnUnif𝔖nY_{n}\sim\operatorname{\mathrm{Unif}}_{{\mathfrak{S}}_{n}}, and define

Bn(Yn)={iAn:Yn(i)=i}.B_{n}(Y_{n})=\left\{i\in A_{n}{\;:\;}Y_{n}(i)=i\right\}. (7.7)

Assume that |An|/nα|A_{n}|/n\to\alpha as nn\to\infty. Then

|Bn(Yn)|n(d)Poiss(α).|B_{n}(Y_{n})|\xrightarrow[n\to\infty]{(d)}\operatorname{Poiss}(\alpha). (7.8)
Proof.

First, if α=1\alpha=1, then a consequence of Lemma 7.1 is that, with high probability, all fixed points of YnY_{n} are in AnA_{n}. The result follows. The same way, if α=0\alpha=0 then (|Bn(Yn)|=0)1\mathbb{P}(|B_{n}(Y_{n})|=0)\to 1 as nn\rightarrow\infty. Now suppose 0<α<10<\alpha<1. Consider the set F(Yn)F(Y_{n}) of fixed points of YnY_{n}, so that Bn(Yn)=F(Yn)AnB_{n}(Y_{n})=F(Y_{n})\cap A_{n}. Fix two integers jk0j\geq k\geq 0. We have, using the fact that YnY_{n} is uniform over 𝔖n{\mathfrak{S}}_{n} and splitting over |F(Yn)||F(Y_{n})|:

pk,j:=(|Bn(Yn)|=k,|F(Yn)|=j)=(|F(Yn)|=j)(|Bn(Yn)|=k||F(Yn)|=j)=(|F(Yn)|=j)(|An|k)(n|An|jk)(nj).\begin{split}p_{k,j}:=\mathbb{P}\left(|B_{n}(Y_{n})|=k,|F(Y_{n})|=j\right)&=\mathbb{P}(|F(Y_{n})|=j)\mathbb{P}\Big{(}|B_{n}(Y_{n})|=k\Big{|}|F(Y_{n})|=j\Big{)}\\ &=\mathbb{P}(|F(Y_{n})|=j)\frac{\binom{|A_{n}|}{k}\binom{n-|A_{n}|}{j-k}}{\binom{n}{j}}.\end{split} (7.9)

Moreover, as nn\to\infty we have (|F(Yn)|=j)e1j!\mathbb{P}(|F(Y_{n})|=j)\to\frac{e^{-1}}{j!} from Lemma 7.1, and since for rr fixed (nr)nrr!\binom{n}{r}\sim\frac{n^{r}}{r!}, we also have as nn\rightarrow\infty:

(|An|k)(n|An|jk)(nj)|An|kk!(n|An|)jk(jk)!njj!j!k!(jk)!αk(1α)jk.\frac{\binom{|A_{n}|}{k}\binom{n-|A_{n}|}{j-k}}{\binom{n}{j}}\sim\frac{\frac{|A_{n}|^{k}}{k!}\frac{(n-|A_{n}|)^{j-k}}{(j-k)!}}{\frac{n^{j}}{j!}}\sim\frac{j!}{k!(j-k)!}\alpha^{k}(1-\alpha)^{j-k}. (7.10)

We deduce that

pk,jne1k!(jk)!αk(1α)jk.p_{k,j}\xrightarrow[n\to\infty]{}\frac{e^{-1}}{k!(j-k)!}\alpha^{k}(1-\alpha)^{j-k}. (7.11)

Observe now that, making the change of indices i=jki=j-k, we have

jk0e1k!(jk)!αk(1α)jk=e1k0αkk!i0(1α)ii!=e1eαe1α=1.\sum_{j\geq k\geq 0}\frac{e^{-1}}{k!(j-k)!}\alpha^{k}(1-\alpha)^{j-k}=e^{-1}\sum_{k\geq 0}\frac{\alpha^{k}}{k!}\sum_{i\geq 0}\frac{(1-\alpha)^{i}}{i!}=e^{-1}e^{\alpha}e^{1-\alpha}=1. (7.12)

Hence, there is no loss of mass and (7.11) holds uniformly for all jk0j\geq k\geq 0. We deduce that for each k0k\geq 0,

(|Bn(Yn)|=k)=jkpk,jnjke1k!(jk)!αk(1α)jk.\mathbb{P}(|B_{n}(Y_{n})|=k)=\sum_{j\geq k}p_{k,j}\xrightarrow[n\to\infty]{}\sum_{j\geq k}\frac{e^{-1}}{k!(j-k)!}\alpha^{k}(1-\alpha)^{j-k}. (7.13)

Factoring out e1αkk!\frac{e^{-1}\alpha^{k}}{k!} and making again the change of indices i=jki=j-k, this rewrites as

e1αkk!i0(1α)ii!=e1αkk!e1α=eααkk!,\frac{e^{-1}\alpha^{k}}{k!}\sum_{i\geq 0}\frac{(1-\alpha)^{i}}{i!}=\frac{e^{-1}\alpha^{k}}{k!}e^{1-\alpha}=\frac{e^{-\alpha}\alpha^{k}}{k!}, (7.14)

which is the probability that a Poiss(α)\operatorname{Poiss}(\alpha) random variable is equal to kk. This concludes the proof. ∎

Let us now show that the cancellations of transpositions after two steps leads to many fixed points. The use of unseparated pairs in the next proof is inspired from [DEG14].

Lemma 7.4.

Let 0<α<1/20<\alpha<1/2 and m0m\geq 0. There exists n1=n1(α,m)n_{1}=n_{1}(\alpha,m) such that for every nn1n\geq n_{1} and every conjugacy class 𝒞{\mathcal{C}} of 𝔖n\mathfrak{S}_{n} such that f2(𝒞)αnf_{2}({\mathcal{C}})\geq\alpha n, we have

Unif𝒞2(Em)eα2αm/2(m/2)!\operatorname{\mathrm{Unif}}_{{\mathcal{C}}}^{*2}(E_{m})\geq\frac{e^{-\alpha}}{2}\frac{\alpha^{m/2}}{(m/2)!} (7.15)
Proof.

Let nn be a (large) integer, 𝒞{\mathcal{C}} be a conjugacy class of 𝔖n{\mathfrak{S}}_{n}, and x,yUnif𝒞x,y\sim\operatorname{\mathrm{Unif}}_{{\mathcal{C}}} be independent random variables. Let q:=2αn2q:=2\left\lfloor\frac{\alpha n}{2}\right\rfloor. Without loss of generality we can assume that

x=(1  2)(3  4)(2q1  2q)x,x=(1\;\;2)(3\;\;4)...(2q-1\;\;2q)x^{\prime}, (7.16)

where xx^{\prime} is a permutation of the integers from 2q+12q+1 to nn.

A way to obtain yy from xx is to consider a permutation zUnif𝔖nz\sim\operatorname{\mathrm{Unif}}_{{\mathfrak{S}}_{n}} and define y=zxz1y=zxz^{-1}. This corresponds to filling the cycles as follows:

y=(z(1)z(2))(z(3)z(4))(z(2q1)z(2q))y.y=(z(1)\;\;z(2))(z(3)\;\;z(4))...(z(2q-1)\;\;z(2q))y^{\prime}. (7.17)

The transpositions of yy that are the same as that of xx then include the transpositions (z(i)z(i+1))(z(i)\;\;z(i+1)) such that z(i)z(i) and z(i+1)z(i+1) are two consecutive integers that are between 1 and 2q2q, a subset of which is

Sn,2q(z):={i[2q]:i is odd,z(i)[2q],z(i) is odd, and z(i+1)=z(i)+1}.S_{n,2q}(z):=\left\{i\in[2q]{\;:\;}i\text{ is odd},z(i)\in[2q],z(i)\text{ is odd, and }z(i+1)=z(i)+1\right\}. (7.18)

Set w=z(n(n1) 2 1)w=z(n\;(n-1)\;...\;2\;1). Then the elements of Sn,2q(z)S_{n,2q}(z) are the odd fixed points of ww that are between 1 and 2q2q, which is a subset of [n][n] of size qαnq\sim\alpha n. We deduce from Lemma 7.3 that

for nn large enough we have, letting XPoiss(α)X\sim\operatorname{Poiss}(\alpha),

(|Sn,2q(z)|m/2)12(Xm/2)12(X=m/2)=12eααm/2(m/2)!.{\mathbb{P}}(|S_{n,2q}(z)|\geq m/2)\geq\frac{1}{2}{\mathbb{P}}(X\geq m/2)\geq\frac{1}{2}{\mathbb{P}}(X=m/2)=\frac{1}{2}e^{-\alpha}\frac{\alpha^{m/2}}{(m/2)!}. (7.19)

Now recall that each such fixed point of ww corresponds to a consecutive pair of zz, which is a transposition of yy that is the same as in xx. Hence, each element of Sn,2q(z)S_{n,2q}(z) corresponds to 2 fixed points in the product xyxy. We conclude that

(f1(xy)m)(|Sn,2q(z)|m/2)eα2αm/2(m/2)!.{\mathbb{P}}(f_{1}(xy)\geq m)\geq{\mathbb{P}}(|S_{n,2q}(z)|\geq m/2)\geq\frac{e^{-\alpha}}{2}\frac{\alpha^{m/2}}{(m/2)!}. (7.20)

We can now use this result to prove that EmE_{m} is a splitting event.

Proposition 7.5.

Let 0<α<1/20<\alpha<1/2. There exists c=c(α)>0c=c(\alpha)>0 and n2=n2(α)n_{2}=n_{2}(\alpha) such that for every nn2n\geq n_{2}, and every conjugacy class 𝒞{\mathcal{C}} of 𝔖n{\mathfrak{S}}_{n} such that f2(𝒞)αnf_{2}({\mathcal{C}})\geq\alpha n, we have

Unif𝒞2(Em)Unif𝔄n(Em)c(α),\operatorname{\mathrm{Unif}}_{{\mathcal{C}}}^{*2}(E_{m})-\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}(E_{m})\geq c(\alpha), (7.21)

and in particular

dTV(Unif𝒞(n)2,Unif𝔄n)c(α).\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right)\geq c(\alpha). (7.22)
Proof.

First observe that for any even integer m2m\geq 2 we have, setting β=eαα2\beta=\frac{e^{-\alpha}\alpha}{2},

m!(eα2αm/2(m/2)!)=mm/2eα2αm/2βm/2mm/2βm/2(m/2)m/2=(βm2)m/2.m!\left(\frac{e^{-\alpha}}{2}\frac{\alpha^{m/2}}{(m/2)!}\right)=m^{\downarrow m/2}\frac{e^{-\alpha}}{2}\alpha^{m/2}\geq\beta^{m/2}m^{\downarrow m/2}\geq\beta^{m/2}(m/2)^{m/2}=\left(\frac{\beta m}{2}\right)^{m/2}. (7.23)

We now fix m(α):=10βm(\alpha):=\left\lceil\frac{10}{\beta}\right\rceil. Then

m(α)!(eα2αm(α)/2(m(α)/2)!)(βm(α)2)m(α)/2βm(α)25.m(\alpha)!\left(\frac{e^{-\alpha}}{2}\frac{\alpha^{m(\alpha)/2}}{(m(\alpha)/2)!}\right)\geq\left(\frac{\beta m(\alpha)}{2}\right)^{m(\alpha)/2}\geq\frac{\beta m(\alpha)}{2}\geq 5. (7.24)

We deduce from Lemma 7.4 that for nn large enough

Unif𝒞2(Em(α))5m(α)!.\operatorname{\mathrm{Unif}}_{{\mathcal{C}}}^{*2}(E_{m(\alpha)})\geq\frac{5}{m(\alpha)!}. (7.25)

We conclude using Lemma 7.2, that for nn large enough

Unif𝒞2(Em(α))Unif𝔄n(Em(α))5m(α)!3m(α)!=2m(α)!=:c(α)>0.\operatorname{\mathrm{Unif}}_{{\mathcal{C}}}^{*2}(E_{m(\alpha)})-\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}(E_{m}(\alpha))\geq\frac{5}{m(\alpha)!}-\frac{3}{m(\alpha)!}=\frac{2}{m(\alpha)!}=:c(\alpha)>0. (7.26)

7.3 Upper bound on characters

We now obtain an upper bound on characters applied to fixed-point free permutations.

Lemma 7.6.

Let n2n\geq 2 and σ𝔖n\sigma\in{\mathfrak{S}}_{n} be a fixed-point free permutation. Then

|χλ(σ)|:=|chλ(σ)|dλdλ1/2dλ16lng(σ)lnn.|\chi^{\lambda}(\sigma)|:=\frac{|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|}{d_{\lambda}}\leq d_{\lambda}^{-1/2}d_{\lambda}^{-\frac{1}{6}\frac{\ln g(\sigma)}{\ln n}}. (7.27)

where g(σ)=ne3Cmax(2f2(σ),1)g(\sigma)=\frac{ne^{-3C}}{\max(2f_{2}(\sigma),1)}. In particular, we have

dλ|χλ(σ)|2dλ13lng(σ)lnn.d_{\lambda}|\chi^{\lambda}(\sigma)|^{2}\leq d_{\lambda}^{-\frac{1}{3}\frac{\ln g(\sigma)}{\ln n}}. (7.28)
Proof.

First, we have ne2=max(Σ2,1)n^{e_{2}}=\max(\Sigma_{2},1) i.e. e2=lnmax(Σ2,1)lnne_{2}=\frac{\ln\max(\Sigma_{2},1)}{\ln n}, so that

E(σ)12e2+13(1e2)=1216(1e2)=1216ln(n/max(Σ2,1))lnn.E(\sigma)\leq\frac{1}{2}e_{2}+\frac{1}{3}(1-e_{2})=\frac{1}{2}-\frac{1}{6}(1-e_{2})=\frac{1}{2}-\frac{1}{6}\frac{\ln(n/\max(\Sigma_{2},1))}{\ln n}. (7.29)

Therefore we have for all C0C\geq 0:

(1+Clnn)E(σ)12+12Clnn16ln(n/max(Σ2,1))lnn=1216ln(ne3C/max(Σ2,1))lnn.\left(1+\frac{C}{\ln n}\right)E(\sigma)\leq\frac{1}{2}+\frac{1}{2}\frac{C}{\ln n}-\frac{1}{6}\frac{\ln(n/\max(\Sigma_{2},1))}{\ln n}=\frac{1}{2}-\frac{1}{6}\frac{\ln(ne^{-3C}/\max(\Sigma_{2},1))}{\ln n}. (7.30)

We deduce from Theorem 1.6 that

|chλ(σ)|dλ(1+Clnn)E(σ)dλ1/2dλ16ln(ne3Cmax(Σ2,1))lnn.|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|\leq d_{\lambda}^{\left(1+\frac{C}{\ln n}\right)E(\sigma)}\leq d_{\lambda}^{1/2}d_{\lambda}^{-\frac{1}{6}\frac{\ln\left(\frac{ne^{-3C}}{\max(\Sigma_{2},1)}\right)}{\ln n}}. (7.31)

This concludes the proof since Σ2=f1+2f2=2f2\Sigma_{2}=f_{1}+2f_{2}=2f_{2}. ∎

Remark 7.1.

Lemma 7.6 shows in particular that if [ne3Cn\geq e^{3C}, f1(σ)=0f_{1}(\sigma)=0, and f2(σ)n2e3Cf_{2}(\sigma)\leq\frac{n}{2e^{3C}}] then |chλ(σ)|dλ|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|\leq\sqrt{d_{\lambda}}.

7.4 Proof of Theorem 1.9

We finally have all the ingredients to prove Theorem 1.9.

Proof of Theorem 1.9.

First, by Proposition 7.5, if there exists some α>0\alpha>0 such that f2(𝒞(n))αnf_{2}({\mathcal{C}}^{(n)})\geq\alpha n (for nn large enough) then dTV(Unif𝒞(n)2,Unif𝔄n)\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right) does not converge to 0. Assume now that f2(𝒞(n))=o(n)f_{2}({\mathcal{C}}^{(n)})=o(n) and let us prove that dTV(Unif𝒞(n)2,Unif𝔄n)n0\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right)\xrightarrow[n\to\infty]{}0.

By the Diaconis–Shahshahani upper bound lemma ([Dia88, Lemma 1 on Page 24 in Chapter 3B] but we refer here to Equation (4) in [Hou16] since we are considering the distance to the uniform measure on 𝔄n{\mathfrak{A}}_{n}), we have

4dTV(Unif𝒞(n)2,Unif𝔄n)2λ𝔖n^(dλ|χλ(𝒞(n))|2)2.4\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right)^{2}\leq\sum_{\lambda\in\widehat{\mathfrak{S}_{n}}^{**}}\left(d_{\lambda}|\chi^{\lambda}({\mathcal{C}}^{(n)})|^{2}\right)^{2}. (7.32)

Set g(𝒞(n))=ne3Cmax(2f2(𝒞(n)),1)g({\mathcal{C}}^{(n)})=\frac{ne^{-3C}}{\max(2f_{2}({\mathcal{C}}^{(n)}),1)} as in Lemma 7.6 and set also sn(𝒞(n)):=23lng(𝒞(n))lnns_{n}({\mathcal{C}}^{(n)}):=\frac{2}{3}\frac{\ln g({\mathcal{C}}^{(n)})}{\ln n}.

Then applying Lemma 7.6 to the right hand side of (7.32) we obtain

4dTV(Unif𝒞(n)2,Unif𝔄n)2λ𝔖n^(dλ13lng(σ)lnn)2=λ𝔖n^1(dλ)sn(𝒞(n)).4\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right)^{2}\leq\sum_{\lambda\in\widehat{\mathfrak{S}_{n}}^{**}}\left(d_{\lambda}^{-\frac{1}{3}\frac{\ln g(\sigma)}{\ln n}}\right)^{2}=\sum_{\lambda\in\widehat{\mathfrak{S}_{n}}^{**}}\frac{1}{(d_{\lambda})^{s_{n}({\mathcal{C}}^{(n)})}}. (7.33)

Finally, we have f2(𝒞(n))=o(n)f_{2}({\mathcal{C}}^{(n)})=o(n) by assumption, which is equivalent to sn(𝒞(n))1lnns_{n}({\mathcal{C}}^{(n)})\ggg\frac{1}{\ln n}. It then follows from Proposition 6.3 that

dTV(Unif𝒞(n)2,Unif𝔄n)n0,\operatorname{d_{TV}}\left(\operatorname{\mathrm{Unif}}_{{\mathcal{C}}^{(n)}}^{*2},\operatorname{\mathrm{Unif}}_{{\mathfrak{A}}_{n}}\right)\xrightarrow[n\to\infty]{}0, (7.34)

which concludes the proof of Theorem 1.9. ∎

8 Further mixing time estimates

This last section consists in the proofs of Theorems 1.10 and 1.12 about mixing time estimates for specific conjugacy classes.

8.1 Mixing time of permutations with few cycles

We prove here Theorem 1.10, which states that conjugacy classes of permutations with o(n)o(\sqrt{n}) cycles mix in 2 steps.

Proof of Theorem 1.10.

Let ε(n)\varepsilon(n) be such that cyc(𝒞(n))=ε(n)n\operatorname{\mathrm{cyc}}({\mathcal{C}}^{(n)})=\varepsilon(n)\sqrt{n}. Then by Theorem 1.7 there exists a constant CC^{\prime} such that for all n2n\geq 2 we have

|chλ(σ)|dλln(ε(n)n)lnn+Clnn=dλ12+ln(ε(n))+Clnn,|\operatorname{\mathrm{ch}}^{\lambda}(\sigma)|\leq d_{\lambda}^{\frac{\ln\left(\varepsilon(n)\sqrt{n}\right)}{\ln n}+\frac{C^{\prime}}{\ln n}}=d_{\lambda}^{\frac{1}{2}+\frac{\ln(\varepsilon(n))+C^{\prime}}{\ln n}}, (8.1)

so that dλ|χλ(σ)|2dλ2ln(ε(n))+Clnn=:dλsnd_{\lambda}|\chi^{\lambda}(\sigma)|^{2}\leq d_{\lambda}^{2\frac{\ln(\varepsilon(n))+C^{\prime}}{\ln n}}=:d_{\lambda}^{-s_{n}}. Since ε(n)0\varepsilon(n)\to 0, we have that snlnns_{n}\ln n\to\infty. We conclude as in the proof of Theorem 1.9 using the Diaconis–Shahshahani upper bound lemma and the bound on the Witten zeta function from Proposition 6.3. ∎

8.2 Cutoff for conjugacy classes with large support and no short cycles

Here we prove Theorem 1.12, that is, cutoff for conjugacy classes with no short cycles.

Let us first prove another bound on E(σ)E(\sigma), this time depending on the length of the smallest non-trivial cycle of σ\sigma.

For n2n\geq 2 and σ𝔖n\{Id}\sigma\in\mathfrak{S}_{n}\backslash\{\operatorname{\mathrm{Id}}\}, let ibis:=ibis(σ)=min{i2:fi1}\operatorname{i_{\mathrm{bis}}}:=\operatorname{i_{\mathrm{bis}}}(\sigma)=\min\left\{i\geq 2{\;:\;}f_{i}\geq 1\right\} be the length of the smallest non-trivial cycle of σ\sigma. We also denote by cyci(σ)\operatorname{\mathrm{cyc}}_{\geq i}(\sigma) the number of cycles of σ\sigma of length at least ii.

Proposition 8.1.

Let n2n\geq 2 and σ𝔖n\{Id}\sigma\in\mathfrak{S}_{n}\backslash\left\{\operatorname{\mathrm{Id}}\right\} with at least one fixed point. Then

E(σ)1ln(n/f1)lnn(11ibis).E(\sigma)-1\leq-\frac{\ln(n/f_{1})}{\ln n}\left(1-\frac{1}{\operatorname{i_{\mathrm{bis}}}}\right). (8.2)
Proof.

For convenience, in this proof, set =ibis(σ)\ell=\operatorname{i_{\mathrm{bis}}}(\sigma). We define, for ii\geq\ell,

Gi=f1+jifj.G_{i}=f_{1}+\ell\sum_{\ell\leq j\leq i}f_{j}. (8.3)

A straightforward adaptation of Lemma 5.3 shows that e1=lnf1lnne_{1}=\frac{\ln f_{1}}{\ln n}, that ei=0e_{i}=0 for 2i<2\leq i<\ell, and that e=ln(1+ff1)lnn=ln(Gf1)lnne_{\ell}=\frac{\ln\left(1+\frac{\ell f_{\ell}}{f_{1}}\right)}{\ln n}=\frac{\ln\left(\frac{G_{\ell}}{f_{1}}\right)}{\ln n}. Then, using Lemma 5.3 and the fact that (1+x)a1+ax(1+x)^{a}\leq 1+ax for x0x\geq 0 and 0a10\leq a\leq 1, we get for i>i>\ell:

iei=ln((1+ifiΣi1)/i)lnnln(1+fiΣi1)lnnln(1+fiGi1)lnn=ln(Gi/Gi1)lnn,\frac{\ell}{i}e_{i}=\frac{\ln\left(\left(1+\frac{if_{i}}{\Sigma_{i-1}}\right)^{\ell/i}\right)}{\ln n}\leq\frac{\ln\left(1+\frac{\ell f_{i}}{\Sigma_{i-1}}\right)}{\ln n}\leq\frac{\ln\left(1+\frac{\ell f_{i}}{G_{i-1}}\right)}{\ln n}=\frac{\ln\left(G_{i}/G_{i-1}\right)}{\ln n}, (8.4)

where we used in the last inequality that

Σi1=f1+ji1jfjf1+ji1fj=Gi1.\Sigma_{i-1}=f_{1}+\sum_{\ell\leq j\leq i-1}jf_{j}\geq f_{1}+\ell\sum_{\ell\leq j\leq i-1}f_{j}=G_{i-1}. (8.5)

The terms then compensate as in the proof of Proposition 5.4 and we get

E(σ)=e1+e+i>eii=lnf1lnn+1ln(G/f1)lnn+1i>ieilnf1lnn+1ln(Gn/f1)lnn.\begin{split}E(\sigma)=e_{1}+\frac{e_{\ell}}{\ell}+\sum_{i>\ell}\frac{e_{i}}{i}=\frac{\ln f_{1}}{\ln n}+\frac{1}{\ell}\frac{\ln(G_{\ell}/f_{1})}{\ln n}+\frac{1}{\ell}\sum_{i>\ell}\frac{\ell}{i}e_{i}\leq\frac{\ln f_{1}}{\ln n}+\frac{1}{\ell}\frac{\ln(G_{n}/f_{1})}{\ln n}.\end{split} (8.6)

We conclude that

E(σ)lnf1lnn+1ln(n/f1)lnn=lnf1lnn(11)+1=1ln(n/f1)lnn(11).E(\sigma)\leq\frac{\ln f_{1}}{\ln n}+\frac{1}{\ell}\frac{\ln(n/f_{1})}{\ln n}=\frac{\ln f_{1}}{\ln n}\left(1-\frac{1}{\ell}\right)+\frac{1}{\ell}=1-\frac{\ln(n/f_{1})}{\ln n}\left(1-\frac{1}{\ell}\right). (8.7)

Remark 8.1.

If σ\sigma consists only of fixed points and ibis\operatorname{i_{\mathrm{bis}}}-cycles, then E(σ)=e1+1e1ibis=e1(11ibis)+1ibisE(\sigma)=e_{1}+\frac{1-e_{1}}{\operatorname{i_{\mathrm{bis}}}}=e_{1}\left(1-\frac{1}{\operatorname{i_{\mathrm{bis}}}}\right)+\frac{1}{\operatorname{i_{\mathrm{bis}}}}. The bound from Proposition 8.1 is therefore sharp in this case.

We now turn to the proof of Theorem 1.12.

Proof of Theorem 1.12.

Let ε>0\varepsilon>0. First assume that f1=0f_{1}=0. Recall that d(n)(0)=1o(1)\mathrm{d}^{(n)}(0)=1-o(1) since the walk starts from the identity permutation, and d(n)(1)=1o(1)\mathrm{d}^{(n)}(1)=1-o(1) since the walk after 1 step is concentrated of 𝒞(n){\mathcal{C}}^{(n)}. The result to prove is therefore that d(n)(2)0\mathrm{d}^{(n)}(2)\to 0.

In this case we have min{i2:fi(𝒞(n))1}=imin\min\left\{i\geq 2{\;:\;}f_{i}({\mathcal{C}}^{(n)})\geq 1\right\}=\operatorname{i_{\mathrm{min}}}, which by assumption tends to infinity, so by Proposition 5.4 we have E(𝒞(n))1imin=o(1)E({\mathcal{C}}^{(n)})\leq\frac{1}{\operatorname{i_{\mathrm{min}}}}=o(1). It follows from Theorem 1.3 that |χλ(𝒞(n))|dλo(1)|\chi^{\lambda}({\mathcal{C}}^{(n)})|\leq d_{\lambda}^{o(1)}, so dλ|χλ(𝒞(n))|2=dλ1+o(1)d_{\lambda}|\chi^{\lambda}({\mathcal{C}}^{(n)})|^{2}=d_{\lambda}^{-1+o(1)}, and we conclude by the Diaconis–Shahshahani upper bound lemma, Theorem 1.6 and bounds on the Witten zeta function that d(n)(2)0\mathrm{d}^{(n)}(2)\to 0.

From now on we assume that f1(𝒞(n))1f_{1}({\mathcal{C}}^{(n)})\geq 1. The lower bound is a classical counting argument that can be done with coupon collector methods. We briefly recall it. A second moment argument shows that, if one multiplies lnnln(n/f)(1ε)\left\lfloor\frac{\ln n}{\ln(n/f)}(1-\varepsilon)\right\rfloor permutations (i.i.d., uniform in a conjugacy class) that have support of size nfn-f, then, with probability 1o(1)1-o(1), at least nε/2n^{\varepsilon}/2 are fixed points of all these permutations and therefore are fixed points of their product. Since uniform permutations (also on 𝔄n{\mathfrak{A}}_{n} and 𝔖n\𝔄n{\mathfrak{S}}_{n}\backslash{\mathfrak{A}}_{n}) have less than lnn\ln n fixed points with probability 1o(1)1-o(1), we deduce that d(n)(lnnln(n/f)(1ε))n1\mathrm{d}^{(n)}\left(\left\lfloor\frac{\ln n}{\ln(n/f)}(1-\varepsilon)\right\rfloor\right)\xrightarrow[n\to\infty]{}1.

Let us now show the cutoff upper bound. The proof is similar to that of Theorem 1.9. Set for the rest of the proof t:=lnnln(n/f)(1+ε)t:=\left\lceil\frac{\ln n}{\ln(n/f)}(1+\varepsilon)\right\rceil, 𝔈n=𝔈(𝒞(n),t){\mathfrak{E}}_{n}={\mathfrak{E}}\left({\mathcal{C}}^{(n)},t\right), and =ibis=min{i2:fi(𝒞(n))1}\ell=\operatorname{i_{\mathrm{bis}}}=\min\left\{i\geq 2{\;:\;}f_{i}({\mathcal{C}}^{(n)})\geq 1\right\}.

Let CC be the constant appearing in Theorems 1.5 and 1.6. First, by Proposition 8.1, and the facts that \ell\to\infty, and f=o(n)f=o(n), we have

(1+Clnn)E(𝒞(n))1=(1+Clnn)(E(𝒞(n))1)+Clnn(1+Clnn)ln(n/f)lnn(11)+Clnn=ln(n/f)lnn(1+o(1)).\begin{split}\left(1+\frac{C}{\ln n}\right)E({\mathcal{C}}^{(n)})-1&=\left(1+\frac{C}{\ln n}\right)\left(E({\mathcal{C}}^{(n)})-1\right)+\frac{C}{\ln n}\\ &\leq-\left(1+\frac{C}{\ln n}\right)\frac{\ln(n/f)}{\ln n}\left(1-\frac{1}{\ell}\right)+\frac{C}{\ln n}\\ &=-\frac{\ln(n/f)}{\ln n}(1+o(1)).\end{split} (8.8)

In particular, it is negative for nn large enough.

We therefore have as nn\rightarrow\infty:

1+t((1+Clnn)E(𝒞(n))1)=1+((1+ε)lnnln(n/f))(ln(n/f)lnn(1+o(1)))=ε+o(1).\begin{split}1+t\left(\left(1+\frac{C}{\ln n}\right)E({\mathcal{C}}^{(n)})-1\right)&=1+\left((1+\varepsilon)\frac{\ln n}{\ln(n/f)}\right)\left(-\frac{\ln(n/f)}{\ln n}(1+o(1))\right)\\ &=-\varepsilon+o(1).\end{split} (8.9)

We deduce from the Diaconis–Shahshahani upper bound lemma, Theorem 1.6, and bounds on the Witten zeta function,

4d(n)(t)λ𝔖n^(dλ|χλ(𝒞(n))|t)2λ𝔖n^dλ2ε+o(1)n0,4\mathrm{d}^{(n)}(t)\leq\sum_{\lambda\in\widehat{\mathfrak{S}_{n}}^{**}}\left(d_{\lambda}|\chi^{\lambda}({\mathcal{C}}^{(n)})|^{t}\right)^{2}\leq\sum_{\lambda\in\widehat{\mathfrak{S}_{n}}^{**}}d_{\lambda}^{-2\varepsilon+o(1)}\xrightarrow[n\to\infty]{}0, (8.10)

which concludes the proof. ∎

Acknowledgements

We are grateful to Nathanaël Berestycki, Persi Diaconis, Valentin Féray and Sam Olesker-Taylor for interesting conversations and helpful comments on early versions of the paper.

L.T. was supported by the Pacific Institute for the Mathematical Sciences, the Centre national de la recherche scientifique, and the Simons fundation, via a PIMS-CNRS-Simons postdoctoral fellowship. L.T. and P.T. were supported by the Austrian Science Fund (FWF) under grants 10.55776/P33083 and SFB F 1002.

References

  • [AD86] David Aldous and Persi Diaconis. Shuffling cards and stopping times. Amer. Math. Monthly, 93(5):333–348, 1986.
  • [Ada14] Victor S. Adamchik. Contributions to the theory of the Barnes function. Int. J. Math. Comput. Sci., 9(1):11–30, 2014.
  • [Ald83] David Aldous. Random walks on finite groups and rapidly mixing Markov chains. In Seminar on probability, XVII, volume 986 of Lecture Notes in Math., pages 243–297. Springer, Berlin, 1983.
  • [BD92] Dave Bayer and Persi Diaconis. Trailing the dovetail shuffle to its lair. Ann. Appl. Probab., 2(2):294–313, 1992.
  • [BŞ19] Nathanaël Berestycki and Batı Şengül. Cutoff for conjugacy-invariant random walks on the permutation group. Probab. Theory Related Fields, 173(3-4):1197–1241, 2019.
  • [BSZ11] Nathanaël Berestycki, Oded Schramm, and Ofer Zeitouni. Mixing times for random kk-cycles and coalescence-fragmentation chains. Ann. Probab., 39(5):1815–1843, 2011.
  • [DEG14] Persi Diaconis, Steven N. Evans, and Ron Graham. Unseparated pairs and fixed points in random permutations. Adv. in Appl. Math., 61:102–124, 2014.
  • [DF23] Persi Diaconis and Jason Fulman. The mathematics of shuffling cards. American Mathematical Society, Providence, RI, [2023] ©2023.
  • [Dia88] Persi Diaconis. Group representations in probability and statistics, volume 11 of Institute of Mathematical Statistics Lecture Notes—Monograph Series. Institute of Mathematical Statistics, Hayward, CA, 1988.
  • [dM13] Pierre Rémond de Montmort. Essay d’analyse sur les jeux de hazard. Chelsea Publishing Co., New York, second edition, 1980, original version from 1713.
  • [DM23] Persi Diaconis and Laurent Miclo. On a Markov construction of couplings. Arxiv preprint, 2023.
  • [DS81] Persi Diaconis and Mehrdad Shahshahani. Generating a random permutation with random transpositions. Z. Wahrsch. Verw. Gebiete, 57(2):159–179, 1981.
  • [FRT54] J. S. Frame, G. de B. Robinson, and R. M. Thrall. The hook graphs of the symmetric groups. Canad. J. Math., 6:316–324, 1954.
  • [FS11] Valentin Féray and Piotr Śniady. Asymptotics of characters of symmetric groups related to Stanley character formula. Ann. of Math. (2), 173(2):887–906, 2011.
  • [Ful24] Jason Fulman. Fixed points of non-uniform permutations and representation theory of the symmetric group. Arxiv preprint, 2024.
  • [Fé10] Valentin Féray. Stanley’s formula for characters of the symmetric group. Ann. Comb., 13(4):453–461, 2010.
  • [GLT20] Robert M. Guralnick, Michael Larsen, and Pham Huu Tiep. Character levels and character bounds. Forum Math. Pi, 8:e2, 81, 2020.
  • [GLT24] Robert M. Guralnick, Michael Larsen, and Pham Huu Tiep. Character levels and character bounds for finite classical groups. Invent. Math., 235(1):151–210, 2024.
  • [Gon42] VL Goncharov. On the distribution of cycles in permutations. In Dokl. Acad. Nauk SSSR, volume 35, pages 299–301, 1942.
  • [Hou16] Bob Hough. The random kk cycle walk on the symmetric group. Probab. Theory Related Fields, 165(1-2):447–482, 2016.
  • [JS24] Vishesh Jain and Mehtaab Sawhney. Hitting time mixing for the random transposition walk. Arxiv preprint, 2024+.
  • [KLS24] Nathan Keller, Noam Lifshitz, and Ohad Sheinfeld. Improved covering results for conjugacy classes of symmetric groups via hypercontractivity. Forum Math. Sigma, 12:e85, 2024.
  • [Lie17] Martin W. Liebeck. Character ratios for finite groups of Lie type, and applications. In Finite simple groups: thirty years of the atlas and beyond, volume 694 of Contemp. Math., pages 193–208. Amer. Math. Soc., Providence, RI, 2017.
  • [LM23] Noam Lifschitz and Avichai Marmor. Bounds for characters of the symmetric group: A hypercontractive approach. Arxiv preprint, 2023.
  • [LOST10] Martin W. Liebeck, E. A. O’Brien, Aner Shalev, and Pham Huu Tiep. The Ore conjecture. J. Eur. Math. Soc. (JEMS), 12(4):939–1008, 2010.
  • [LP02] Nathan Lulov and Igor Pak. Rapidly mixing random walks and bounds on characters of the symmetric group. J. Algebraic Combin., 16(2):151–163, 2002.
  • [LP17] David A. Levin and Yuval Peres. Markov chains and mixing times. American Mathematical Society, Providence, RI, 2017. Second edition of [ MR2466937], With contributions by Elizabeth L. Wilmer, With a chapter on “Coupling from the past” by James G. Propp and David B. Wilson.
  • [LS04] Martin W. Liebeck and Aner Shalev. Fuchsian groups, coverings of Riemann surfaces, subgroup growth, random quotients and random walks. J. Algebra, 276(2):552–601, 2004.
  • [LS05] Martin W. Liebeck and Aner Shalev. Fuchsian groups, finite simple groups and representation varieties. Invent. Math., 159(2):317–367, 2005.
  • [LS08] Michael Larsen and Aner Shalev. Characters of symmetric groups: sharp bounds and applications. Invent. Math., 174(3):645–687, 2008.
  • [LS09] Michael Larsen and Aner Shalev. Word maps and Waring type problems. J. Amer. Math. Soc., 22(2):437–466, 2009.
  • [LST24] Michael Larsen, Aner Shalev, and Pham Huu Tiep. Characteristic covering numbers of finite simple groups. Math. Ann., 388(1):167–189, 2024.
  • [LT24] Michael Larsen and Pham Huu Tiep. Uniform character bounds for finite classical groups. Ann. of Math. (2), 200(1):1–70, 2024.
  • [Lul96] Nathan Anton Mikerin Lulov. Random walks on the symmetric group generated by conjugacy classes. ProQuest LLC, Ann Arbor, MI, 1996. Thesis (Ph.D.)–Harvard University.
  • [MSP07] Thomas W. Müller and Jan-Christoph Schlage-Puchta. Character theory of symmetric groups, subgroup growth of Fuchsian groups, and random walks. Adv. Math., 213(2):919–982, 2007.
  • [Mé17] Pierre-Loïc Méliot. Representation Theory of Symmetric Groups (1st ed.). Chapman and Hall/CRC., 2017.
  • [NOT21] E. Nestoridi and S. Olesker-Taylor. Limit profiles for reversible Markov chains. Probab. Theory Related Fields, 2021.
  • [Roi96] Yuval Roichman. Upper bound on the characters of the symmetric groups. Invent. Math., 125(3):451–485, 1996.
  • [Sta06] Richard P. Stanley. A conjectured combinatorial interpretation of the normalized irreducible character values of the symmetric group. Arxiv preprint, 2006.
  • [Tey20] Lucas Teyssier. Limit profile for random transpositions. Ann. Probab., 48(5):2323–2343, 2020.
  • [Wit91] Edward Witten. On quantum gauge theories in two dimensions. Comm. Math. Phys., 141(1):153–209, 1991.