On the Moments of the Moments of the Characteristic Polynomials of Random Unitary Matrices

Abstract

Denoting by $P_N(A,\theta )=\det (I-Ae^{-i\theta })$ the characteristic polynomial on the unit circle in the complex plane of an $N\times N$ random unitary matrix A, we calculate the kth moment, defined with respect to an average over $A\in U(N)$, of the random variable corresponding to the $2\beta$th moment of $P_N(A,\theta )$ with respect to the uniform measure $\frac{d\theta }{2\pi }$, for all $k, \beta \in \mathbb {N}$ . These moments of moments have played an important role in recent investigations of the extreme value statistics of characteristic polynomials and their connections with log-correlated Gaussian fields. Our approach is based on a new combinatorial representation of the moments using the theory of symmetric functions, and an analysis of a second representation in terms of multiple contour integrals. Our main result is that the moments of moments are polynomials in N of degree $k^2\beta ^2-k+1$. This resolves a conjecture of Fyodorov and Keating (Philos Trans R Soc A 372(2007):20120503, 2014) concerning the scaling of the moments with N as $N\rightarrow \infty$, for $k,\beta \in \mathbb {N}$. Indeed, it goes further in that we give a method for computing these polynomials explicitly and obtain a general formula for the leading coefficient.

Appendix

1.1 Examples

Here we give explicit examples of the polynomials ${{\,\mathrm{MoM}\,}}_N(k,\beta )$ for small values of $k,\beta$. The formulae we record extend the results of preliminary calculations due to Keating and Scott [32] (c.f. [29]), which formed the basis for some of the numerical computations in [21].

First, the general technique is described and then explicit forms of the polynomials are given in the cases of $\beta =1$, $k\in \{1,2,3,4\}$ and $\beta =2$, $k\in \{1,2\}$. We should remark that the moment formula of Keating and Snaith [33] gives the full polynomials for the case $k=1$, $\beta \in \mathbb {N}$; see (10).

The technique we use is in a slightly more general form than is needed here, because we see it as having other potential applications; we then specialise back to the actual formula required for our calculations. The more general form we start with was first derived by Conrey, Farmer and Zirnbauer [13], and later by Bump and Gamburd using symmetric function theory [7]. Note that we used a special case of this result to prove Theorem 1.3 in Sect. 4. First, define for finite sets A, B, C, D,

$$\begin{aligned} R(A,B;C,D)\,{:=}\,\int _{U(N)}\frac{\prod _{\alpha \in A}\det (I-X^*e^{-\alpha })\prod _{\beta \in B}\det (I-Xe^{-\beta })}{\prod _{\gamma \in C}\det (I-X^*e^{-\gamma })\prod _{\delta \in D}\det (I-Xe^{-\delta })}dX. \end{aligned}$$

Further if

$$\begin{aligned} Z(A,B)\,{:=}\, \prod _{\begin{array}{c} \alpha \in A,\\ \beta \in B \end{array}}\frac{1}{(1-e^{-(\alpha +\beta )})}, \end{aligned}$$

then define

$$\begin{aligned} Z(A,B;C,D)\,{:=}\,\frac{Z(A,B)Z(C,D)}{Z(A,D)Z(B,C)}. \end{aligned}$$

Finally, if $S\subset A$ and $T\subset B$ then $\overline{S}=A-S$, $\overline{T}=B-T$, $S^-=\{-\hat{\alpha }:\hat{\alpha }\in S\}$ and similarly for T. Note that here we are using the notation $U+V$, $U-V$ (to be interpreted as $U\cup V$ and $U\backslash V$ respectively for sets U, V) to be consistent with the statement of the Theorem in [14].

Theorem

([11]). With $N\ge 0$ and ${{\,\mathrm{Re}\,}}(\gamma )>0, {{\,\mathrm{Re}\,}}(\delta )>0$ for $\gamma \in C$, $\delta \in D$, $|C|\le |A|+N$, $|D|\le |B|+N$, we have

$$\begin{aligned} R(A,B;C,D)=\sum _{\begin{array}{c} S\subset A,T\subset B\\ |S|=|T| \end{array}}e^{-N(\sum _{\hat{\alpha }\in S}\hat{\alpha }+\sum _{\hat{\beta }\in T}\hat{\beta })}Z(\overline{S}+T^-,\overline{T}+S^-;C,D), \end{aligned}$$

where $A=S+\overline{S}$ and $B=T+\overline{T}$.

To see how this is used to give the full polynomials for ${{\,\mathrm{MoM}\,}}_N(k,\beta )$, we outline the simplest case with $k=\beta =1$. We note that

$$\begin{aligned} {{\,\mathrm{MoM}\,}}_N(1,1)&=\frac{1}{2\pi }\int _0^{2\pi }\mathbb {E}_{A\in U(N)}\left( |P_N(A,\theta )|^2\right) d\theta \\&=\frac{1}{2\pi }\int _0^{2\pi }\int _{U(N)}P_N(A,\theta )P_N(A^*,-\theta )dAd\theta , \end{aligned}$$

so we apply the above theorem with $A=\{i\alpha \}, B=\{i\beta \}$, $C,D=\emptyset$. This gives us that

$$\begin{aligned} {{\,\mathrm{MoM}\,}}_N(1,1)&=\frac{1}{2\pi }\int _0^{2\pi }\lim _{\beta \rightarrow -\alpha }Z(A,B)+e^{-iN(\alpha +\beta )}Z(B^-,A^-)d\alpha \\&=\frac{1}{2\pi }\int _0^{2\pi }\lim _{\beta \rightarrow -\alpha }\sum _{m=0}^Ne^{-im(\alpha +\beta )}d\alpha \\&=N+1. \end{aligned}$$

Higher values of $k, \beta$ clearly result in bigger sets A, B, and hence many more choices for S, T. Nevertheless, small cases of ${{\,\mathrm{MoM}\,}}_N(k,\beta )$ can be computed in the same way. For example

$$\begin{aligned} {{\,\mathrm{MoM}\,}}_N(1,1)&=N+1\\ {{\,\mathrm{MoM}\,}}_N(2,1)&=\frac{1}{6}(N+3)(N+2)(N+1)\\ {{\,\mathrm{MoM}\,}}_N(3,1)&=\frac{1}{2520}(N+5)(N+4)(N+3)(N+2)(N+1)(N^2+6N+21)\\ {{\,\mathrm{MoM}\,}}_N(4,1)&=\frac{1}{778377600}(N+7)(N+6)(N+5)(N+4)(N+3)(N+2)(N+1)\\&\quad \times (7N^6+168N^5+1804N^4+10944N^3+41893N^2+99624N+154440)\\ {{\,\mathrm{MoM}\,}}_N(1,2)&=\frac{1}{12}(N+1)(N+2)^2(N+3)\\ {{\,\mathrm{MoM}\,}}_N(2,2)&=\frac{1}{163459296000}(N+7)(N+6)(N+5)(N+4)(N+3)(N+2)(N+1)\\&\quad \times (298N^8+9536N^7+134071N^6+1081640N^5+5494237 N^4+18102224N^3\\&\quad +38466354N^2+50225040N+32432400). \end{aligned}$$

It is worth noting explicitly that this method gives exact information about the moments of the moments at the freezing transition $\beta =1$.

1.2 Vandermonde determinant coefficients

Recall that we are interested in determining the coefficient of terms of the form $(x_1\cdots x_n)^{n-1}$ in the square of the Vandermonde determinant,

$$\begin{aligned} \Delta (x_1,\dots ,x_n)^2=\sum _{\sigma ,\tau \in S_n}{{\,\mathrm{sgn}\,}}(\sigma ){{\,\mathrm{sgn}\,}}(\tau )\prod _{i=1}^nx_i^{\sigma (i)+\tau (i)-2}. \end{aligned}$$

(121)

Thus, we require that $\sigma (i)+\tau (i)=n+1$ for all $i\in \{1,\dots ,n\}$ and in particular we want to show that this coefficient is strictly positive.

Immediately, we see that there will be n! terms of the required form since fixing $\sigma (i)$ completely determines $\tau (i)$. Consider the bijection

$$\begin{aligned} \phi :\{1,\dots ,n\}\rightarrow \{1,\dots ,n\}, \quad i\mapsto n+1-i. \end{aligned}$$

The order of $\phi$ is 2 and if n is even, it has no fixed point, whereas if n is odd there is a unique fixed point $(n+1)/2$. Thus, $\phi \in S_n$ and it consists of n / 2 transpositions if n is even, and $(n-1)/2$ transpositions if n is odd. Now set $\tau =\phi \circ \sigma$, so $\tau \in S_n$, and $\tau (i)=n+1-\sigma (i)$. Given $\sigma$, we have found our unique permutation. To determine the sign of $\tau$, note that ${{\,\mathrm{sgn}\,}}(\tau )={{\,\mathrm{sgn}\,}}(\phi ){{\,\mathrm{sgn}\,}}(\sigma )$, and

$$\begin{aligned} {{\,\mathrm{sgn}\,}}(\phi )=(-1)^{\left\lfloor \tfrac{n}{2}\right\rfloor }={\left\{ \begin{array}{ll}+1&{}\text {if }n\equiv 0,1\mod 4\\ -1&{}\text {if }n\equiv 2,3\mod 4.\end{array}\right. } \end{aligned}$$

Thus, the coefficient of $(x_1\cdots x_n)^{n-1}$ in $\Delta (x_1,\dots ,x_n)^2$ is ${{\,\mathrm{sgn}\,}}(\phi )n!$. It now follows that the coefficient of

$$\begin{aligned} \prod _{i=1}^{l_1}v_i^{l_1-1}\prod _{i=l_1+1}^{l_1+l_2}v_i^{l_2-1}\cdots \prod _{i=\sum _{j=1}^{k-1}l_j+1}^{k\beta }v_i^{l_k-1}\prod _{i=k\beta +1}^{\sum _{j=1}^{k-1}l_j+2\beta }v_i^{2\beta -l_k-1}\cdots \prod _{i=2(k-1)\beta +1+l_1}^{2k\beta }v_i^{2\beta -l_1-1} \end{aligned}$$

in

$$\begin{aligned} \prod _{n=1}^{k}\Delta (v_{\sum _{j=1}^{n-1}l_j+1},\dots ,v_{\sum _{j=1}^{n}l_j})^2 \prod _{n=1}^{k}\Delta (v_{\sum _{j=1}^{n}l_j+2(k-n)\beta +1},\dots ,v_{\sum _{j=1}^{n-1}l_j+2(k-(n-1))\beta })^2 \end{aligned}$$

is given by

$$\begin{aligned} (-1)^{\sum _{j=1}^k\left( \big \lfloor \tfrac{l_j}{2}\big \rfloor +\big \lfloor \tfrac{2\beta -l_j}{2}\big \rfloor \right) }\prod _{j=1}^kl_j!(2\beta -l_j)!&=(-1)^{k\beta +\sum _{j=1}^k\left( \big \lfloor \tfrac{l_j}{2}\big \rfloor +\big \lfloor \tfrac{-l_j}{2}\big \rfloor \right) }\prod _{j=1}^kl_j!(2\beta -l_j)!\\&=(-1)^{k\beta }(-1)^{\sum _{j=1}^k\delta _{\{l_j\text { is odd\}}}}\prod _{j=1}^kl_j!(2\beta -l_j)!\\&=(-1)^{k\beta }(-1)^{\#\{j:l_j\text { is odd}\}}\prod _{j=1}^kl_j!(2\beta -l_j)!. \end{aligned}$$

This proves the result since the parity of $\#\{j:l_j\text { is odd}\}$ is the same as the parity of $k\beta$ as $\sum _{j=1}^kl_j= k\beta$.

1.3 Uniqueness of construction

When trying to construct the term of the form $(v_1\cdots v_{2k\beta })^{2\beta -1}$ in

$$\begin{aligned}&\prod _{q=1}^{k}\Delta (v_{\sum _{j=1}^{q-1}l_j+1},\dots ,v_{\sum _{j=1}^{q}l_j})^2 \prod _{q=1}^{k}\Delta (v_{\sum _{j=1}^{q}l_j+2(k-q)\beta +1},\dots ,v_{\sum _{j=1}^{q-1}l_j+2(k-(q-1))\beta })^2\\&\prod _{\begin{array}{c} m\le k\beta <n\\ \alpha _m=\alpha _n \end{array}}\left( {v_n-v_m}\right) , \end{aligned}$$

first note that the variables $v_m$, for $m\in \{1,\dots ,k\beta \}$ only appear in the Vandermonde determinants and the products

$$\begin{aligned} \prod _{\begin{array}{c} m\le k\beta <n\\ \alpha _m=\alpha _n \end{array}}\left( {v_n-v_m}\right) =\prod _{\begin{array}{c} m\in \{1,\dots ,l_1\}\\ n\in \{2(k-1)\beta +1+l_1,\dots ,2k\beta \} \end{array}}(v_n-v_m)\cdots \prod _{\begin{array}{c} m\in \{\sum _{j=1}^{k-1}l_j+1,\dots ,k\beta \}\\ n\in \{k\beta +1,\dots ,\sum _{j=1}^{k-1}l_j+2\beta \} \end{array}}(v_n-v_m). \end{aligned}$$

In particular, after fixing $q\in \{1,\dots ,k\}$ take $v_j$ with $j\in \{\sum _{i=1}^{q-1}l_i+1,\dots ,\sum _{i=1}^ql_i\}$. Then $v_j$ only appears in the following two terms:

$$\begin{aligned} \Delta (v_{\sum _{i=1}^{q-1}l_i+1},\dots ,v_{\sum _{i=1}^ql_i})^2\text { and }\prod _{\begin{array}{c} m\in \{\sum _{i=1}^{q-1}l_i+1,\dots ,\sum _{i=1}^ql_i\}\\ n\in \{2k\beta -\sum _{i=1}^{q}(2\beta -l_i)+1,\dots ,2k\beta -\sum _{i=1}^{q-1}(2\beta -l_i)\} \end{array}}(v_n-v_m). \end{aligned}$$

In particular these are both homogeneous polynomials: the former of degree $l_q(l_q-1)$ in $l_q$ variables and the latter is of degree $l_q(2\beta -l_q)$ in $2\beta$ variables. We will show that the only way to construct a term of the form $(v_1\cdots v_{2k\beta })^{2\beta -1}$ is as described following Eq. (106). Without loss of generality, we will set $q=1$ and assume $l_1\ge 2$. From the above discussion, the square of the Vandermonde determinant consists of terms of the form

$$\begin{aligned} v_1^{a_1}\cdots v_{l_1}^{a_{l_1}},\text { with }\sum _{i=1}^{l_1}a_i=l_1(l_1-1). \end{aligned}$$

(122)

Similarly, the product term is built of elements of the form

$$\begin{aligned} v_1^{b_1}\cdots v_{l_1}^{b_{l_1}}v_{2(k-1)\beta +1+l_1}^{b_{l_1+1}}\cdots v_{2k\beta }^{b_{2\beta }},\text { with }\sum _{i=1}^{2\beta }b_i=l_1(2\beta -l_1),\ 0\le b_i\le 2\beta -l_1.\nonumber \\ \end{aligned}$$

(123)

Hence, each term of

$$\begin{aligned} \Delta (v_1,\dots ,v_{l_1})^2\prod _{\begin{array}{c} m\in \{1,\dots ,l_1\}\\ n\in \{2(k-1)\beta +1+l_1,\dots ,2k\beta \} \end{array}}(v_n-v_m) \end{aligned}$$

is of the form

$$\begin{aligned} v_1^{a_1+b_1}\cdots v_{l_1}^{a_{l_1}+b_{l_1}}v_{2(k-1)\beta +1+l_1}^{b_{l_1+1}}\cdots v_{2k\beta }^{b_{2\beta }}, \end{aligned}$$

with $a_i,b_i$ satisfying the homogenous conditions. To reach our goal, we need to find all possibilities for $a_i, 1\le i\le l_1$ and $b_i, 1\le i\le 2\beta$ that $a_i+b_i=2\beta -1$ for $i\in \{1,\dots ,l_1\}$. This implies that we need $\sum _{i=1}^{l_1}(a_i+b_i)=l_1(2\beta -1)$. Now note that the ‘homogeneous conditions’ in Eqs. (122) and (123) together mean that

$$\begin{aligned} \sum _{i=1}^{l_1}(a_i+b_i)+\sum _{l_1+1}^{2\beta }b_i=\sum _{i=1}^{l_1}a_i+\sum _{i=1}^{2\beta }b_i=l_1(2\beta -1). \end{aligned}$$

Thus, we must set $b_{l_1+1},\dots ,b_{2\beta }=0$ if we want to construct the required term. This leaves us with finding all $a_i,b_i$$1\le i\le l_1$ such that all the following are satisfied,

$$\begin{aligned} a_i+b_i&=2\beta -1,\\ \sum _{i=1}^{l_1}a_i&=l_1(l_1-1),\\ \sum _{i=1}^{l_1}b_i&=l_1(2\beta -l_1),\\ 0\le&b_i\le 2\beta -l_1. \end{aligned}$$

However, the latter two conditions imply that we must have $b_i=2\beta -l_1$ for all $1\le i\le l_1$ which in turn gives us that $a_i=l_1-1$ for all $1\le i\le l_1$, and these are the only possible choices. This exactly matches the construction described following Eq. (106). The case for $q\in \{2,\dots ,k\}$ follows similarly.