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On the Number of Eigenvalues of Modified Permutation Matrices in Mesoscopic Intervals

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Abstract

We are interested in two random matrix ensembles related to permutations: the ensemble of permutation matrices following Ewens’ distribution of a given parameter \(\theta >0\) and its modification where entries equal to 1 in the matrices are replaced by independent random variables uniformly distributed on the unit circle. For the elements of each ensemble, we focus on the random numbers of eigenvalues lying in some specified arcs of the unit circle. We show that for a finite number of fixed arcs, the fluctuation of the numbers of eigenvalues belonging to them is asymptotically Gaussian. Moreover, for a single arc, we extend this result to the case where the length goes to zero sufficiently slowly when the size of the matrix goes to infinity. Finally, we investigate the behavior of the largest and smallest spacings between two distinct consecutive eigenvalues.

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Acknowledgements

The author wishes to thank his Ph.D. advisor Joseph Najnudel for suggesting the problem and helpful comments.

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Correspondence to Valentin Bahier.

Appendices

Appendix A

We give a proof of Lemma 4 using Cesàro means. Our proof provides a slightly better upper-bound than the one given in [2]. We begin with the following lemma:

Lemma 10

For all \(1\le j\le n\),

$$\begin{aligned} \sum _{p=j}^{n-1} \frac{A_{p-j}^{\theta -1}}{p A_p^\theta } = \varPsi _n (j) \left( \frac{1}{j} - \frac{1}{n} \right) . \end{aligned}$$
(34)

Proof

With the same notation as in the proof of Proposition 2, we observe that

$$\begin{aligned} J_n(Ms)_n&= \sum _{p=1}^{n-1} J_n M_{n,p} s_p + s_n = \sum _{p=1}^{n-1} \frac{\theta }{\theta +p} \frac{1}{p} \sum _{j=1}^{p} \varPsi _p (j) w_j + \frac{1}{n} \sum _{j=1}^n \varPsi _n (j) w_j \\&= \sum _{j=1}^{n-1} w_j \left[ \sum _{p=j}^{n-1} \frac{\theta }{\theta +p} \frac{\varPsi _p (j)}{p} + \frac{\varPsi _n (j)}{n} \right] \\&\quad +\,\frac{\varPsi _n (n)}{n} w_n \end{aligned}$$

and

$$\begin{aligned} J_n t_n = \sum _{j=1}^{n-1} w_j \frac{\varPsi _n (j)}{j} + \frac{\varPsi _n (n)}{n} w_n. \end{aligned}$$

Therefore, by identification, for all \(j\in [\![ 1 , n-1 ]\!]\),

$$\begin{aligned} \frac{\varPsi _n (j)}{j} = \sum _{p=j}^{n-1} \frac{\theta }{\theta +p} \frac{\varPsi _p (j)}{p} + \frac{\varPsi _n (j)}{n}. \end{aligned}$$

Finally, using the definition of Cesàro numbers it is clear that \(\frac{\theta }{\theta +p} \frac{\varPsi _p (j)}{p} = \frac{A_{p-j}^{\theta -1}}{p A_p^\theta }\) for all \(1\le j \le p\). \(\square \)

We are ready to prove Lemma 4.

We introduce independent Bernoulli variables \(\xi _r\), \(r\ge 1\), defined as

$$\begin{aligned} \mathbb {P} (\xi _r = 1)= \frac{\theta }{\theta + r -1} \quad , \quad \mathbb {P} (\xi _r =0) = \frac{r-1}{\theta +r-1}. \end{aligned}$$

The Feller Coupling characterizes the variables \(a_{n,j}\) and \(W_j\) on the same probability space in function of the \(\xi _r\), by the following equalities: For all \(1 \le j \le n\),

$$\begin{aligned} a_{n,j}&= \# \{j-\text {spacings between two consecutive 1 in the word } (1 \ \xi _2 \ \cdots \ \xi _n \ 1) \} \\&= \sum _{k=1}^{n-j} \xi _k (1-\xi _{k+1}) \cdots (1-\xi _{k+j-1})\xi _{k+j} + \xi _{n-j+1} (1- \xi _{n-j+1} ) \cdots (1- \xi _n ), \end{aligned}$$

and for all \(j \in \mathbb {N}^*\),

$$\begin{aligned} W_j = \sum _{k=1}^{+\infty } \xi _k (1-\xi _{k+1}) \cdots (1-\xi _{k+j-1})\xi _{k+j}. \end{aligned}$$

To begin with, it is easy to notice that

$$\begin{aligned} \left| a_{n,j} - W_j \right| \le \mathbb {1}_{J_n=j} + W_{j,n} + \mathbb {1}_{J_n+K_n = j+1} \end{aligned}$$

where

$$\begin{aligned} \left\{ \begin{array}{l} J_n := \min \{ k\ge 1 : \ \xi _{n-k+1} =1 \} \\ K_n:= \min \{ k\ge 1 : \ \xi _{n+k} =1\} \\ W_{j,n} := \sum _{k=n+1}^{+\infty } \xi _k (1-\xi _{k+1}) \cdots (1-\xi _{k+j-1})\xi _{k+j} \end{array} \right. \end{aligned}$$

therefore

$$\begin{aligned} \mathbb {E} \left| a_{n,j} - W_j \right| \le \mathbb {P} (J_n=j ) + \mathbb {E} (W_{j,n}) + \mathbb {P} (J_n+K_n = j+1 ). \end{aligned}$$

We look separately at the three right-hand side terms in this inequality.

$$\begin{aligned} \mathbb {P} (J_n = j) = \frac{\theta }{\theta + n - j} \frac{n-j+1}{\theta + n -j +1} \cdots \frac{n-1}{\theta + n-1} = \frac{\theta }{n} \varPsi _n (j). \end{aligned}$$

Hence, \(\sum \limits _{j=1}^n \mathbb {P} (J_n = j) = \frac{\theta }{n} \sum \limits _{j=1}^n \varPsi _n (j) = 1\).

$$\begin{aligned} \mathbb {E} (W_{j,n})&= \sum \limits _{k=n+1}^{+\infty } \frac{\theta }{\theta + k -1} \frac{k}{\theta + k} \cdots \frac{k+j-2}{\theta + k+j-2} \frac{\theta }{\theta + k +j -1} \\&= \sum \limits _{p=n+j+1}^{+\infty } \frac{\theta }{\theta + p-j -1} \frac{p-j}{\theta + p-j} \cdots \frac{p-2}{\theta + p-2} \frac{\theta }{\theta + p -1}. \end{aligned}$$
$$\begin{aligned}&\mathbb {P} (J_n+K_n = j+1 ) \\&\qquad = \sum \limits _{k=1}^j \frac{\theta }{\theta + n +k-j-1} \frac{n+k-j}{\theta + n+k-j} \cdots \frac{n+k-2}{\theta + n+k-2} \frac{\theta }{\theta + n + k-1} \\&\qquad = \sum \limits _{p=n+1}^{n+j} \frac{\theta }{\theta + p-j -1} \frac{p-j}{\theta + p-j} \cdots \frac{p-2}{\theta + p-2} \frac{\theta }{\theta + p -1}. \end{aligned}$$

In particular,

$$\begin{aligned}&\mathbb {E} (W_{j,n}) + \mathbb {P} (J_n+K_n = j+1 ) \\&\qquad = \sum \limits _{p=n+1}^{+\infty } \frac{\theta }{\theta + p-j -1} \frac{p-j}{\theta + p-j} \cdots \frac{p-2}{\theta + p-2} \frac{\theta }{\theta + p -1} \\&\qquad = \theta \sum _{p=n}^{+\infty } \frac{1}{p} \frac{A_{p-j}^{\theta -1}}{A_p^\theta } = \theta \left[ \sum _{p=j}^{+\infty } \frac{1}{p} \frac{A_{p-j}^{\theta -1}}{A_p^\theta } - \sum _{p=j}^{n-1} \frac{1}{p} \frac{A_{p-j}^{\theta -1}}{A_p^\theta } \right] . \end{aligned}$$

Since, by Lemma 10,

$$\begin{aligned} \sum _{p=j}^{n-1} \frac{1}{p} \frac{A_{p-j}^{\theta -1}}{A_p^\theta } = \varPsi _n (j) \left( \frac{1}{j} - \frac{1}{n} \right) \underset{n\rightarrow \infty }{\longrightarrow } \frac{1}{j} \end{aligned}$$

it follows

$$\begin{aligned} \mathbb {E} (W_{j,n}) + \mathbb {P} (J_n+K_n = j+1 ) = \theta \left[ \frac{1}{j} - \varPsi _n (j) \left( \frac{1}{j} - \frac{1}{n} \right) \right] . \end{aligned}$$

Then,

$$\begin{aligned} \sum _{j=1}^n \mathbb {E} (W_{j,n}) + \mathbb {P} (J_n+K_n = j+1 )&= \theta \sum _{j=1}^n \frac{1}{j} - \theta \sum _{j=1}^n \frac{\varPsi _n (j)}{j} + \frac{\theta }{n} \sum _{j=1}^n \varPsi _n (j) \\&= \theta \sum _{j=1}^n \frac{1}{j} - \theta \sum _{j=1}^n \frac{1}{\theta + j -1} \ + \ 1 \\&= \theta \sum _{j=1}^n \frac{\theta -1}{j(\theta + j -1)} \ + \ 1 \end{aligned}$$

using Lemma 3 for the second equality. We deduce

$$\begin{aligned} \begin{aligned} \mathbb {E} \left( \sum _{j=1}^n \left| a_{n,j} - W_j \right| \right)&\le \,2 + \theta (\theta -1) \sum _{j=1}^n \frac{1}{j(\theta + j-1)} \\&\le \,1 + \theta \mathbb {1}_{0<\theta <1} + (1 + \theta (\gamma + \psi (\theta )))\mathbb {1}_{\theta \ge 1} \end{aligned} \end{aligned}$$
(35)

where \(\gamma \) is the Euler–Mascheroni constant, and \(\psi \) is the digamma function. As a particular consequence,

$$\begin{aligned} \limsup \limits _{\theta \rightarrow 0^+} \ \sup \limits _{n \ge 1} \mathbb {E} \left( \sum _{j=1}^n \left| a_{n,j} - W_j \right| \right) \le 1. \end{aligned}$$
(36)

Appendix B

The following proposition gives the whole possible values for the constant \(c_2\) appearing in Proposition 4, in function of \(\alpha \) and \(\beta \).

Proposition 11

Let pr be integers and qs positive integers with p and q relatively prime, r and s relatively prime, \(r\ne 0\). Then

$$\begin{aligned}&c_2 \\&\quad = \left\{ \begin{array}{l@{\quad }l} \frac{1}{6} &{} \mathrm{{if}} \,\,\alpha \,\, \mathrm{{and}} \,\, \beta \,\, \mathrm{{are}}\, \mathrm{{irrational}}\, \mathrm{{and}} \\ &{} \mathrm{{linearly}} \,\, \mathrm{{independent}}\, \mathrm{{over}}\, \mathbb {Q} \\ \frac{1}{6} + \frac{1}{6q^2} &{} \mathrm{{if}} \,\, \alpha = \frac{p}{q}\,\, \mathrm{and} \,\,\beta \,\, \mathrm{{is}}\,\, \mathrm{{irrational }} \\ \frac{1}{6} + \frac{1}{6s^2} &{} \mathrm{{if}} \,\, \alpha \,\, \mathrm{{is}}\,\, \mathrm{{irrational}} \,\, \mathrm{{and }} \,\,\beta = \frac{r}{s} \\ \frac{(2q-1)(q-1)}{6q^2} + \frac{(2s-1)(s-1)}{6s^2} - \frac{2}{qs} \sum _{j=1}^{qs} \lbrace \frac{jp}{q} \rbrace \lbrace \frac{jr}{s} \rbrace &{} \mathrm{{if}}\,\, \alpha = \frac{p}{q} \,\, \mathrm{{and}} \,\,\beta = \frac{r}{s} \\ \frac{1}{6} - \frac{\gcd (s,q)^2 }{6srq^2} &{} \mathrm{{if}} \,\, \alpha \,\, \mathrm{{is}} \,\, \mathrm{{irrational}}~\mathrm{{and}} \\ &{} \beta = \frac{p}{q}+\frac{r}{s}\alpha . \end{array}\right. \end{aligned}$$

The four first cases have been shown by Wieand, as well as the last case for \(s=1\) by the same author in [13]. We complete her work in this appendix, treating the case s arbitrary.

Therefore, suppose \(\alpha \) and \(\beta \) irrational numbers which are linearly dependent over \(\mathbb {Q}\), say \(\beta = \frac{p}{q} + \frac{r}{s} \alpha \), with \(s\ge 2\). Let us recall that \(c_2 := \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n (\{j\beta \} - \{j\alpha \})^2\). Since \(\lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n \{jx \}^2\) is easy to compute for all real numbers x, it remains to study what was denoted by \(s_3\) in [13], defined as

$$\begin{aligned} s_3 (\alpha , \beta ) := \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n \{j\alpha \} \{j \beta \}. \end{aligned}$$
(37)

We first assume that we have \(r>0\). We are going to proceed in two steps.

\(\bullet \,\, {\hbox {Computation of} \,s_{3} (\alpha , \frac{r}{s}\alpha )}\) : Before starting the calculation, it is good to notice that if \(\psi := \frac{\alpha }{s}\) then \(\psi \) is still irrational and consequently the sequence \((n\psi )_{n\in \mathbb {N}^*}\) is still equidistributed. Thus \(s_3 (s\psi , r\psi ) = s_3 (s\alpha , r\alpha )\), i.e., \(s_3 (\alpha , \frac{r}{s}\alpha ) = s_3 (s\alpha , r\alpha ) \). Indeed, it is a direct consequence of Theorem 9 given in [13] that we recall here:

Theorem 7

Let f be a Riemann integrable function on [0, 1], and let \(t\in \mathbb {R}\setminus \mathbb {Q}\). Then for all real numbers b,

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} \sum _{j=1}^n f(\{jt + b \} ) = \int _0^1 f(x) \mathrm {d}x. \end{aligned}$$

(We take \(b=0\) and \(f(x)=\{sx\} \{rx\}\).) The expression of \(s_3 (s\alpha , r\alpha )\) is more convenient to handle in practice. We decompose for \(j\ge 1\),

$$\begin{aligned} \lbrace j s\alpha \rbrace \lbrace j r\alpha \rbrace&= \left( s \lbrace j \alpha \rbrace - \lfloor s \lbrace j \alpha \rbrace \rfloor \right) \left( r \lbrace j \alpha \rbrace - \lfloor r \lbrace j \alpha \rbrace \rfloor \right) \\&=\,rs \lbrace j \alpha \rbrace ^2 - r \sum _{l =1}^{s-1} \lbrace j \alpha \rbrace \mathbb {1}_{ \lbrace j \alpha \rbrace \ge \frac{l}{s}} - s \sum _{k=1}^{r-1} \lbrace j \alpha \rbrace \mathbb {1}_{ \lbrace j \alpha \rbrace \ge \frac{k}{r}} \\&\qquad + \,\sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \mathbb {1}_{ \lbrace j \alpha \rbrace \ge \max \left( \frac{l}{s} , \frac{k}{r} \right) } \end{aligned}$$

The three first terms will not pose any difficulty since the limits of their respective means have already been evaluated in [13]. The novelty holds in the fourth term. For \(k \in [\![ 1 , r-1 ]\!]\) and \(l \in [\![ 1 , s-1 ]\!]\), let \(f_{k,l} (x) :=\mathbb {1}_{ x \ge \max \left( \frac{l}{s} , \frac{k}{r} \right) } \). These functions are clearly Riemann integrable on [0, 1], thus

$$\begin{aligned} \frac{1}{n} \sum _{j=1}^n \sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \mathbb {1}_{ \lbrace j \alpha \rbrace \ge \max \left( \frac{l}{s} , \frac{k}{r} \right) }&\underset{n\rightarrow \infty }{\longrightarrow } \sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \int _0^1 f_{k,l} (x) \mathrm {d}x \\&=\, (r-1)(s-1) - \sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \max \left( \frac{l}{s} , \frac{k}{r} \right) \end{aligned}$$

with

$$\begin{aligned}&\sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \max \left( \frac{l}{s} , \frac{k}{r} \right) \\&\qquad = \sum _{k=1}^{r-1} \left( \left( \sum _{l =1}^{\left\lfloor \frac{sk}{r}\right\rfloor } \frac{k}{r} \right) + \left( \sum _{l = \left\lfloor \frac{sk}{r}\right\rfloor + 1 }^{s-1} \frac{l}{s} \right) \right) \\&\qquad = \sum _{k=1}^{r-1} \left( \left( \frac{k}{r}\left\lfloor \frac{sk}{r}\right\rfloor \right) + \left( \frac{s-1}{2} -\frac{\left\lfloor \frac{sk}{r}\right\rfloor \left( \left\lfloor \frac{sk}{r}\right\rfloor +1\right) }{2s} \right) \right) \\&\qquad = \frac{(r-1)(s-1)}{2} + \frac{1}{2s} \sum _{k=1}^{r-1} \left( \frac{sk}{r} \right) ^2 - \frac{sk}{r} - \left\{ \frac{sk}{r} \right\} ^2 + \left\{ \frac{sk}{r} \right\} \\&\qquad = \frac{(r-1)(s-1)}{2} + \frac{1}{2s} \left( \frac{s^2-1}{r^2} \frac{r(r-1)(2r-1)}{6} - \frac{s-1}{r} \frac{r(r-1)}{2} \right) . \end{aligned}$$

Once expanded, we get

$$\begin{aligned} (r-1)(s-1) - \sum \limits _{\begin{array}{c} 1\le k \le r-1 \\ 1\le l \le s-1 \end{array}} \max \left( \frac{l}{s} , \frac{k}{r} \right) = \frac{1}{4} + \frac{rs}{3} - \frac{r}{4} - \frac{s}{4} - \frac{s}{12r} - \frac{r}{12s} + \frac{1}{12sr}. \end{aligned}$$

We deduce

$$\begin{aligned} s_3 (s\alpha , r\alpha )&= \frac{rs}{3} - r \left( \frac{s}{3} - \frac{1}{4} - \frac{1}{12s} \right) - s \left( \frac{r}{3} - \frac{1}{4} - \frac{1}{12r} \right) \\&\qquad + \,\left( \frac{1}{4} + \frac{rs}{3} - \frac{r}{4} - \frac{s}{4} - \frac{s}{12r} - \frac{r}{12s} + \frac{1}{12sr} \right) \\&=\, \frac{1}{4} + \frac{1}{12sr} \end{aligned}$$

\(\bullet \)   Computation of \({s_3 (\alpha , \frac{p}{q} + \frac{r}{s}\alpha )}\) : First of all, the particular case \(q=1\) is trivial since \(s_3 (\alpha , p + \frac{r}{s}\alpha ) = s_3 (\alpha ,\frac{r}{s}\alpha )\) and then \(c_2= \frac{2}{3} - 2 \left( \frac{1}{4} + \frac{1}{12sr} \right) = \frac{1}{6} - \frac{\gcd (s,1)^2}{6sr\times 1^2}\). Thus, we assume in all the following that \(q\ge 2\). In the same manner as at the beginning of the first step, we notice that \(s_3 (\alpha , \frac{p}{q} + \frac{r}{s}\alpha ) = s_3 (s\alpha , \frac{p}{q} + r\alpha )\) (using Theorem 7 with \(b=\frac{p}{q}\) and for f a complicated expression that we do not precise here, which is bounded and piecewise continuous so Riemann integrable).

For all \(j \ge 1\),

$$\begin{aligned} \left\{ j s \alpha \right\} \left\{ j \frac{p}{q} + j r \alpha \right\} = \left\{ j s \alpha \right\} \left\{ j \frac{p}{q} \right\} + \lbrace j s \alpha \rbrace \lbrace j r \alpha \rbrace - \lbrace j s \alpha \rbrace \mathbb {1}_{ \lbrace j \frac{p}{q} \rbrace + \lbrace j r \alpha \rbrace \ge 1 }. \end{aligned}$$

The mean of the first term tends to \(s_3 (s \alpha , \frac{p}{q}) = \frac{1}{4} - \frac{1}{4q}\). The mean of the second term tends to \(s_3 (s\alpha , r\alpha ) = \frac{1}{4} + \frac{1}{12sr}\) by the first step. It remains to study the third term. In a similar way to what is done in [13], it is easy to check that

$$\begin{aligned} \frac{1}{q \left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=1}^{q \left\lfloor \frac{n}{q} \right\rfloor } \lbrace j s \alpha \rbrace \mathbb {1}_{ \lbrace j \frac{p}{q} \rbrace + \lbrace j r \alpha \rbrace \ge 1 } \underset{n\rightarrow \infty }{=} \frac{1}{n} \sum _{j=1}^n \lbrace j s \alpha \rbrace \mathbb {1}_{ \lbrace j \frac{p}{q} \rbrace + \lbrace j r \alpha \rbrace \ge 1 } \ + \ o(1). \end{aligned}$$

Let \(n\in \mathbb {N}^*\). We use the periodicity of the sequence \((\lbrace \frac{kp}{q} \rbrace )\) to write that the left-hand side expression is equal to

$$\begin{aligned}&\frac{1}{q \left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=0}^{\left\lfloor \frac{n}{q} \right\rfloor - 1} \sum _{k=1}^q \lbrace (k+jq)s\alpha \rbrace \mathbb {1}_{ \lbrace (k+jq)r\alpha \rbrace \ge 1 - \left\{ \frac{kp}{q} \right\} } \\&\qquad = \,s \cdot \frac{1}{q} \sum _{k=1}^q \frac{1}{\left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=0}^{\left\lfloor \frac{n}{q} \right\rfloor - 1} \lbrace (k+jq)\alpha \rbrace \mathbb {1}_{ \lbrace (k+jq)r\alpha \rbrace \ge 1 - \left\{ \frac{kp}{q} \right\} } \\&\qquad \quad - \,\frac{1}{q} \sum _{k=1}^q \frac{1}{\left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=0}^{\left\lfloor \frac{n}{q} \right\rfloor - 1} \sum _{m=1}^{s-1} \mathbb {1}_{ \left( \lbrace (k+jq) \alpha \rbrace \ge \frac{m}{s} \right) \cap \left( \lbrace (k+jq) r \alpha \rbrace \ge 1 - \left\{ \frac{kp}{q} \right\} \right) } \end{aligned}$$

with, (see [13]),

$$\begin{aligned} \frac{1}{q} \sum _{k=1}^q \frac{1}{\left\lfloor \frac{n}{q} \right\rfloor } \sum _{j=0}^{\left\lfloor \frac{n}{q} \right\rfloor - 1} \lbrace (k+jq)\alpha \rbrace \mathbb {1}_{ \lbrace (k+jq)r\alpha \rbrace \ge 1 - \left\{ \frac{kp}{q} \right\} } \underset{n\rightarrow \infty }{\longrightarrow } \frac{1}{4} + \frac{1}{12r} - \frac{1}{4q} - \frac{1}{12rq^2}. \end{aligned}$$

For \(k \in [\![ 1 , q]\!]\) and \(m \in [\![ 1 , s-1 ]\!]\), let

$$\begin{aligned} f_{k,m} (x) = \mathbb {1}_{ \left( x \ge \frac{m}{s} \right) \cap \left( rx - \sum _{l =1}^{r-1} \mathbb {1}_{x \ge \frac{l}{r} } \ge 1 - \left\{ \frac{kp}{q} \right\} \right) }. \end{aligned}$$

We compute its integral between 0 and 1:

$$\begin{aligned} \int _0^1 f_{k,m} (x) \mathrm {d}x&= \sum _{l = 0}^{r-1} \int _{\frac{1 - \left\{ \frac{kp}{q} \right\} + l}{r}}^{\frac{l+1}{r}} \mathbb {1}_{ \left( x \ge \frac{m}{s} \right) } \mathrm {d}x \\&= \,0 + \left( \frac{\left\lfloor \frac{rm}{s} \right\rfloor +1}{r} - \max \left( \frac{m}{s} , \frac{1 - \left\{ \frac{kp}{q} \right\} + \left\lfloor \frac{rm}{s} \right\rfloor }{r} \right) \right) \\&\quad + \,\sum _{l = \left\lfloor \frac{rm}{s} \right\rfloor +1 }^{r-1} \frac{l +1}{r} - \frac{1 - \left\{ \frac{kp}{q} \right\} + l}{r} \\&= \frac{1- \left\{ \frac{rm}{s} \right\} }{r} - \frac{1- \left\{ \frac{kp}{q} \right\} - \left\{ \frac{rm}{s} \right\} }{r} \mathbb {1}_{1- \left\{ \frac{kp}{q} \right\} - \left\{ \frac{rm}{s} \right\} \ge 0} \\&\quad + \,\left\{ \frac{kp}{q} \right\} \left( 1 - \frac{1}{r} - \frac{m}{s} + \frac{1}{r} \left\{ \frac{rm}{s} \right\} \right) \\&= \,\frac{1- \left\{ \frac{kp}{q} \right\} - \left\{ \frac{rm}{s} \right\} }{r} \mathbb {1}_{1- \left\{ \frac{kp}{q} \right\} - \left\{ \frac{rm}{s} \right\} < 0} \\&\quad + \left\{ \frac{kp}{q} \right\} \left( 1 - \frac{m}{s}+ \,\frac{1}{r} \left\{ \frac{rm}{s} \right\} \right) . \end{aligned}$$

Thus,

$$\begin{aligned} \sum _{k=1}^q \sum _{m=1}^{s-1} \int _0^1 f_{k,m} (x) \mathrm {d}x&= - \,\sum _{k=1}^q \left\{ \frac{kp}{q} \right\} + \sum _{k=1}^q \sum _{m=0}^{s-1} \int _0^1 f_{k,m} (x) \mathrm {d}x \\&=\, -\,\sum _{k=0}^{q-1} \frac{k}{q} + \sum _{k=0}^{q-1} \sum _{m=0}^{s-1} \frac{1 - \frac{k}{q} - \frac{m}{s}}{r} \mathbb {1}_{\frac{m}{s}> 1 - \frac{k}{q} }\\&\quad + \,\frac{k}{q} \left( 1- \frac{m}{s} + \frac{m}{sr} \right) \\&=\, \frac{q-1}{2} \left( s-1 - \frac{s-1}{2} + \frac{s-1}{2r} \right) \\&\quad +\, \sum _{k=0}^{q-1} \sum _{m=0}^{s-1} \frac{1 - \frac{k}{q} - \frac{m}{s}}{r} \mathbb {1}_{\frac{m}{s} > 1 - \frac{k}{q} } \end{aligned}$$

where we get the second equality noting that the numbers \(\left\{ \frac{kp}{q} \right\} \) (resp. \(\left\{ \frac{mr}{s} \right\} \) ) cycle through some rearrangement of the numbers \(0 , \frac{1}{q} ,\ldots , \frac{q-1}{q}\) (resp. \(0 , \frac{1}{s} ,\ldots , \frac{s-1}{s}\)). Now, it just remains to compute the second term in the last expression, which is equal to

$$\begin{aligned} \frac{1}{r} \sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \sum _{m = \left\lfloor s- \frac{ks}{q} \right\rfloor + 1}^{s-1} \left( 1 - \frac{k}{q} - \frac{m}{s} \right) . \end{aligned}$$

Note that the condition \(\left\lfloor \frac{q}{s} \right\rfloor + 1 \le q-1\) is always satisfied when \(q\ge 3\) since \(s\ge 2 > \frac{3}{2} \ge \frac{q}{q-1}\). When \(q=2\), it only fails for \(s=2\) (in this case we have an empty sum). For now, assume we are not in the case \(q=s=2\). We will easily treat this remaining case at the end. We have:

$$\begin{aligned} \sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \sum _{m = \left\lfloor s- \frac{ks}{q} \right\rfloor + 1}^{s-1} \left( 1 - \frac{k}{q} - \frac{m}{s} \right)&=\, \sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \left( (s-1)- \left\lfloor s- \frac{ks}{q} \right\rfloor \right) \left( 1 - \frac{k}{q} \right) \\&\quad - \,\frac{1}{s} \sum _{m = \left\lfloor s- \frac{ks}{q} \right\rfloor + 1}^{s-1} m \\&= \,\sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \frac{1}{2} \left( \frac{k}{q} - s \left( \frac{k}{q} \right) ^2 \right) \\&\quad + \,\frac{1}{2s} \left( \left\{ \frac{-ks}{q} \right\} ^2 - \left\{ \frac{-ks}{q} \right\} \right) . \end{aligned}$$

In addition, using the facts that for all \(x \in \mathbb {R}\), \(\{-x\}(1-\{-x\})= \{ x\}(1-\{ x\})\), and for all \(k \in [\![ 0 , \left\lfloor \frac{q}{s} \right\rfloor ]\!]\), \(0\le \frac{ks}{q} <1\), it comes

$$\begin{aligned}&\sum _{k=0}^{\left\lfloor \frac{q}{s} \right\rfloor } \frac{1}{2} \left( \frac{k}{q} - s \left( \frac{k}{q} \right) ^2 \right) + \frac{1}{2s} \left( \left\{ \frac{-ks}{q} \right\} ^2 - \left\{ \frac{-ks}{q} \right\} \right) \\&\qquad \qquad = \sum _{k=0}^{\left\lfloor \frac{q}{s} \right\rfloor } \frac{1}{2} \left( \frac{k}{q} - s \left( \frac{k}{q} \right) ^2 \right) + \frac{1}{2s} \left( \left( \frac{ks}{q} \right) ^2 - \frac{ks}{q} \right) = 0. \end{aligned}$$

Furthermore, denoting \(d:=\gcd (s,q)\) and \(s=d\tilde{s}\), \(q=d \tilde{q}\),

$$\begin{aligned} \sum _{k=0}^{q-1} \left\{ \frac{ks}{q} \right\} \left( 1 - \left\{ \frac{ks}{q} \right\} \right)&=\, d \sum _{k=0}^{\tilde{q}-1} \left\{ \frac{k\tilde{s}}{\tilde{q}} \right\} \left( 1 - \left\{ \frac{k\tilde{s}}{\tilde{q}} \right\} \right) \\&=\, d \sum _{k=0}^{\tilde{q}-1} \frac{k}{\tilde{q}} \left( 1 - \frac{k}{\tilde{q}} \right) \\&= \frac{q}{6} - \frac{d^2}{6q}. \end{aligned}$$

Thus,

$$\begin{aligned} \sum _{k= \left\lfloor \frac{q}{s} \right\rfloor + 1}^{q-1} \sum _{m = \left\lfloor s- \frac{ks}{q} \right\rfloor + 1}^{s-1} \left( 1 - \frac{k}{q} - \frac{m}{s} \right)&= \,-\, \frac{q}{12s} + \frac{d^2}{12qs} + \frac{1}{2} \sum _{k=0}^{q-1} \left( \frac{k}{q} - s \left( \frac{k}{q} \right) ^2 \right) . \end{aligned}$$

Expanding the expressions it follows

$$\begin{aligned} \frac{1}{n} \sum _{j=1}^n \lbrace js \alpha \rbrace \mathbb {1}_{ \lbrace j \frac{p}{q} \rbrace + \lbrace j r \alpha \rbrace \ge 1 }&\underset{n\rightarrow \infty }{\longrightarrow } s \left( \frac{1}{4} + \frac{1}{12r} - \frac{1}{4q} - \frac{1}{12rq^2} \right) \\&\qquad - \,\frac{1}{q} \left( \frac{1}{4} + \frac{qs}{4} - \frac{q}{4} + \frac{qs}{4r} - \frac{q}{4r} - \frac{s}{4} - \frac{s}{4r} + \frac{1}{4r} \right) \\&\qquad - \,\frac{1}{qr} \left( - \frac{q}{12s} + \frac{d^2}{12sq} -\frac{qs}{6} + \frac{s}{4} - \frac{s}{12q} + \frac{q}{4} - \frac{1}{4} \right) \\&\quad = \frac{1}{4} - \frac{1}{4q} + \frac{1}{12sr} - \frac{d^2}{12srq^2} \end{aligned}$$

Consequently,

$$\begin{aligned} s_3 \left( \alpha , \frac{p}{q} + \frac{r}{s} \alpha \right)&= \left( \frac{1}{4} - \frac{1}{4q} \right) + \left( \frac{1}{4} + \frac{1}{12sr} \right) - \left( \frac{1}{4} - \frac{1}{4q} + \frac{1}{12sr} - \frac{d^2}{12srq^2} \right) \\&= \frac{1}{4} + \frac{d^2}{12srq^2}. \end{aligned}$$

It is easy to check that this formula remains true for \(q=s=2\), since in this case we have

$$\begin{aligned} \sum _{k=1}^q \sum _{m=1}^{s-1} \int _0^1 f_{k,m} (x) \mathrm {d}x = \frac{1}{4}+ \frac{1}{4r} \end{aligned}$$

and thus

$$\begin{aligned} s_3 \left( \alpha , \frac{p}{q} + \frac{r}{s} \alpha \right)&= \left( \frac{1}{4} - \frac{1}{4q} \right) + \left( \frac{1}{4} + \frac{1}{12sr} \right) \\&\quad - \,s \left( \frac{1}{4} + \frac{1}{12sr} - \frac{1}{4q} - \frac{1}{12rq^2} \right) + \frac{1}{q} \left( \frac{1}{4}+ \frac{1}{4r} \right) \\&= \frac{1}{4} + \frac{1}{24r} = \frac{1}{4} + \frac{d^2}{12srq^2}. \end{aligned}$$

Finally, we assume \(r<0\).

As we have seen above, the result of \(s_3 \left( \alpha , \frac{p}{q} + \frac{r}{s} \alpha \right) \) is not depending on the choice of the irrational \(\alpha \); thus, one can replace it by \((-s)\alpha \), and then

$$\begin{aligned} s_3 \left( \alpha , \frac{p}{q} + \frac{r}{s} \alpha \right)&= s_3 \left( s (-\alpha ) , \frac{p}{q} + \vert r \vert \alpha \right) \\&= \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n (1- \{ js\alpha \} ) \left\{ j \left( \frac{p}{q} + \vert r \vert \alpha \right) \right\} \\&= \lim _{n \rightarrow \infty } \frac{1}{n} \sum _{j=1}^n \left\{ j \left( \frac{p}{q} + \vert r \vert \alpha \right) \right\} - s_3 \left( s\alpha , \frac{p}{q} + \vert r \vert \alpha \right) \\&= \frac{1}{2} - \left( \frac{1}{4} + \frac{d^2}{12s \vert r \vert q^2} \right) \\&= \frac{1}{4} + \frac{d^2}{12srq^2}. \end{aligned}$$

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Bahier, V. On the Number of Eigenvalues of Modified Permutation Matrices in Mesoscopic Intervals. J Theor Probab 32, 974–1022 (2019). https://doi.org/10.1007/s10959-017-0798-5

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