# Factoring determinants and applications to number theory

Estelle Basor  American Institute of Mathematics, San Jose, USA
<EMAIL_ADDRESS>and  Brian Conrey  American Institute of Mathematics, San
Jose, USA<EMAIL_ADDRESS>
###### Abstract.

Products of shifted characteristic polynomials, and ratios of such products,
averaged over the classical compact groups are of great interest to number
theorists as they model similar averages of L-functions in families with the
same symmetry type as the compact group. We use Toeplitz and Toeplitz plus
Hankel operators and the identities of Borodin - Okounkov, Case - Geronimo,
and Basor - Erhardt to prove that, in certain cases, these unitary averages
factor as polynomials in the matrix size into averages over the symplectic
group and the orthogonal group. Building on these identities we present new
proofs of the exact formulas for these averages where the “swap” terms that
are characteristic of the number theoretic averages occur from the Fredholm
expansions of the determinants of the appropriate Hankel operator. This is the
fourth different proof of the formula for the averages of ratios of products
of shifted characteristic polynomials; the other proofs are based on
supersymmetry; symmetric function theory, and orthogonal polynomial methods
from Random Matrix Theory.

##  1\. introduction

The statistical study of L-functions is the backbone of modern analytic number
theory.The quantities of primary interest are the zeros and their
distributions and the values both large and small. One approaches these
statistics by calculating averages (moments) of products of shifted
L-functions and averages of ratios of such. The averages are over a family of
L-functions with conductor up to some bound.

Our understanding of moments and ratios of L-functions in families has
increased dramatically in the last 25 years stimulated initially by the work
of Keating and Snaith [KeSn] showing that Random Matrix Theory provides a good
model for these moments and by the work of Katz and Sarnak [KaSa] that showed
how to associate a symmetry type to each family of L-functions. The symmetry
type is one of Unitary, Symplectic or Orthogonal and governs which classical
group with Haar measure one should use in trying to model a given family of
L-functions.

Two conjectures about these averages have come into prominence: the recipe
tells how to calculate moments in a family, and the ratios conjecture tells
how to average ratios of products of L-functions. Each of these has a
corresponding theorem in Random Matrix Theory (RMT), (see [CFKRS, I and II],
[BG], [CFZ], [CFS]. RMT has become an indispensable tool for predicting the
answers for questions about L-functions. For example, if one wants to know the
distribution of the real parts of the zeros of $\zeta^{\prime}(s)$ one would
might start by asking the same question about the radial distribution of zeros
of $\Lambda_{U}^{\prime}(z)$ where $U\in U(N)$, the group of $N\times N$
unitary matrices, and $\Lambda_{U}(x)=\det(I-xU^{*})$ is the characteristic
polynomial of $U$. If one could evaluate

$\int_{U(N)}|\Lambda_{U}^{\prime}(x)|^{2k}dU$

for small positive real $k$ and $x$ near 1 one could answer this (unsolved)
problem and the answer, we believe, would be the same distribution (with
appropriate scaling) as the question for zeros of $\zeta^{\prime}(s)$.

In general, moments of L-functions give information about the value
distribution (in particular large values) of L-functions, while averages of
ratios of products of L-functions give us information about the distribution
of small values, in particular zeros, of L-functions (see [CFKRS], [CS07], and
[CS08] for some examples).

Examples of families of L-functions include:

  * •

$\\{\zeta(1/2+it):t\in\mathbb{R}\\}$ is a unitary family. In fact any
L-function regarded in $t$ aspect is a family parametrized by the (continuous
variable) $t$. All other families are discrete.

  * •

$\\{L(1/2,\chi_{d}):\chi_{d}\mbox{ is a real primitive Dirichlet
character}\\}$ is a symplectic family. These characters are extensions of the
Legendre symbol $\chi_{p}(n)=\big{(}\frac{n}{p}\big{)}$ which is 1 if $n$ is a
square modulo $p$ and $-1$ if $n$ is not a square.

  * •

$\\{L_{f}(1/2):f\mbox{ is a weight 2, level $q$ newform}\\}$ is an orthogonal
family. Examples include the L-functions associated with elliptic curves. This
is a family of degree 2 L-functions.

The symbiotic relationship between Analytic Number Theory (ANT) and RMT reaps
many rewards. Conjectures we make in ANT have counterparts that are expected
to be proveable in RMT. These are often of a different nature than random
matrix theorists might typically think about. And theorems in RMT lead us to
wonder whether the analogue in ANT is true and so to conjectures in ANT that
guide the way as we try to understand L-functions.

In this paper we prove a theorem in RMT that would be very surprising in ANT.
It has to do with moments. For example we might consider

$\int_{0}^{\infty}\psi(\frac{t}{T})\zeta(1/2+it+\alpha)\zeta(1/2-it+\beta)~{}dt$

where $\psi$ is (a) compactly supported (test-function) on $[1,2]$ The RMT
analogue of this simplest shifted moment problem would be

$\int_{U(N)}\det(I-a~{}U)\det(I-b~{}U^{*})dU$

where we integrate with respect to Haar measure $dU$ and where
$a=e^{-\alpha}$. In ANT we can prove that

$\displaystyle\int_{0}^{\infty}\psi(\frac{t}{T})\zeta(1/2+it+\alpha)\zeta(1/2-it+\beta)~{}dt$
$\displaystyle\quad=\int_{0}^{\infty}\psi(\frac{t}{T})\bigg{(}\zeta(1+\alpha+\beta)+\big{(}\frac{t}{2\pi}\big{)}^{-\alpha-\beta}\zeta(1-\alpha-\beta)\bigg{)}~{}dt+O(T^{\frac{1}{3}})$

for $\alpha,\beta\ll(\log T)^{-1}$ and in RMT we can prove that

$\int_{U(N)}\det(I-e^{-\alpha}U)\det(I-e^{-\beta}U^{*})~{}dU=z(\alpha+\beta)+e^{-N(\alpha+\beta)}z(-\alpha-\beta)$

for any $\alpha$ and $\beta$ where

$z(x)=\frac{1}{1-e^{-x}}.$

The surprising theorem that we prove is that in certain cases the moments of a
product of shifted characteristic polynomials over the unitary group factors
into a moment over the symplectic group times a moment over the orthogonal
group. We do not expect this to be literally true for moments of families of
L-functions but it is intriguing to try to speculate what an analogue might
look like.

There are examples of similar phenomena that have been previously observed in
RMT. For example, see section 7.6 of [KaSa] and the many formulas in [BR].
Also, equations (6.10) and (6.11) of [Ke] and equation (2.3) of [Faz].

This factorization also works for unitary averages of ratios of products of
shifted characteristic polynomials.

The proof depends on some remarkable identities. The first is known as BOCG
which holds for for unitary averages. The second is a non-trivial extension to
symplectic and orthogonal averages discovered and proven by Basor and Erhardt.

We give some examples of factorizations: We have

$\displaystyle\int_{U(3)}\Lambda_{U}(a)\Lambda_{U}(b)\Lambda_{U^{*}}(a)\Lambda_{U^{*}}(b)~{}dU$
$\displaystyle=(1+a^{2}+ab+b^{2}+a^{2}b^{2})$
$\displaystyle\qquad\times(1+a^{4}+ab+a^{3}b+2a^{2}b^{2}+ab^{3}+a^{3}b^{3}+b^{4}+a^{4}b^{4})$

while

$\displaystyle\int_{USp(2)}\Lambda_{U}(a)\Lambda_{U}(b)dU=1+a^{2}+ab+b^{2}+a^{2}b^{2}$

and

$\displaystyle\int_{O^{+}(4)}\Lambda_{U}(a)\Lambda_{U}(b)dU$ $\displaystyle=$
$\displaystyle 1+a^{4}+ab+a^{3}b+2a^{2}b^{2}$
$\displaystyle\qquad+ab^{3}+a^{3}b^{3}+b^{4}+a^{4}b^{4}.$

As an example of a ratios calculations we mention

$\displaystyle\int_{U(N)}\frac{\Lambda_{U}(a)\Lambda_{U^{*}}(a)}{\Lambda_{U}(c)\Lambda_{U^{*}}(c)}dU$
$\displaystyle\qquad=\frac{(1+a^{N+1}-ac-a^{N}c)(1-a^{N+1}-ac+a^{N}c)}{(1-a)(1+a)(1-c)(1+c)}$

while for even $N$

$\displaystyle\int_{USp(N)}\frac{\Lambda_{U}(a)}{\Lambda_{U}(c)}~{}dU=\frac{(1-a^{N+2}-ac+a^{N+1}c)}{(1-a)(1+a)}$

and

$\displaystyle\int_{O^{(}+N)}\frac{\Lambda_{U}(a)}{\Lambda_{U}(c)}~{}dU$
$\displaystyle=$ $\displaystyle\frac{(1+a^{N}-ac-a^{N-1}c)}{(1-c)(1+c)}.$

Thus, for $N$ odd, we have

$\displaystyle\int_{U(N)}\frac{\Lambda_{U}(a)\Lambda_{U^{*}}(a)}{\Lambda_{U}(c)\Lambda_{U^{*}}(c)}dU=\left(\int_{USp(N-1)}\frac{\Lambda_{U}(a)}{\Lambda_{U}(c)}dU\right)\left(\int_{O^{+}(N+1)}\frac{\Lambda_{U}(a)}{\Lambda_{U}(c)}~{}dU\right).$

In the process of recording the proof of this factorization, we discovered a
new proof for the ratios theorem. This is the fourth known proof. The first
was [CFZ] motivated by physics and uses supersymmetry; the second was [BG]
using symmetric function theory and representation theory. The third was [CFS]
using orthogonal polynomial methods from RMT. The new one uses operator theory
and has the nice feature that the idea is very simple to explain. The unitary
moment can be expressed via BOCG as a manifestation of the Szego-Widom theorem
for Toeplitz matrices. One piece of this identity involves a finite rank
operator $K$ whose Fredholm expansion gives the desired swap terms. To handle
the formulas for symplectic and orthogonal one proceeds in the same way using
the Basor-Erhart identity which relates to operators that are “Toeplitz plus
Hankel.” The first proof works for all matrix size $N\geq 1$, whereas the
second and third proofs have a condition that $N$ must be bigger than the
number of factors in the average. Our new proof works for all $N\geq 1$.

Work of the second author was supported in part by a Focused Research Grant
DMS 1854398 from the NSF.

##  2\. Paper organization

In the next section we outline all the necessary results from operator theory
that we need to prove the factorization theorem and to provide the new proof
of the ratios formulas. This section requires very little background knowledge
of operator theory and is nearly self-contained. The basic properties of
Toeplitz operators are presented along with a proof of the Borodin-Okounkov-
Case-Geronino (BOCG) identity.

The next section describes the connections with the averages over the
classical groups and determinants of finite sections of either Toeplitz or
“Toeplitz plus Hankel”. This is followed by the proofs of the factorizations
of the averages illustrated by the previous examples. Then Section 6 contains
the new proofs of the ratio formulas using Fredholm determinant expansions.
Finally, more examples can be found in the Appendix.

##  3\. Toeplitz Preliminaries

We begin with a function $\phi$ defined on the unit circle ${\mathbb{T}}$ with
Fourier coefficients

$\phi_{k}=\frac{1}{2\pi}\int_{0}^{2\pi}\phi(e^{i\theta})\,e^{-ik\theta}\,\,d\theta,$
$\phi(e^{i\theta})=\sum_{-\infty}^{\infty}\phi_{k}\,e^{i\,k\,\theta}=\sum_{-\infty}^{\infty}\phi_{k}\,z^{k},$

($z=e^{i\theta}$) and consider the matrix

$T_{n}(\phi)\,=\,(\phi_{j-k})_{j,\,k\,=\,0,\,\cdots,\,n-1}.$

The matrix $T_{n}(\phi)$ is referred to as the finite Toeplitz matrix with
symbol $\phi.$

This matrix has the form

$\left[\begin{array}[]{ccccc}\phi_{0}&\phi_{-1}&\phi_{-2}&\cdots&\phi_{-(n-1)}\\\
\phi_{1}&\phi_{0}&\phi_{-1}&\cdots&\phi_{-(n-2)}\\\
\phi_{2}&\phi_{1}&\phi_{0}&\cdots&\phi_{-(n-3)}\\\
\cdots&\cdots&\cdots&\cdots&\cdots\\\ \cdots&\cdots&\cdots&\cdots&\cdots\\\
\phi_{n-1}&\phi_{n-2}&\phi_{n-3}&\cdots&\phi_{0}\end{array}\right]$

Since the 1950s there has been considerable interest in the asymptotics of the
determinants of Toeplitz matrices as $n\rightarrow\infty.$ The interest was
motivated by the study of the two-dimensional Ising model correlations, but
other important connections are related to many other areas of applied
mathematics, such as entanglement computations in statistical physics or
problems in random matrix theory, See for example [IJK, IMM, JK] for some
applications and [BS99] and [BS06] for more history and general facts about
Toeplitz operators.

For a certain class of “nice” symbols the asymptotics are described by the
Szegö-Widom Limit Theorem. This theorem states that if the symbol $\phi$
defined on the unit circle ${\mathbb{T}}$ has a sufficiently well-behaved
logarithm then the determinant of the Toeplitz matrix

$T_{n}(\phi)=(\phi_{j-k})_{\,\,j,\,k\,=\,0,\,\cdots,\,n-1}$

has the asymptotic behavior

$D_{n}(\phi):=\det T_{n}(\phi)\,\sim G(\phi)^{n}\,E(\phi)\ \ \ {\rm as}\
n\to\infty.$

Here are the constants:

$G(\phi)=e^{(\log\phi)_{0}}$

and

$E(\phi)=\det\left(T(\phi)T(\phi^{-1})\right)$

where $\phi^{-1}=1/\phi$ and

$T(\phi)=(\phi_{j-k})\,\,\,\,\,\,\,\,0\leq j,k<\infty$

is the Toeplitz operator defined on $\ell^{2},$ the space of complex sequences
$\\{x_{k}\\}_{k=0}^{\infty},$ with the usual $2$-norm.

If $\phi$ is in $L^{\infty}({\mathbb{T}})$ then it is well known that
$T(\phi)$ is a bounded operator on $\ell^{2}.$ More about the properties of
Toeplitz operators will be described later.

$E(\phi)$ is often referred to as Widom’s Constant since Widom discovered this
form of the constant when he extended the original Szegö theorem to the block
matrix case.

To make sense of Widom’s constant, that is the term
$\det\left(T(\phi)T(\phi^{-1})\right)$, we need to think about operators of
the form $I+K$ where $K$ is trace class. A compact operator $K$ is trace class
if the eigenvalues $s_{i}$ of $(K^{*}K)^{1/2}$ satisfy
$\sum_{i=0}^{\infty}s_{i}<\infty.$ When this is finite, we define the trace
norm of $K$ as

$||K||_{1}=\sum_{i=0}^{\infty}s_{i}.$

This set of operators is a closed ideal in the set of all bounded operators.

For trace class operators with discrete eigenvalues $\lambda_{i}$ one can show
that

$\sum_{i=0}^{\infty}|\lambda_{i}|<\infty$

and thus

$\det(I+K)=\prod_{i=0}^{\infty}(1+\lambda_{i})$

is well defined. Proofs of the above statements can be found in [GK].

We will later show that the operator $T(\phi)T(\phi^{-1})$ is of this form.

To state the necessary conditions for the Szegö-Widom theorem to hold, it is
useful to consider a certain Banach algebra (see [BS06], section 1.10).

Let $B$ stand for the set of all function $\phi$ such that the Fourier
coefficients satisfy

$\|\phi\|_{B}:=\sum_{k=-\infty}^{\infty}|\phi_{k}|+\Big{(}\sum_{k=-\infty}^{\infty}|k|\cdot|\phi_{k}|^{2}\Big{)}^{1/2}<\infty.$

With this norm and pointwise defined algebraic operations on $T$ the set $B$
becomes a Banach algebra of continuous functions on the unit circle.

The Szegö-Widom Limit Theorem holds provided that $\phi\in B$ and the function
$\phi$ does not vanish on ${\mathbb{T}}$ and has winding number zero. These
conditions imply that both $\phi^{-1}$ and $\log\phi$ (any continuous choice
of the logarithm will suffice) are defined and in $B.$

The most direct way to prove the Szegö-Widom theorem is to prove an identity
for the determinants, an identity called the Borodin-Okounkov-Case-Geronimo
(BOCG) identity. We should point out here that while the identity easily
yields the asymptotics, the asymptotics were known long before the identity.

In addition to the Toeplitz operator, we also define a Hankel operator

$\displaystyle T(\phi)$ $\displaystyle=$
$\displaystyle(\phi_{j-k}),\,\,\,\,\,\,\,\,\,\,\,\,\,0\leq j,k<\infty,$
$\displaystyle H(\phi)$ $\displaystyle=$
$\displaystyle(\phi_{j+k+1}),\,\,\,\,\,\,\,\,0\leq j,k<\infty.$

It is useful to think of the Toepltiz operator in matrix form, but it is also
worth pointing out that it can be defined as

$T(\phi):\ell^{2}\rightarrow\ell^{2}\,\,\,\,\quad T(\phi)f=P(\phi f)$

where $P$ is the Riesz projection onto $\ell^{2},$

$P:\sum_{k=-\infty}^{\infty}\phi_{k}e^{ik\theta}\rightarrow\sum_{k=0}^{\infty}\phi_{k}e^{ik\theta},$

and $f$ is the function identified by its Fourier coefficients
$\\{f_{n}\\}_{n=0}^{\infty}.$ In other words, start with the sequence
$\\{f_{n}\\}_{n=0}^{\infty}$ in $\ell^{2}$ and then let

$f(\theta)=\sum_{n=0}^{\infty}f_{n}e^{in\theta};$

multiply pointwise by the function $\phi$ and then project by chopping off all
negative Fourier coefficients of $\phi f$ to obtain a sequence back in
$\ell^{2}.$

For $\phi,\psi\in L^{\infty}(\mathbb{T})$ the well-known identities

$\displaystyle T(\phi\psi)$ $\displaystyle=$ $\displaystyle
T(\phi)T(\psi)+H(\phi)H({\tilde{\psi}})$ (3.1) $\displaystyle H(\phi\psi)$
$\displaystyle=$ $\displaystyle T(\phi)H(\psi)+H(\phi)T({\tilde{\psi}})$ (3.2)

where $\tilde{\phi}(e^{i\,\theta})=\phi(e^{-i\,\theta})$ are the basis of
almost everything we do later on. They can be proved by simply analyzing the
matrix entries of the operators and using convolution to find the Fourier
coefficients of products. They are also found in [BS06], Proposition 2.14.

It follows from these identities that if $\psi_{-}$ and $\psi_{+}$ have the
property that all their Fourier coefficients vanish for $k>0$ and $k<0$,
respectively, then

$\displaystyle T(\psi_{-}\phi\psi_{+})$ $\displaystyle=$ $\displaystyle
T(\psi_{-})T(\phi)T(\psi_{+}),$ $\displaystyle
H(\psi_{-}\phi\tilde{\psi}_{+})$ $\displaystyle=$ $\displaystyle
T(\psi_{-})H(\phi)T(\psi_{+}).$

Now suppose that the conditions of the Szegö-Widom theorem hold. Then there
exists a well-defined logarithm of $\phi$ that is in $B$. We can then write
$\log\phi$ as a sum of two functions, one with Fourier coefficients that
vanish for positive indices, and one with Fourier coefficients that vanish for
negative indices. If we then exponentiate these two functions, the factors
(and their inverses) have the same Fourier properties and are in $B$. Thus we
see that

$\phi=\phi_{-}\phi_{+},\,\,\,\phi^{-1}=\phi_{-}^{-1}\phi_{+}^{-1}.$

This factorization is known as a Wiener-Hopf factorization.

The BOCG identity can now be stated as follows.

######  Theorem 3.1.

Suppose $\phi$ satisfies the conditions of the Szegö-Widom theorem. Then

$\displaystyle D_{n}(\phi)$ $\displaystyle=$ $\displaystyle
G(\phi)^{n}\,\,E(\phi)\cdot\det\left(I-H(z^{-n}\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+}z^{-n})\right)$

In the above $z=e^{i\theta}.$

###### Proof.

We let $P_{n}$ be the projection defined on $\ell^{2}$ that maps

$\\{x_{k}\\}_{k=0}^{\infty}\,\,\,\mbox{to}\,\,\,\\{x_{0},\dots,x_{n-1},0,0,\dots\\}$

identifying a function with its Fourier coefficients, and we define
$Q_{n}=I-P_{n}.$

It is easy to check that

$Q_{n}=T(z^{n})T(z^{-n}),\,\,\,I=T(z^{-n})T(z^{n}).$

In what follows and for the rest of the paper, determinants are always defined
on the appropriate space. Thus for example, $\det P_{n}AP_{n}$ is the
determinant defined on the image of $P_{n}.$

Next we make some simple observations:

1) If $\phi_{k}=0$ for $k<0$ or $k>0,$ then the Toeplitz matrices are
triangular and $D_{n}(\phi)=\det T_{n}(\phi)=(\phi_{0})^{n}.$

2) $T_{n}(\phi)=P_{n}T(\phi)P_{n}$ i.e. $T_{n}(\phi)$ is the upper left corner
of the Toeplitz operator $T(\phi).$

The projection $P_{n}$ has the nice property that if $U$ is an operator whose
matrix representation has an upper triangular form, then

$P_{n}UP_{n}=UP_{n}.$

And if $L$ is an operator whose matrix representation has a lower triangular
form, then

$P_{n}LP_{n}=P_{n}L.$

So if we had an operator of the form $LU,$ then

$P_{n}LUP_{n}=P_{n}LP_{n}UP_{n}$

and the corresponding determinants would be easy to compute.

What happens for Toeplitz operators is the opposite.

Toeplitz operators generally do not factor as $L\,U$ but rather as $U\,L.$

To see this, we use a Wiener-Hopf factorization. We have

$\phi=\phi_{-}\,\phi_{+},\,\,\,\,\ \phi^{-1}=\phi_{-}^{-1}\phi_{+}^{-1}.$

Recall this means that $(\phi_{+})_{k}=0$ for $k<0$ and thus $T(\phi_{+})$ is
lower triangular and that $(\phi_{-})_{k}=0$ for $k>0$ and $T(\phi_{-})$ is
upper triangular. We also know from the algebra property (3.1) that the
factorization of $\phi$ shows that

$T(\phi)=T(\phi_{-})\,T(\phi_{+}).$

One final observation, the algebra property (3.1) inplies that

$T(\phi_{+}^{-1})T(\phi_{+})=T(\phi_{+})T(\phi_{+}^{-1})=I$

and that

$T(\phi_{-}^{-1})T(\phi_{-})=T(\phi_{-})T(\phi_{-}^{-1})=I.$

We write

$\displaystyle P_{n}T(\phi)P_{n}$ $\displaystyle=$ $\displaystyle
P_{n}T(\phi_{-})\,T(\phi_{+})P_{n}$ $\displaystyle=$ $\displaystyle
P_{n}T(\phi_{+})T(\phi_{+}^{-1})T(\phi_{-})\,T(\phi_{+})T(\phi_{-}^{-1})T(\phi_{-})P_{n}$
$\displaystyle=$ $\displaystyle
P_{n}T(\phi_{+})P_{n}T(\phi_{+}^{-1})T(\phi_{-})\,T(\phi_{+})T(\phi_{-}^{-1})P_{n}T(\phi_{-})P_{n}.$

Taking determinants, we have

$\det P_{n}T(\phi_{+})P_{n}=((\phi_{+})_{0})^{n}$

and

$\det P_{n}T(\phi_{-})P_{n}=((\phi_{-})_{0})^{n}.$

So at this point we have

$D_{n}(\phi)=G(\phi)^{n}\det
P_{n}T(\phi_{+}^{-1})T(\phi_{-})\,T(\phi_{+})T(\phi_{-}^{-1})P_{n}.$

The term

$P_{n}T(\phi_{+}^{-1})T(\phi_{-})\,T(\phi_{+})T(\phi_{-}^{-1})P_{n}$

is of the form

$P_{n}(I+K)P_{n}$

where $K$ is trace class. To see this we note that an operator is Hilbert-
Schmidt if, in matrix form, its entries satisfy

$\sum_{i,j\geq 0}|a_{i,j}|^{2}<\infty,$

and that it is well-known that the product of two Hilbert-Schmidt operators is
trace class. For a good reference for this fact and general properties of
trace class and Hilbert-Schmidt operators, please see [GK].

Now the operator

$T(\phi_{+}^{-1})T(\phi_{-})=T(\phi_{+}^{-1}\phi_{-})-H(\phi_{+}^{-1})H(\tilde{\phi_{-}})$

and the Banach algebra norm condition means that each of the above Hankel
operators are Hilbert-Schmidt and thus their product is trace class. Thus

$T(\phi_{+}^{-1})T(\phi_{-})=T(\phi_{+}^{-1}\phi_{-})+K_{1}$

where $K_{1}$ is trace class. Applying the same argument to
$T(\phi_{+})T(\phi_{-}^{-1})$ we have

$T(\phi_{+})T(\phi_{-}^{-1})=T(\phi_{+}\phi_{-}^{-1})+K_{2}$

where $K_{2}$ is trace class. Hence

$T(\phi_{+}^{-1})T(\phi_{-})T(\phi_{+})T(\phi_{-}^{-1})=(T(\phi_{+}^{-1}\phi_{-})+K_{1})(T(\phi_{+}\phi_{-}^{-1})+K_{2})$
$=T(\phi_{+}^{-1}\phi_{-})T(\phi_{+}\phi_{-}^{-1})+T(\phi_{+}^{-1}\phi_{-})K_{2}+K_{1}T(\phi_{+}\phi_{-}^{-1})+K_{1}K_{2}.$

This yields for a trace class $K_{3}$

$T(1)+K_{3}+T(\phi_{+}^{-1}\phi_{-})K_{2}+K_{1}T(\phi_{+}\phi_{-}^{-1})+K_{1}K_{2}.$
$=I+K$

since $T(1)=I$ and trace class operators form an ideal.

To obtain the next step we use the Jacobi identity

$\det P_{n}AP_{n}=\det A\cdot\det(Q_{n}A^{-1}Q_{n})$

in which $Q_{n}=I-P_{n}$ and $A$ is an invertible operator of the form
identity plus trace class.

This holds because

$\displaystyle\det P_{n}AP_{n}$ $\displaystyle=$
$\displaystyle\det(Q_{n}+P_{n}AP_{n})=\det(Q_{n}+AP_{n})$ $\displaystyle=$
$\displaystyle\det A\cdot\det(A^{-1}Q_{n}+P_{n})=\det
A\cdot\det(Q_{n}A^{-1}Q_{n}).$

We apply this to $A=T(\phi_{+}^{-1})T(\phi_{-})\,T(\phi_{+})T(\phi_{-}^{-1})$.
It follows that

$\displaystyle\det
P_{n}T(\phi_{+}^{-1})T(\phi_{-})\,T(\phi_{+})T(\phi_{-}^{-1})P_{n}$
$\displaystyle=$ $\displaystyle\det
T(\phi_{+}^{-1})T(\phi_{-})\,T(\phi_{+})T(\phi_{-}^{-1})$
$\displaystyle\qquad\times\det
Q_{n}T(\phi_{-})T(\phi_{+}^{-1})\,T(\phi_{-}^{-1})T(\phi_{+})Q_{n}.$

Using the fact that $\det AB=\det BA$ the first term on the right is easily
seen to be $\det T(\phi)T(\phi^{-1})=E(\phi).$ Thus we now have

$D_{n}(\phi)=G(\phi)^{n}E(\phi)\det
Q_{n}T(\phi_{-})T(\phi_{+}^{-1})\,T(\phi_{-}^{-1})T(\phi_{+})Q_{n}.$

Now

$\displaystyle\det Q_{n}A^{-1}Q_{n}$ $\displaystyle=\det
Q_{n}T(\phi_{-})T(\phi_{+}^{-1})\,T(\phi_{-}^{-1})T(\phi_{+})Q_{n}$
$\displaystyle=\det(P_{n}+Q_{n}T(\phi_{-}\phi_{+}^{-1})\,T(\phi_{-}^{-1}\phi_{+})Q_{n})$
$\displaystyle=\det(P_{n}+Q_{n}(I-H(\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+}))Q_{n})$
$\displaystyle=\det(I-Q_{n}H(\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+})Q_{n})$
$\displaystyle=\det(I-T(z^{n})T(z^{-n})H(\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+})T(z^{n})T(z^{-n}))$
$\displaystyle=\det(I-T(z^{-n})H(\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+})T(z^{n}))$
$\displaystyle=\det(I-H(z^{-n}\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+}z^{-n}))$

and the identity is proved. For other proofs (including two original ones) we
refer the reader to [BW, BO, Bot, GC].

From this BOCG identity we have an instant proof of the Szegö-Widom theorem.
Notice that $Q_{n}f$ tends to zero for any fixed $f$ as $n\rightarrow\infty$
and this allows one to say that

$Q_{n}\,H(\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+})\,Q_{n}$

tends to zero in the trace norm. Thus the determinant

$\det(I-Q_{n}\,H(\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+})\,Q_{n})$

tends to one and

$D_{n}(\phi)=\det T_{n}(\phi)\,\sim G(\phi)^{n}\,E(\phi)\ \ \ {\rm as}\
n\to\infty.$

For more details see [BW].

The constant $E(\phi)$ has a nice concrete description.

If we have a Wiener-Hopf factorization for $\phi=\phi_{-}\phi_{+}$, then

$T(\phi)T(\phi^{-1})=T(\phi_{-})T(\phi_{+})T^{-1}(\phi_{-})T^{-1}(\phi_{+})$

and this is of the form

$e^{\,A}e^{\,B}e^{-A}e^{-B}$

where

$A=T(\log(\phi_{-})),\,\,\,\,B=T(\log(\phi_{+})).$

From this we can use a formula for determinants of multiplicative commutators
of this form,

$\det\,(e^{A}e^{B}e^{-A}e^{-B})=\exp\,\,(\operatorname{trace}(AB-BA)).$

It is tempting to think that the above formula is trivial, that is, that the
left and right sides should be $1$ since the trace of $AB$ should be the trace
of $BA$ and that the operators on the left side should be inverses of each
other. But this is not the case. For our $A$ and $B,$ $AB$ is not trace class
(indeed not even compact), but from our Banach algebra conditions what is true
is that $AB-BA$ is trace class. Moreover, the operators $e^{A}$ and $e^{B}$ do
not commute. In this situation the formula above yields a non-trivial answer.
This was first used in this context in [Wi76].

Now $AB-BA$ is the same as

$T(\log(\phi_{-}))T(\log(\phi_{+}))-T(\log(\phi_{+}))T(\log(\phi_{-}))$
$=T(\log(\phi_{-})(\log(\phi_{+}))-T(\log(\phi_{+}))T(\log(\phi_{-}))$

which by 3.1 is

$H(\log(\phi_{+}))H(\widetilde{\log(\phi_{-})}).$

After computing the trace and exponentiating we arrive at

$\det
T(\phi)T(\phi^{-1})=\exp\left(\sum_{k=1}^{\infty}k\,(\log\phi)_{k}\,(\log\phi)_{-k}\right).$

Now we turn to a different class of structured operators.

##  4\. Toeplitz plus Hankel operators

Many other classes of structured matrices arise in applications other than
finite Toeplitz matrices. For example, one may want to find asymptotics for
the determinants of matrices generated by the Fourier coefficients of $\phi$
with $j,k$ entry

$\phi_{j-k}+\phi_{j+k+1}.$

The purpose of this section is to state identities similar to the BOCG type
proved in the previous section. The proofs of these are similar (but not
exactly the same) as the one done in the previous section and the proofs and
additional details can be found in [BE4].

Let $\mathcal{S}$ stand for a unital Banach algebra of functions on the unit
circle continuously embedded into $L^{\infty}({\mathbb{T}})$ with norm
$||\cdot||_{S}$ and with the property that $\phi\in\mathcal{S}$ implies that
$\tilde{\phi}\in\mathcal{S}$ and $P\phi\in\mathcal{S}$. Moreover, define

$\displaystyle\mathcal{S}_{-}$ $\displaystyle=$
$\displaystyle\Big{\\{}\phi\in\mathcal{S}\;:\;\phi_{n}=0\mbox{ for all
}n>0\Big{\\}},$ $\displaystyle\mathcal{S}_{0}$ $\displaystyle=$
$\displaystyle\Big{\\{}\phi\in\mathcal{S}\;:\;\phi=\tilde{\phi}\Big{\\}}.$

In the above $\tilde{\phi}(e^{i\theta})=\phi(e^{-i\theta})$ and $P$ is the
Riesz projection.

Assume that $M:\phi\in L^{\infty}\mapsto M(\phi)\in\mathcal{L}(\ell^{2})$ (the
set of bounded operators on $\ell^{2}$) is a continuous linear map such that:

  1. (a)

If $\phi\in\mathcal{S}$, then $M(\phi)-T(\phi)$ is trace class and there
exists some constant $C$ such that

$\|M(\phi)-T(\phi)\|_{1}\leq C\,\|\phi\|_{\mathcal{S}}.$

  2. (b)

If $\psi\in\mathcal{S}_{-}$, $\phi\in\mathcal{S}$, $\gamma\in\mathcal{S}_{0}$,
then

$M(\psi\phi\gamma)=T(\psi)M(\phi)M(\gamma).$

  3. (c)

$M(1)=I$.

Then we say $M$ and $\mathcal{S}$ are compatible pairs.

All of the following can be realized as compatible pairs.

  1. (I)

$M(\phi)=T(\phi)+H(\phi)$,

  2. (II)

$M(\phi)=T(\phi)-H(\phi)$,

  3. (III)

$M(\phi)=T(\phi)-H(z^{-1}\phi)$

  4. (IV)

$M(\phi)=T(\phi)+H(z\,\phi)Q_{1}$

The matrix representations of the operators are of the form

$(\phi_{j-k}\pm\phi_{j+k-\varkappa+1})$

with $\varkappa=0,1,-1.$

A Banach algebra that satisfies the compatible pair conditions for all four of
the above cases is the set of all $\phi$ such that

$|\phi_{0}|+||H(\phi)||_{1}+||H(\tilde{\phi})||_{1}<\infty.$

We are interested in the determinants (where the matrices or operators are
always thought of as acting on the image of the projection of the appropriate
space) of

$P_{n}\,M(\phi)\,P_{n},$

or for our examples, determinants of finite Toeplitz plus finite Hankel
matrices.

As mentioned earlier the analog of the BOCG for the determinants of
$P_{n}M(\phi)P_{n}$ was proved in [BE4]. Here is the statement of the identity
for even functions ($\phi=\tilde{\phi}$), but can be made more general. Note
that in the theorem below $E(\phi)$ always denotes the constant in the
identity and is different depending on the compatible pair.

######  Theorem 4.1.

Let $M$ and $\mathcal{S}$ be a compatible pair, and let
$\beta_{+}\in\mathcal{S}_{+}$. Let $\phi=\phi_{+}\tilde{\phi}_{+}=\exp(\beta)$
with $\phi_{+}=\exp(\beta_{+})$, $\beta=\beta_{+}+\tilde{\beta}_{+}$. Then

$\det P_{n}M(\phi)P_{n}=G(\phi)^{n}E(\phi)\det(I+Q_{n}\,K\,Q_{n}),$

where

$E(\phi)=\exp\Big{(}\operatorname{trace}(M(\beta)-T(\beta))+\frac{1}{2}\operatorname{trace}\,H(\beta)^{2}\Big{)},$

and $K=M(\phi_{+}^{-1})T(\phi_{+})-I.$

For case (I), $(\phi_{j-k}+\phi_{j+k+1}),$

$K=M(\phi_{+}^{-1})T(\phi_{+})-I=H(\phi_{+}^{-1})T(\phi_{+})=H(\tilde{\phi}_{+}/\phi_{+}).$

For case (II), $(\phi_{j-k}-\phi_{j+k+1}),$

$K=M(\phi_{+}^{-1})T(\phi_{+})-I=-H(\phi_{+}^{-1})T(\phi_{+})=-H(\tilde{\phi}_{+}/\phi_{+}).$

For case (III), $(\phi_{j-k}-\phi_{j+k+2}),$

$K=M(\phi_{{+}}^{-1})T(\phi_{+})-I,$ $M(\phi)=T(\phi)-H(z^{-1}\phi),$

we have

$\displaystyle K$ $\displaystyle=$
$\displaystyle(T(\phi_{+}^{-1})-H(z^{-1}\phi_{+}^{-1}))T(\phi_{+})-I$
$\displaystyle=$ $\displaystyle
T(\phi_{+}^{-1})T(\phi_{+})-H(z^{-1}\phi_{+}^{-1}))T(\phi_{+})-I$
$\displaystyle=$ $\displaystyle-H(z^{-1}\phi_{+}^{-1}))T(\phi_{+})$
$\displaystyle=$ $\displaystyle-H(z^{-1}\phi_{+}^{-1}\tilde{\phi_{+}}).$

For case (IV)

$K=M(\phi_{+}^{-1})T(\phi_{+})-I,$ $M(\phi)=T(\phi)+H(z\phi)Q_{1},$

we have

$\displaystyle K$ $\displaystyle=$
$\displaystyle(T(\phi_{+}^{-1})+H(z\,\phi_{+}^{-1})Q_{1})T(\phi_{+})-I$
$\displaystyle=$ $\displaystyle
T(\phi_{+}^{-1})T(\phi_{+})+H(z\,\phi_{+}^{-1})Q_{1}T(\phi_{+})-I$
$\displaystyle=$ $\displaystyle H(z\,\phi_{+}^{-1})Q_{1}T(\phi_{+}).$

Note for case (IV),

$M(\phi)\,R=(\phi_{j-k}+\phi_{j+k})$

where $R=2P_{1}+Q_{1},$ that is, a diagonal matrix with $2$ in the upper-left
entry and ones otherwise.

Note also that

$\det P_{n}\,M(\phi)\,R\,P_{n}=2\det P_{n}\,M(\phi)\,P_{n}.$

The need for formulas for the above determinants and their asymptotics arise
in various applications. One example is the connection to averages over
certain groups or cosets of groups. We will now state the connections. The
measure in all cases is normalized Haar measure. All of the results are found
in Section 2 of [BR].

######  Proposition 4.2.

  1. (1)

Let $I_{U(N)}(\phi)$ denote the average over the unitary group $U(N)$ of
$\det\phi(U).$ Then

$I_{U(N)}(\phi)=D_{N}(\phi).$

  2. (2)

Let $N=2m,\phi=g\tilde{g}$. Then

$I_{O^{-}(N+1)}(g)=g(-1)\det P_{m}(\phi_{j-k}+\phi_{j+k+1})P_{m}$

averaging $\det g(U)$ over the coset of orthogonal matrices with determinant
$-1.$

  3. (3)

Let $N=2m,\phi=g\tilde{g}$.Then

$I_{O^{+}(N+1)}(g)=g(1)\det P_{m}(\phi_{j-k}-\phi_{j+k+1})P_{m}$

averaging $\det g(U)$ over the subgroup of matrices with determinant $1.$

  4. (4)

Let $N=2m,\phi=g\tilde{g}$ Then

$I_{USp(N)}(g)=\det P_{m}(\phi_{j-k}-\phi_{j+k+2})P_{m}$

averaging $\det g(U)$ over the symplectic group.

  5. (5)

Let $N=2m,\phi=g\tilde{g}.$ Then

$I_{O^{+}(N)}(g)=\frac{1}{2}\det P_{m}(\phi_{j-k}+\phi_{j+k})P_{m}$

averaging $\det g(U)$ over the subgroup of matrices with determinant $1.$

  6. (6)

Let $N=2m,\phi=g\tilde{g}.$ Then

$I_{O^{-}(N)}(g)=g(1)g(-1)\det P_{m-1}(\phi_{j-k}-\phi_{j+k+2})P_{m-1}$

averaging $\det g(U)$ over the coset of matrices with determinant $-1.$

In the following sections we shall see how the above formulas and the
determinant identities of this section arise in number theory calculations.

##  5\. Factoring determinants

The example given in the introduction showed a factorization of the averages
over the unitary groups for the $\det\phi$ when $\phi$ was a certain rational
function. It turns out that is always the case for even functions.

The following theorem describes the factorization of the Toeplitz determinant
when $N$ is even.

######  Theorem 5.1.

Suppose $\phi=g\tilde{g}$ satisfies the conditions of Theorem 4.1. Let $N=2m.$
Then

$I_{U(N)}(g\tilde{g})=(g(1)g(-1))^{-1}I_{O^{+}(N+1)}(g)I_{O^{-}(N+1)}(g),$

or in the operator language $D_{N}(\phi)$ equals

$\det P_{m}(T(\phi)+H(\phi))P_{m}\times\det P_{m}(T(\phi)-H(\phi))P_{m},$

Here $g(\pm 1)$ means substitute $\pm 1$ for $e^{i\theta}$ in the formula for
$g$. So for example, if $g=1-ae^{i\theta}$ then $g(1)=1-a$ and $g(-1)=1+a$.

###### Proof.

The proof of this follows directly from the formulas of Theorems 3.1 and 4.1.
It is easy to see that both the geometric mean part and the constant term part
of the determinants on the right-hand side of the above multiply to give the
correct term for the Toeplitz or Unitary case. For the Fredholm determinant
part of the identity we have

$\det(I-Q_{N}\,K\,Q_{N})$

where

$K=H(\tilde{\phi_{+}}/\phi_{+})^{2}.$

Now

$Q_{N}H(\tilde{\phi_{+}}/\phi_{+})^{2}Q_{N}=T(z^{2m})T(z^{-2m})H(\tilde{\phi_{+}}/\phi_{+})^{2}T(z^{2m})T(z^{-2m}).$

This is the same as

$T(z^{2m})T(z^{-m})H(z^{-m}\tilde{\phi_{+}}/\phi_{+})H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{m})T(z^{-2m})$

or

$\displaystyle T(z^{2m})T(z^{-m})H(z^{-m}\tilde{\phi_{+}}/\phi_{+})T(z^{-m})$
$\displaystyle\qquad\times
T(z^{m})H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{m})T(z^{-2m}),$

which equals

$T(z^{2m})T(z^{-m})H(\tilde{\phi_{+}}/\phi_{+}))T(z^{m})T(z^{-m})H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-2m})$

Hence the determinant of $I-Q_{N}\,K_{u}\,Q_{N}$ is the same as

$\det(I-T(z^{m})Q_{m}H(\tilde{\phi_{+}}/\phi_{+})Q_{m}H(\tilde{\phi_{+}}/\phi_{+})Q_{m}T(z^{-m}))$

or

$\displaystyle\det(I-Q_{m}H(\tilde{\phi_{+}}/\phi_{+})Q_{m}H(\tilde{\phi_{+}}/\phi_{+})Q_{m})$
$\displaystyle\qquad=\det(I-Q_{m}H(\tilde{\phi_{+}}/\phi_{+})Q_{m})\det(I+Q_{m}H(\tilde{\phi_{+}}/\phi_{+})Q_{m})$

and the result follows. ∎

Next we have the case when $N$ is odd.

######  Theorem 5.2.

Suppose $\phi=g\tilde{g}$ satisfies the conditions of Theorem 4.1. Let
$N-1=2m,$

$I_{U(N)}(g\tilde{g})=I_{USp(N-1)}(g)I_{O^{+}(N+1)}(g),$

or in the operator language $D_{N}(\phi)$ equals

$\det P_{m}(T(\phi)-H(z^{-1}\phi))P_{m}\times\frac{1}{2}\det
P_{m+1}(T(\phi)+H(z\phi)Q_{1})R\,P_{m+1}$ $=\det
P_{m}(T(\phi)-H(z^{-1}\phi))P_{m}\times\det
P_{m+1}(T(\phi)+H(z\phi)Q_{1})\,P_{m+1}.$

As before, each formula has a $G(a)$ part, a constant part, and a Fredholm
determinant part.

For $I_{U(N)},$ the exponent of $G(a)$ is $N=2m+1.$ For $I_{USp(N-1)}$ the
exponent is $m$ and for $I_{O^{+}(N+1)},$ $m+1.$

We write
$\phi=\phi_{+}\tilde{\phi_{+}}=\exp\beta=\exp(\beta_{+}+\tilde{\beta_{+}}).$
For $I_{U(N)},$ we have constant term
$E(\phi)=\exp\,(\operatorname{trace}\,H(\beta)^{2})$.

For $I_{USp(N-1)}$ the constant term is

$\exp\operatorname{trace}(-H(z^{-1}\beta)+\frac{1}{2}H(\beta)^{2}).$

For $I_{O^{+}(N+1)}$ it is

$\frac{1}{2}\times
2\exp\operatorname{trace}(H(z\,\beta)Q_{1}+\frac{1}{2}H(\beta)^{2}).$

But then the product of the last two is

$\exp(-\sum_{j=1}^{\infty}\beta_{2j}+\sum_{j=1}^{\infty}\beta_{2j}+\operatorname{trace}\,H(\beta)^{2}).$

Now we compare the Fredholm determinants. For the unitary case, we have

$\det(I-Q_{N}\,K_{u}\,Q_{N})$

where

$K_{u}=(H(\tilde{\phi_{+}}/\phi_{+}))^{2}.$

For the symplectic case we have

$\det(I-Q_{m}\,K_{s}\,Q_{m})$

with

$K_{s}=-H(z^{-1}\tilde{\phi_{+}}/\phi_{+}).$

And for the orthogonal case we have

$\det(I-Q_{m+1}\,K_{o}\,Q_{m+1})$

with

$K_{o}=H(z/\phi_{+})Q_{1}T(\phi_{+}).$

Thus we consider

$\det(I+Q_{m}\,H(z^{-1}\tilde{\phi_{+}}/\phi_{+})\,Q_{m})\det(I-Q_{m+1}\,H(z/\phi_{+})Q_{1}T(\phi_{+})\,Q_{m+1}).$

The determinant

$\det(I+Q_{m}\,H(z^{-1}\tilde{\phi_{+}}/\phi_{+})\,Q_{m})$

is the same as

$\det(I+T(z^{m})T(z^{-m})T(z^{-1})H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-m}))$

or

$\det(I+T(z^{m+1})T(z^{-m-1})H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-m-1}))$
$\det(I+Q_{m+1}H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-m-1})).$

The determinant

$\det(I-Q_{m+1}\,H(z/\phi_{+})Q_{1}T(\phi_{+})\,Q_{m+1})$

is the same as

$\det(I-Q_{m+1}H(z\,/\phi_{+})T(z)T(z^{-1})T(\phi_{+})T(z^{m+1})T(z^{-m-1}))$

or

$\det(I-Q_{m+1}H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-m-1})).$

Thus for the product we have $\det(I-L)$ where $L$ is

$Q_{m+1}H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-m-1})Q_{m+1}H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-m-1})$
$Q_{m+1}H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-m-1})H(\tilde{\phi_{+}}/\phi_{+})T(z^{m})T(z^{-m-1})$
$T(z^{-m})Q_{2m+1}H(\tilde{\phi_{+}}/\phi_{+})H(\tilde{\phi_{+}}/\phi_{+})T(z^{2m+1})T(z^{-m-1}),$

and

$\det(I-T(z^{-m})Q_{2m+1}H(\tilde{\phi_{+}}/\phi_{+})H(\tilde{\phi_{+}}/\phi_{+})T(z^{2m+1})T(z^{-m-1}))$
$=\det(I-Q_{2m+1}H(\tilde{\phi_{+}}/\phi_{+})H(\tilde{\phi_{+}}/\phi_{+})Q_{2m+1})$
$=\det(I-Q_{N}\,K_{u}\,Q_{N}).$

Thus we have shown that when $N$ is odd, $D_{N}(\phi)$ is the same as

$\det P_{m}(T(\phi)-H(z^{-1}\phi))P_{m}\times\det
P_{m+1}(T(\phi)+H(z\phi)Q_{1})\,P_{m+1}$

which is our desired result.

##  6\. New proof of ratios formula

We now use the identities described in the previous sections, along with an
expansion for a Fredholm determinant to prove the ratios formulas mentioned in
the introduction. In this section we use the following notation.

For sets of complex numbers $A$ and $B$ let

$Z(A,B)=\prod_{a\in A\atop b\in B}(1-ab)^{-1}.$

Further, let

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

Also, define

$Z_{S}(A;C)=\frac{Z_{S}(A)Z_{O}(C)}{Z(A,C)}$

where, if $A=\\{a_{1},\dots,a_{k}\\}$ then

$Z_{S}(A)=\prod_{1\leq i\leq j\leq k}(1-a_{i}a_{j})^{-1}$

and

$Z_{O}(A)=\prod_{1\leq i<j\leq k}(1-a_{i}a_{j})^{-1}$

For the set $A$ with $U\subset A,$ the set $A-U+U^{-1}$ is the set $A$ with
the elements of $U$ removed and with the inverses of the elements of $U$ put
back in. Finally, $U^{N}=\prod_{a_{i}\in U}a_{I}^{N}.$

######  Theorem 6.1.

Let

$g=\prod_{i=1}^{k}\frac{(1-a_{i}e^{i\theta})}{(1-c_{i}e^{i\theta})},$

and let $A=\\{a_{1},a_{2},\dots,a_{k}\\}$. Suppose that $|c_{i}|<1$ for all
$i.$ Then

$I_{Sp(2m)}(g)=\sum_{U\subset A}U^{N}Z_{S}(A-U+U^{-1};C)$

###### Proof.

We first note that the function $g$ satisfies the conditions of Theorem 4.1
providing that $|a_{i}|<1$ for all $i$. We make that assumption in our
computation below, but then by analytic continuation the result holds for all
$a_{i}.$

We use the identity in Theorem 4.1 to compute the determinant corresponding to
the symplectic case exactly. Our symbol is

$\phi(\theta)=g\tilde{g}=\prod_{i=1}^{k}\frac{(1-a_{i}e^{i\theta})(1-a_{i}e^{-i\theta})}{(1-c_{i}e^{i\theta})(1-c_{i}e^{-i\theta})}.$

We see for this symbol the geometric mean $G(\phi)=1.$ To compute $E(\phi)$ is
straightforward and is seen to be

$\frac{\prod_{i,j}(1-a_{i}c_{j})}{\prod_{1\leq i\leq j\leq
k}(1-a_{i}a_{j})\prod_{1\leq i<j\leq k}(1-c_{i}c_{j})}.$

Now for the Fredholm determinant part of the identity, we use the standard
Fredholm expansion. Recall that if we have a discrete kernel $K(i,j),$ for the
operator $K,$ then the determinant of $I-K$ is given by a sum

$1+\sum_{l=1}^{\infty}\frac{(-1)^{l}}{l!}\left(\sum_{i_{1},i_{2},\cdots,i_{l}}\det
K(i_{j},i_{k})_{1\leq j,k\leq l}\right).$

The first term here is simply the trace of the operator $K$. The second term
is a two-by-two determinant, and so on.

In our case we need to apply this to the Hankel operator

$K_{s}=Q_{m}H(z^{-1}\tilde{\phi_{+}}/\phi_{+})Q_{m}.$

So the first thing we need are the positive Fourier coefficients of the
function, $\tilde{\phi_{+}}/\phi_{+}.$ These are the Fourier coefficients of

$\prod_{i=1}^{k}\frac{(1-a_{i}e^{-i\theta})(1-c_{i}e^{i\theta})}{(1-a_{i}e^{i\theta})(1-c_{i}e^{-i\theta})}.$

The Fourier $l$th coefficient is given by

$\sum_{j=1}^{k}\alpha_{j}a_{j}^{l-1},\,\,\,\,\,\,\alpha_{j}=\prod_{i\neq
j}\frac{1}{(a_{j}-a_{i})}\prod_{i}\frac{(1-a_{i}a_{j})(a_{j}-c_{i})}{(1-c_{i}a_{j})}.$

This is found by writing the usual integral for the Fourier coefficients, then
letting $z=e^{i\theta}$ on the contour $|z|=1$ and computing the residues at
infinity and the points $z=1/a_{i}.$ Notice first that any term in the
Fredholm expansion with index greater than $k$ must vanish. This is easy to
see since if we expand the determinants in columns one would need to pick two
columns with the same $a_{j}$’s and thus the determinant vanishes.

Notice also the effect of the $Q_{m}$ operators is to simply cut off the first
$m$ rows and columns of the Hankel matrix. Now we will compare the terms of
the sum with the terms in the ratio expansion. Letting $U$ be the empty set,
the zero’th term in the ratio expansion is $Z_{S}(A,C)$ which is exactly the
$E(\phi)$ term

$\frac{\prod_{i,j}(1-a_{i}c_{j})}{\prod_{1\leq i\leq j\leq
k}(1-a_{i}a_{j})\prod_{1\leq i<j\leq k}(1-c_{i}c_{j})}.$

Now suppose that the set $U$ has cardinality one. Let $U=\\{a_{1}\\}.$
Consider the first term in the expansion of our determinant with
$K=Q_{m}H(z^{-1}\tilde{\phi_{+}}/\phi_{+})Q_{m}.$ This is the trace of the
operator $K.$ The operator $K$ is a sum of rank one operators (one for each
$a_{j}$). One of these when $j=1$ has trace

$\alpha_{1}a_{1}^{2m+1}/(1-a_{1}^{2})=a_{1}^{2m+1}/(1-a_{1}^{2})\prod_{i\neq
1}\frac{1}{(a_{1}-a_{i})}\prod_{i}\frac{(1-a_{i}a_{1})(a_{1}-c_{i})}{(1-c_{i}a_{1})}.$

After multiplying through by the $E(\phi)$ term, we have

$-a_{1}^{2m}\frac{\prod_{i\neq
1,j}(1-a_{i}c_{j})\prod_{i}(1-c_{i}/a_{1})}{(1-1/a_{1}^{2})\prod_{\begin{subarray}{c}j\neq
1\\\ 1\leq i\leq j\leq k\end{subarray}}(1-a_{i}a_{j})\prod_{1\leq i<j\leq
k}(1-c_{i}c_{j})\prod_{i\neq 1}(1-a_{i}/a_{1})}.$

After adjusting for the minus sign, this is exactly the answer given by the
ratios formula for $Sp(2m).$ And thus the trace term in the Fredholm expansion
yields the sum of such terms, one for each set of cardinality one.

Now for the general term in the Fredholm expansion, we let $n=l$ and suppose
that $l\leq k.$ We need to compute the term

$\sum_{i_{1},i_{2},\cdots,i_{l}}\det K(i_{j},i_{h})_{1\leq j,h\leq l}.$

To do this, first recall that each column of $K(i_{j},i_{h})_{1\leq j,h\leq
l}$ is a sum of the terms $\alpha_{j}a_{j}^{i_{j}+i_{h}+1}.$ We can expand the
determinant by the columns. Thus when two of the $a_{j}$’s are the same the
determinant vanishes since each is a multiple of the other. It follows that we
only get a contribution when the $a_{j}$’s are distinct.

Now suppose that we wish to compute the determinant for the set of $a_{j}$’s
with $j=1,2,\dots,l.$ First, we factor out the product

$\prod_{j}\alpha_{j}=\prod_{j}\left(\prod_{i\neq
j}\frac{1}{(a_{j}-a_{i})}\prod_{i}\frac{(1-a_{i}a_{j})(a_{j}-c_{i})}{(1-c_{i}a_{j})}\right)$

or

$\prod_{\begin{subarray}{c}i\neq j\\\ 1\leq j\leq l\\\ 1\leq i\leq
k\end{subarray}}\frac{1}{(a_{j}-a_{i})}\prod_{\begin{subarray}{c}1\leq j\leq
l\\\ 1\leq i\leq
k\end{subarray}}\frac{(1-a_{i}a_{j})(a_{j}-c_{i})}{(1-c_{i}a_{j})}.$

What is left is columns with entries of the form $a_{j}^{i_{g}+i_{h}+1}.$ We
can also factor out the product of the $a_{j}$’s to have columns that are
simply of the from $a_{j}^{i_{g}+i_{h}}.$ We next expand this using the
permutation definition of the determinants for some fixed choice of the
columns, To begin we assume the $a_{j}$’s are in ascending order in the
columns. ($\sigma$ denotes a permutation of the set $\\{1,2,\cdots,l\\}.)$

$\sum_{\sigma}sgn(\sigma)a_{1}^{i_{\sigma(1)}+i_{1}}a_{2}^{i_{\sigma(2)}+i_{2}}\cdots
a_{l}^{i_{\sigma(l)}+i_{l}}$

Now pair each $a_{j}^{i_{j}}$ with the corresponding $a_{j^{\prime}}$ which
also has $i_{j}$ as its exponent and then sum. The result is

$\sum_{\sigma}sgn(\sigma)(a_{1}a_{2}\cdots
a_{l})^{2m}/((1-a_{1}a_{\sigma(1)})(1-a_{2}a_{\sigma(2)})\dots(1-a_{l}a_{\sigma(l)})).$

Now note that this answer is independent of the order of the columns, so we
could have started with any permutation of the $a_{j}$’s. There are $l!$ such
permutations and each one yields the same answer.

If we factor out the term $(a_{1}a_{2}\cdots a_{l})^{2m}$ we see this itself
is another determinant of the matrix with entries

$\frac{1}{1-a_{i}a_{j}}\,\,\,\,\,1\leq i,j,\leq l,$

and this a Cauchy determinant after factoring out the $a_{j}$’s from each
column.

The result is that we have

$(a_{1}a_{2}\cdots
a_{l})^{2m}\frac{\prod_{i=2}^{l}\prod_{j=1}^{i-1}(1/a_{i}-1/a_{j})(a_{i}-a_{j})}{\prod_{i=1}^{l}\prod_{j=1}^{l}(1/a_{j}-a_{i})}$

We now just need to collect our answers and remember that we must multiply
through by $E(\phi)$. Here is what we have so far:

  1. (1)

$E(\phi)=\frac{\prod_{i,j}(1-a_{i}c_{j})}{\prod_{1\leq i\leq j\leq
k}(1-a_{i,}a_{j})\prod_{1\leq i<j\leq k}(1-c_{i}c_{j})}$

  2. (2)

The product of the $\alpha_{j}$ terms is

$\prod\alpha_{j}=\prod_{\begin{subarray}{c}i\neq j\\\ 1\leq j\leq l\\\ 1\leq
i\leq k\end{subarray}}\frac{1}{(a_{j}-a_{i})}\prod_{\begin{subarray}{c}1\leq
j\leq l\\\ 1\leq i\leq
k\end{subarray}}\frac{(1-a_{i}a_{j})(a_{j}-c_{i})}{(1-c_{i}a_{j})}$

  3. (3)

The answer from the determinant part

$(a_{1}a_{2}\cdots
a_{l})^{2m}\frac{\prod_{i=2}^{l}\prod_{j=1}^{i-1}(1/a_{i}-1/a_{j})(a_{i}-a_{j})}{\prod_{i=1}^{l}\prod_{j=1}^{l}(1/a_{j}-a_{i})}$

Note that we factored out the product of the $a_{j}$’s from both the numerator
and the denominator in the matrix columns for the computation in (3).

We now simplify the product of the above by considering three types of terms.

  1. (1)

The terms with only $c_{j}$’s. This comes only from the $E(\phi)$ term and is

$\frac{1}{\prod_{1\leq i<j\leq k}(1-c_{i}c_{j})}$

  2. (2)

The terms with $c_{j}$’s and $a_{j}$’s. This comes from (1) and (2) above and
is

$\prod_{i,j}(1-a_{i}c_{j})\prod_{\begin{subarray}{c}1\leq j\leq l\\\ 1\leq
i\leq
k\end{subarray}}\frac{(a_{j}-c_{i})}{(1-c_{i}a_{j})}=\prod_{\begin{subarray}{c}1\leq
i\leq k\\\ l+1\leq j\leq
k\end{subarray}}(1-c_{i}a_{j})\prod_{\begin{subarray}{c}1\leq j\leq l\\\ 1\leq
i\leq k\end{subarray}}(a_{j}-c_{i})$

It is probably useful to point out that in the above (1) is $Z_{O}(C)$ and
that (2) is

$\frac{\prod a_{j}^{k}}{Z_{S}(A-U+U^{-1},C)}$

where $U=\\{a_{1},a_{2},\cdots,a_{l}\\}$

All the rest of the factors only involve the terms with $a_{j}$’s and we turn
to these now. First notice that

$\frac{\prod_{\begin{subarray}{c}1\leq j\leq l\\\ 1\leq i\leq
k\end{subarray}}(1-a_{i}a_{j})}{\prod_{1\leq i\leq j\leq
k}(1-a_{i,}a_{j})\prod_{\begin{subarray}{c}1\leq i\leq l\\\ 1\leq j\leq
l\end{subarray}}(1/a_{j}-a_{i})}$

yields when simplified

$\frac{(-1)^{l^{2}}}{\prod_{1}^{l}a_{j}\prod_{l+1\leq i\leq j\leq
k}(1-a_{i}a_{j})\prod_{1\leq i\leq j\leq l}(1-1/a_{i}a_{j})}$

Next the remaining factors are

$\frac{\prod_{i=2}^{l}\prod_{j=1}^{i-1}(1/a_{i}-1/a_{j})(a_{i}-a_{j})}{\prod_{\begin{subarray}{c}i\neq
j\\\ 1\leq j\leq l\\\ 1\leq i\leq k\end{subarray}}(a_{j}-a_{i})}.$

This is the same as

$\frac{\prod_{\begin{subarray}{c}i\neq j\\\ 1\leq i\leq l\\\ 1\leq j\leq
l\end{subarray}}(a_{i}-a_{j})}{\prod_{j}a_{j}^{l-1}\prod_{\begin{subarray}{c}i\neq
j\\\ 1\leq j\leq l\\\ 1\leq i\leq k\end{subarray}}(a_{j}-a_{i})}$

or

$\frac{1}{\prod_{j}a_{j}^{l-1}\prod_{\begin{subarray}{c}1\leq j\leq l\\\
l+1\leq i\leq
k\end{subarray}}(a_{j}-a_{i})}=\frac{1}{\prod_{j}a_{j}^{l-1}a_{j}^{k-l}\prod_{\begin{subarray}{c}1\leq
j\leq l\\\ l+1\leq i\leq k\end{subarray}}(1-a_{i}/a_{j})}.$

Putting this all together we have

$\frac{(-1)^{l^{2}}}{\prod a_{j}^{k}\prod_{l+1\leq i\leq j\leq
k}(1-a_{i}a_{j})\prod_{1\leq i\leq j\leq
l}(1-1/a_{i}a_{j})\prod_{\begin{subarray}{c}1\leq j\leq l\\\ l+1\leq i\leq
k\end{subarray}}(1-a_{i}/a_{j})},$

which is exactly

$(-1)^{l}Z_{S}(A-U+U^{-1})/\prod a_{j}^{k}$

and this shows that for this set $U$ the contribution of the Fredholm
determinant yields the same answer as in the ratios formula.

We have shown agreement with the coefficient of the term
$a_{1}^{N}a_{2}^{N}\cdots a_{l}^{N}.$ The indexing is not important and hence
this works for any subset $U$. Notice we have equated the $l\times l$
determinant in the Fredholm expansion with the sum over all subsets of size
$l$ on the right-hand side of our formula in Theorem 6.1.

One final comment is that the in the proof we assumed that the number of
factors in the numerator and denominator of $g$ were the same. But a moment’s
thought shows that if there are fewer or more $c_{i}$ terms than $a_{i}$
terms, the proof is easily modified to hold in those cases.

This same proof with minor changes works in the orthogonal cases as well. Here
is the statement in that case.

######  Theorem 6.2.

Let $A=\\{a_{1},a_{2},\dots,a_{k}\\}$ and let
$C=\\{c_{1},a_{2},\dots,c_{k}\\}$ where $|c_{i}|<1$. Let

$g(\theta)=\prod_{i=1}^{k}\frac{(1-a_{i}e^{i\theta})}{(1-c_{i}e^{i\theta})}.$

For a matrix $U$ with eigenvalues $\\{e^{i\theta_{j}}:1\leq j\leq N\\}$ we
define

$\det g(U)=\prod_{1\leq i\leq k\atop 1\leq j\leq
N}\frac{(1-a_{i}e^{i\theta_{j}})}{(1-c_{i}e^{i\theta_{j}})}.$

Further, define

$Z_{O}(A;C)=\frac{Z_{O}(A)Z_{S}(C)}{Z_{U}(A,C)}.$

  1. (1)

Let $N=2m+1.$ Then

$\int_{O^{+}(N)}\det g(U)~{}dU=\sum_{U\subset
A}(-1)^{|U|}U^{N}Z_{O}(A-U+U^{-1};C).$

  2. (2)

Let $N=2m+1.$ Then

$\int_{O^{-}(N)}\det g(U)~{}dU=\sum_{U\subset A}U^{N}Z_{O}(A-U+U^{-1};C).$

  3. (3)

Let $N=2m.$ Then

$\int_{O^{-}(N)}\det g(U)~{}dU=\sum_{U\subset
A}(-1)^{|U|}U^{N}Z_{O}(A-U+U^{-1};C).$

  4. (4)

Let $N=2m.$ Then

$\int_{O^{+}(N)}\det g(U)~{}dU=\sum_{U\subset A}U^{N}Z_{O}(A-U+U^{-1};C).$

We can do the same for the unitary case, although this is a bit harder. Before
we begin the proof of the identity, we need three lemmas and some notation.

Let $\\{a_{i},a_{2},\dots a_{l}\\}$ and $\\{b_{i},b_{2},\dots b_{l}\\}$ be any
two ordered sets of complex numbers and define the $l\times l$ determinant by

$D(\\{a_{i}\\},\\{b_{j}\\})=\det(a_{h}^{i_{g}}\,b_{h}^{i_{h}})_{1\leq g,h\leq
l}$

Here is the $3\times 3$ example for sets $\\{a_{1},a_{2},a_{3}\\}$ and
$\\{b_{1},b_{2},b_{3}\\}$.

$D(\\{a_{i}\\},\\{b_{j}\\})=\left|\begin{array}[]{ccc}a_{1}^{i_{1}}b_{1}^{i_{1}}&a_{2}^{i_{1}}b_{2}^{i_{2}}&a_{3}^{i_{1}}b_{3}^{i_{3}}\\\
a_{1}^{i_{2}}b_{1}^{i_{1}}&a_{2}^{i_{2}}b_{2}^{i_{2}}&a_{3}^{i_{2}}b_{3}^{i_{3}}\\\
a_{1}^{i_{3}}b_{1}^{i_{1}}&a_{2}^{i_{3}}b_{2}^{i_{2}}&a_{3}^{i_{3}}b_{3}^{i_{3}}\end{array}\right|.$

The reason for defining the above is that it will occur in our Fredholm
expansion.

######  Lemma 6.3.

If any $a_{i}=a_{j},i\neq j,$ then $D(\\{a_{i}\\},\\{b_{j}\\})=0.$

This is easy to see since we can factor out $b_{h}^{i_{h}}$ from any column
and the resulting determinant has two equal columns.

######  Lemma 6.4.

If any $b_{i}=b_{j},i\neq j$ then $\sum_{i_{i},\dots,i_{l}\geq
m}D(\\{a_{i}\\},\\{b_{j}\\})=0.$

###### Proof.

It suffices to assume that $b_{1}=b_{2}$. Factor first the $b_{h}^{i_{h}}$s
from all the columns and then use the permutation definition of a determinant
and consider the product of $a_{g}^{i_{h}}$ in each summand. There will be
exactly one $a_{g}^{i_{h}}$ term with $i_{1}$ as an exponent and exactly one
other $a_{g}^{i_{h}}$ term with $i_{2}$ as an exponent. This is because there
can only be one element from each column. There is also a permutation with
these exponents reversed and whose permutation sign is the opposite of the
other. For this second permutation, if we change variables in the sum
switching $i_{1}$ and $i_{2}$, we see that after summing the two terms cancel.

Notice the $D(\\{a_{i}\\},\\{b_{j}\\})$ depends on the order of the
$\\{a_{i}\\},\\{b_{j}\\}$ in the determinant.

######  Lemma 6.5.

Suppose there is a permutation of the $\\{b_{j}\\}$s in the columns by the
permutation $\sigma.$ Then

$\sum_{i_{i},\dots,i_{l}\geq m}D(\\{a_{i}\\},\\{b_{j}\\})$

is changed by the sign of the permutation.

###### Proof.

Change both the $a_{j}$’s and $b_{j}$’s by the given permutation. After
summing the answer does not change. Permuting the $a_{j}$s back just permutes
the rows which changes the answer by the permutation sign and leaves the
$b_{j}$’s still permuted.

######  Theorem 6.6.

Let

$\phi=\prod_{i=1}^{k}\frac{(1-a_{i}e^{-i\theta})(1-b_{i}e^{i\theta})}{(1-c_{i}e^{-i\theta})(1-d_{i}e^{i\theta})},$

and let
$A=\\{a_{1},a_{2},\dots,a_{k}\\},B=\\{b_{1},b_{2},\dots,b_{k}\\},C=\\{c_{1},c_{2},\dots,c_{k}\\},D=\\{d_{1},d_{2},\dots,d_{k}\\}$.
Then

$I_{U(N)}\phi=\sum_{S\subset A,T\subset
B\atop|S|=|T|}S^{N}T^{N}Z(A-S+T^{-1},B-T+S^{-1};C,D).$

###### Proof.

We use the BOCG identity. First we note that by direct computation
$E(\phi)=Z(A,B,C,D),$ which corresponds exactly to the empty set in the above.

Referring to the identity the operator which occurs in the Fredholm
determinant part of the identity is a product given by

$Q_{N}H(\phi_{-}\phi_{+}^{-1})H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+})Q_{N}.$

The Fourier coefficients of $\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+}$ are given
by

$\sum_{j=1}^{k}\alpha_{j}a_{j}^{l-1},\,\,\,\,\,\,\alpha_{j}=\prod_{i\neq
j}\frac{1}{(a_{j}-a_{i})}\prod_{i}\frac{(1-b_{i}a_{j})(a_{j}-c_{i})}{(1-d_{i}a_{j})}.$

The Fourier coefficients of $\phi_{-}\phi_{+}^{-1}$ are given by

$\sum_{j=1}^{k}\beta_{j}b_{j}^{l-1},\,\,\,\,\,\,\beta_{j}=\prod_{i\neq
j}\frac{1}{(b_{j}-b_{i})}\prod_{i}\frac{(1-a_{i}b_{j})(b_{j}-d_{i})}{(1-c_{i}b_{j})}.$

Each $j,k,$ entry in $H(\tilde{\phi}_{-}^{-1}\tilde{\phi}_{+})$ is a sum of
terms of the form $\alpha_{j}a_{j}^{j+k}$ and each $j,k$ entry in
$H(\phi_{-}\phi_{+}^{-1})$ is of the form $\beta_{j}b_{j}^{j+k}.$ Thus when
squaring the operator we find that the $j,k$ entry of the product is given by
a sum of terms

$\frac{\beta_{g}\alpha_{h}b_{g}^{j}a_{h}^{k}}{(1-b_{g}a_{h})}.$

Next consider the $l$th term in the Fredholm expansion,

$\sum_{i_{1},i_{2},\cdots,i_{l}}\det K(i_{j},i_{k})_{1\leq j,k\leq l}.$

Since $K$ is a sum of columns of the above form, we can expand
$K(i_{j},i_{k})$ in columns over all possible choices. Let us begin with the
choice $\\{a_{1},a_{2},\dots,a_{l}\\}$ and $\\{b_{1},b_{2},\dots,b_{l}\\},$
where the $b_{i}$ represents a subset of $B$. By our lemmas we have no
contribution if any of the $a_{i}$’s overlap or if any of the $b_{j}$’s
overlap. A moment’s thought shows that $\det K(i_{j},i_{k})$ is a sum of
determinants, each of the form

$\frac{\prod_{j=1}^{l}\alpha_{j}\beta_{j}}{\prod_{j=1}^{l}(1-a_{j}b_{j})}D(\\{b_{i}\\}\\{a_{j}\\}).$

We fix the order of the $b_{j}$’s. Then the contribution from all possible
permutations of the $a_{j}$’s is

$\prod_{j=1}^{l}\alpha_{j}\beta_{j}\,\,(\tilde{D}(\\{b_{i}\\}\\{a_{j}\\})\left(\sum_{\sigma}sgn(\sigma)\frac{1}{\prod_{j=1}^{l}(1-b_{j}a_{\sigma(j)})}\right).$

Here $\tilde{D}$ represents the determinant $D(\\{b_{i}\\}\\{a_{j}\\})$ with
the elements in the standard order. This expression on the right is a Cauchy
determinant and after evaluating yields

$\prod_{j=1}^{l}\alpha_{j}\beta_{j}\,\,(\tilde{D}(\\{b_{i}\\}\\{a_{j}\\})\frac{\prod_{1\leq
i<j\leq l}(1/a_{i}-1/a_{j})\prod_{1\leq i<j\leq
l}(b_{i}-b_{j})}{\prod_{i=1}^{i}a_{i}\prod_{i,j}(1/a_{i}-b_{j})}.$

After summing as before $\tilde{D}$ is also Cauchy and the result is that the
above is

$\prod_{j=1}^{l}a_{j}^{N}b_{j}^{N}\prod_{j=1}^{l}\alpha_{j}\beta_{j}\,\,\left(\frac{\prod_{1\leq
i<j\leq l}(1/a_{i}-1/a_{j})\prod_{1\leq i<j\leq
l}(b_{i}-b_{j})}{\prod_{i=1}^{i}a_{i}\prod_{i,j}(1/a_{i}-b_{j})}\right)^{2}.$

Notice that this expression is independent of a permutation of the $b_{j}$’s.
Hence we have $l!$ of these which cancels with $1/l!$ in the Fredholm
expression.

Using the notation of the theorem we think of the set $S$ as
$\\{a_{1},a_{2},\dots,a_{l}\\}$ and $T$ as $\\{b_{1},b_{2},\dots,b_{l}\\}.$
Now we just need to gather terms as before, and as before we have four types
of terms.

  1. (1)

The terms with only $c_{j}$’s and $d_{j}$’s occurs only in $E(\phi)$ and is
easily seen to be $Z(C,D).$

  2. (2)

The terms with both $a_{j}$’s, $b_{j}$’s and $d_{j}$’s is

$\prod_{i,j}(1-a_{i}d_{j})\prod_{\begin{subarray}{c}1\leq j\leq l\\\ 1\leq
i\leq k\end{subarray}}\frac{(b_{j}-d_{i})}{(1-d_{i}a_{j})},$

which is easily seen to be

$\frac{\prod_{i=1}^{l}b_{i}^{k}}{Z(A-S+T^{-1},D)}.$

  3. (3)

The terms with both $a_{j}$’s, $b_{j}$’s and $c_{j}$’s is

$\prod_{i,j}(1-b_{i}c_{j})\prod_{\begin{subarray}{c}1\leq j\leq l\\\ 1\leq
i\leq k\end{subarray}}\frac{(a_{j}-c_{i})}{(1-b_{i}c_{j})},$

or

$\frac{\prod_{i=1}^{l}a_{i}^{k}}{Z(B-T+S^{-1},D)}.$

Hence the only thing left to consider is the terms with only $a_{j}$’s and
$b_{j}$’s. This is

$Z(A,B)\prod_{\begin{subarray}{c}i\neq j\\\ 1\leq j\leq l\\\ 1\leq i\leq
k\end{subarray}}\frac{1}{(a_{j}-a_{i})}\prod_{\begin{subarray}{c}1\leq j\leq
l\\\ 1\leq i\leq k\end{subarray}}(1-b_{i}a_{j})\prod_{\begin{subarray}{c}i\neq
j\\\ 1\leq j\leq l\\\ 1\leq i\leq
k\end{subarray}}\frac{1}{(b_{j}-b_{i})}\prod_{\begin{subarray}{c}1\leq j\leq
l\\\ 1\leq i\leq k\end{subarray}}(1-b_{j}a_{i})\\\ $
$\times\left(\frac{\prod_{1\leq i<j\leq l}(1/a_{i}-1/a_{j})\prod_{1\leq
i<j\leq
l}(b_{i}-b_{j})}{\prod_{i=1}^{i}a_{i}\prod_{i,j}(1/a_{i}-b_{j})}\right)^{2}.$

After simplifying the above, we find that it becomes

$(-1)^{l}Z(A-S+T^{-1},B-T+S^{-1})/\prod_{i=1}^{l}(a_{i}b_{i})^{k}$

and after combining with the other terms yields the correct answer.

## References

  * [BR] Baik, J.and Rains, E. M. Algebraic aspects of increasing subsequences Duke Math. J. 109 (2001), no. 1, 1–65.
  * [BW] E.L. Basor and H. Widom, On a Toeplitz determinant identity of Borodin and Okounkov. Int. Eqs. Operator Th. 37(4), 397–401 (2000).
  * [BE4] E.L. Basor, T. Ehrhardt, Determinant computations for some classes of Toeplitz-Hankel matrices, Oper. Matrices 3, no. 2 (2009), 167–186.
  * [BO] A. Borodin, A., Okounkov, A., A Fredholm determinant formula for Toeplitz determinants. Int. Eqs. Operator Th. 37, 386–396 (2000)
  * [BS99] A. Böttcher, B. Silbermann, Introduction to large truncated Toeplitz matrices, Universitext. Springer-Verlag, New York, 1999.
  * [Bot] A. Böttcher, One more proof of the Borodin-Okounkov formula for Toeplitz determinants. Int. Eqs. Operator Th. 41(1), 123–125 (2001)
  * [BS06] A. Böttcher, B. Silbermann, Analysis of Toeplitz operators, 2nd edition. Prepared jointly with A. Karlovich. Springer Monographs in Mathematics, Springer, Berlin, 2006.
  * [BG] D. Bump, A. Gamburd, On the averages of characteristic polynomials from classical groups. Comm. Math. Phys. 265 (2006), no. 1, 227–274.
  * [CFS] J. B. Conrey, P. J. Forrester, N. C. Snaith, Averages of ratios of characteristic polynomials for the compact classical groups. Int. Math. Res. Not. (2005), no. 7, 397–431.
  * [CFZ] J. B. Conrey, D. W. Farmer, M. R. Zirnbauer, Autocorrelation of ratios of L-functions. arXiv:0711.0718
  * [CFKRS] J. B. Conrey, D. W. Farmer, J. P. Keating, M. O. Rubinstein, N. C. Snaith, Integral moments of L-functions. Proc. London Math. Soc. (3) 91 (2005), no. 1, 33–104.
  * [CS07] J. B. Conrey, N. C. Snaith, Applications of the L-functions ratios conjectures. Proc. Lond. Math. Soc. (3) 94 (2007), no. 3, 594–646.
  * [CS08] J. B. Conrey, N. C. Snaith, Correlations of eigenvalues and Riemann zeros. Commun. Number Theory Phys. 2 (2008), no. 3, 477–536.
  * [DIK13] P. Deift, A. R. Its, I. Krasovsky, Toeplitz matrices and Toeplitz determinants under the impetus of the Ising model: some history and some recent results, Comm. Pure Appl. Math. 66 (2013), no. 9, 1360–1438.
  * [Faz] A. Fazzari, A weighted one-level density of families of L-functions. arXiv:2109.07244
  * [GC] J. F. Geronimo and K. M. Case, Scattering theory and polynomials orthogonal on the unit circle. J. Math. Phys. 20, 299–310 (1979)
  * [GK] I.C. Gohberg and M. G. Kreĭn, Introduction to the theory of linear nonselfadjoint operators Translations of Mathematical Monographs, Vol. 18, American Mathematical Society, Providence, 1969.
  * [IJK] A. R. Its, B. Q. Jin and V. E. Korepin, Entanglement in the XY spin chain. J. Phys. A 38 (2005), 2975-2990.
  * [KaSa] N. Katz and P. Sarnak, Random matrices, Frobenius eigenvalues, and monodromy. American Mathematical Society Colloquium Publications, 45, American Mathematical Society, Providence, 1999.
  * [Ke] J. P. Keating, L-functions and the Characteristic Polynomials of Random Matrices. In: Recent Perspectives on Random Matrix Theory and Number Theory. London Math. Soc. Lecture Note Ser., vol. 322, Cambridge University Press, Cambridge, 2005, pp. 251–277
  * [KeSn] J. P. Keating and N. Snaith, Random matrix theory and $\zeta(1/2+it)$. Comm. Math. Phys. 214 (2000), no. 1, 57-89.
  * [IMM] A. R. Its, F. Mezzadri, M. Y. Mo, Entanglement entropy in quantum spin chains with finite range interaction. Comm. Math. Phys. 284 (2008), no. 1, 117–185.
  * [JK] B.-Q. Jin, V. E. Korepin, Quantum spin chain, Toeplitz determinants and the Fisher-Hartwig conjecture. J. Statist. Phys. 116 (2004), no. 1-4, 79–95.
  * [Wi74] H. Widom, Asymptotic behavior of block Toeplitz matrices and determinants. Advances in Math.13 (1974), 284–322.
  * [Wi76] H. Widom, Asymptotic behavior of block Toeplitz matrices and determinants. II. Advances in Math.21 (1976), no. 1, 1–29.

##  7\. Appendix

Here we give some additional computations.

As the simplest interesting example of a ratios factorization, suppose that
$A=\\{a\\},B=\\{b\\},C=\\{c\\},D=\\{d\\}$ where $|c|,|d|<1.$ By Theorem 6.6,

$\displaystyle\int_{U(N)}\frac{\Lambda_{U}(a)\Lambda_{U^{*}}(b)}{\Lambda_{U}(c)\Lambda_{U^{*}}(d)}dU$
$\displaystyle=$
$\displaystyle\frac{(1-ad)(1-bc)}{(1-ab)(1-cd)}+a^{N}b^{N}\frac{(1-\frac{d}{b})(1-\frac{c}{a})}{(1-\frac{1}{ab})(1-cd)}.$

We are interested in the case that $a=b$ and $c=d$; we assume this from now
on. We have

$\displaystyle R_{U(N)}(a;c)$ $\displaystyle:=$
$\displaystyle\int_{U(N)}\frac{\Lambda_{U}(a)\Lambda_{U^{*}}(a)}{\Lambda_{U}(c)\Lambda_{U^{*}}(c)}dU$
$\displaystyle=$
$\displaystyle\frac{(1+a^{(1+N)}-ac-a^{N}c)(1-a^{(1+N)}-ac+a^{N}c)}{(1-a)(1+a)(1-c)(1+c)}.$

Thus,

$\displaystyle\begin{array}[]{|c|l|}\hline\cr N&R_{U}(N)\\\ \hline\cr
1&\frac{1-2ac+a^{2}}{1-c^{2}}\\\ \hline\cr
2&\frac{(1-a+a^{2}-ac)(1+a+a^{2}-ac)}{1-c^{2}}\\\ \hline\cr
3&\frac{(1+a^{2}-ac)(1+a^{4}-ac-a^{3}c)}{1-c^{2}}\\\ \hline\cr
4&\frac{(1+a+a^{2}+a^{3}+a^{4}-ac-a^{2}c-a^{3}c)(1-a+a^{2}-a^{3}+a^{4}-ac+a^{2}c-a^{3}c)}{1-c^{2}}\\\
\hline\cr\end{array}$

For the symplectic group we have by Theorem 6.1,

$\displaystyle R_{S(N)}$ $\displaystyle:=$
$\displaystyle\int_{USp(N)}\frac{\Lambda_{U}(a)}{\Lambda_{U}(c)}~{}dU$
$\displaystyle=$
$\displaystyle\frac{(1-ac)}{(1-a^{2})}+a^{N}\frac{(1-c/a)}{(1-a^{-2})}.$

Thus

$\displaystyle\begin{array}[]{|c|l|}\hline\cr N&R_{S}(N)\\\ \hline\cr
2&1+a^{2}-ac\\\ \hline\cr 4&1+a^{2}+a^{4}-ac-a^{3}c\\\ \hline\cr
6&1+a^{2}+a^{4}-+a^{6}-ac-a^{3}c-a^{5}c\\\ \hline\cr
8&1+a^{2}+a^{4}+a^{6}+a^{8}-ac-a^{3}c-a^{5}c-a^{7}c\\\ \hline\cr\end{array}$

Using Theorem 6.2 and an obvious notation for the orthogonal averages we have

$\displaystyle\begin{array}[]{|c|l|}\hline\cr N&R_{O^{+}(N)}\\\ \hline\cr
1&\frac{1-a}{1-c}\\\ \hline\cr 2&\frac{1+a^{2}-2ac}{1-c^{2}}\\\ \hline\cr
3&\frac{1-ac-a^{3}+a^{2}c}{1-c^{2}}\\\ \hline\cr
4&\frac{1-ac+a^{4}-a^{3}c}{1-c^{2}}\\\ \hline\cr
5&\frac{1-ac-a^{5}+a^{4}c}{1-c^{2}}\\\ \hline\cr
6&\frac{1-ac+a^{6}-a^{5}c}{1-c^{2}}\\\ \hline\cr\end{array}$

and

$\displaystyle\begin{array}[]{|c|l|}\hline\cr N&R_{O^{-}(N)}\\\ \hline\cr
1&\frac{1+a}{1+c}\\\ \hline\cr 2&\frac{1-a^{2}}{1-c^{2}}\\\ \hline\cr
3&\frac{1-ac+a^{3}-a^{2}c}{1-c^{2}}\\\ \hline\cr
4&\frac{1-ac-a^{4}+a^{3}c}{1-c^{2}}\\\ \hline\cr
5&\frac{1-ac+a^{5}-a^{4}c}{1-c^{2}}\\\ \hline\cr
6&\frac{1-ac-a^{6}+a^{5}c}{1-c^{2}}\\\ \hline\cr\end{array}$

From these we can see the verification of our factorization formulas.

$\displaystyle R_{U(1)}$ $\displaystyle=$ $\displaystyle R_{S(0)}R_{O^{+}(2)}$
$\displaystyle R_{U(2)}$ $\displaystyle=$
$\displaystyle(g(1)g(-1))^{-1}R_{O^{+}(3)}R_{O^{-}(3)}$ $\displaystyle
R_{U(3)}$ $\displaystyle=$ $\displaystyle R_{S(2)}R_{O^{+}(4)}$ $\displaystyle
R_{U(4)}$ $\displaystyle=$
$\displaystyle(g(1)g(-1))^{-1}R_{O^{+}(5)}R_{O^{-}(5)}$ $\displaystyle
R_{U(5)}$ $\displaystyle=$ $\displaystyle R_{S(4)}R_{O^{+}(6)}.$

Here $g(\pm 1)$ means substitute $\pm 1$ for $e^{i\theta}$ in the formula for
$g$. So for example, if $g=\frac{1-ae^{i\theta}}{1-ce^{i\theta}}$ then
$g(1)=\frac{1-a}{1-c}$.

We now consider the slightly more complicated case of unitary moments with two
numerator parameters $a$ and $b$

$I_{U(N)}(a,b):=\int_{U(M)}\Lambda_{U}(a)\Lambda_{U}(b)\Lambda_{U^{*}}(a)\Lambda_{U^{*}}(b)~{}dU.$

By Theorem 6.6 with $A=B=\\{a,b\\}$ and $C=D=\\{\\}$, the empty set, we have

$\displaystyle I_{U(1)}(a,b)$ $\displaystyle=$ $\displaystyle
1+a^{2}+2ab+b^{2}+a^{2}b^{2}$ $\displaystyle I_{U(2)}(a,b)$ $\displaystyle=$
$\displaystyle(1-a+a^{2}-b+2ab-a^{2}b+b^{2}-ab^{2}+a^{2}b^{2})$
$\displaystyle\quad\times(1+a+a^{2}+b+2ab+a^{2}b+b^{2}+ab^{2}+a^{2}b^{2})$
$\displaystyle I_{U(3)}(a,b)$ $\displaystyle=$
$\displaystyle(1+a^{2}+ab+b^{2}+a^{2}b^{2})$
$\displaystyle\quad\times(1+a^{4}+ab+a^{3}b+2a^{2}b^{2}+ab^{3}+a^{3}b^{3}+b^{4}+a^{4}b^{4})$
$\displaystyle I_{U(4)}(a,b)$ $\displaystyle=$
$\displaystyle(1-a+a^{2}-a^{3}+a^{4}-b+2ab-2a^{2}b+2a^{3}b-a^{4}b$
$\displaystyle\qquad+b^{2}-2ab^{2}+3a^{2}b^{2}-2a^{3}b^{2}+a^{4}b^{2}b^{3}+2ab^{3}-2a^{2}b^{3}$
$\displaystyle\qquad+2a^{3}b^{3}-a^{4}b^{3}+b^{4}-ab^{4}+a^{2}b^{4}-a^{3}b^{4}+a^{4}b^{4})$
$\displaystyle\quad\times(1+a+a^{2}+a^{3}+a^{4}+b+2ab+2a^{2}b+2a^{3}b$
$\displaystyle\qquad+a^{4}b+b^{2}+2ab^{2}+3a^{2}b^{2}+2a^{3}b^{2}+a^{4}b^{2}+b^{3}$
$\displaystyle\qquad+2ab^{3}+2a^{2}b^{3}+2a^{3}b^{3}+a^{4}b^{3}+b^{4}+ab^{4}+a^{2}b^{4}$
$\displaystyle\qquad+a^{3}b^{4}+a^{4}b^{4})$

For the symplectic moments we have

$\displaystyle I_{USp(2)}(a,b)$ $\displaystyle=$ $\displaystyle
1+a^{2}+ab+b^{2}+a^{2}b^{2}$ $\displaystyle I_{USp(4)}(a,b)$ $\displaystyle=$
$\displaystyle 1+a^{2}+a^{4}+ab+a^{3}b+b^{2}+2a^{2}b^{2}+a^{4}b^{2}+ab^{3}$
$\displaystyle\qquad+a^{3}b^{3}+b^{4}+a^{2}b^{4}+a^{4}b^{4}$

For the orthogonal moment we have,

$\displaystyle I_{O^{-}(3)}(a,b)$ $\displaystyle=$
$\displaystyle(1+a)(1+b)(1-a+a^{2}-b+2ab-a^{2}b$
$\displaystyle\qquad+b^{2}-ab^{2}+a^{2}b^{2})$ $\displaystyle
I_{O^{+}(3)}(a,b)$ $\displaystyle=$
$\displaystyle(1-a)(1-b)(1+a+a^{2}+b+2ab+a^{2}b$
$\displaystyle\qquad+b^{2}+ab^{2}+a^{2}b^{2})$ $\displaystyle
I_{O^{-}(5)}(a,b)$ $\displaystyle=$
$\displaystyle(1+a)(1+b)(1-a+a^{2}-a^{3}+a^{4}-b+2ab$
$\displaystyle\qquad-2a^{2}b+2a^{3}b-a^{4}b+b^{2}-2ab^{2}+3a^{2}b^{2}$
$\displaystyle\qquad-2a^{3}b^{2}+a^{4}b^{2}b^{3}+2ab^{3}-2a^{2}b^{3}+2a^{3}b^{3}$
$\displaystyle\qquad\quad-a^{4}b^{3}+b^{4}-ab^{4}+a^{2}b^{4}-a^{3}b^{4}+a^{4}b^{4})$
$\displaystyle I_{O^{+}(5)}(a,b)$ $\displaystyle=$
$\displaystyle(1-a)(1-b)(1+a+a^{2}+a^{3}+a^{4}+b+2ab$
$\displaystyle\qquad+2a^{2}b+2a^{3}b+a^{4}b+b^{2}+2ab^{2}+3a^{2}b^{2}$
$\displaystyle\qquad+2a^{3}b^{2}+a^{4}b^{2}+b^{3}+2ab^{3}+2a^{2}b^{3}$
$\displaystyle\qquad+2a^{3}b^{3}+a^{4}b^{3}+b^{4}+ab^{4}+a^{2}b^{4}+a^{3}b^{4}+a^{4}b^{4})$
$\displaystyle I_{O^{+}(4)}(a,b)$ $\displaystyle=$ $\displaystyle
1+a^{4}+ab+a^{3}b+2a^{2}b^{2}+ab^{3}+a^{3}b^{3}+b^{4}+a^{4}b^{4}.$

Thus, we have the following identities:

$\displaystyle I_{U(1)}(a,b)$ $\displaystyle=$ $\displaystyle
I_{O^{+}(2)}(a,b)$ $\displaystyle I_{U(2)}(a,b)$ $\displaystyle=$
$\displaystyle(g(1)g(-1))^{-1}I_{O^{-}(3)}(a,b)I_{O^{+}(3)}(a,b)$
$\displaystyle I_{U(3)}(a,b)$ $\displaystyle=$ $\displaystyle
I_{USp(2)}(a,b)I_{O^{+}(4)}(a,b)$ $\displaystyle I_{U(4)}(a,b)$
$\displaystyle=$
$\displaystyle(g(1)g(-1))^{-1}I_{O^{-}(5))}(a,b)I_{O^{+}(5)}(a,b).$