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Large deviations for the \({\hbox {Sine}}_\beta \) and \(\hbox {Sch}_\tau \) processes

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Abstract

The \({\hbox {Sine}}_\beta \) process is the bulk point process limit of the Gaussian \(\beta \)-ensemble. For \(\beta = 1, 2\) and 4 this process gives the limit of the GOE, GUE and GSE random matrix models. The \(\hbox {Sch}_\tau \) process is obtained similarly as the bulk scaling limit of the spectrum of certain discrete one-dimensional random Schrödinger operators. Both point processes have asymptotically constant average density, in our chosen normalization one expects close to \(\tfrac{1}{2\pi } \lambda \) points in a large interval of length \(\lambda \). We prove large deviation principles for the average densities of the processes, identifying the rate function in both cases. Our approach is based on the representation of the counting functions using coupled systems of stochastic differential equations. Our techniques work for the full range of parameter values. The results are novel even in the classical \(\beta = 1, 2\) and 4 cases for the \({\hbox {Sine}}_\beta \) process. They are consistent with the existing rigorous results on large gap probabilities and confirm the physical predictions made using log-gas arguments.

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Acknowledgments

B. Valkó was partially supported by the National Science Foundation CAREER award DMS-1053280. The authors would like to thank P. J. Forrester for helpful comments and additional references.

Author information

Correspondence to Benedek Valkó.

Appendices

Appendix A: All about \(\mathcal {I}\)

In this section we prove the needed estimates about the function \({\mathcal {I}}\).

Proposition 16

The function \({\mathcal {I}}(x)\) is strictly convex and continuous on \((0,\infty )\). It has an absolute minimum at \(x=1\) where it is equal to 0.

Proof

\(K\) is strictly increasing on \((-\infty ,1)\) which shows that \({\mathcal {I}}(x)\) is well-defined (and differentiable) on \((0,\infty )\). By differentiating (11) and using the identities

$$\begin{aligned} K'(x)=\frac{E(x)-(1-x)K(x)}{2(1-x)x}, \qquad E'(x)=\frac{E(x)-K(x)}{2x} \end{aligned}$$

we can compute that

$$\begin{aligned} {\mathcal {I}}'(x)=\frac{1}{x}{\mathcal {I}}(x)- \frac{1}{8 x}K^{-1}\left( \frac{\pi }{2x}\right) , \end{aligned}$$
(85)

and

$$\begin{aligned} {\mathcal {I}}''(x)=\frac{\pi }{16 x^3} \frac{1}{K'(K^{-1}(\frac{\pi }{2x}))}. \end{aligned}$$
(86)

Observe that \(K'(y)> 0\) for \(y<1\) which gives \({\mathcal {I}}''(x)> 0\) for \(x>0\) and the strict convexity of \({\mathcal {I}}\).

Using \(K(0)=E(0)=\tfrac{\pi }{2}\) we get \({\mathcal {I}}(1)={\mathcal {I}}'(1)=0\) which (by the strict convexity) proves the second half of the proposition. \(\square \)

Proposition 17

We have \(\lim \nolimits _{x\rightarrow 0^{+}}{\mathcal {I}}(x)=\tfrac{1}{8}\) and \( \lim \nolimits _{x\rightarrow 0^{+}}{\mathcal {I}}'(x)=-\tfrac{1}{2\pi }\). There is a constant \(c_1>0\) so that

$$\begin{aligned}&{\mathcal {I}}(x)\ge c_1 (x-1)^2, \qquad \mathrm{for}\, \mathrm{all}\,\, x, \mathrm{and}\end{aligned}$$
(87)
$$\begin{aligned}&\frac{{\mathcal {H}}(-x)}{\sqrt{x} \log x}, \, \frac{{\mathcal {I}}(x)}{x^2\log ^2 x} \mathrm{and }\frac{-K^{-1}(1/x)}{x^2 \log ^2 x} \mathrm{are \,\, bounded \,\, away \,\, from \,\, 0 \,\, and} \, \infty \,\, \mathrm{for }\,\, x>2.\nonumber \\ \end{aligned}$$
(88)

Proof

The following asymptotics can be readily derived from the definitions of elliptic integrals (or by the existing more sophisticated expansions c.f. [6, 15]). There is a constant \(c>0\) so that

$$\begin{aligned}&\left| K(-a)-\frac{1}{2\sqrt{a}}\log (16 a)\right| \le \frac{c}{a^{3/2}}\log (a), \nonumber \\&\quad \left| E(-a)-\sqrt{a} \right| \le \frac{c}{a^{1/2}}\log (a), \quad \hbox {for } a>2. \end{aligned}$$
(89)

From this it is easy to check that

$$\begin{aligned}&\lim \limits _{x\rightarrow \infty } \frac{{\mathcal {H}}(-x)}{\sqrt{x} \log x}=\frac{1}{2}, \qquad \lim \limits _{x\rightarrow \infty } \frac{-K^{-1}(1/x)}{x^2\log ^2 x}=1, \quad \hbox {and} \nonumber \\&\qquad \lim \limits _{x\rightarrow \infty } \frac{{\mathcal {I}}(x)}{ x^2 \log ^2 x}=\frac{1}{2\pi ^2}. \end{aligned}$$
(90)

This gives (88). Note, that together with (85) this also gives

$$\begin{aligned} \lim \limits _{x\rightarrow \infty } \frac{{\mathcal {I}}'(x)}{x\log ^2 x}=\frac{1}{\pi ^2}. \end{aligned}$$
(91)

Using the functional identities

$$\begin{aligned} E(z)=\sqrt{1-z} E\left( \frac{z}{z-1}\right) , \qquad K(z)=\frac{1}{\sqrt{1-z}} K\left( \frac{z}{z-1}\right) , \qquad z\in (0,1) \end{aligned}$$

the asymptotics of (89) can be transformed into

$$\begin{aligned} K(a)\sim -\tfrac{1}{2}\log (1-a), \qquad E(a)\sim 1, \qquad \hbox { as }a\rightarrow 1^{-1} \end{aligned}$$

with explicit error bounds for \(2/3<a<1\). From this we can obtain \(\lim \nolimits _{x\rightarrow 0^{+}} {\mathcal {I}}(x)=\tfrac{1}{8}\) and \( \lim \nolimits _{x\rightarrow 0^{+}}{\mathcal {I}}'(x)=-\tfrac{1}{2\pi }\). Using (88) with the continuity of \({\mathcal {I}}\) and the fact that \({\mathcal {I}}(1)=0\) is an absolute minimum with \({\mathcal {I}}''(1)>0\) gives (87). \(\square \)

The following two lemmas help to consolidate error terms that appear in the proofs Theorems 1 and 2.

Lemma 18

There exists an absolute constant \(c\) such that for any \(t, q>0\) we have

$$\begin{aligned} |{\mathcal {H}}(a)|+|a|t/2 \le c(t+1)( \mathcal {I}(q)+1) \end{aligned}$$
(92)

where \(a=a(q)=K^{-1}(\pi /(2q))\).

Proof

Using (89) with the definition (11) we get that there is a constant \(c_2\) so that

$$\begin{aligned} c_{2}^{-1} a(q)\le {\mathcal {I}}(q)\le c_2 a(q), \qquad \hbox {if }q>2, \end{aligned}$$
(93)

and the same bounds also give

$$\begin{aligned} |{\mathcal {H}}(a(q))|\le c_3 \sqrt{|a(q)|} \log |a(q)| \end{aligned}$$
(94)

for some constant \(c_3\) in the same region. This shows the existence of a constant \(A\) with

$$\begin{aligned} |{\mathcal {H}}(a)| \le A \, \mathcal {I}(q), \quad \text { and } \qquad a(q) \le A \mathcal {I}(q), \qquad \hbox {for }q>2. \end{aligned}$$

Since for \(0<q<2\) both \(a(q)\) and \({\mathcal {H}}(a(q))\) are bounded the lemma follows.

Lemma 19

For any \(0\le \varepsilon < 1/2\) there exists an absolute constant \(c\), so that

$$\begin{aligned} \mathcal {I}(x+\varepsilon ) \le (1+\varepsilon )I(x)+ c \varepsilon \end{aligned}$$
(95)

Proof

Since \({\mathcal {I}}\) is convex, we have

$$\begin{aligned} \mathcal {I}(x+ \varepsilon )\le \mathcal {I}(x)+ \varepsilon \mathcal {I}'(x+\epsilon ). \end{aligned}$$

Since \({\mathcal {I}}(x)\) is decreasing on \([0,\pi /2]\), the bound (95) follows immediately for \(x\in [0,\pi /2-\varepsilon ]\) with any \(c\ge 0\). Using (85) we get

$$\begin{aligned} \mathcal {I}(x+ \varepsilon )&\le \mathcal {I}(x)+ \frac{\varepsilon }{x+\varepsilon }\left( \mathcal {I}(x+\varepsilon )- \frac{1}{8} K^{-1}\left( \frac{\pi }{2(x+\varepsilon )} \right) \right) . \end{aligned}$$
(96)

From (90) it follows that there exists an \(x_0>0\) such that

$$\begin{aligned} \frac{\varepsilon }{x+\varepsilon }\left( \mathcal {I}(x+\varepsilon )- \frac{1}{8} K^{-1}\left( \frac{\pi }{2(x+\varepsilon )}\right) \right) \le \varepsilon {\mathcal {I}}(x), \qquad \hbox {if }x\ge x_0 \end{aligned}$$

uniformly in \(\varepsilon \in [0,1/2]\). Therefore, for \(x>x_0\) we have that \(\mathcal {I}(x+\varepsilon ) \le (1+\varepsilon )\mathcal {I}(x).\) We can assume \(x_0>\pi /2\). By choosing

$$\begin{aligned} c= \sup _{x\in [\pi /2, x_0+1/2]} {\mathcal {I}}'(x)={\mathcal {I}}'(x_0+1/2) \end{aligned}$$

we get \(\mathcal {I}(x+\varepsilon ) \le \mathcal {I}(x)+c \varepsilon \) on \([\pi /2-\varepsilon , x_0]\) with any \(0\le \varepsilon <1/2\) and the lemma follows. \(\square \)

Appendix B: Properties of \(\hbox {I}_{\mathrm{Sine}}\)

In the final section of the appendix we describe the behavior of the function \(I_{\mathrm{Sine}}(\rho )\) near \(\rho =\tfrac{1}{2\pi }\) and \(\rho \rightarrow \infty \).

Proposition 20

The functions \(\gamma (\nu )\) and \(I_{\mathrm{Sine}}(\rho )\) satisfy the following.

  1. 1.

    The function \(\gamma (\nu )\) defined in (7) is continuous and strictly decreasing. It satisfies the differential equation \(4x(1-x)\gamma ''(x)=\gamma (x)\) on \((-\infty ,0)\) and on \((0,1)\) with boundary behavior \(\lim \nolimits _{x\rightarrow 0^{\pm }} \gamma (x)=\tfrac{1}{2\pi }\), \( \gamma (1)=0\) and \(\lim \nolimits _{x\rightarrow -\infty } \frac{\gamma (x)}{\sqrt{|x|}}=\tfrac{1}{4}\). These limits and the differential equation identify \(\gamma (x)\) uniquely on \((-\infty ,0)\cup (0,1]\).

  2. 2.

    We have \(I_{\mathrm{Sine}}(0)=\tfrac{1}{64}\), \(I_{\mathrm{Sine}}(\tfrac{1}{2\pi })=0\), and \(I_{\mathrm{Sine}}''(x)>0\) for \(x\ne \tfrac{1}{2\pi }\). Moreover, we have the following limits:

    $$\begin{aligned} \lim \limits _{x\rightarrow 0} \frac{I_{\mathrm{Sine}}(\tfrac{1}{2\pi }+x)}{\tfrac{x^2}{\log (1/ |x|)}}=\frac{\pi ^2}{4} , \qquad \hbox {and} \qquad \lim \limits _{\rho \rightarrow \infty } \frac{I_{\mathrm{Sine}}(\rho )}{\rho ^2\log \rho }=\frac{1}{2}. \end{aligned}$$

Proof

Recall the function \(\gamma (\nu )\) given in (7). Using the asymptotics (90) proved in Proposition 17 it is easy to see that \(\gamma (\nu )\) is well-defined and positive in \((-\infty ,0)\cup (0,1)\) with \(\lim \nolimits _{\nu \rightarrow 1^{-}} \gamma (\nu )=0=\gamma (1)\) and \(\lim \nolimits _{x\rightarrow -\infty } \frac{\gamma (x)}{\sqrt{|x|}}=\tfrac{1}{4}\). We also get that \(\gamma (x){\mathcal {H}}(x)\) blows up as \(\nu \rightarrow 0^-\) or \(0^+\).

Differentiating (7) and using the definition (5) lead to

$$\begin{aligned} {\mathcal {H}}(x)\gamma '(x)= \frac{1}{8}+ {\mathcal {H}}'(x)\gamma (x), \qquad \frac{\gamma ''(x)}{\gamma (x)}= \frac{{\mathcal {H}}''(x)}{{\mathcal {H}}(x)}=\frac{1}{4x(1-x)}, \end{aligned}$$
(97)

for \(x\in (-\infty ,0)\cup (0,1)\). We have \({\mathcal {H}}'(x)=-\frac{K(x)}{2}<0\) for \(x<1\) and \({\mathcal {H}}(0)=0\). Thus from the second identity we get that \(\gamma '(\nu )\) is strictly decreasing in \((-\infty ,0)\) and strictly increasing in \((0,1)\). From the asymptotics (90) of Proposition 17 it is not hard to check that \(\lim \nolimits _{\nu \rightarrow -\infty } \gamma '(\nu )=0\) and \(\lim \nolimits _{\nu \rightarrow 1^{-}} \gamma '(\nu )=-\tfrac{1}{8}\). This, together with the previous statement, proves that \(\gamma (\nu )\) is decreasing on \((-\infty ,0)\) and also on \((0,1]\).

Since \((\gamma (x) {\mathcal {H}}(x)^{-1})'=\tfrac{1}{8}{\mathcal {H}}(x)^{-2}\), L’Hospital’s rule gives

$$\begin{aligned} \lim \limits _{\nu \rightarrow 0} \gamma (\nu )=\lim \limits _{\nu \rightarrow 0}\frac{\gamma (\nu ) {\mathcal {H}}(\nu )^{-1}}{{\mathcal {H}}(\nu )^{-1}}=-\tfrac{1}{8}H'(0)=\frac{1}{2\pi }=\gamma (0). \end{aligned}$$

Then from (97) it follows that

$$\begin{aligned} \lim \limits _{\nu \rightarrow 0} \gamma ''(\nu )\nu =\frac{1}{8\pi }, \qquad \lim \limits _{\nu \rightarrow 0} \frac{\gamma '(\nu )}{\log |\nu |}=\frac{1}{8\pi }, \end{aligned}$$

and also that

$$\begin{aligned} \lim \limits _{x\rightarrow 0} \frac{\gamma ^{(-1)}(\tfrac{1}{2\pi }+x)}{8\pi \tfrac{x}{\log |x|}}=1. \end{aligned}$$
(98)

This shows that \(\gamma (\nu )\) is continuous and strictly decreasing on \((-\infty ,1]\). We have \(\gamma ^{(-1)}(1)=0\) and \(I_{\mathrm{Sine}}(0)=\tfrac{1}{64}\).

Note, the fact that \(\gamma (x)\) solves \(4x(1-x) \gamma ''(x)=\gamma (x)\) on \((-\infty ,0)\cup (0,1)\) and has the proven asymptotics at \(-\infty , 0\) and \(1\) uniquely identifies it. The equation \(4x(1-x) y''(x)=y(x)\) has two linearly independent solutions on both \((-\infty ,0)\) and \((0,1)\). The function \({\mathcal {H}}(x)\) also solves the equation (on both intervals), but with \({\mathcal {H}}(0)=0\), \(\lim \nolimits _{x\rightarrow 1^{-}}{\mathcal {H}}(x)=-1\) and \(\lim \nolimits _{x\rightarrow -\infty } \frac{{\mathcal {H}}(x)}{\sqrt{|x|} \log |x|}=\tfrac{1}{2}\). This shows that any solution on \((-\infty ,0)\) or \((0,1)\) can be expressed as \(c_1 \gamma +c_2 {\mathcal {H}}\) with some constants \(c_1, c_2\), and the values of the constants are determined by the behavior of the solution at the end of the interval.

Using (97) together with (6) we can also compute that

$$\begin{aligned} I_{\mathrm{Sine}}'(\rho )&= \frac{1}{8} \left[ \frac{1}{8\gamma '(\nu )}+{\mathcal {H}}(\nu )+ \frac{\gamma (\nu ){\mathcal {H}}'(\nu )}{\gamma '(\nu )}\right] = \frac{1}{4} {\mathcal {H}}(\nu ), \hbox { and}\\ I_{\mathrm{Sine}}''(\rho )&= \frac{1}{4} \frac{{\mathcal {H}}'(\nu )}{\gamma '(\nu )}= - \frac{1}{8} \frac{K(\nu )}{\gamma '(\nu )}, \end{aligned}$$

where \(\nu \) is short for \(\gamma ^{-1}(\rho )\). From this \(I_{\mathrm{Sine}}(\tfrac{1}{2\pi })=0\) follows, together with \(I_{\mathrm{Sine}}(x)>0\) for \(x\ne \tfrac{1}{2\pi }\). The asymptotics of \(I_{\mathrm{Sine}}(\tfrac{1}{2\pi }+x)\) as \(x\rightarrow 0\) can be obtained from the definition (6), the asymptotics (98), and the fact that \({\mathcal {H}}(0)=0, {\mathcal {H}}'(0)=-\tfrac{\pi }{4}\).

Lastly we can look at the asymptotics of \(I_{\mathrm{Sine}}(\rho )\) as \(\rho \rightarrow \infty \). Recalling again (90) we get

$$\begin{aligned} {\mathcal {H}}(-x) \sim \tfrac{1}{2} \sqrt{x} \log x, \qquad \gamma (-x)&\sim \frac{\sqrt{x}}{4}, \qquad \gamma ^{(-1)}(x)\sim 16 x^2\qquad \hbox { as }x\rightarrow \infty , \end{aligned}$$

from which \(I_{\mathrm{Sine}}(\rho )\sim \tfrac{1}{2} \rho ^2 \log \rho \) follows for \(\rho \rightarrow \infty \).

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Holcomb, D., Valkó, B. Large deviations for the \({\hbox {Sine}}_\beta \) and \(\hbox {Sch}_\tau \) processes. Probab. Theory Relat. Fields 163, 339–378 (2015). https://doi.org/10.1007/s00440-014-0594-4

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Mathematics Subject Classification

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  • 60F10
  • 15B52