The many faces of the stochastic zeta function

Abstract

We introduce a framework to study the random entire function $\zeta _\beta$ whose zeros are given by the Sine$_\beta$ process, the bulk limit of beta ensembles. We present several equivalent characterizations, including an explicit power series representation built from Brownian motion. We study related distributions using stochastic differential equations. Our function is a uniform limit of characteristic polynomials in the circular beta ensemble; we give upper bounds on the rate of convergence. Most of our results are new even for classical values of $\beta$. We provide explicit moment formulas for $\zeta$ and its variants, and we show that the Borodin–Strahov moment formulas hold for all $\beta$ both in the limit and for circular beta ensembles. We show a uniqueness theorem for $\zeta$ in the Cartwright class, and deduce some product identities between conjugate values of $\beta$. The proofs rely on the structure of the $\mathtt {Sine}_{\beta }$ operator to express $\zeta$ in terms of a regularized determinant.

Appendices

Proof of the Taylor Expansion of $\zeta _{{\varvec{\tau }}}$

Proof of Proposition 14

We start by adapting Lemma 15 to integral operators with matrix valued kernels. We can embed the space of ${{\mathbb {R}}}^2$-valued $L^2(0, \sigma ]$ functions into the space of real valued $L^2(0,2 \sigma ]$ functions using the following invertible isometry:

$$\begin{aligned} g=[g_1,g_2]^{\dagger }, \quad {\mathcal {J}}g(t)={\left\{ \begin{array}{ll} g_1(s), &{} 0<s\le \sigma \\ g_2(s- \sigma ), &{} \sigma <s\le 2 \sigma . \end{array}\right. } \end{aligned}$$

(189)

If B is a Hilbert–Schmidt integral operator acting on ${{\mathbb {R}}}^2$-valued functions on $(0, \sigma ]$ with kernel $K=\left( \begin{array}{cc} K_{11} &{} K_{12} \\ K_{21} &{} K_{22} \\ \end{array} \right)$ then $A={\mathcal {J}} B {\mathcal {J}}^{-1}$ is a Hilbert–Schmidt integral operator acting on scalar functions on $(0,2 \sigma ]$, and the integral kernel is

$$\begin{aligned} k(s,t)=K_{1+{\mathbf {1}}(s\ge \sigma ), 1+{\mathbf {1}}(t\ge \sigma )}(s,t). \end{aligned}$$

(190)

For a matrix M denote by $\Vert M\Vert _2$ the Frobenius norm $\Vert M\Vert _2^2={{\text {Tr}}}\,(M M^\dagger )$.

We have $\int _0^{ \sigma }\int _0^{ \sigma } \Vert K(s,t)\Vert ^2_2 ds\, dt=\int _0^{2 \sigma }\int _0^{2 \sigma } |k(s,t)|^2 ds\, dt$ and $\int _0^{2 \sigma } k(s,s)ds=\int _0^{ \sigma } {{\text {Tr}}}\, K(s,s) ds$. Assuming that both of these integrals are finite we may apply Lemma 15 to the integral operator A. Since the spectrum of A is the same as the spectrum of B, we have ${\det }_2(I-z B)={\det }_2(I-z A)$ and hence

$$\begin{aligned} {\det }_2(I-z B) e^{-z \int _0^{ \sigma } {{\text {Tr}}}\,K(s,s)}=1+\sum _{n=1}^\infty (-1)^n \frac{z^n}{n!} \int _{(0,2 \sigma ]^n}\det \left[ k(t_i,t_j) \right] _{i,j=1}^n dt_1\cdots dt_n. \end{aligned}$$

(191)

From (190) we have

$$\begin{aligned}&\int _{(0,2 \sigma ]^n}\det \left[ k(t_i,t_j) \right] _{i,j=1}^n dt_1\cdots dt_n\\&\quad =\sum _{i_1=1}^2 \cdots \sum _{i_n=1}^2 \int _{(0, \sigma ]^n} \det (K_{i_a, i_b}(s_a,s_b))_{a,b=1}^n ds_1\cdots ds_n, \end{aligned}$$

Now set $B={\mathtt {r}\,}{\varvec{\tau }}$. From (17) we get that the entries $K_{ij}$ of the matrix valued kernel are given by

$$\begin{aligned} K_{i,j}(s,t)=\tfrac{1}{2}\left( a_i(s) c_j(t){\mathbf {1}}(s<t)+c_i(s)a_j(t){\mathbf {1}}(t\le s)\right) . \end{aligned}$$

(192)

Note that $K(s,t)^{\dagger }=K(t,s)$ and thus $K_{i,j}(s,t)=K_{j,i}(t,s)$. Because of this we have

$$\begin{aligned}&\frac{1}{n!}\sum _{i_1=1}^2 \cdots \sum _{i_n=1}^2 \int _{(0, \sigma ]^n} \det (K_{i_j, i_\ell }(s_j,s_\ell ))_{j,\ell =1}^n ds_1\cdots ds_n \\&\quad =\sum _{i_1=1}^2 \cdots \sum _{i_n=1}^2\, \, \idotsint \limits _{0<s_1<\cdots <s_n\le \sigma } \det (K_{i_j, i_\ell }(s_j,s_\ell ))_{j,\ell =1}^n ds_1\cdots ds_n. \end{aligned}$$

Fix $i_1, \ldots , i_n\in \{1,2\}$ and $0<s_1<\cdots s_n\le \sigma$. Introduce the temporary notation

$$\begin{aligned} p_k=a_{i_k}(s_k),\quad q_k=c_{i_k}(s_k), \end{aligned}$$

then by (192) we have $2 K_{i_j, i_\ell }(s_j,s_\ell )= p_{\min (j,\ell )} \cdot q_{\max (j,\ell )}$. For example, for $n=3$ we have

$$\begin{aligned} (2K_{i_j, i_\ell }(s_j,s_\ell ))_{j,\ell =1}^n=\left( \begin{array}{ccc} p_1 q_1 &{}\quad p_1 q_2 &{}\quad p_1 q_3 \\ p_1 q_2 &{}\quad p_2 q_2 &{}\quad p_2 q_3 \\ p_1 q_3 &{}\quad p_2 q_3 &{}\quad p_3 q_3 \end{array} \right) . \end{aligned}$$

We show that

$$\begin{aligned} \det ( p_{\min (j,\ell )} \cdot q_{\max (j,\ell )})_{j,\ell =1}^n=p_1 q_n\prod _{j=1}^{n-1} (p_{j+1} q_{j}-p_{j}q_{j+1}). \end{aligned}$$

(193)

Subtract row $n-1$ times $q_{n}/q_{n-1}$ from row n. Then the last row becomes

$$\begin{aligned}{}[0,\ldots ,0, p_nq_n-p_{n-1} q_n^2/q_{n-1}]. \end{aligned}$$

The identity (193) now follows by induction.

Note that $p_{j+1} q_{j}-p_{j}q_{j+1}=[p_j,q_j]J [p_{j+1},q_{j+1}]^{\dagger }$, hence, with $v_{i_k}(s_k)=[p_k,q_k]^{\dagger }$ we have

$$\begin{aligned}&\det ( p_{\min (j,\ell )} \cdot q_{\max (j,\ell )})_{j,\ell =1}^n\nonumber \\&\quad =\left[ v_{i_1}(s_1) v_{i_1}(s_1)^{\dagger } J v_{i_2}(s_2) v_{i_2}(s_2)^{\dagger } J\cdots v_{i_n}(s_n) v_{i_n}(s_n)^{\dagger } J\right] _{1,1}. \end{aligned}$$

(194)

Note that

$$\begin{aligned} v_{1}(s)v_1(s)^{\dagger }+v_2(s)v_2(s)^{\dagger }=2U^{\dagger } R(s)U, \quad U=[{\mathfrak {u}}_0 ,{\mathfrak {u}}_1 ]. \end{aligned}$$

Summing (194) for all choices of $i_1, \ldots , i_n$ gives

$$\begin{aligned} 2^n\left[ U^{\dagger } R(s_1) U J U^{\dagger } R(s_2) U J\cdots U^{\dagger } R(s_n) U J\right] _{1,1}=-(-2)^n {\mathfrak {u}}_0 ^{\dagger } R(s_1)JR(s_2)J\cdots R(s_n){\mathfrak {u}}_1 . \end{aligned}$$

In the last step we used $U J U^{\dagger }=-J$, which is equivalent to the assumption (14).

The statement (28) of the proposition now follows from (191). $\square$

Law of Iterated Logarithm for Brownian Integrals

Theorem 81

For every $a>2, b<1/2$ there is a constant c so that

$$\begin{aligned} P\Big (\sup _{t>0} \left( B(t)^2/t-a \log (1+|\log t|)\right) >y\Big )\le c e^{-by} \quad \text {for all}\quad y\ge 0. \end{aligned}$$

(195)

This is an effective small and large-time version of the upper bound in the law of iterated logarithm. By setting $t=1$ inside the $\sup$ we get $B(1)^2$, which shows that the rate of the exponential decay cannot be more than 1/2. The lower bound 2 on the parameter a is sharp by the usual law of iterated logarithm.

Proof

Let $f,g:(-1,\infty )\rightarrow {\mathbb {R}}$ be non-decreasing functions and $\varepsilon \in (0,1)$. If $f(t)>g(t)$ for $t\ge 0$ then $f(s)\ge g(s-\varepsilon )$ for $s\in [t, t+\varepsilon ]$. Hence

$$\begin{aligned} 1(f(t)\ge g(t) \text{ for } \text{ some } t\ge 0)\le \tfrac{1}{\varepsilon } \int _0^\infty 1(f(s)\ge g(s-\varepsilon ))ds. \end{aligned}$$

(196)

Let ${\bar{B}}_t=\max _{0\le s\le t} |B_s|$, and apply (196) to $f(t)={\bar{B}}(e^{t})^2$ and $g(t)=e^{t}(y+a\log (1+t))$. The expectation of the resulting inequality bounds $P(\sup _{t\ge 1} B_t^2/t-a \log (1+\log t)>y)$ above as

$$\begin{aligned} P({\bar{B}}(e^{t})^2\ge g(t)\hbox { for some}\ t\ge 0) \le \tfrac{1}{\varepsilon } \int _0^\infty P({\bar{B}}(e^s)^2\ge e^{s-\varepsilon }(y+a\log (1+s-\varepsilon ))ds. \end{aligned}$$

(197)

Since $\max _{0\le r\le 1} B(r)$ is distributed as |B(1)|, union and Gaussian tail bounds yield

$$\begin{aligned} P({\bar{B}}(e^s)^2\ge e^s x)=P({\bar{B}}(1)^2\ge x)\le 2P(B(1)^2\ge x)\le 2 e^{-x/2}. \end{aligned}$$

Thus the right hand side of (197) is bounded above by

$$\begin{aligned} \tfrac{2}{\varepsilon } \int _0^\infty e^{- e^{-\varepsilon }(y+a\log (1+s-\varepsilon ))/2} ds=\frac{ 4(1-\varepsilon )^{1-e^{-\varepsilon } a/2}e^{-e^{-\varepsilon } y/2}}{(a e^{-\varepsilon }-2)\varepsilon }. \end{aligned}$$

To make the last step valid and to get the required bound we need $a e^{-\varepsilon }>2$ and $e^{-\varepsilon }/2>b$, so we choose $\varepsilon <\min (1, \log (a/2),\log (1/(2b))$. Time inversion $B_t \rightarrow tB(1/t)$ gives the same bound for the supremum for $0< t\le 1$, from which (195) follows. $\square$

We apply Theorem 81 to estimate the growth of Brownian integrals.

Proposition 82

Suppose that B is two-sided Brownian motion and $x_u, u\le 0$ is adapted to the filtration generated by its increments. Assume further that there is a random variable C and a constant $a>0$ so that $|X_u|\le C e^{a u}$ for $u\le 0$. Then

$$\begin{aligned}&|\int _{-\infty }^u X_u dB |\\&\quad \le \frac{2C}{\sqrt{a}} e^{a u} \left( Z+ \log (1+2a|u|)+|\log a| +\log (1+2|\log C |)\right) \quad \text {for all } u\le 0, \end{aligned}$$

so the left hand side is well defined. With an absolute constant c, the random variable Z satisfies

$$\begin{aligned} P(Z>y)\le ce^{- y}, \quad y\ge 0. \end{aligned}$$

(198)

Proof

We have

$$\begin{aligned} \int _{-\infty }^u X_s^2\, ds \le \int _{-\infty }^u C^2 e^{2a s} ds=\frac{C^2}{2a} e^{2a u}<\infty , \end{aligned}$$

so the process $M_u=\int _{-\infty }^u X_u dB$ is well defined. Moreover,

$$\begin{aligned}{}[M]_u\le \frac{C^2}{2a} e^{2a u}, \quad \text {for all}\quad u\le 0. \end{aligned}$$

(199)

By the Dubins–Schwarz theorem there is a Brownian motion $W(x), x\ge 0$ so that $M_u=W([M]_u)$. Let

$$\begin{aligned} Z=\tfrac{1}{3} \sup _{x>0} (W(x)^2/x-3 \log (1+|\quad \log x|)). \end{aligned}$$

By Theorem 81 this random variable satisfies (198), and for $u\le 0$ we have

$$\begin{aligned} M_u^2\le 3 [M]_u Z+3 [M]_u (1+\log (1+|\quad \log [M]_u|)). \end{aligned}$$

The function $x(1+\log (1+| \log x|))$ is increasing, so by (199) we also get

$$\begin{aligned} M_u^2\le \frac{3C^2}{2a} e^{2a u} \left( Z+1+ \log \left( 1+\left| \log \frac{C^2}{2a} e^{2a u}\right| \right) \right) . \end{aligned}$$

(200)

For $x,y>0$ we have $| \log (xy)|\le | \log x|+| \log y|$, $\log (1+x+y)\le \log (1+x)+\log (1+y)$, and $\log (1+x)\le x$. Using these bounds repeatedly we get

$$\begin{aligned} \log (1+| \log \frac{C^2}{2a} e^{2a u}|)&\le \log (1+2 | \log C|+ | \log 2a|+2a |u|) \\&\le \log (1+2a |u|)+\log 2+| \log a| +\log (1+2| \log C |). \end{aligned}$$

Take square roots in (200) and use the inequality $\sqrt{1+y}\le 1+y$ for $y>0$. For q fixed, $Z+q$ satisfies the same tail bound as Z with a different c. This implies the claim. $\square$

Moment Bounds for an Almost Linear SDE

The following lemma is used when calculating moments of ratios of $\zeta$.

Lemma 83

Consider the diffusion

$$\begin{aligned} dX=Y dt+Z dW, \end{aligned}$$

(201)

with $|Y|\le a |X|$, $|Z|\le b |X|$, and a, b and $E|X_0^2|$ finite. Then for any $t\ge 0$

$$\begin{aligned} E|X_t|^2\le E|X_0|^2 e^{(2 a+2b^2)t}, \quad EX_t=EX_0+\int _0^{t} E Y_s ds. \end{aligned}$$

In particular, if $Y=\eta X$ for $\eta \in {{\mathbb {C}}}$ then $EX_t=EX_0 e^{\eta t}$.

Proof

Let $\tau _c$ be the first time $|X_t|\ge c$. By Itô’s formula

$$\begin{aligned} \int _0^{\tau _c\wedge s}d|X|^2=\int _0^{\tau _c\wedge s}2 \Re ({\bar{X}} Z dW)+(2 \Re X {\bar{Y}} +2|Z|^2)dt. \end{aligned}$$

In the interval $[0,\tau _c\wedge s]$ the quadratic variation is bounded, so

$$\begin{aligned} M_t=|X_{t\wedge \tau _c}|^2-\int _0^{t\wedge \tau _c} (2 \Re X_s {\bar{Y}}_s+2Z_s^2)\, ds \end{aligned}$$

is a martingale, and

$$\begin{aligned} E|X_{t\wedge \tau _c}^2|=E\int _0^{t\wedge \tau _c} (2 \Re X_s {\bar{Y}}_s+2Z_s^2)\, ds \le (2a+2b^2)\, E\int _0^{t\wedge \tau _c} |X|^2 ds. \end{aligned}$$

(202)

Since

$$\begin{aligned} \int _0^{t\wedge \tau _c} |X|^2 ds=\int _0^{t} |X|^2 {\mathbf {1}}(s\le \tau _c) ds \le \int _0^{t} |X|_{s\wedge \tau _c}^2 ds \end{aligned}$$

we have $E|X_{t\wedge \tau _c}^2|\le E|X_0^2|e^{(2a+2b^2) t}$ by Gronwall’s inequality. Fatou’s lemma gives

$$\begin{aligned} E|X_t^2|\le E|X_0^2|e^{(2a+2b^2) t} \end{aligned}$$

as well. From this we see that the quadratic variation of $\int _0^{t} Z dW$ has finite expectation, so it is a martingale. Thus we can take expectations in (201) which gives (202). The last claim follows from solving the equation $EX_t=EX_0+\eta \int _0^{t} EX_s ds.$ $\square$