1 Introduction: general setting and several reformulations of the problem

1.1 General setting

In this article, we are interested in the computation of Toeplitz integrals of the form:

$$\begin{aligned} Z_n({\mathcal {I}})=\frac{1}{(2\pi )^nn!}\int _{{\mathcal {I}}^n} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _n \prod _{1\le p<q\le n}\left| e^{i \theta _p}-e^{i\theta _q}\right| ^2 \end{aligned}$$
(1.1)

where \({\mathcal {I}}\) is a union of d (\(d\ge 1\)) intervals in \([-\pi ,\pi ]\):

$$\begin{aligned} {\mathcal {I}}=\bigcup _{j=1}^d \left[ \alpha _j,\beta _j\right] \text { with } -\pi \le \alpha _j\le \beta _j\le \pi \end{aligned}$$
(1.2)

Note that in the case of a full support \({\mathcal {I}}=[-\pi ,\pi ]\) it is well known that (see [1]):

$$\begin{aligned} Z_n([-\pi ,\pi ])=1 \end{aligned}$$
(1.3)

Thus, in the rest of the article, we assume that\(\varvec{{\mathcal {I}}}\varvec{\ne } \varvec{[}\varvec{-}\varvec{\pi }\varvec{,}\varvec{\pi }\varvec{]}\). Since the integral is obviously invariant under a global angular translation (\(\varvec{\theta }\mapsto \varvec{\theta }-\text {Cste }{\mathbf {1}}\)), we may assume that no interval\(\left( \varvec{[}\varvec{\alpha }_{\mathbf {j}},\varvec{\beta }_{\mathbf {j}}\varvec{]}\right) _{{\mathbf {1}}\varvec{\le }{\mathbf {j}}\varvec{\le }{\mathbf {d}}}\)contains\(\varvec{\pm }\varvec{\pi }\). Integrals of type (1.1) can also be understood as the partition functions of a gas of particles restricted to the set \({\mathcal {T}}=\{e^{it}, t\in {\mathcal {I}}\}\) with interactions given by \(\mathop {\prod }\nolimits _{1\le p<q\le n}\left| e^{i \theta _p}-e^{i\theta _q}\right| ^2\). Moreover, it is also well known that integrals of type (1.1) are Toeplitz integrals that can be reformulated as the determinant of a \(n\times n\) Toeplitz matrix with a non-vanishing symbol (defined below) on some arc intervals of the unit circle. Asymptotic expansions of Toeplitz determinants and integrals of the form (1.1) have been studied for a long time, and many results already exist in the literature using different strategies. For example, studies using orthogonal polynomials on the unit circle, properties of powers of random unitary matrices, Fredholm determinants, Riemann–Hilbert problems, etc., have been used to tackle the problem. A non-exhaustive list of articles on the subject is [2,3,4,5,6,7,8,9,10,11,12,13,14,15].

The purpose of this article is to provide a rigorous large n expansion of Toeplitz integrals of the form (1.1) to all order in \(\frac{{\mathbf {1}}}{{\mathbf {n}}}\) using the Eynard–Orantin topological recursion defined in [16]. In particular, we shall prove the following results:

  • A complete large n expansion when the support is restricted to only one interval (i.e., \(d=1\)) in Sect. 2.

  • Results up to O(1) when the support is composed of \(d=2r+1\ge 3\) intervals of the form \([\alpha _j,\beta _j]=\left[ \frac{2\pi j}{2r+1}-\frac{\pi \epsilon }{2r+1},\frac{2\pi j}{2r+1}+\frac{\pi \epsilon }{2r+1}\right] \) with \(-r\le j\le r\) and \(0<\epsilon <1\) in Sect. 3.3.

  • Results up to O(1) when the support is composed of \(d=2s\ge 2\) intervals of the form \([\alpha _j,\beta _j]=\left[ \frac{\pi \left( j-\frac{1}{2}\right) }{s}-\frac{\pi \epsilon }{2s},\frac{\pi \left( j-\frac{1}{2}\right) }{s}+\frac{\pi \epsilon }{2s}\right] \) with \(-(s-1)\le j\le s\) and \(0<\epsilon <1\) in Sect. 3.6.

The strategy used in this article is to reformulate the Toeplitz integrals in terms of some Hermitian matrix integrals (with some restrictions on the eigenvalues support). Then we compute the associated spectral curve and the corresponding limiting eigenvalues density. Using the theory developed in [17,18,19], we are able to rigorously prove the general form of the large n expansions of the correlators and of the partition function, as well as relate them with quantities computed from the topological recursion. In the case of a single interval, we finally use a Selberg integral to fix the normalization issues of the partition function and thus provide the complete large n expansion of the Toeplitz integrals. We eventually compare our theoretical predictions with numeric simulations performed on the Toeplitz determinant reformulation (which is very convenient for numeric computations) up to \(o\left( \frac{1}{n^4}\right) \).

1.2 Various reformulations of the problem

There are several useful rewritings of the integral (1.1). We list them in the following theorem:

Lemma 1.1

(Various reformulations of the problem) Defining \({\mathcal {I}}=\mathop {\bigcup }\nolimits _{j=1}^{d} [\alpha _j,\beta _j] \subset (-\pi ,\pi )\), \({\mathcal {T}}=\mathop {\bigcup }\nolimits _{j=1}^{d} \left\{ e^{it}\,,\,t\in [\alpha _j,\beta _j]\right\} \) and \({\mathcal {J}}=\mathop {\bigcup }\nolimits _{j=1}^{d} [\tan \frac{\alpha _j}{2},\tan \frac{\beta _j}{2}]\), the following quantities are equal to each other:

  1. 1.

    A Toeplitz integral with symbol \(f=\mathbb {1}_{{\mathcal {T}}}\):

    $$\begin{aligned} Z_n({\mathcal {I}})= & {} \frac{1}{(2\pi )^nn!}\int _{[-\pi ,\pi ]^n} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _n \left( \prod _{k=1}^n f(e^{i\theta _k})\right) \prod _{1\le p<q\le n}\left| e^{i \theta _p}-e^{i\theta _q}\right| ^2 \nonumber \\= & {} \frac{1}{(2\pi )^nn!}\int _{{\mathcal {I}}^n} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _n \prod _{1\le p<q\le n}\left| e^{i \theta _p}-e^{i\theta _q}\right| ^2 \end{aligned}$$
    (1.4)
  2. 2.

    The determinant of a \(n\times n\) Toeplitz matrix:

    $$\begin{aligned} Z_n({\mathcal {I}})= \det \left( T_{p,q}=t_{p-q}\right) _{1\le p,q\le n} \end{aligned}$$
    (1.5)

    with discrete Fourier coefficients given by:

    $$\begin{aligned} t_0= & {} \frac{1}{2\pi }\sum _{j=1}^d(\beta _j-\alpha _j)=\frac{|{\mathcal {I}}|}{2\pi } \nonumber \\ t_k= & {} \frac{1}{2\pi }\sum _{j=1}^d e^{ik\frac{\alpha _j+\beta _j}{2}}(\beta _j-\alpha _j)\sin _c\frac{k(\beta _j-\alpha _j)}{2} \,\,\,\,,\,\,\,\, \forall \, k\ne 0 \end{aligned}$$
    (1.6)

    where we denoted \(\sin _c(x)=\frac{\sin x}{x}\) the cardinal sine function.

  3. 3.

    A real n-dimensional integral with logarithmic potential and Vandermonde interactions:

    $$\begin{aligned} Z_n({\mathcal {I}})= & {} \frac{2^{n(n-1)}}{(2\pi )^n n!}\int _{{\mathcal {I}}^n} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _n \prod _{1\le p<q\le n}\sin ^2\left( \frac{\theta _p-\theta _q}{2}\right) \nonumber \\= & {} \frac{2^{n^2}}{(2\pi )^n n!}\int _{{\mathcal {J}}^n} \mathrm{d}t_1\ldots \mathrm{d}t_n \, \Delta (t_1,\ldots ,t_n)^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+t_k^2)} \end{aligned}$$
    (1.7)

    where \(\Delta (t_1,\ldots ,t_n)\) is the usual Vandermonde determinant \(\Delta ({\mathbf {t}})=\mathop {\prod }\nolimits _{1\le p<q\le n}(t_p-t_q)^2\).

  4. 4.

    A Hermitian matrix integral with prescribed eigenvalues support:

    $$\begin{aligned} Z_n({\mathcal {I}})= c_n\int _{{\mathcal {N}}_n({\mathcal {J}})} \frac{\mathrm{d}M_n}{\left( \det (I_n+M_n^2)\right) ^n} \end{aligned}$$
    (1.8)

    where \({\mathcal {N}}_n\) is the set of Hermitian matrices with eigenvalues belonging to \({\mathcal {J}}\). The normalizing constant \(c_n\) is related to the volume of the unitary group:

    $$\begin{aligned} c_n=\frac{1}{(2\pi )^nn!} \frac{1}{\text {Vol } {\mathcal {U}}_n}=\frac{1}{(2\pi )^nn!} \frac{(n!)\mathop {\prod }\nolimits _{j=1}^{n-1}j! }{\pi ^{\frac{n(n-1)}{2}}}= \frac{1}{2^n\pi ^{\frac{n(n+1)}{2}}}\mathop {\prod }\limits _{j=1}^{n-1}j! \end{aligned}$$
  5. 5.

    A complex n-dimensional integral over some segments of the unit circle with Vandermonde interactions:

    $$\begin{aligned} Z_n({\mathcal {I}})=(-1)^{\frac{n(n+1)}{2}}i^n \int _{{\mathcal {T}}^n} {\mathrm {d}}u_1\ldots {\mathrm{d}} u_n\, \Delta (u_1,\ldots ,u_n)^2 e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln u_k} \end{aligned}$$
    (1.9)

Proof

The proof of the previous lemma is rather elementary. The reformulation in terms of Toeplitz determinant is standard [3, 4]. Indeed, it is well known [4] that for a function (usually called “symbol” in the context of Toeplitz integrals) f measurable on the unit circle we have:

$$\begin{aligned} I_n(f)= & {} \frac{1}{(2\pi )^n n!}\int _{{\mathcal {I}}^n} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _n \left( \prod _{k=1}^n f(e^{i\theta _k})\right) \prod _{1\le p<q\le n}\left| e^{i \theta _p}-e^{i\theta _q}\right| ^2\\= & {} \det \left( T_{p,q}=t_{p-q}\right) _{1\le p,q\le n} \text { with } t_k\\= & {} \frac{1}{2\pi }\int _0^{2\pi } f(e^{i\theta })e^{ik\theta }\mathrm{d}\theta \,\,,\,\, \forall \, \, k\in \llbracket -(n-1),n-1\rrbracket \end{aligned}$$

Thus, equality between (1.4) and (1.5) corresponds to the application of the last identity with \(f=\mathbb {1}_{{\mathcal {T}}}\). Equality between (1.4) and (1.7) follows from the change of variables \(\theta _j=\tan \frac{t_j}{2}\) which is allowed since the support of the angles is included into \((-\pi ,\pi )\). With this change of variables, we get:

$$\begin{aligned} \left| e^{i \theta _p}-e^{i\theta _q}\right| ^2=\frac{4\left( \tan \frac{\theta _p}{2}-\tan \frac{\theta _q}{2}\right) ^2}{\left( 1+\tan ^2\frac{\theta _p}{2}\right) \left( 1+\tan ^2\frac{\theta _q}{2}\right) } \end{aligned}$$

Observing that \(\mathrm{d}\theta _k\) provides a factor \(\mathrm{d}\theta _k=\frac{2}{1+t_k^2}\mathrm{d}t_k\) and that we have:

$$\begin{aligned} \prod _{1\le p<q\le n}\frac{1}{\left( 1+t_p^2\right) \left( 1+t_q^2\right) }=\prod _{k=1}^n\frac{1}{(1+t_k^2)^{n-1}} \end{aligned}$$

immediately gives (1.7). Reformulating the real integral (1.7) in terms of a Hermitian matrix integral is standard (see [1]) from diagonalization \(M=U\Lambda U^\dagger \) of normal matrices. We only note here that the support of eigenvalues is prescribed to \({\mathcal {J}}\). Eventually the volume of the unitary group can be found in [20] and equality between (1.4) and (1.9) is straightforward from the change of variables \(u_j=e^{i\theta _j}\). \(\square \)

As presented in the previous theorem, the complex (1.9) and real (1.7) integral reformulations of the problem share the important point that the interactions are given by a Vandermonde determinant \(\Delta (x)^2\). We stress that this situation is rather exceptional since a (non-affine) change of variables in such integrals does not generally preserve the form of the interactions.

Remark 1.1

We inform the reader that part of the results proven in this article have already been presented in [21] using the reformulation in terms of the complex integrals (1.9) in the context of return times for the eigenvalues of a random unitary matrix. Indeed, as one can obviously see from (1.7) and (1.9), the complex and real integral reformulations share the crucial fact that the interactions between the eigenvalues are of Vandermonde type: \(\Delta (\mathbf{x })^2\). This implies that correlation functions of both reformulations satisfy some loop equations and that the topological recursion may be applied to both models. This leads to the fact that the spectral curves found in this article are related by some symplectic transformations to the ones presented in [21] and thus that they provide the same sets of free energies (that reconstruct up to some constants \(\ln Z_n({\mathcal {I}})\)). However, we stress that several practical and theoretical issues were disregarded in [21] that can be solved using the real integral reformulation:

  1. 1.

    In [21], results from [19] were used to justify the form of the large n expansions of the correlators and partition functions. However, results from [19] have only been proved for integrals with support included in \({\mathbb {R}}\), but not in the case of a generic closed curve in \({\mathbb {C}}\) as would require the unit circle. Though it is believed that the tools developed in [19] should remain valid for certain non-real domains of integration, the mathematical proof is still missing.

  2. 2.

    In [21], normalization issues related to the partition function were completely disregarded. Consequently, the large n asymptotic expansions of the Toeplitz determinants presented in [21] lack some constant terms that we provide in this paper. Though the normalization issues were not particularly important for the physics problem studied in [21], they become essential when one wants to compute exact probabilities and compare them with numeric simulations. It turns out that the reformulation (1.7) offers, at least in the one-interval case, a simple way to deal with the normalization issues by connecting them to a Selberg integral. In the complex setting the connection to a known integral is less obvious and thus the normalization issues were disregarded in [21].

  3. 3.

    Stronger results than those developed for general interactions [19] regarding the large n expansion and the reconstruction by the topological recursion are available in the case of real integrals with Vandermonde interactions like (1.7) in [17,18,19]. In particular, these results allow a proper rigorous mathematical derivation of the full asymptotic expansion of the Toeplitz integrals (1.1) and provide the explicit expressions of the first orders including the proper normalizing factors.

  4. 4.

    Numerical simulations in [21] were only performed to leading order \(O(n^2)\), while we are able to match the theoretical results with numeric simulations up to \(O\left( n^{-6}\right) \) in this article.

2 Study of the one-interval case

2.1 Known results

When \(d=1\), we can use the rotation invariance and take \(\beta _1=-\alpha _1=\pi \epsilon \) with \(0<\epsilon <1\). For simplicity we shall denote in this section\({\mathbf {a}}\varvec{=}\mathbf{tan } \frac{\varvec{\pi }\varvec{\epsilon }}{{\mathbf {2}}}\) and a function of \(\epsilon \) will equivalently be seen as a function of a and vice versa depending on the relevance of the parameter in the discussion. In this section, we want to compute the large expansion of the Toeplitz integral:

$$\begin{aligned} Z_n(a)\overset{\text {notation}}{\equiv } Z_n(\epsilon )=\frac{1}{(2\pi )^nn!}\int _{[-\pi \epsilon ,\pi \epsilon ]^n} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _n \prod _{1\le p<q\le n}\left| e^{i \theta _p}-e^{i\theta _q}\right| ^2\nonumber \\ \end{aligned}$$
(2.1)

The corresponding Toeplitz determinant reformulation is particularly easy:

$$\begin{aligned} Z_n(\epsilon )=\det \left( T_{p,q}=t_{p-q}\right) _{1\le p,q\le n} \end{aligned}$$
(2.2)

with the discrete Fourier coefficients given by:

$$\begin{aligned} t_k=\epsilon \sin _c(k\pi \epsilon ) ,\quad \forall \, k \in \llbracket -(n-1),n-1\rrbracket \end{aligned}$$
(2.3)

The reformulation in terms of a n-fold integral corresponding to the diagonalized form of a Hermitian matrix integral is given by:

$$\begin{aligned} Z_n(a)=\frac{2^{n^2}}{(2\pi )^n n!}\int _{[-a,a]^n} \mathrm{d}t_1\ldots \mathrm{d}t_n \Delta (t_1,\ldots ,t_n)^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+t_k^2)} \end{aligned}$$
(2.4)

Such integrals have been studied by Widom in [3, 29] and do not fall in the standard theory of Toeplitz integrals developed by Szegö. Indeed, the standard case corresponds to a symbol f which is strictly positive and continuous on the unit circle. In that case, the standard theory of Toeplitz determinants can be applied and one would obtain the strong Szegö theorem [2]:

$$\begin{aligned} \frac{1}{n}\ln \det \left( T_n(f)\right) \overset{n\rightarrow \infty }{\rightarrow }\frac{1}{2\pi }\int _0^{2\pi } \ln (f(e^{i\theta }))\mathrm{d}\theta \end{aligned}$$
(2.5)

When the symbol is discontinuous but remains strictly positive, adaptations of the theory, known as Fisher–Hartwig singularities, have been developed and the convergence of \(\frac{1}{n}\ln \det \left( T_n(f)\right) \) is still obtained though formulas get more involved. However, in our present case, the symbol is vanishing on several intervals of the unit circle and the convergence of \(\frac{1}{n}\ln \det \left( T_n(f)\right) \) does no longer hold. Indeed, in this case, Widom proved the following proposition [3]:

Theorem 2.1

(Widom’s result) Let \(0<\theta _0<\pi \) and define \({\mathcal {T}}(\theta _0)=\{e^{it},t\in [-\theta _0,\theta _0]\}\). Then we have:

$$\begin{aligned} \ln \det \left( T_n(\mathbb {1}_{{\mathcal {T}}(\theta _0)})\right)= & {} n^2\ln \left( \sin \frac{\theta _0}{2}\right) -\frac{1}{4}\ln n-\frac{1}{4}\ln \left( \cos \frac{\theta _0}{2}\right) \\&+3\,\zeta '(-1)+\frac{1}{12}\ln 2+o(1) \end{aligned}$$

where \(\zeta \) denotes the Riemann zeta function.

We propose in this section to improve Widom’s result by providing a mathematical proof of the form of the large n expansion of \(\ln Z_n(\epsilon )\) as well as a general way to compute all sub-leading corrections.

2.2 Spectral curve and limiting eigenvalues density

Integral (2.4) may be seen as a gas of eigenvalues with Vandermonde interactions and evolving in a potential \(V(x)=\ln (1+x^2)\). Consequently, it falls into the category of integrals studied in [17,18,19]. In particular, the potential is analytic on \({\mathbb {R}}\) and has only one minimum at \(x=0\). Moreover, since the support of the integration is restricted to a compact set \([-a,a]\), the convergence issues are trivial. Under such conditions, it is proved in [22, 23] that the empirical eigenvalues density \(\delta _n\) converges almost surely toward an absolutely continuous limiting eigenvalues density:

$$\begin{aligned} \delta _n=\frac{1}{n}\sum _{k=1}^n\delta (x-t_k)\mathop {\rightarrow }\limits _{n\rightarrow \infty } \mathrm{d}\mu _\infty (x)=\nu (x)\mathrm{d}x \end{aligned}$$
(2.6)

whose support is a finite union of intervals in \([-a,a]\). Determining the limiting eigenvalues density can be done in many ways, but since we plan to use the topological recursion, it seems appropriate to derive the limiting eigenvalues density using the loop equations method. We define the following correlation functions:

Definition 2.1

(Correlation functions) We define the correlation functions by:

$$\begin{aligned} W_{1,a}(x)= & {} \left\langle \sum _{k=1}^n \frac{1}{x-t_k}\right\rangle \\ W_{p,a}^{n.c.}(x_1,\ldots ,x_p)= & {} \left\langle \sum _{i_1,\ldots ,i_p=1}^n \frac{1}{x_1-t_{i_1}}\ldots \frac{1}{x_p-t_{i_p}}\right\rangle \\ W_{p,a}(x_1,\ldots ,x_p)= & {} \left\langle \sum _{i_1,\ldots ,i_p=1}^n \frac{1}{x_1-t_{i_1}}\ldots \frac{1}{x_p-t_{i_p}}\right\rangle _c \end{aligned}$$

where the average of a function of the eigenvalues is defined by:

$$\begin{aligned} \left\langle g(t_1,\ldots ,t_n)\right\rangle =\frac{2^{n^2}}{(2\pi )^n n! Z_n(a)}\int _{[-a,a]^n} \mathrm{d}t_1\ldots \mathrm{d}t_n \, g(t_1,\ldots ,t_n)\Delta ({\mathbf {t}})^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+t_k^2)} \end{aligned}$$

The index \(\,_c\) stands for “connected” or “cumulant” in the sense that:

$$\begin{aligned} W_{2,a}(x_1,x_2)= & {} W_{2,a}^{n.c.}(x_1,x_2)-W_{1,a}(x_1)W_{1,a}(x_2)\\ W_{3,a}(x_1,x_2,x_3)= & {} W_{3,a}^{n.c.}(x_1,x_2,x_3)-W_{1,a}(x_1)W_{2,a}^{n.c.}(x_2,x_3)-W_{1,a}(x_2)W_{2,a}^{n.c.}(x_1,x_3)\\&-W_{1,a}(x_3)W_{2,a}^{n.c.}(x_1,x_2)+2W_{1,a}(x_1)W_{1,a}(x_2)W_{1,a}(x_3)\\&\quad {\text {etc.}} \end{aligned}$$

or in a more general way by the inverse relation:

$$\begin{aligned} W_{p,a}^{n.c.}(x_1,\ldots ,x_p)= \sum _{\mu \,\vdash \{x_1,\ldots ,x_p\}}\prod _{i=1}^{\text {l}(\mu )} W_{|\mu _i|,a}(\mu _i) \end{aligned}$$

Integral (2.4) is a Hermitian matrix integral with hard edges at \(t=\pm a\). Loop equations for Hermitian matrix integrals with hard edges have been written in many places [24,25,26]. They can also be easily obtained with the integral:

$$\begin{aligned} \frac{1}{(2\pi )^n(n!) Z_n(a)}\int _{[-a,a]^n}\mathrm{d}t_1\ldots d t_n \sum _{j=1}^n\frac{d}{\mathrm{d}t_j}\left( \frac{1}{x-t_j} \Delta (t_1,\ldots ,t_n)^2 e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+t_k^2)}\right) \nonumber \\ \end{aligned}$$
(2.7)

Indeed, the last integral is equivalent to:

$$\begin{aligned} W_{1,a}^2(x)+W_{2,a}(x,x)-\frac{2nx}{1+x^2}W_{1,a}(x)+n\left\langle \sum _{k=1}^n\frac{V'(x)-V'(t_k)}{x-t_k}\right\rangle =\frac{c_1(n)}{x-a}-\frac{c_2(n)}{x+a}\nonumber \\ \end{aligned}$$
(2.8)

Equation (2.8) is exact, and the coefficients \(c_1(n)\) and \(c_2(n)\) are given by:

$$\begin{aligned} c_1(n)= & {} \frac{e^{-n\ln (1+a^2)}}{(2\pi )^n(n!) Z_n(a)}\nonumber \\&\sum _{j=1}^n\int _{[-a,a]^{n-1}} \mathrm{d}t_1\ldots \mathrm{d}t_{j-1} \mathrm{d}t_{j+1} \ldots d t_n\,\Delta (t_1,\ldots ,t_{j-1},a,t_{j+1},\ldots , t_n)^2 e^{-n\mathop {\sum }\limits _{k\ne j}^{n} \ln (1+t_k^2)} \nonumber \\ c_2(n)= & {} \frac{e^{-n\ln (1+a^2)}}{(2\pi )^n(n!) Z_n(a)}\nonumber \\&\sum _{j=1}^n\int _{[-a,a]^{n-1}} \mathrm{d}t_1\ldots \mathrm{d}t_{j-1} \mathrm{d}t_{j+1} \ldots d t_n\,\Delta (t_1,\ldots ,t_{j-1},-a,t_{j+1},\ldots , t_n)^2 e^{-n\mathop {\sum }\limits _{k\ne j}^{n} \ln (1+t_k^2)}\nonumber \\ \end{aligned}$$
(2.9)

Note that since the integral (2.4) is invariant under the change \({\mathbf {t}}\rightarrow -{\mathbf {t}}\) we automatically get \(c_1(n)=c_2(n)\). In order to obtain the spectral curve of the problem (which is the Stieltjes transform of the limiting eigenvalues density), we take the leading order in n of Eq. (2.8). Results from [17,18,19] show that \(W_{1,a}(x)\mathop {\sim }\nolimits _{n\rightarrow \infty } nW_{1,a}^{(0)}(x)\) and \(W_{2,a}(x_1,x_2)\mathop {=}\nolimits _{n\rightarrow \infty } O(1)\). Defining

$$\begin{aligned} y(x)=W_{1,a}^{(0)}(x)-\frac{1}{2}V'(x)=W_{1,a}^{(0)}(x)-\frac{x}{1+x^2} \end{aligned}$$
(2.10)

we end up with:

$$\begin{aligned} y(x)^2= & {} \frac{x^2}{(1+x^2)^2}-\frac{2}{1+x^2}\left( \mathop {\lim }\limits _{n\rightarrow \infty }\left\langle \frac{1}{n}\sum _{k=1}^n\frac{1}{1+t_k^2}\right\rangle -x\mathop {\lim }\limits _{n\rightarrow \infty }\left\langle \frac{1}{n}\sum _{k=1}^n\frac{t_k}{1+t_k^2}\right\rangle \right) \nonumber \\&+\frac{c(a)}{x-a}-\frac{c(a)}{x+a} \end{aligned}$$
(2.11)

where the constant c(a) is given by \(c(a)=\mathop {\lim }\nolimits _{n\rightarrow \infty }\frac{1}{n}c_1(n)\). Since the integral (2.4) is invariant under \({\mathbf {t}}\rightarrow -{\mathbf {t}}\), we get that \(\left\langle \mathop {\sum }\nolimits _{k=1}^{n}\frac{t_k}{1+t_k^2}\right\rangle =0\) so that:

$$\begin{aligned} y(x)^2=\frac{x^2}{(1+x^2)^2}-\frac{2d(a)}{1+x^2}+\frac{c(a)}{x-a}-\frac{c(a)}{x+a} \end{aligned}$$
(2.12)

where d(a) and c(a) are so far undetermined constants (i.e., independent of x). Moreover, note that by definition we must have at large x:

$$\begin{aligned} W_{1,a}^{(0)}(x)= & {} \frac{1}{x}-\mathop {\lim }\limits _{n\rightarrow \infty }\left\langle \frac{1}{n}\sum _{k=1}^nt_k\right\rangle \frac{1}{x^2}+O\left( \frac{1}{x^3}\right) \nonumber \\= & {} \frac{1}{x}+O\left( \frac{1}{x^3}\right) \,\,\, \Rightarrow \,\,\,y(x)=O\left( \frac{1}{x^3}\right) \end{aligned}$$
(2.13)

since the integral is invariant under \({\mathbf {t}}\rightarrow -{\mathbf {t}}\). Using the fact that \(y^2(x)=O\left( \frac{1}{x^6}\right) \) in (2.12) provides two independent equations satisfied by (c(a), d(a)) that can be explicitly solved. We find:

$$\begin{aligned} c(a)=\frac{1}{2a(1+a^2)} \text { and } d(a)=\frac{2+a^2}{2(1+a^2)} \end{aligned}$$
(2.14)

Finally, we get:

$$\begin{aligned} y^2(x)=\frac{1+a^2}{(1+x^2)^2(x^2-a^2)}=\frac{1}{\cos ^2\left( \frac{\pi \epsilon }{2}\right) (1+x^2)^2\left( x^2-\tan ^2\left( \frac{\pi \epsilon }{2}\right) \right) } \end{aligned}$$
(2.15)

In other words since \(\cos (\frac{\pi \epsilon }{2})>0\) for \(\epsilon \in (0,1)\):

$$\begin{aligned} y(x)=\frac{1}{\cos \left( \frac{\pi \epsilon }{2}\right) (1+x^2)\sqrt{x^2-\tan ^2\left( \frac{\pi \epsilon }{2}\right) }} \end{aligned}$$
(2.16)

This is equivalent to say that the limiting eigenvalues density is given by:

$$\begin{aligned} \mathrm{d}\mu _\infty (x)= \frac{\mathrm{d}x}{\pi \cos \left( \frac{\pi \epsilon }{2}\right) (1+x^2)\sqrt{\tan ^2\left( \frac{\pi \epsilon }{2}\right) -x^2}}\mathbb {1}_{ \left[ -\tan \frac{\pi \epsilon }{2},\tan \frac{\pi \epsilon }{2}\right] }(x) \end{aligned}$$
(2.17)

The last density can be verified numerically using Monte–Carlo simulations (Fig. 1):

Fig. 1
figure 1

Empirical eigenvalues density obtained from 100 independent Monte–Carlo simulations of the integral (2.4) in the case \(\epsilon =\frac{1}{7}\) and \(n=20\). The black curve is the theoretical curve corresponding to (2.17)

The limiting eigenvalues density is supported on the whole interval \([-\tan \frac{\pi \epsilon }{2},\tan \frac{\pi \epsilon }{2}]\) and exhibits inverse square-root behavior near the edges of the support:

$$\begin{aligned} \mathrm{d}\mu _\infty (x)\overset{x\rightarrow \tan \frac{\pi \epsilon }{2}}{=} O\left( \frac{\mathrm{d}x}{\sqrt{x-\tan \frac{\pi \epsilon }{2}}}\right) \text { and } \mathrm{d}\mu _\infty (x)\overset{x\rightarrow -\tan \frac{\pi \epsilon }{2}}{=} O\left( \frac{\mathrm{d}x}{\sqrt{x+\tan \frac{\pi \epsilon }{2}}}\right) \nonumber \\ \end{aligned}$$
(2.18)

Moreover, it is strictly positive inside \([-\tan \frac{\pi \epsilon }{2},\tan \frac{\pi \epsilon }{2}]\).

Remark 2.1

The spectral curve (2.16) is related to the one found in [21] (eq. C.14) using the complex integral reformulation rather than the real integral reformulation. Indeed, in [21], the spectral curve in the one-interval case is \({\tilde{y}}^2=\frac{({\tilde{x}}+1)^2}{4{\tilde{x}}^2({\tilde{x}}-e^{i\pi \epsilon })({\tilde{x}}+e^{-i\pi \epsilon })}\). In fact, both curves are equivalent up to the symplectic transformation:

$$\begin{aligned} {\tilde{x}}=e^{2i\text {Arctan}(x)}\overset{def}{=}f(x) \text { and } {\tilde{y}}=\frac{1}{f'(x)}y=\frac{1+x^2}{2i}e^{-2i\text {Arctan}(x)}y \end{aligned}$$
(2.19)

This transformation follows from the combination of \(x=\tan \frac{\theta }{2}\) and \({\tilde{x}}=e^{i\theta }\) corresponding to a simple reparametrization of the x function (which is a trivial case of the symplectic invariance). In particular, as explained in “Appendix B,” both curves provide the same set of “symplectic invariants” \(\left( F^{(g)}\right) _{g\ge 0}\) that reconstruct the large n expansion of \(\ln Z_n(a)\). However, the Eynard–Orantin differentials attached to both curves differ which is coherent with the fact that the correlation functions are different in both settings.

2.3 General form of the large n expansions

As soon as the limiting eigenvalue density (2.17) is determined, we may apply the main results of [17,18,19] to obtain the general form of the large n expansions of the correlators and of the partition function. However, we first need to prove that Hypothesis 1.1 of [17] is satisfied so that we may apply the main results of [17]. We have:

Proposition 2.1

The following conditions (Hypothesis 1.1 of [17]) are met for integral (2.4) (note that in our case, the potential V does not depend on n so that some conditions of [17] are trivially verified):

  • (Regularity): The potential V is continuous on the integration domain \([b_-,b_+]\).

  • (Confinement of the potential): If \(\pm \infty \) belong to the integration domain, then the potential is assumed to be decaying sufficiently fast:

    $$\begin{aligned} \liminf _{x\rightarrow \pm \infty } \frac{V(x)}{2\ln |x|}>1 \end{aligned}$$
  • (One-cut regime): The support of the limiting eigenvalues density is a single interval \([\alpha _-,\alpha _+]\) not reduced to a point.

  • (Control of large deviations): The function \(x\mapsto \frac{1}{2}V(x)+\int _{{\mathbb {R}}} |x-\xi |\mathrm{d}\mu _\infty (\xi )\) defined on \([b_-,b_+]\setminus (\alpha _-,\alpha _+)\) achieves its minimum only in \(\alpha _-\) or \(\alpha _+\).

  • (Off-Criticality): The limiting eigenvalues density is off-critical in the sense that it is strictly positive inside the interior of its support and behaves like \(O\left( \frac{1}{\sqrt{x-b_\pm }}\right) \) if \(b_\pm \) is a hard edge or like \(O\left( \sqrt{x-\alpha _\pm }\right) \) if \(\alpha _\pm \) is a soft edge.

  • (Analyticity): V can be extended to an analytic function inside a neighborhood of \([\alpha _-,\alpha _+]\).

Proof

In our case, most of the points required are easily verified:

  • (Regularity): \(x\mapsto \ln (1+x^2)\) is obviously continuous on \([b_-,b_+]=[-a,a]\).

  • (Confinement of the potential): No confinement is required since the support is a compact set of \({\mathbb {R}}\).

  • (One-cut regime): This condition directly follows from Eq. (2.17).

  • (Control of large deviations): Since \(\alpha _-=b_-=-a\) and \(\alpha _+=b_+=a\) (the limiting density is supported on the whole integration domain) then \([b_-,b_+]\setminus (\alpha _-,\alpha _+)=\{\alpha _-,\alpha _+\}\) so the condition is trivially realized.

  • (Off-criticality): We only have two hard edges and Eq. (2.18) provides the correct behavior. Moreover, we directly observe from its expression that \(\mathrm{d}\mu _\infty (x)\) is strictly positive inside its support.

  • (Analyticity): \(x\mapsto \ln (1+x^2)\) is trivially analytic in a neighborhood of \([-a,a]\).

\(\square \)

Therefore, we can apply the main result of [17] for \(\beta =2\) (as well as Theorems 1.3 and 1.4 of [18] or results of [19] for the partition function with hard edges) and we obtain that:

Theorem 2.2

(Large n expansions) The correlators and the partition functions \(Z_n(a)\) admit a large n expansion (usually called “topological expansion”) of the form:

$$\begin{aligned} W_{p,a}(x_1,\ldots ,x_p)= & {} \sum _{g=0}^{\infty } W_{p,a}^{\{2-p-2g\}}(x_1,\ldots ,x_p)n^{2-p-2g} \nonumber \\ Z_n(a)= & {} \frac{n^{n+\frac{1}{4}}}{n!}\,\text {exp}\left( \sum _{k=-2}^\infty {\tilde{F}}^{\{k\}}(a)n^{-k} \right) \nonumber \\ \ln Z_n(a)= & {} -\frac{1}{4}\ln n+\sum _{k=-2}^\infty F^{\{k\}}(a)n^{-k} \end{aligned}$$
(2.20)

The previous large n asymptotic expansions have the precise meaning that \(\forall \, K\ge 0\):

$$\begin{aligned} W_{p,a}(x_1,\ldots ,x_p)= & {} \sum _{g=0}^K W_{p,a}^{\{2-p-2g\}}(x_1,\ldots ,x_p)n^{2-p-2g} + o\left( n^{2-p-2K}\right) \nonumber \\ \ln Z_n(a)= & {} -\frac{1}{4}\ln n+\sum _{k=-2}^K F^{\{k\}}(a)n^{-k} + o\left( n^{-K}\right) \end{aligned}$$
(2.21)

where the \(o\left( n^{2-p-2K}\right) \) and \(o\left( n^{-K}\right) \) are uniform for \(x_1,\ldots ,x_n\) in any compact set of \([-a,a]\) but are not uniform in n nor K. In the rest of the article, series like \(\mathop {\sum }\nolimits _{k=-2}^{\infty } f_k n^{-k}\) will always be considered as asymptotic expansions in the sense presented above.

Notice that the asymptotic expansion of a given correlation function only involves powers of n with the same parity and that the series expansion of \(W_{p,a}\) starts at \(\left( \frac{1}{n^{p-2}}\right) \). On the contrary, the large n expansion of the partition function \(Z_n(a)\) may involve all powers of n and has an extra factor \(n^{n+\frac{1}{4}}\) and n!. Indeed, the results of [18, 19] providing the r.h.s. of (2.20) only apply directly to \(\frac{(2\pi )^n n!}{2^{n^2}}Z_n(a)\). While the factors \((2\pi )^n\) and \(2^{n^2}\) may be absorbed in the definition of the constants \(F^{\{k\}}(a)\), the term n! may not. A direct corollary of the previous theorem is that the coefficients \(W_{p,a}^{\{2-p-2g\}}\) of the correlators are obtained from the topological recursion:

Corollary 2.1

(Reconstruction of the correlators via the topological recursion) For all \(p\ge 1\) and \(g\ge 0\), we have:

$$\begin{aligned} W_{p,a}^{^{\{2-p-2g\}}}(x_1,\ldots ,x_p)\mathrm{d}x_1\ldots \mathrm{d}x_n=\omega _p^{(g)}(x_1,\ldots ,x_p) \end{aligned}$$
(2.22)

where \(\omega _p^{(g)}(x_1,\ldots ,x_p)\) is the (pg) Eynard–Orantin differential (see “Appendix B” or [16]) computed from the application of the topological recursion to the (genus zero) spectral curve (2.16).

Details about the topological recursion are presented for completeness in “Appendix B” and more can be found in [16] and [27]. In particular, since the spectral curve (2.16) is of genus zero, it can be parametrized globally on \(\overline{{\mathbb {C}}}={\mathbb {C}}\cup \{\infty \}\) via:

$$\begin{aligned} x(z)= & {} \frac{1}{2}\tan \left( \frac{\pi \epsilon }{2}\right) \left( z+\frac{1}{z}\right) \nonumber \\ y(z)= & {} \frac{2}{\sin (\frac{\pi \epsilon }{2})\left( 1+\frac{1}{4}\tan ^2(\frac{\pi \epsilon }{2})\left( z+\frac{1}{z}\right) \right) \left( z-\frac{1}{z}\right) } \end{aligned}$$
(2.23)

Moreover, the normalized bi-differential \(\omega _2^{(0)}(z_1,z_2)\) required to initialize the topological recursion is \(\omega _2^{(0)}(z_1,z_2)=\frac{dz_1dz_2}{(z_1-z_2)^2}\) for genus zero curves. The proof of corollary 2.1 is standard. Indeed, by construction the correlation functions \(W_{p,a}(x_1,\ldots ,x_p)\) satisfy the loop equations arising in Hermitian matrix models. Moreover, Theorem 2.2 ensures that they have a topological expansion and by definition these correlators may only have singularities at the branchpoints or at the edges. These properties are also satisfied by the Eynard–Orantin differentials (see [16]) and thus both sets must match since, under these conditions (topological expansion and location of the singularities), the loop equations admit a unique solution.

The situation is more complicated for the partition function \(Z_n(a)\). Indeed, as discussed in [18] and [19], only \(\frac{\partial \ln Z_n(a)}{\partial a}\) can be matched with \(\frac{\partial \ln \tau (a)}{\partial a}\) where \(\ln \tau \) is the free energies generating series associated with the spectral curve (2.16) defined by:

$$\begin{aligned} \ln \tau (a)=-\sum _{g=0}^\infty F_{\text {Top. Rec.}}^{(g)}(a)\, n^{2-2g} \end{aligned}$$
(2.24)

The coefficients \(\left( F_{\text {Top. Rec.}}^{(g)}(a)\right) _{g\ge 0}\) are computed from the topological recursion and are commonly called “symplectic invariants” or “free energies” (see “Appendix B” or [16] for the formulas). Note that there is a change of convention regarding the indexes between the indexes of the topological recursion (noted \(\,^{(g)}\) and corresponding to \(n^{2-2g}\)) and the indexes of the asymptotic expansion of \(Z_n(a)\) (noted \(\,^{\{k\}}\) and corresponding to \(n^{-k}\)). We keep the notation of [16] for the topological recursion side to avoid confusion (hence, the notation is \(F_{\text {Top. Rec.}}^{(0)}\), \(F_{\text {Top. Rec.}}^{(1)}\) for powers \(n^{2}\), \(n^0\) and so on). Thus, the coefficients \(\left( F^{\{k\}}(a)\right) _{k\ge -2}\) match with \(\left( F_{\text {Top. Rec.}}^{(g)}(a)\right) _{g\ge 0}\) only up to some constants terms (in the sense independent of a). For applications in topological string theory and integrable systems, these constants are generally disregarded because normalization issues are mostly irrelevant. But in the context of Toeplitz determinants and probability one needs to find a way to compute them. Otherwise, one may only consider ratio of Toeplitz determinants as studied in [15]. In our case, we may compute these constants from the knowledge of the asymptotic expansion of \(\mathop {\lim }\nolimits _{a_0\rightarrow 0}\ln Z_n(a_0)\) and using results of [28] as we will see below.

2.4 Normalization and limit \(\epsilon \rightarrow 0\)

From [18] we have for all \((a,a_0)\in {\mathbb {R}}_+^*\times {\mathbb {R}}_+^*\):

$$\begin{aligned} \ln Z_n(a)-\ln Z_n(a_0)= & {} -\sum _{k=-2}^{+\infty } F^{\{k\}}(a)n^{-k}+\sum _{k=-2}^{+\infty } F^{\{k\}}(a_0)n^{-k} \nonumber \\= & {} -\sum _{g=0}^{+\infty } F_{\text {Top. Rec.}}^{(g)}(a)\, n^{2-2g}+\sum _{g=0}^{+\infty } F_{\text {Top. Rec.}}^{(g)}(a_0)\, n^{2-2g}\nonumber \\ \end{aligned}$$
(2.25)

Note that when a and \(a_0\) are positive numbers, the corresponding potentials satisfy conditions of Proposition 2.1 (in particular, the limiting eigenvalues distributions remain one-cut) thus allowing the use of results of [18] that directly gives the previous result. In order to have convergent quantities when \(a_0\rightarrow 0\), we then isolate a trivial divergent term for \(g=0\):

$$\begin{aligned} \ln Z_n(a)= & {} -\sum _{g=0}^{+\infty } F_{\text {Top. Rec.}}^{(g)}(a)\, n^{2-2g}+\left( \ln \left( Z_n(a_0)\right) -n^2\ln a_0\right) \nonumber \\&+\sum _{g=0}^{+\infty } \left( F_{\text {Top. Rec.}}^{(g)}(a_0)+\delta _{g=0}\ln a_0\right) \, n^{2-2g} \end{aligned}$$
(2.26)

In “Appendix A.3” we prove that \(\left( \ln \left( Z_n(a_0)\right) -n^2\ln a_0\right) \) has a limit when \(a_0\rightarrow 0\) which is given by a Selberg integral whose asymptotic expansion can be computed exactly. More precisely, we find:

$$\begin{aligned}&\ln Z_n(a_0)-n^2\ln (a_0)\mathop {\rightarrow }\limits _{a_0\rightarrow 0}4\ln (G(n+1))-\ln (G(2n+1))+2n^2\ln 2-n\ln (2\pi )\nonumber \\&\quad =-\frac{1}{4}\ln n +3\zeta '(-1)+\frac{1}{12}\ln 2 +\sum _{g=1}^\infty \frac{(4-2^{-2g})B_{2g+2}}{2g(2g+2) n^{2g}} \end{aligned}$$
(2.27)

where G is the G-Barnes function, \(\zeta \) is the Riemann zeta function and \(\left( B_k\right) _{k\ge 0}\) are the Bernoulli numbers. Moreover, using invariance of the free energies under symplectic transformations (we only need a trivial symplectic transformation but not the difficult and controversial \((x,y)\leftrightarrow (y,-x)\) exchange), we prove in “Appendix A.4” that the limits when \(a_0\rightarrow 0\) of the free energies \( \left( F_{\text {Top. Rec.}}^{(g)}(a_0)+\delta _{g=0}n^2\ln a_0\right) _{g\ge 0}\) are given by the free energies of Legendre’s spectral curve \(y=\frac{1}{\sqrt{x^2-1}}\) whose values have been recently computed in [28] for all \(g\ge 0\). This term by term convergence is extended to a large n asymptotic expansion in “Appendix A.5” using results of [18]. We find:

$$\begin{aligned} \sum _{g=0}^{+\infty }\left( F_{\text {Top. Rec.}}^{(g)}(a_0) +\delta _{g=0}\ln a_0 \right) \, n^{2-2g}\overset{a_0\rightarrow 0}{\rightarrow }n^2\ln 2-\sum _{g=2}^{+\infty }\frac{B_{2g}\left( 4-2^{-(2g-2)}\right) }{2g(2g-2)n^{2g-2}}\nonumber \\ \end{aligned}$$
(2.28)

Note in particular that we have the simplification (to be understood again in the sense of large n asymptotic expansions):

$$\begin{aligned} \left( \ln \left( Z_n(a_0)\right) -n^2\ln a_0\right) +\sum _{g=0}^{+\infty } \left( F_{\text {Top. Rec.}}^{(g0)}(a_0)+\delta _{g=0}\ln a_0\right) \, n^{2-2g} \overset{a_0\rightarrow 0}{\rightarrow } n^2\ln 2-\frac{1}{4}\ln n +3\zeta '(-1)+\frac{1}{12}\ln 2\nonumber \\ \end{aligned}$$
(2.29)

Thus, we get from (2.25) the following final result:

$$\begin{aligned} \ln Z_n(a)=-\sum _{g=0}^{+\infty } F_{\text {Top. Rec.}}^{(g)}(a)\, n^{2-2g}+n^2\ln 2-\frac{1}{4}\ln n +3\zeta '(-1)+\frac{1}{12}\ln 2\nonumber \\ \end{aligned}$$
(2.30)

Application of the topological recursion to the spectral curve (2.16) is standard and is presented in “Appendix A.1” with details for the specific computations for \(\left( F_{\text {Top. Rec.}}^{(0)}(\epsilon ),F_{\text {Top. Rec.}}^{(1)}(\epsilon )\right) \) given in “Appendix A.2” (equations (A.9) and (A.11)). We remind here that \(a=\tan \frac{\pi \epsilon }{2}\). We obtain the first orders:

$$\begin{aligned} F_{\text {Top. Rec.}}^{(0)}(\epsilon )= & {} \ln 2-\ln \left( \sin \frac{\pi \epsilon }{2}\right) \nonumber \\ F_{\text {Top. Rec.}}^{(1)}(\epsilon )= & {} \frac{1}{4}\ln \left( \cos \left( \frac{\pi \epsilon }{2}\right) \right) \nonumber \\ F_{\text {Top. Rec.}}^{(2)}(\epsilon )= & {} \frac{1}{64}-\frac{1}{32}\tan ^2\left( \frac{\pi \epsilon }{2}\right) \nonumber \\ F_{\text {Top. Rec.}}^{(3)}(\epsilon )= & {} -\frac{1}{256}-\frac{1}{128}\tan ^2\left( \frac{\pi \epsilon }{2}\right) -\frac{5}{128}\tan ^4\left( \frac{\pi \epsilon }{2}\right) \end{aligned}$$
(2.31)

so that Eq. (2.32) gives:

$$\begin{aligned} \ln Z_n(a)= & {} n^2\ln \left( \sin \left( \frac{\pi \epsilon }{2}\right) \right) -\frac{1}{4}\ln n -\frac{1}{4}\ln \left( \cos \left( \frac{\pi \epsilon }{2}\right) \right) +3\,\zeta '(-1)+\frac{1}{12}\ln 2 \nonumber \\&+\,\frac{1}{64n^2}\left( 2\tan ^2\left( \frac{\pi \epsilon }{2}\right) -1\right) + \frac{1}{256n^4}\left( 1+2\tan ^2\left( \frac{\pi \epsilon }{2}\right) +10\tan ^4\left( \frac{\pi \epsilon }{2}\right) \right) \nonumber \\&+\,O\left( \frac{1}{n^6}\right) \end{aligned}$$
(2.32)

2.5 Final result

Using the rotation invariance of the problem, we may easily generalize the previous result for any interval \([\alpha ,\beta ]\) and we obtain the following theorem:

Theorem 2.3

(Asymptotic expansion of Toeplitz determinants in the one-interval case) For \((\alpha ,\beta )\) such that \(0<|\beta -\alpha |<2\pi \), the Toeplitz determinant with symbol \(f=\mathbb {1}_{{\mathcal {T}}(\alpha ,\beta )}\) where \({\mathcal {T}}(\alpha ,\beta )=\{e^{it}, t\in [\alpha ,\beta ]\}\) admits a large n asymptotic expansion of the form (with the same meaning as the one given in Theorem 2.2):

$$\begin{aligned}&\ln \det T_n(\mathbb {1}_{{\mathcal {T}}(\alpha ,\beta )})=n^2\ln \left( \sin \left( \frac{|\beta -\alpha |}{4}\right) \right) -\frac{1}{4}\ln n -\frac{1}{4}\ln \left( \cos \left( \frac{|\beta -\alpha |}{4}\right) \right) \\&\quad +3\,\zeta '(-1)+\frac{1}{12}\ln 2- \sum _{g=2}^\infty F^{(g)}(a)n^{2-2g} \end{aligned}$$

where \(a=\tan \left( \frac{|\beta -\alpha |}{4}\right) \) and the coefficients \(\left( F^{(g)}(a)\right) _{g\ge 2}\) are the Eynard–Orantin free energies (also called symplectic invariants) associated with the spectral curve

$$\begin{aligned} y^2(x)=\frac{1}{\cos ^2\left( \frac{|\beta -\alpha |}{4}\right) (1+x^2)^2\left( x^2-\tan ^2\left( \frac{|\beta -\alpha |}{4}\right) \right) } \end{aligned}$$

The previous large n asymptotic expansion has the precise meaning given in Theorem 2.2. In particular, the first orders of the expansion are given by:

$$\begin{aligned} \ln \det T_n(\mathbb {1}_{{\mathcal {T}}(\alpha ,\beta )})= & {} n^2\ln \left( \sin \left( \frac{|\beta -\alpha |}{4}\right) \right) -\frac{1}{4}\ln n-\frac{1}{4}\ln \left( \cos \left( \frac{|\beta -\alpha |}{4}\right) \right) \\&+\,3\,\zeta '(-1)+\frac{1}{12}\ln 2+\frac{1}{64n^2}\left( 2\tan ^2\left( \frac{|\beta -\alpha |}{4}\right) -1\right) \\&+\,\frac{1}{256n^4}\left( 1+2\tan ^2\left( \frac{|\beta -\alpha |}{4}\right) +10\tan ^4\left( \frac{|\beta -\alpha |}{4}\right) \right) +O\left( \frac{1}{n^6}\right) \end{aligned}$$

2.6 Numerical study

We can efficiently compute the Toeplitz determinants (2.2) numerically up to \(n=35\). This allows to compare the theoretical formula (2.32) with the numeric simulations up to order \(o\left( \frac{1}{n^4}\right) \). We obtain the following picture (Fig. 2):

We obviously see on the last figure that the numeric simulations are compatible with the theoretical results up to order \(o\left( \frac{1}{n^4}\right) \). This provides additional credit for the general formulas proved in Theorem 2.3 and the reconstruction of the expansion from the topological recursion. Note that the reformulation of \(Z_n(a)\) in terms of the determinant of a symmetric Toeplitz matrix is particularly useful since it allows fast computations of \(Z_n(a)\) even for relatively large values of n. We performed the computations using Maple software and could compute the values of \(Z_n(a)\) from \(n=2\) to \(n=35\) in no more than a few minutes on a standard laptop.

3 Toeplitz determinants with a discrete rotational symmetry

Let \(r\ge 0\) be a given integer and let \(\epsilon \in (0,1)\) be a given number. In this section, we consider the Toeplitz determinants:

$$\begin{aligned} Z_n({\mathcal {I}}_r)= & {} \frac{1}{(2\pi )^n n!} \int _{\left( {\mathcal {I}}_r\right) ^n} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _n \prod _{1\le p<q\le n}\left| e^{i\theta _p}-e^{i\theta _q}\right| ^2 \text { with } \nonumber \\ {\mathcal {I}}_r= & {} \mathop {\bigcup }\limits _{k=-r}^{r}\left[ \frac{2\pi k}{2r+1}-\frac{\pi \epsilon }{2r+1}, \frac{2\pi k}{2r+1}+\frac{\pi \epsilon }{2r+1}\right] \end{aligned}$$
(3.1)

For simplicity, we denote \(\alpha ^{(r)}_k=\frac{2\pi k}{2r+1}-\frac{\pi \epsilon }{2r+1}\), \(\beta ^{(r)}_k=\frac{2\pi k}{2r+1}+\frac{\pi \epsilon }{2r+1}\) and \(\gamma ^{(r)}_k=\frac{2\pi k}{2r+1}\) for \(-r\le k\le r\). We also define:

$$\begin{aligned} a_k^{(r)}=\tan \left( \frac{\alpha _k^{(r)}}{2}\right) \text { and } b_k^{(r)}=\tan \left( \frac{\beta _k^{(r)}}{2}\right) \text { for } -r\le k\le r \end{aligned}$$

Note that \(Z_n({\mathcal {I}}_r)\) can also be interpreted as the probability to obtain all angles \((\theta _j)_{1\le j\le n}\) in \({\mathcal {I}}_r\). Similarly to the last section, we also introduce the sets:

Fig. 2
figure 2

Computations (from (2.2)) of the Toeplitz determinants \(\theta \mapsto \ln Z_n(\frac{\theta }{\pi })\) with \(0<\theta <\pi \) for \(2\le n\le 35\) with subtraction of the first coefficients of the large n expansion (2.32) (colored dots: starting from orange to yellow, green and purple as n increases). The black curves are the theoretical predictions given by (from top to bottom and from left to right): \(\theta \mapsto \ln \left( \sin \left( \frac{\theta }{2}\right) \right) \), \(\theta \mapsto -\frac{1}{4}\ln \left( \cos \left( \frac{\theta }{2}\right) \right) +3\,\zeta '(-1)+\frac{1}{12}\ln 2\), \(\theta \mapsto \frac{1}{64}\left( 2\tan ^2\left( \frac{\theta }{2}\right) -1\right) \) and \(\theta \mapsto \frac{1}{256}\left( 1+2\tan ^2\left( \frac{\theta }{2}\right) +10\tan ^4\left( \frac{\theta }{2}\right) \right) \) (color figure online)

$$\begin{aligned} {\mathcal {J}}_r= & {} \mathop {\bigcup }\limits _{k=-r}^{r}\left[ a_k^{(r)},b_k^{(r)}\right] =\mathop {\bigcup }\limits _{k=-r}^{r} \left[ \tan \left( \frac{\pi k}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) ,\tan \left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right] \\ {\mathcal {T}}_r= & {} \left\{ e^{i\theta },\theta \in {\mathcal {I}}_r\right\} \end{aligned}$$

Therefore, from Lemma 1.1, \(Z_n({\mathcal {I}}_r)\) can be reformulated as follow:

  1. 1.

    \(Z_n({\mathcal {I}}_r)=\det T_n^{(r)}\) with \(\left( T_n^{(r)}\right) _{i,j}=t_{i-j}\) the \(n\times n\) Toeplitz matrix given by:

    $$\begin{aligned} t_0= & {} \frac{|{\mathcal {I}}_r|}{2\pi }=\epsilon \nonumber \\ t_k= & {} \epsilon \sin _c\left( \frac{\pi \epsilon k}{2r+1}\right) \delta _{k\,\equiv \, 0\,[2r+1]} \,\text { for }k\ne 0 \end{aligned}$$
    (3.2)

    Note in particular that the Toeplitz matrices are mostly empty since only bands with indexes multiple of \(2r+1\) are nonzero.

  2. 2.

    A real integral with Vandermonde interactions:

    $$\begin{aligned} Z_n({\mathcal {I}}_r)=\frac{2^{n^2}}{(2\pi )^n n!}\int _{\left( {\mathcal {J}}_r\right) ^n} \mathrm{d}t_1\ldots \mathrm{d}t_n\, \Delta (t_1,\ldots ,t_n)^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+t_k^2)} \end{aligned}$$
    (3.3)
  3. 3.

    A complex integral:

    $$\begin{aligned} Z_n({\mathcal {I}}_r)=(-1)^{\frac{n(n+1)}{2}}i^n \int _{\left( {\mathcal {T}}_r\right) ^n} \mathrm{d}u_1\ldots d u_n\, \Delta (u_1,\ldots ,u_n)^2 e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln u_k} \end{aligned}$$
    (3.4)

The situation can be illustrated as follows (Fig. 3):

3.1 Computation of the spectral curve

We want to compute the spectral curve associated with the integral (3.3). This integral corresponds to a Hermitian matrix integral with hard edges at \(\left( a_k^{(r)}\right) _{-r\le k\le r}\) and \(\left( b_k^{(r)}\right) _{-r\le k\le r}\). Following the same method as in Sect. 2.2, we define \(\left\langle g({\mathbf {t}})\right\rangle _r\) as the average of the function \(g({\mathbf {t}})\) relatively to the measure induced by (3.3). With the same arguments as in Sect. 2.2 , we get a spectral curve of the form:

$$\begin{aligned} y(x)^2= & {} \frac{x^2}{(1+x^2)^2}-\frac{2}{1+x^2}\left( \mathop {\lim }\limits _{n\rightarrow \infty }\left\langle \frac{1}{n}\sum _{j=1}^n\frac{1}{1+t_j^2}\right\rangle _r-x\mathop {\lim }\limits _{n\rightarrow \infty }\left\langle \frac{1}{n}\sum _{j=1}^n\frac{t_j}{1+t_j^2}\right\rangle _r\right) \nonumber \\&\quad +\sum _{j=-r}^{r}\left( \frac{A_k}{x-a_k^{(r)}}+\frac{B_k}{x-b_k^{(r)}}\right) \end{aligned}$$
(3.5)

where the constants \(A_k\) (resp. \(B_k\)) are given by \(A_k=-\mathop {\lim }\nolimits _{n\rightarrow \infty }\frac{1}{n}A_{k,n}\) and \(B_k=-\mathop {\lim }\nolimits _{n\rightarrow \infty }\frac{1}{n}B_{k,n}\) with:

$$\begin{aligned} A_{k,n}= & {} -\frac{e^{-n\ln (1+(a_k^{(r)})^2)}}{(2\pi )^n(n!) Z_n({\mathcal {I}}_r)}\nonumber \\&\sum _{j=1}^n\int _{\left( {\mathcal {J}}_r\right) ^{n-1}} \frac{\mathrm{d}t_1\ldots \mathrm{d}t_{j-1} \mathrm{d}t_{j+1} \ldots d t_n}{a_k^{(r)}-t_j}\Delta (t_1,\ldots ,t_{j-1},a_k^{(r)},t_{j+1},\ldots , t_n)^2 e^{-n\mathop {\sum }\limits _{q\ne j}^{n} \ln (1+t_q^2)} \nonumber \\ B_{k,n}= & {} \frac{e^{-n\ln (1+(b_k^{(r)})^2)}}{(2\pi )^n(n!) Z_n({\mathcal {I}}_r)}\nonumber \\&\sum _{j=1}^n\int _{\left( {\mathcal {J}}_r\right) ^{n-1}} \frac{\mathrm{d}t_1\ldots \mathrm{d}t_{j-1} \mathrm{d}t_{j+1} \ldots d t_n}{b_k^{(r)}-t_j}\Delta (t_1,\ldots ,t_{j-1},b_k^{(r)},t_{j+1},\ldots , t_n)^2 e^{-n\mathop {\sum }\limits _{q\ne j}^{n} \ln (1+t_q^2)}\nonumber \\ \end{aligned}$$
(3.6)
Fig. 3
figure 3

Illustration (in green) of the set \({\mathcal {T}}_{r=2}\) for \(\epsilon =\frac{1}{5}\) (color figure online)

Note that the integral (3.3) is invariant under \({\mathbf {t}}\rightarrow -{\mathbf {t}}\); thus, we get that \(\left\langle \mathop {\sum }\nolimits _{j=1}^{n}\frac{t_j}{1+t_j^2}\right\rangle _r=0\). Therefore, we end up with:

$$\begin{aligned} y(x)^2=\frac{x^2}{(1+x^2)^2}-\frac{2d}{1+x^2}+\sum _{j=-r}^{r}\left( \frac{A_k}{x-a_k^{(r)}}+\frac{B_k}{x-b_k^{(r)}}\right) \end{aligned}$$
(3.7)

where the coefficients d and \((A_k,B_k)_{-(r-1)\le k\le r-1}\) are so far undetermined and independent of x. Observe now that the choice of \({\mathcal {I}}_r\) implies that we have:

$$\begin{aligned} a_{-k}^{(r)}=-b_k^{(r)} \text { for all } 0\le k\le r \end{aligned}$$

Using the invariance of the integral relatively to \({\mathbf {t}}\mapsto -{\mathbf {t}}\), we obtain that:

$$\begin{aligned} A_{-k}=-B_k \text { and } B_{-k}=-A_k \text { for all } 0\le k\le r \end{aligned}$$
(3.8)

Consequently, we can reduce the spectral curve to:

$$\begin{aligned} y(x)^2=\frac{x^2}{(1+x^2)^2}-\frac{2d}{1+x^2}+ \sum _{j=-r}^{r}\frac{2b_k^{(r)}B_k}{x^2-(b_k^{(r)})^2} \end{aligned}$$
(3.9)

Similarly to the one-interval case, the study of large x implies that \(y^2(x)=O\left( \frac{1}{x^6}\right) \) giving some algebraic relations between the \(2r+2\) undetermined coefficients \((d,B_{-r},\ldots ,B_{r})\). However, unlike the one-interval case, when \(r\ge 1\), these relations are not sufficient to determine completely the coefficients. Indeed, when \(r\ge 1\), the number of unknown coefficients is strictly larger than the number of relations obtained from the fact that \(y^2(x)=O\left( \frac{1}{x^6}\right) \). Therefore, some additional relations are required. In order to obtain them, we perform the following change of variables (denoting \({\mathbf {1}}=(1,\ldots ,1)^t \in {\mathbb {R}}^n\), \({\mathbf {t}}=(t_1,\ldots ,t_n)^t\) and \({\tilde{{\mathbf {t}}}}=({\tilde{t}}_1,\ldots ,{\tilde{t}}_n)^t\)) in the integral (3.3). Let \(-r\le j_0\le r\):

$$\begin{aligned} {\mathbf {t}}= & {} \tan \left( \text {Arctan}({\tilde{{\mathbf {t}}}})+\frac{\pi j_0}{2r+1}{\mathbf {1}}\right) =\frac{{\tilde{{\mathbf {t}}}}+\tan \left( \frac{\pi j_0}{2r+1}\right) }{1-{\tilde{{\mathbf {t}}}} \tan \left( \frac{\pi j_0}{2r+1}\right) }\,\,\Leftrightarrow \nonumber \\ {\tilde{{\mathbf {t}}}}= & {} \tan \left( \text {Arctan}({\mathbf {t}})-\frac{\pi j_0}{2r+1}{\mathbf {1}}\right) =\frac{{\mathbf {t}}-\tan \left( \frac{\pi j_0}{2r+1}\right) }{1+{\mathbf {t}} \tan \left( \frac{\pi j_0}{2r+1}\right) } \end{aligned}$$
(3.10)

Note that the domain of integration \({\mathcal {J}}_r\) is invariant under the former change of variables since any interval \(\left[ a_k^{(r)},b_k^{(r)}\right] \) is mapped to \(\left[ a_{k-j_0}^{(r)},b_{k-j_0}^{(r)}\right] \) where the indexes \(k-j_0\) are to be understood modulo \(2r+1\). Then, straightforward computations show that:

$$\begin{aligned} \mathrm{d}t_p= & {} \frac{1+\tan ^2\left( \frac{\pi j_0}{2r+1}\right) }{\left( 1-{\tilde{t}}_p \tan \left( \frac{\pi j_0}{2r+1}\right) \right) ^2}d{\tilde{t}}_p \nonumber \\ 1+t_p^2= & {} \frac{\left( 1+\tan ^2\left( \frac{\pi j_0}{2r+1}\right) \right) (1+{\tilde{t}}_p^2)}{\left( 1-{\tilde{t}}_p \tan \left( \frac{\pi j_0}{2r+1}\right) \right) ^2} \nonumber \\ (t_p-t_q)^2= & {} \frac{\left( 1+\tan ^2\left( \frac{\pi j_0}{2r+1}\right) \right) ^2({\tilde{t}}_p-{\tilde{t}}_q)^2}{\left( 1-{\tilde{t}}_p \tan \left( \frac{\pi j_0}{2r+1}\right) \right) ^2\left( 1-{\tilde{t}}_q \tan \left( \frac{\pi j_0}{2r+1}\right) \right) ^2} \end{aligned}$$
(3.11)

Therefore, the general form of the integral remains invariant:

$$\begin{aligned} Z_n({\mathcal {I}}_r)= & {} \frac{2^{n^2}}{(2\pi )^n n!}\int _{\left( {\mathcal {J}}_r\right) ^n} \mathrm{d}t_1\ldots \mathrm{d}t_n \Delta (t_1,\ldots ,t_n)^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+t_k^2)}\\= & {} \frac{2^{n^2}}{(2\pi )^n n!}\int _{\left( {\mathcal {J}}_r\right) ^n} d{\tilde{t}}_1\ldots d{\tilde{t}}_n \Delta ({\tilde{t}}_1,\ldots ,{\tilde{t}}_n)^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+{\tilde{t}}_k^2)} \end{aligned}$$

and thus the function \(W_1(x)\) is given by:

$$\begin{aligned}&W_1(x)=\frac{2^{n^2}}{(2\pi )^n (n!)Z_n({\mathcal {I}}_r)}\int _{\left( {\mathcal {J}}_r\right) ^n} d{\mathbf {t}}\left( \sum _{k=1}^n\frac{1}{x-t_k}\right) \Delta ({\mathbf {t}})^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+t_k^2)}=\frac{2^{n^2}}{(2\pi )^n (n!)Z_n({\mathcal {I}}_r)} \nonumber \\&\int _{\left( {\mathcal {J}}_r\right) ^n} d{\tilde{{\mathbf {t}}}}\left( \sum _{k=1}^n\frac{1-{\tilde{t}}_k\tan \left( \frac{\pi j_0}{2r+1}\right) }{x-\tan \left( \frac{\pi j_0}{2r+1}\right) -{\tilde{t}}_k\left( 1+x\tan \left( \frac{\pi j_0}{2r+1}\right) \right) }\right) \Delta ({\tilde{{\mathbf {t}}}})^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+{\tilde{t}}_k^2)} \end{aligned}$$
(3.12)

We now observe that the equation:

$$\begin{aligned} \frac{1-at}{x-a-t(1+ax)}=\mu +\frac{\nu }{x-t}+\frac{\rho \, t}{x^2-t^2} \end{aligned}$$
(3.13)

where \((\mu ,\nu ,\rho )\) are independent of t admits the following solution:

$$\begin{aligned} (\mu ,\nu ,\rho ,a)=\left( \frac{a}{1+ax},\frac{x(1+a^2)}{(x-a)(1+ax)},-\frac{(1+a^2)(1+x^2)}{(x-a)(1+ax)^2},-\frac{2x}{x^2-1}\right) \nonumber \\ \end{aligned}$$
(3.14)

In particular, we always have \(\frac{\mu }{1-\nu }=\frac{x}{1+x^2}=\frac{1}{2}V'(x)\). Inserting these results with \(a=\tan \left( \frac{\pi j_0}{2r+1}\right) \), \(t={\tilde{t}}_k\) and:

$$\begin{aligned} \left\{ \begin{array}{ccc} x_{j_0} &{}=&{}\tan \left( \frac{\pi i_0}{2r+1}-\frac{\pi }{2(2r+1)}\right) \,\text { if } \,j_0=2i_0 \text { is even} \\ x_{j_0} &{}=&{}\tan \left( \frac{\pi i_0}{2r+1}+\frac{\pi }{2(2r+1)}\right) \,\text { if } \,j_0=2i_0+1 \text { is odd} \end{array} \right. \end{aligned}$$
(3.15)

that are solutions of \(a(x^2-1)+2x=0\) located outside of \({\mathcal {J}}_r\), we end up in (3.12) with:

$$\begin{aligned} W_1(x_{j_0})=\mu +\nu W_1(x_{j_0}) +\rho \left\langle \mathop {\sum }\nolimits _{k=1}^{n}\frac{t}{x_{j_0}^2-t^2}\right\rangle _r \end{aligned}$$
(3.16)

Using the fact that \(\left\langle \mathop {\sum }\nolimits _{k=1}^{n}\frac{t}{x^2-t^2}\right\rangle _r=0\) from the symmetry \({\mathbf {t}}\rightarrow -{\mathbf {t}}\), we get to:

$$\begin{aligned} W_1(x_{j_0})=\mu +\nu W_1(x_{j_0}) \Rightarrow W_1(x_{j_0})=\frac{\mu }{1-\nu }=\frac{x_{j_0}}{1+x_{j_0}^2} \,\,\, \Rightarrow y(x_{j_0})=0\nonumber \\ \end{aligned}$$
(3.17)

Hence, the values \(\left( \tan \left( \frac{\pi j}{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) _{-r\le j\le r-1}\) (that are identical to \(\left( \tan \left( \frac{\pi j}{2r+1}-\frac{\pi }{2(2r+1)}\right) \right) _{-(r-1)\le j\le r}\))) are 2r distinct zeros of the function \(x\mapsto y(x)\) located outside \({\mathcal {J}}_r\). This provides 2r distinct double zeros for \(y^2(x)\). From (3.9) and the fact that \(y^2(x)\mathop {=}\nolimits _{x\rightarrow \infty }O\left( \frac{1}{x^6}\right) \), we get that the spectral curve must be of the form:

$$\begin{aligned} y^2(x)=\frac{P_{4r+4}(x)}{(1+x^2)^2 \mathop {\prod }\nolimits _{k=-r}^{r}(x^2-(b_k^{(r)})^2)} \text { with } P_{4r+4} \text { a polynomial of degree } 4r\nonumber \\ \end{aligned}$$
(3.18)

Since we have found 4r zeros (counted with their multiplicities), we have:

$$\begin{aligned} y^2(x)=\lambda _r \frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \left( x^2-\tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) ^2}{(1+x^2)^2\mathop {\prod }\nolimits _{k=-r}^{r}\left( x^2-\tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right) } \end{aligned}$$
(3.19)

with the convention that for \(r=0\) we take empty products (like \(\mathop {\prod }\nolimits _{k=0}^{-1}\left( x^2-\tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) ^2\)) equal to 1. The constant \(\lambda _r\) can be determined using the behavior of \(y^2(x)\) around \(\pm i\). Indeed, by definition the function \(z\mapsto W_1(z)\) is analytic in a neighborhood of \(\pm i\). Consequently, the poles of \(y^2(x)\) at \(x=\pm i\) only come from the shift by \(-\frac{1}{2}V'(x)=\frac{x}{1+x^2}\). Therefore, we should get \(y^2(x)\mathop {\sim }\nolimits _{x\rightarrow i} \frac{1}{4(x-i)^2}\). We get:

$$\begin{aligned} \frac{1}{4}=\lambda _r\frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \left( 1+\tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) ^2}{4\mathop {\prod }\nolimits _{k=-r}^{r}\left( 1+\tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right) } \,\Leftrightarrow \, \lambda _r=\frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \cos ^4\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) }{\mathop {\prod }\nolimits _{k=-r}^{r}\cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) }\nonumber \\ \end{aligned}$$
(3.20)

Finally, we get:

$$\begin{aligned} y^2(x)= & {} \frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \cos ^4\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) }{\mathop {\prod }\nolimits _{k=-r}^{r}\cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) } \frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \left( x^2-\tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) ^2}{(1+x^2)^2\mathop {\prod }\nolimits _{k=-r}^{r}\left( x^2-\tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right) } \nonumber \\= & {} \frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \left( x^2\cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) -\sin ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) ^2}{(1+x^2)^2\mathop {\prod }\limits _{k=-r}^{r}\left( x^2\cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) -\sin ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right) } \nonumber \\= & {} \frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \left( (x^2+1)\cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) -1\right) ^2}{(1+x^2)^2\mathop {\prod }\limits _{k=-r}^{r}\left( (x^2+1)\cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) -1\right) } \end{aligned}$$
(3.21)

The corresponding limiting eigenvalues density is therefore given by:

$$\begin{aligned} \mathrm{d}\mu _\infty (x)= & {} \frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) }{\mathop {\prod }\nolimits _{k=-r}^{r}\cos \left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) } \frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \left| \tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) -x^2\right| }{\pi (1+x^2)\mathop {\prod }\nolimits _{k=-r}^{r}\sqrt{\left| \tan ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) -x^2\right| }} \mathbb {1}_{{\mathcal {J}}_r}(x)\mathrm{d}x \nonumber \\= & {} \frac{\mathop {\prod }\nolimits _{k=0}^{r-1} \left| (1+x^2)\cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi }{2(2r+1)}\right) -1\right| }{\pi (1+x^2)\mathop {\prod }\nolimits _{k=-r}^{r}\sqrt{\left| (1+x^2)\cos ^2\left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) -1\right| }}\mathbb {1}_{{\mathcal {J}}_r}(x)\mathrm{d}x \end{aligned}$$
(3.22)

In particular, one can verify that it is properly normalized: \(\int _{{\mathcal {J}}_r} \mathrm{d}\mu _\infty (x)=1\). Note that in the case \(r=0\), we recover the spectral curve for a single interval (2.16). We can verify numerically the last limiting eigenvalues density with Monte–Carlo simulations (Fig. 4).

Fig. 4
figure 4

Histogram of 50 independent simulations of the eigenvalues density induced by (3.3) in the case \(r=1\), \(\epsilon =\frac{1}{10}\) and \(n=60\). The black curve is the theoretical limiting eigenvalues density computed in Eq. (3.22)

For a clearer view, it is also interesting to zoom on each interval (Fig. 5).

Fig. 5
figure 5

Zoom of the previous histogram in each of the three intervals. The black curve is the theoretical limiting eigenvalues density computed in Eq. (3.22)

3.2 Filling fractions

We want to determine the proportion of eigenvalues in each of the \(2r+1\) intervals of the limiting eigenvalues density. We start with the previous result:

$$\begin{aligned} y\mathrm{d}x=\lambda _r \frac{\mathop {\prod }\nolimits _{j=-r}^{r-1} \left( x-\tan \left( \frac{j\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) }{(1+x^2)\mathop {\prod }\nolimits _{k=-r}^{r} \sqrt{\left( x-\tan \left( \frac{\pi k}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) \right) \left( x-\tan \left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right) }}\mathrm{d}x\nonumber \\ \end{aligned}$$
(3.23)

and observe that the filling fraction corresponding to the \(k_0{\text {th}}\) interval, with \(-r\le k_0\le r\), is given by:

$$\begin{aligned} \varvec{\epsilon }_{k_0+r+1}\overset{\text {def}}{=}\frac{1}{i\pi }\int _{\tan \left( \frac{\pi k_0}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) }^{\tan \left( \frac{\pi k_0}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) } \mathrm{d}\mu _\infty (x) \end{aligned}$$
(3.24)

Let \(-r\le j_0\le r\) be a given integer with \(j_0\ne 0\). We perform the change of variables \(x=\frac{{\tilde{x}}-\tan \frac{\pi j_0}{2r+1}}{1+{\tilde{x}}\tan \frac{\pi j_0}{2r+1}}\) described above in the previous integral. Observe in particular that we have \(\frac{\mathrm{d}x}{1+x^2}=\frac{d{\tilde{x}}}{1+{\tilde{x}}^2}\). Moreover, under this change of variable, the interval \(x\in \left[ \tan \left( \frac{\pi k}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) , \tan \left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right] \) is directly mapped to the interval \({\tilde{x}}\in \left[ \tan \left( \frac{\pi (k+j_0)}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) , \tan \left( \frac{\pi (k+j_0)}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right] \). We also have the identities:

$$\begin{aligned}&\left( x-\tan \left( \frac{\pi k}{2r+1}\pm \frac{\pi \epsilon }{2(2r+1)}\right) \right) \nonumber \\&\quad = \frac{\left( {\tilde{x}}-\tan \left( \frac{\pi (k+j_0)}{2r+1}\pm \frac{\pi \epsilon }{2(2r+1)}\right) \right) \left( 1- \tan \left( \frac{\pi k}{2r+1}\pm \frac{\pi \epsilon }{2(2r+1)}\right) \tan \left( \frac{\pi j_0}{2r+1}\right) \right) }{1+{\tilde{x}}\tan \left( \frac{\pi j_0}{2r+1}\right) } \nonumber \\&\left( x-\tan \left( \frac{i_0\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) \nonumber \\&\quad =\frac{\left( {\tilde{x}}-\tan \left( \frac{(i_0+j_0)\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) \left( 1-\tan \left( \frac{\pi i_0}{2r+1}+\frac{\pi }{2(2r+1)}\right) \tan \left( \frac{\pi j_0}{2r+1}\right) \right) }{1+{\tilde{x}}\tan \left( \frac{\pi j_0}{2r+1}\right) } \nonumber \\&\qquad \text { if } i_0+j_0\not \equiv r[2r+1] \end{aligned}$$
(3.25)

When \(i_0+j_0\equiv r[2r+1]\), then we get instead:

$$\begin{aligned} \left( x-\tan \left( \frac{i_0\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right)= & {} \frac{-\tan \left( \frac{\pi j_0}{2r+1}\right) -\tan \left( \frac{i_0\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) }{1+{\tilde{x}}\tan \left( \frac{\pi j_0}{2r+1}\right) }\nonumber \\= & {} \frac{-\tan \left( \frac{\pi j_0}{2r+1}\right) -\frac{1}{\tan \left( \frac{\pi j_0}{2r+1}\right) }}{1+{\tilde{x}}\tan \left( \frac{\pi j_0}{2r+1}\right) }\nonumber \\= & {} \frac{-1}{\cos \left( \frac{\pi j_0}{2r+1}\right) \sin \left( \frac{\pi j_0}{2r+1}\right) \left( 1+{\tilde{x}}\tan \left( \frac{\pi j_0}{2r+1}\right) \right) }\nonumber \\ \end{aligned}$$
(3.26)

Observe now that the powers of \(1+{\tilde{x}}\tan \left( \frac{\pi j_0}{2r+1}\right) \) produce a factor with power 1 at the numerator and that we can express it like:

$$\begin{aligned} 1+{\tilde{x}}\tan \left( \frac{\pi j_0}{2r+1}\right) =\tan \left( \frac{\pi j_0}{2r+1}\right) \left( {\tilde{x}}-\tan \left( \frac{(r+j_0)\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) \nonumber \\ \end{aligned}$$
(3.27)

Therefore, collecting the terms depending on \({\tilde{x}}\) we get:

$$\begin{aligned}&\prod _{k=-r}^{r}\sqrt{\left( {\tilde{x}}-\tan \left( \frac{\pi (k+j_0)}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) \right) \left( {\tilde{x}}-\tan \left( \frac{\pi (k+j_0)}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right) } \nonumber \\&\quad =\prod _{k=-r}^{r}\sqrt{\left( {\tilde{x}}-\tan \left( \frac{\pi k}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) \right) \left( {\tilde{x}}-\tan \left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \right) } \nonumber \\&\left( {\tilde{x}}-\tan \left( \frac{(r+j_0)\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) \prod _{i_0=-r, i_0+j_0\not \equiv r [2r+1] }^{r-1}\left( {\tilde{x}}-\tan \left( \frac{(i_0+j_0)\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) \nonumber \\&\quad =\prod _{i_0=-r, i+j_0\not \equiv r [2r+1] }^{r}\left( {\tilde{x}}-\tan \left( \frac{(i_0+j_0)\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) \nonumber \\&\quad =\mathop {\prod }\limits _{i_0=-r}^{r-1} \left( {\tilde{x}}-\tan \left( \frac{i_0\pi }{2r+1}+\frac{\pi }{2(2r+1)}\right) \right) \end{aligned}$$
(3.28)

Thus, we obtain:

$$\begin{aligned} \varvec{\epsilon }_{k_0+r+1}=C_{j_0}\varvec{\epsilon }_{k_0+j_0+r+1} \end{aligned}$$
(3.29)

where the constant \(C_{j_0}\) is given by:

$$\begin{aligned} C_{j_0}= & {} - \frac{{\mathop {\prod }\nolimits _{i_0=-r, i_0+j_0\not \equiv r [2r+1] }^{r-1}} \left( 1-\tan \left( \frac{\pi i_0}{2r+1}+\frac{\pi }{2(2r+1)}\right) \tan \left( \frac{\pi j_0}{2r+1}\right) \right) }{\cos ^2\left( \frac{\pi j_0}{2r+1}\right) \mathop {\prod }\nolimits _{k=-r}^{r}\sqrt{\left( 1- \tan \left( \frac{\pi k}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) \tan \left( \frac{\pi j_0}{2r+1}\right) \right) \left( 1- \tan \left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \tan \left( \frac{\pi j_0}{2r+1}\right) \right) }} \nonumber \\= & {} -\frac{{\mathop {\prod }\nolimits _{i_0=-r, i_0+j_0\not \equiv r [2r+1] }^{r-1}} \frac{\cos \left( \frac{\pi (i_0+j_0)}{2r+1}+\frac{\pi }{2(2r+1)}\right) }{\cos \left( \frac{\pi i_0}{2r+1}\right) \cos \left( \frac{\pi j_0}{2r+1}\right) }}{\cos ^2\left( \frac{\pi j_0}{2r+1}\right) \mathop {\prod }\nolimits _{k=-r}^{r}\sqrt{\frac{\cos \left( \frac{\pi (k+j_0)}{2r+1}+\frac{\pi \epsilon }{2r+1}\right) \cos \left( \frac{\pi (k+j_0)}{2r+1}-\frac{\pi \epsilon }{2r+1}\right) }{\cos \left( \frac{\pi k}{2r+1}-\frac{\pi \epsilon }{2r+1}\right) \cos \left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2r+1}\right) \cos ^2\left( \frac{\pi j_0}{2r+1}\right) }} }=\left( \cos \left( \frac{\pi j_0}{2r+1}\right) \right) ^{-2-(2r-1)+(2r+1)} \nonumber \\= & {} 1 \end{aligned}$$
(3.30)

Hence, we have just proved that all filling fractions are equal:

Proposition 3.1

The filling fractions are the same in each interval:

$$\begin{aligned} \forall \, 1\le k\le 2r+1\,\,:\,\,\varvec{\epsilon }_k=\frac{1}{2r+1} \end{aligned}$$

Remark 3.1

This result is compatible with the one found in [21] where the filling fractions were also proved to be equal.

3.3 General form of the large n asymptotic

After determining the limiting eigenvalues density, the next step is to determine the general form of the asymptotic of the correlation functions and of the partition function. However, since the spectral curve is of strictly positive genus (\({{\mathfrak {g}}}=2r\)), the general form of the asymptotic is more complicated than in the one-interval case. Using the main result of [18] or [19], we can prove the following:

Theorem 3.1

We have the following large n expansion:

(3.31)

where \(\Theta \) is the Siegel theta function:

$$\begin{aligned} \Theta _{\varvec{\gamma }}(\varvec{\nu },{\mathbf {T}})=\sum _{{\mathbf {m}}\in {\mathbb {Z}}^g} \text {exp}\left( -\frac{1}{2}({\mathbf {m}}+\varvec{\gamma })^t\,\cdot {\mathbf {T}}\cdot ({\mathbf {m}}+\varvec{\gamma })+\varvec{\nu }^t\,\cdot ({\mathbf {m}}+\varvec{\gamma })\right) \end{aligned}$$

and \(F^{\{2k\},(l)}_{\varvec{\epsilon }}\) are defined as the \(l{\text {th}}\) derivative of the coefficient \(F_{\varvec{\epsilon }}^{\{2k\}}\) relatively to the filling fractions \(\varvec{\epsilon }=(\epsilon _1,\ldots ,\epsilon _{2r+1})^t \in \{{\mathbf {u}}\in ({\mathbb {Q}}_+)^{2r+1}\,/ \, \mathop {\sum }\nolimits _{i=1}^{n}u_i=1\}\). The “optimal” filling fractions \(\varvec{\epsilon }^\star \) correspond to the vector of filling fractions (i.e., the proportion of eigenvalues in each interval of the support of the limiting eigenvalues density) of the limiting eigenvalues density:

$$\begin{aligned} \varvec{\epsilon }^\star =\left( \frac{1}{2r+1},\ldots ,\ldots ,\frac{1}{2r+1}\right) ^t \end{aligned}$$

Again large n asymptotic expansions presented in this theorem are to be understood as asymptotic expansions up to any arbitrary large negative power of n as in Theorem 2.2.

The proof of the last theorem consists in verifying the conditions required to apply the main theorem of [18] and [19]. The conditions are very similar to the one-interval case, but for completeness we summarize them here:

  • (Regularity): The potential V is continuous on the integration domain \(\mathop {\bigcup }\limits _{k=-r}^{r}[b_-^{(k)},b_+^{(k)}]\). In our case \(x\mapsto \ln (1+x^2)\) is obviously continuous on \({\mathbb {R}}\).

  • (Confinement of the potential): Not required since the integration domain is a compact set of \({\mathbb {R}}\).

  • (Genus 2r regime): The support of the limiting eigenvalues density is given by the union of \(2r+1\) single intervals \([\alpha _-^{(k)},\alpha _+^{(k)}]\) not reduced to a point. This is trivial from the explicit expression (3.22).

  • (Control of large deviations): The function \(x\mapsto \frac{1}{2}V(x)+\int _{{\mathbb {R}}} |x-\xi |\mathrm{d}\mu _\infty (\xi )\) defined on \(\left( \mathop {\bigcup }\nolimits _{k=-r}^{r}[b_-^{(k)},b_+^{(k)}]\right) \setminus \left( \mathop {\bigcup }\nolimits _{k=-r}^{r}(\alpha _-^{(k)},\alpha _+^{(k)})\right) \) achieves its minimum only at the endpoints \(\alpha _-^{(k)}\) or \(\alpha _+^{(k)}\). In our case this condition is trivial since the limiting eigenvalues density spans the whole integration domain.

  • (Off-criticality): The limiting eigenvalues density is off-critical in the sense that it is strictly positive inside the interior of its support and behaves like \(O\left( \frac{1}{\sqrt{x-b_\pm }}\right) \) if \(b_\pm \) is a hard edge or like \(O\left( \sqrt{x-\alpha _\pm }\right) \) if \(\alpha _\pm \) is a soft edge. In our case, we have \(2(2r+1)\) hard edges and the explicit expression (3.22) provides the correct behavior. Moreover, it is obvious from (3.22) that \(\mathrm{d}\mu _\infty (x)\) is strictly positive inside its support.

  • (Analyticity): V can be extended into an analytic function inside a neighborhood of the integration domain. In our case, \(x\mapsto \ln (1+x^2)\) is obviously analytic in a neighborhood of any compact set of \({\mathbb {R}}\).

As one can see, the general form of the large n expansion in the multi-cut regime is much more complicated than in the one-cut regime since eigenvalues may move from an interval to another and thus a summation on the filling fractions is necessary and adds new terms (given by the Siegel theta function). In particular, orders O(1) and beyond exhibit a far more complex structure than in the one-cut case. Indeed, Eq. 3.31 and the fact that the present situation has a discrete symmetry (in particular, \(\varvec{\epsilon }^\star =\left( \frac{1}{2r+1},\ldots ,\frac{1}{2r+1}\right) ^t\)) imply that these orders depend on the remainder of the Euclidean division of n by \(2r+1\).

The term \(\text {exp}\left( \mathop {\sum }\nolimits _{k=-1}^{\infty } n^{-2k}F_{\varvec{\epsilon }^\star }^{\{2k\}}\right) \) is usually called the “perturbative” or “formal” part of the expansion. It is connected to the symplectic invariants computed from the topological recursion by the following proposition (see [18] for details):

Proposition 3.2

The coefficients \(\left( F_{\varvec{\epsilon }^\star }^{\{k\}}\right) _{k\ge -2}\) are related to the symplectic invariants \(\left( F^{(g)}\right) _{g\ge 0}\) computed from the topological recursion applied to the spectral curve (3.21) by:

$$\begin{aligned} \forall \,k\ge -1\,&:\,\, F_{\varvec{\epsilon }^\star }^{\{2k\}}=-F^{(2k+2)}+f_{2k} \text { with } f_{2k} \text { independent of } \epsilon \\ \forall \,k\ge -1\,&:\,\, F_{\varvec{\epsilon }^\star }^{\{2k+1\}}=f_{2k+1} \text { with } f_{2k+1} \text { independent of } \epsilon \end{aligned}$$

Note that the non-perturbative part of (3.31) starts at O(1). The main difficulty of the expansion (3.31) lies in the fact that the non-perturbative part requires the knowledge of the spectral curve in the case where the filling fractions are arbitrarily fixed in order to compute the derivatives relatively to the filling fractions. In our case, if we take arbitrary filling fractions, the symmetries (3.17) are lost and thus the determination of the spectral curve requires to solve fixed filling fractions conditions:

$$\begin{aligned} \varvec{\epsilon }_k=\oint _{{\mathcal {A}}_k} \mathrm{d}\mu _\infty (x) \text { for all } -r\le k\le r \end{aligned}$$

with \({\mathcal {A}}_k\) a closed contour circling the interval \(\big [\tan \left( \frac{\pi k}{2r+1}-\frac{\pi \epsilon }{2(2r+1)}\right) ,\tan \left( \frac{\pi k}{2r+1}+\frac{\pi \epsilon }{2(2r+1)}\right) \big ]\). Unfortunately, solving analytically these conditions remains an open challenge and therefore the non-perturbative part of (3.31) remains mostly out of reach for theoretical computations. Another possibility to bypass this difficulty may be to rewrite derivatives relatively to filling fractions of \(F^{\{2k\},(l)}\) as some integrals over \({\mathcal {B}}\)-cycles of the correlation functions \(\omega ^{(k+1)}_l\) using results of [16]. In this case, there appears to have no need for using arbitrary filling fractions but the strategy requires the computations of all correlation functions. However, this approach seems more convenient than dealing with arbitrary filling fractions. In particular, the case \(m=0\) in (3.31) simply gives:

$$\begin{aligned} \text {Term }m=0 \text { in }(3.31)\,:\, \Theta _{-n\varvec{\epsilon }^\star }\left( {\mathbf {0}}\big |F^{\{-2\},(2)}_{\varvec{\epsilon }^\star }\right) =\Theta _{-n\varvec{\epsilon }^\star }\left( {\mathbf {0}}\big |\,-2i\pi \varvec{\tau }_{|\varvec{\epsilon }=\varvec{\epsilon }^*}\right) \qquad \end{aligned}$$
(3.32)

where \(\varvec{\tau }\) is the Riemann matrix of periods of the spectral curve. In the case \(\varvec{\epsilon }=\varvec{\epsilon }^*=\left( \frac{1}{2r+1},\ldots ,\frac{1}{2r+1}\right) ^t\) the rotational symmetry of the problem may allow the computation of the Riemann matrix of periods even if the expression of the spectral curve is rather complicated (see (3.21)). Since we will have no use of this order, we let it as an open problem.

3.4 Normalization and limit \(\epsilon \rightarrow 0\)

Finally, we also note that even the perturbative part is challenging in the multi-cut regime. Indeed, proposition 3.2 determines the perturbative part up to some constants. The standard way to compute them is to choose the parameters (in our case \(\epsilon \)) in such a way that the initial integral is explicitly connected to a known case. In the one-cut regime, we connected the integral to a Selberg integral after a proper rescaling. In the multi-cut case, we may also compute the limit of \(Z_n({\mathcal {I}}_r)\) when \(\epsilon \rightarrow 0\) since the integrals decouple in this regime. We obtain the following theorem:

Theorem 3.2

(Limit of \(Z_n({\mathcal {I}}_r)\) when \(\epsilon \rightarrow 0\))

For a given \(r\ge 0\) and \(n\ge 1\), we have:

$$\begin{aligned} Z_{n}({\mathcal {I}}_r)&\overset{\epsilon \rightarrow 0}{\sim }(2\pi \epsilon )^{(2r+1)\lfloor \frac{n}{2r+1}\rfloor ^2+\left( 2\lfloor \frac{n}{2r+1}\rfloor +1\right) m_n}\frac{(2\pi )^{-n}\left( {\mathop {\prod }\nolimits _{k=1}^{{\lfloor \frac{n}{2r+1}\rfloor -1}}}k!\right) ^{4(2r+1)} \left( \left( \lfloor \frac{n}{2r+1}\rfloor \right) !\right) ^{4m_n}}{\left( {\mathop {\prod }\nolimits _{k=1}^{2\lfloor \frac{n}{2r+1}\rfloor -1}}k!\right) ^{2r+1}\left( \left( 2\lfloor \frac{n}{2r+1}\rfloor \right) !\right) ^{m_n} \left( \left( 2\lfloor \frac{n}{2r+1}\rfloor +1\right) !\right) ^{m_n}} \nonumber \\&\overset{\epsilon \rightarrow 0}{\sim }(2\pi \epsilon )^{\frac{n^2-m_n^2}{2r+1}+m_n}\frac{(2\pi )^{-n} \left( {\mathop {\prod }\nolimits _{k=1}^{{\frac{n-m_n}{2r+1}-1}}}k!\right) ^{4(2r+1)} \left( \left( \frac{n-m_n}{2r+1}\right) !\right) ^{4m_n}}{\left( {\mathop {\prod }\nolimits _{k=1}^{\frac{2n-2m_n}{2r+1}-1}}k!\right) ^{2r+1}\left( \left( \frac{2n-2m_n}{2r+1}\right) !\right) ^{m_n} \left( \left( \frac{2n-2m_n}{2r+1}+1\right) !\right) ^{m_n}} \end{aligned}$$
(3.33)

This formula may easily be verified numerically by computing the leading order in the \(\epsilon \rightarrow 0\) series expansion of the Toeplitz determinants given by (3.2) for low values of n (up to 25) and low values of r (up to 10). Moreover, the formula reduces for \(r=0\) to (A.13). When \(r>1\), we observe as expected that the result depends both on the value of n but also on the value of \(n \mod (2r+1)\). This is of course in agreement with the asymptotic expansion given by (3.31).

Proof

Let us first decompose the partition function as a sum over all possible partitions of the eigenvalues in the \(2r+1\) intervals:

$$\begin{aligned} Z_n({\mathcal {I}}_r)= & {} \frac{1}{(2\pi )^n n!}\sum _{{\mathbf {n}}\in N_{n,r}}\left( {\begin{array}{c}n\\ n_{-r},\ldots , n_r\end{array}}\right) \int _{[\alpha _{-r}^{(r)},\beta _{-r}^{(r)}]} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _{n_{-r}}\ldots \int _{[\alpha _{r}^{(r)},\beta _{r}^{(r)}]}d{\theta _{n-n_r+1}}\ldots d{\theta _n} \\&\prod _{1\le p<q\le n}\left| e^{i\theta _p}-e^{i\theta _q}\right| ^2\\= & {} \frac{1}{(2\pi )^n n!}\sum _{{\mathbf {n}}\in N_{n,r}}\left( {\begin{array}{c}n\\ n_{-r},\ldots , n_r\end{array}}\right) (2\pi )^{\mathop {\sum }\nolimits _{k=-r}^{r} n_k}(n_{-r})!\,\ldots (n_r)!\, Z_{n_{-r},\ldots ,n_r}({\mathcal {I}}_r)\\= & {} \sum _{{\mathbf {n}}\in N_{n,r}}Z_{n_{-r},\ldots ,n_r}({\mathcal {I}}_r) \end{aligned}$$

where \(N_{n,r}=\{(i_{-r},\ldots ,i_r)\in {\mathbb {N}}^{2r+1}\,/\, \mathop {\sum }\nolimits _{k=-r}^{r}i_k=n\}\), \(\left( {\begin{array}{c}n\\ n_{-r},\ldots , n_r\end{array}}\right) \) stands for the multinomial coefficient and

$$\begin{aligned} Z_{n_{-r},\ldots ,n_r}({\mathcal {I}}_r)= & {} \frac{1}{(2\pi )^{\mathop {\sum }\nolimits _{k=-r}^{r} n_k}\mathop {\prod }\nolimits _{k=-r}^{r} (n_k)!} \,\int _{[\alpha _{-r}^{(r)},\beta _{-r}^{(r)}]} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _{n_{-r}}\ldots \int _{[\alpha _{r}^{(r)},\beta _{r}^{(r)}]}d{\theta _{n-n_r+1}}\ldots d{\theta _n} \\&\prod _{1\le p<q\le n}\left| e^{i\theta _p}-e^{i\theta _q}\right| ^2 \end{aligned}$$

stands for the partition function with filling fractions fixed to \({\mathbf {n}}=(n_{-r},\ldots ,n_r)^t\). Note that the normalization prefactor \(\frac{1}{(2\pi )^n n!}\) in the definition of the partition functions exactly cancels the combinatorial coefficients appearing in the decomposition.

In the limit \(\epsilon \rightarrow 0\), interactions between eigenvalues in two different intervals reduce to interactions between the center of the two intervals. Thus, the integrals decouple and \(Z_{n_{-r},\ldots ,n_r}({\mathcal {I}}_r)\) reduces to a product of one-cut partition functions whose values are given by (A.13) (with \(a_0=\tan \frac{\pi \epsilon }{2(2r+1)}\)):

$$\begin{aligned} Z_{n_{-r},\ldots ,n_r}({\mathcal {I}}_r)\overset{\epsilon \rightarrow 0}{\sim } \mathop {\prod }\nolimits _{k=-r}^{r} Z_{n_k}(\epsilon )=(2\pi \epsilon )^{\mathop {\sum }\nolimits _{k=-r}^{r} n_k^2}(2\pi )^{-n} \frac{\mathop {\prod }\nolimits _{k=-r}^{r}\left( {\mathop {\prod }_{j=1}^{{n_k}}} j!\right) ^4}{\mathop {\prod }\nolimits _{k=-r}^{r} {\mathop {\prod }\limits _{j=1}^{2n_k-1}} j!}\nonumber \\ \end{aligned}$$
(3.34)

We now observe that the exponent in \(\epsilon \) is minimal when the sum \(\mathop {\sum }\nolimits _{k=-r}^{r} n_k^2\) is minimal. Since the sum is constrained to \(\mathop {\sum }\nolimits _{k=-r}^{r}n_k=n\) fixed, it is well known that the previous sum is minimal when all \((n_k)_{k\in \llbracket -r,r\rrbracket }\) are equal. This is also coherent with the fact that the perturbative analysis leads to identical filling fractions (see Proposition 3.1). Indeed, let us define \(f({\mathbf {n}})=\mathop {\sum }\nolimits _{k=-r}^{r} n_k^2\), then

$$\begin{aligned}&f(n_{-r},\ldots ,n_{i_0}+1,\ldots ,n_{j_0}-1,\ldots ,n_r)-f(n_{-r},\ldots ,n_{i_0},\ldots ,n_{j_0},\ldots ,n_r)\\&\quad =2(n_{i_0}-n_{j_0}+1)\left\{ \begin{array}{ll} <0 \text { if } n_{i_0}\le n_{j_0}-2\\ =0\text { if } n_{i_0}=n_{j_0}-1\\ >0 \text { if } n_{i_0}\ge n_{j_0} \end{array} \right. \end{aligned}$$

Therefore, to minimize f on \(N_{n,r}\), one needs to balance the vector \({\mathbf {n}}\) with differences between the components at most 1. This proves that if \(m_n=n\mod (2r+1)\ne 0\) the sum is minimal when we spread the additional \(m_n\) eigenvalues in \(m_n\) distinct intervals, i.e., we put \(\lfloor \frac{n}{2r+1}\rfloor \) eigenvalues in \(2r+1-m_n\) intervals and \(\lfloor \frac{n}{2r+1}\rfloor +1\) eigenvalues in the remaining \(m_n\) intervals and that all other configurations provide a strictly greater value of the exponent. Since the sum contains a finite number of terms (n and r are fixed so \(N_{n,r}\) is a finite set) and because each of them is strictly positive (thus the sum of terms with the same exponent in \(\epsilon \) cannot give of a strictly lower order in \(\epsilon \)), we obtain from (3.34) the result presented in Theorem 3.2. \(\square \)

3.5 Final result and conjecture for \(F^{\{0\}}\)

Note that the knowledge of the equivalent of \(Z_n({\mathcal {I}}_r)\) when \(\epsilon \rightarrow 0\) given by Theorem 3.2 is not sufficient to determine the constants \(\left( f^{\{k\}}\right) _{k\ge -2}\) of Proposition 3.2. Indeed, in order to obtain them, one needs to exchange the limit \(\epsilon \rightarrow 0\) and the limit \(n\rightarrow \infty \) and possible entropy contributions may arise. It is also unclear if arguments similar to the one-cut case may be adapted to this context. The only known cases are \(F^{\{-2\}}\) and \(F^{\{-1\}}\) (given in section 1.4 of [18] with the important reminder that we included a normalization factor \(\frac{2^{n^2}}{(2\pi )^n n!}\) in the definition of the partition function that is absent in [18]) . We get:

$$\begin{aligned} F^{\{-2\}}= & {} \ln 2-F^{(0)} \,\,\,\Leftrightarrow \,\,\, f_{-2}=\ln 2 \\ F^{\{-1\}}= & {} 0 \,\,\,\Leftrightarrow \,\,\, f_{-1}=0 \end{aligned}$$

Remark 3.2

Note that the constants \(f_{-2}=\ln 2\) and \(f_{-1}=0\) differ from those given in [18] because our choice of normalization of the partition function is different from the one used in [18]. Indeed, our partition function is normalized with an additional factor \(\frac{2^{n^2}}{(2\pi )^n n!}\) compared to the one used in [18]. In particular, the numerator \(2^{n^2}\) provides an additional factor \(n^2\ln 2\) in the asymptotic expansion of \(\ln Z_n({\mathcal {I}}_r)\), thus giving \(f_{-2}=\ln 2\). Similarly, the denominator \((2\pi )^nn!\overset{n\rightarrow \infty }{\sim }(2\pi )^nn^ne^{-n}\sqrt{2\pi n}\) provides an additional factor \(n\ln n\) already included in the general form of the asymptotic expansion of \(\ln Z_n({\mathcal {I}}_r)\) given in Proposition 3.1 as well as an additional factor \(-n(\ln (2\pi )-1)\) in the asymptotic expansion of \(\ln Z_n({\mathcal {I}}_r)\) that precisely cancels the one obtained in [18], hence leading to \(f_{-1}=0\).

Thus, we end up with:

Proposition 3.3

(First orders of the asymptotic expansion of \(Z_n({\mathcal {I}}_r)\)) For a given \(r\ge 0\), we have:

$$\begin{aligned} \ln (Z_n({\mathcal {I}}_r))=\frac{n^2}{2r+1}\ln \left( \sin \left( \frac{\pi \epsilon }{2}\right) \right) -\frac{2r+1}{4}\ln n+O(1) \end{aligned}$$

As explained above, the next orders of the large n expansion, and in particular, the O(1) term, exhibit a far more complex structure than in the one-cut case and depend on the remainder of the Euclidean division of n by \(2r+1\). In fact, as discussed in [18], for a given remainder m such that \(0\le m \le 2r\), the sequence \(\left( Z_{(2r+1)p+m}\right) _{p\in {\mathbb {N}}}\) admit a large p expansion taking a similar form as the one-cut case with n replaced by p (the precise form being given in Theorem 2.2) and with coefficients depending on the remainder m. However, because we could compute numerically the Toeplitz determinants up to sufficiently large n, we can propose some conjectures for the coefficient \(F^{\{0\}}\) depending on the value of the remainder m.

Conjecture 3.1

(Conjecture for \(F^{\{0\}}\)) We conjecture the following:

  • In the case \(m=0\), \(F^{\{0\}}\) has a dependence in \(\epsilon \) given by \(-\frac{2r+1}{4}\ln \left( \cos \left( \frac{\pi \epsilon }{2}\right) \right) +C_{0}\) where \(C_{0}\) is a constant. This generalizes the one-cut case that would correspond to \(r=0\).

  • The cases \(m=j\) and \(m=2r-1-j\) for some \(1\le j\le r\) provides the same \(F^{\{0\}}\) coefficient. However, the dependence in \(\epsilon \) is more involved than for \(m=0\) and we conjecture that it is given by:

    $$\begin{aligned} F^{\{0\}}(j)= A_{j}\ln \left( \cos \left( \frac{\pi \epsilon }{2}\right) \right) +B_{j} \ln \left( \tan \left( \frac{\pi \epsilon }{2}\right) \right) +C_j \end{aligned}$$

    with some nonzero constants \((A_j,B_j,C_j)\) depending on the remainder.

  • The constants \((C_j)_{1\le j\le r}\) and \(C_{0}\) are the same and correspond to the normalization issue of the partition function.

The case \(m=0\) exhibits an additional symmetry since the number of eigenvalues is precisely a multiple of the number of intervals so that they can spread evenly in all intervals. In particular, when evaluating derivatives of the free energies at \(\varvec{\epsilon }^\star =\frac{1}{2r+1}{\mathbf {1}}\) cancellations are more likely to happen, thus explaining why the coefficient in front of \(\ln \left( \tan \left( \frac{\pi \epsilon }{2}\right) \right) \) vanishes only in this case. We illustrate our conjecture with numeric simulations (Fig. 6):

Fig. 6
figure 6

Representation of \(\ln (Z_n({\mathcal {I}}_r)) -\frac{1}{2r+1}\ln \left( \sin \left( \frac{\pi \epsilon }{2}\right) \right) +\frac{2r+1}{4}\ln n\) for values of n ranging from 2 to 70 and classified by the remainder m (from left to right: \(m=0\), \(m\in \{j,2r-1-j\}\), with j from 1 to r) of the Euclidean division of n by \(2r+1\). The black curves correspond to the best numerical matches of an affine combination of \(\ln \left( \cos \left( \frac{\pi \epsilon }{2}\right) \right) \) and \(\ln \left( \tan \left( \frac{\pi \epsilon }{2}\right) \right) \) as explained more specifically below

More precisely, the black curves correspond to the best matches with curves of the form:

$$\begin{aligned} f(\epsilon )=\alpha \ln \left( \cos \left( \frac{\pi \epsilon }{2}\right) \right) +\beta \ln \left( \tan \left( \frac{\pi \epsilon }{2}\right) \right) +\gamma \end{aligned}$$
(3.35)

with rational coefficients \((\alpha ,\beta ,\gamma )\) of the form \(\frac{i}{256}\) with \(i\in {\mathbb {Z}}\). Note that we chose to express the coefficients as rational numbers with a specific denominator, but we have no evidence that the coefficients are indeed rational or that the denominator is a power of 2. However, it seems that this particular choice provides very accurate results that are not numerically improved by adding multiplicative powers of 3 or 5 in the denominators or increasing the power of 2. Numerically, we obtain the best matches for the values:

  • Case \(r=1\):

    $$\begin{aligned} m=0&: (\alpha ,\beta ,\gamma )= \left( -\frac{3}{4},0,-\frac{63}{128}\right) \\ m\in \{1,2\}&: (\alpha ,\beta ,\gamma )= \left( -\frac{11}{128},\frac{85}{128},-\frac{63}{128}\right) \end{aligned}$$
  • Case \(r=2\):

    $$\begin{aligned} m=0&: (\alpha ,\beta ,\gamma )= \left( -\frac{5}{4},0,-\frac{3}{16}\right) \\ m\in \{1,4\}&: (\alpha ,\beta ,\gamma )= \left( -\frac{15}{32},\frac{203}{256},-\frac{3}{16}\right) \\ m\in \{2,3\}&: (\alpha ,\beta ,\gamma )= \left( -\frac{1}{16},\frac{6}{5},-\frac{3}{16}\right) \end{aligned}$$
  • Case \(r=3\):

    $$\begin{aligned} m=0&: (\alpha ,\beta ,\gamma )=\left( -\frac{7}{4},0,\frac{85}{256}\right) \\ m\in \{1,6\}&:(\alpha ,\beta ,\gamma )=\left( -\frac{115}{128},\frac{109}{128},\frac{85}{256}\right) \\ m\in \{2,5\}&:(\alpha ,\beta ,\gamma )=\left( -\frac{43}{128},\frac{183}{128},\frac{85}{256}\right) \\ m\in \{3,4\}&:(\alpha ,\beta ,\gamma )=\left( -\frac{7}{128},\frac{219}{128},\frac{85}{256}\right) \end{aligned}$$
  • Case \(r=4\):

    $$\begin{aligned} m=0&: (\alpha ,\beta ,\gamma )=\left( -\frac{9}{4},0,\frac{255}{256}\right) \\ m\in \{1,8\}&:(\alpha ,\beta ,\gamma )=\left( -\frac{355}{256},\frac{227}{256},\frac{255}{256}\right) \\ m\in \{2,7\}&:(\alpha ,\beta ,\gamma )=\left( -\frac{184}{256},\frac{398}{256},\frac{255}{256}\right) \\ m\in \{3,6\}&:(\alpha ,\beta ,\gamma )=\left( -\frac{72}{256},\frac{510}{256},\frac{255}{256}\right) \\ m\in \{4,5\}&:(\alpha ,\beta ,\gamma )=\left( -\frac{17}{256},\frac{565}{256},\frac{255}{256}\right) \end{aligned}$$

3.6 Even number of intervals

The method developed in the last section can be adapted in the case of an even number of intervals. However, we need to be careful since in order to apply \(\theta \mapsto \tan \frac{\theta }{2}\) we need to avoid the angles \(\theta =\pm \pi \). Therefore, we use the invariance \(\varvec{\theta }\rightarrow \varvec{\theta }+\text {Cste }\,{\mathbf {1}}\) of the integral (3.36) to shift the intervals so that they do not contain \(\pm \pi \). We define for \(s\ge 1\):

$$\begin{aligned} \alpha _k^{(s)}= & {} \frac{\pi (k-\frac{1}{2})}{s}- \frac{\pi \epsilon }{2s} \text { , } \beta _k^{(s)}=\frac{\pi (k-\frac{1}{2})}{s}+ \frac{\pi \epsilon }{2s} \text { , } \gamma _k^{(s)}\\= & {} \frac{\pi (k-\frac{1}{2})}{s} \,\,\text { , } \,\, \forall \, -(s-1)\le k\le s\\ {\mathcal {I}}_s= & {} \mathop {\bigcup }\limits _{k=-(s-1)}^{s}\left[ \alpha _k^{(s)},\beta _k^{(s)}\right] \\ {\mathcal {J}}_s= & {} \mathop {\bigcup }\limits _{k=-(s-1)}^{s}\left[ \tan \left( \frac{\alpha _k^{(s)}}{2}\right) ,\tan \left( \frac{\beta _k^{(s)}}{2}\right) \right] \\ {\mathcal {T}}_s= & {} \{e^{it}\,,\,\, t\in {\mathcal {I}}_s\} \end{aligned}$$

and the integral:

$$\begin{aligned} Z_n({\mathcal {I}}_s)=\frac{1}{(2\pi )^n n!} \int _{\left( {\mathcal {I}}_s\right) ^n} \mathrm{d}\theta _1\ldots \mathrm{d}\theta _n \prod _{1\le p<q\le n}\left| e^{i\theta _p}-e^{i\theta _q}\right| ^2 \end{aligned}$$
(3.36)

which is also equal to \(\det T_n^{(s)}\) with \(\left( T_n^{(s)}\right) _{p,q}=t_{p-q}\) the \(n\times n\) Toeplitz matrix given by:

$$\begin{aligned} t_0= & {} \frac{|{\mathcal {I}}_s|}{2\pi }=\epsilon \\ t_k= & {} \epsilon \, \sin _c\left( \frac{k \pi \epsilon }{2s}\right) \delta _{k \, \equiv \, 0\,[2s]} \,\text { for }k\ne 0 \end{aligned}$$

Note again that the Toeplitz matrix is mostly empty since only bands with indexes multiple of 2s are nonzero. Moreover, it is also equal to a Hermitian integral:

$$\begin{aligned} Z_n({\mathcal {I}}_s)=\frac{2^{n^2}}{(2\pi )^n n!}\int _{\left( {\mathcal {J}}_s\right) ^n} \mathrm{d}t_1\ldots \mathrm{d}t_n \Delta (t_1,\ldots ,t_n)^2e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln (1+t_k^2)} \end{aligned}$$
(3.37)

or a complex integral:

$$\begin{aligned} Z_n({\mathcal {I}}_s)=(-1)^{\frac{n(n+1)}{2}}i^n \int _{\left( {\mathcal {T}}_s\right) ^n} \mathrm{d}u_1\ldots d u_n\, \Delta (u_1,\ldots ,u_n)^2 e^{-n\mathop {\sum }\nolimits _{k=1}^{n} \ln u_k} \end{aligned}$$
(3.38)

The situation can be illustrated as follows (Fig. 7).

The method developed in the case of an odd number of intervals can be easily adapted to this case. In particular, we can prove the following:

Fig. 7
figure 7

Illustration (in green) of the set \({\mathcal {T}}_{s=3}\) for \(\epsilon =\frac{1}{5}\)

Theorem 3.3

(Results for an even number of symmetric arc intervals) In the case of an even number of symmetric arc intervals (3.36), we have:

  1. 1.

    The spectral curve attached to integral (3.36) is given by:

    $$\begin{aligned} y^2(x)= & {} \frac{\mathop {\prod }\nolimits _{k=1}^{s-1} \cos ^2\left( \frac{\pi \left( k-\frac{1}{2}\right) }{2s}\right) }{\mathop {\prod }\nolimits _{k=-(s-1)}^{s} \cos ^2\left( \frac{\pi \left( k-\frac{1}{2}\right) }{2s}-\frac{\pi \epsilon }{4s}\right) } \frac{ x^2\mathop {\prod }\nolimits _{k=1}^{s-1}\left( x^2-\tan ^2\left( \frac{\pi k}{2s}\right) \right) ^2}{(1+x^2)^2\mathop {\prod }\nolimits _{k=-(s-1)}^{s}\left( x^2-\tan ^2\left( \frac{\pi \left( k-\frac{1}{2}\right) }{2s}+\frac{\pi \epsilon }{4s}\right) \right) } \end{aligned}$$
  2. 2.

    The corresponding limiting eigenvalues distribution is given by:

    $$\begin{aligned} \mathrm{d}\mu _\infty (x)=\frac{\mathop {\prod }\nolimits _{k=1}^{s-1} \cos \left( \frac{\pi \left( k-\frac{1}{2}\right) }{2s}\right) }{\mathop {\prod }\nolimits _{k=-(s-1)}^{s} \cos \left( \frac{\pi \left( k-\frac{1}{2}\right) }{2s}-\frac{\pi \epsilon }{4s}\right) } \frac{ |x|\mathop {\prod }\nolimits _{k=1}^{s-1}\left| x^2-\tan ^2\left( \frac{\pi k}{2s}\right) \right| }{\pi (1+x^2)\mathop {\prod }\nolimits _{k=-(s-1)}^{s}\sqrt{\left| x^2-\tan ^2\left( \frac{\pi \left( k-\frac{1}{2}\right) }{2s}+\frac{\pi \epsilon }{4s}\right) \right| }}\mathbb {1}_{{\mathcal {J}}_s}(x)\mathrm{d}x \end{aligned}$$
  3. 3.

    The filling fractions are the same in all the 2s intervals:

    $$\begin{aligned} \forall \, k\in \llbracket -(s-1),s\rrbracket \,:\, \varvec{\epsilon }_{k+s}\overset{\text {def}}{=}\int _{\tan \left( \frac{\pi \left( k-\frac{1}{2}\right) }{2s}-\frac{\pi \epsilon }{4s}\right) }^{\tan \left( \frac{\pi \left( k-\frac{1}{2}\right) }{2s}+\frac{\pi \epsilon }{4s}\right) } \mathrm{d}\mu _\infty (x)=\frac{1}{2s} \end{aligned}$$
  4. 4.

    The function \(\ln Z_n({\mathcal {I}}_s)\) admits a large n expansion given by Theorem 3.1 with partial reconstruction by the topological recursion given by proposition 3.2. In particular, we have for \(s\ge 1\):

    $$\begin{aligned} \ln (Z_n({\mathcal {I}}_s))=\frac{1}{2s}\ln \left( \sin \left( \frac{\pi \epsilon }{2}\right) \right) -\frac{2s}{4}\ln n+O(1) \end{aligned}$$

    The O(1) order depends on the remainder \(m_n\) of n modulo 2s. Numerically, its dependence in \(\epsilon \) follows the conjecture 3.1 (with \(2r+1\) replaced by 2s and the special case still given by \(m_n=0\)).

  5. 5.

    The limit \(\epsilon \rightarrow 0\) of \(Z_n({\mathcal {I}}_s)\) with \(n=m_n\mod 2s\) is given by:

    $$\begin{aligned} Z_{n}({\mathcal {I}}_s)\overset{\epsilon \rightarrow 0}{\sim }(2\pi \epsilon )^{2s\lfloor \frac{n}{2s}\rfloor ^2+\left( 2\lfloor \frac{n}{2s}\rfloor +1\right) m_n}\frac{(2\pi )^{-n}\left( {\mathop {\prod }\nolimits _{k=1}^{\lfloor \frac{n}{2s}\rfloor -1}}k!\right) ^{8s} \left( \left( \lfloor \frac{n}{2s}\rfloor \right) !\right) ^{4m_n}}{\left( {\mathop {\prod }\nolimits _{k=1}^{2\lfloor \frac{n}{2s}\rfloor -1}}k!\right) ^{2s}\left( \left( 2\lfloor \frac{n}{2s}\rfloor \right) !\right) ^{m_n} \left( \left( 2\lfloor \frac{n}{2s}\rfloor +1\right) !\right) ^{m_n}}\nonumber \\ \end{aligned}$$
    (3.39)

4 Conclusion and outlooks

In this article, we proposed a rigorous mathematical derivation of the asymptotic expansions of Toeplitz determinants with symbols given by \(f=\mathbb {1}_{{\mathcal {T}}(\alpha ,\beta )}\) where \({\mathcal {T}}(\alpha ,\beta )=\{e^{it},t\in [\alpha ,\beta ]\}\). This generalizes Widom’s result and completes the approach developed in [21]. We also provided numerical simulations up to \(o\left( \frac{1}{n^4}\right) \) to illustrate these results. For symbols \(f=\mathbb {1}_{{\mathcal {T}}_d}\) with \({\mathcal {T}}_d={\mathop {\bigcup }\nolimits _{k=1}^{d}}[\alpha _k,\beta _k]\) and \(d\ge 2\), the situation is more complex, but we were able to provide a rigorous derivation of the large n asymptotic of the corresponding Toeplitz determinants when the arc intervals exhibit a discrete rotational symmetry on the unit circle. We also provided the first terms of the large n expansion up to O(1), and we proposed a conjecture for the O(1) term supported by numeric simulations. Moreover, the results presented in this article raise the following challenges:

  • Prove conjecture 3.1 regarding the O(1) term in the symmetric multi-cut case. In particular, it would be interesting to find a way to obtain the normalization constants. This requires to connect the Toeplitz integral to a known integral for at least one value of the parameter \(\epsilon \) (very likely \(\epsilon \rightarrow 0\)).

  • In the multi-cut cases without discrete symmetry, it would be interesting to prove that, for generic choices of the edges, the corresponding spectral curves are regular and that the hypothesis required to prove the large n asymptotic of Theorem 3.1 are verified. Contrary to the case with a discrete rotational symmetry, it seems unlikely that we obtain an explicit formula for the spectral curves. However, it may be possible to obtain sufficient information on the location of the zeros of y(x) in order to prove that the spectral curves are regular.

  • The method developed in this article could also be tried for more general Toeplitz determinants. In particular, the transition from a symbol supported on a compact set of \(\{e^{it},t\in (-\pi ,\pi )\}\) to a strictly positive symbol on the unit circle deserves some analysis. Indeed, in the case of a strictly positive symbol on the unit circle, Szegö’s theorem implies that \(\frac{1}{n}\ln \det T_n\) admits a non-trivial limit (given by \(\frac{1}{2\pi }\int _0^{2\pi } \ln (f(e^{i\theta }))\mathrm{d}\theta \)) so that the large n expansion proposed in Theorem 3.1 requires some adaptations. However, it may happen that the sub-leading corrections may still be given by Theorem 3.1 and in particular that sub-leading corrections may be reconstructed from the topological recursion. This conjecture is supported by the fact that Toeplitz determinants can also be reformulated as Fredholm determinants (see [8]) that are known to be deeply related with the topological recursion.