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Coulomb Gas, Random Matrices, Matrix Models
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\begin{document}
\title[Matrix models for circular ensembles]{Matrix models for circular ensembles}
\author[R.~Killip and I.~Nenciu]{Rowan Killip and Irina Nenciu}
\address{Rowan Killip\\
UCLA Mathematics Department\\
Box 951555\\
Los Angeles, CA 90095}
\email{killip@math.ucla.edu}
\address{Irina Nenciu\\
Mathematics 253-37\\
Caltech\\
Pasadena, CA 91125}
\email{nenciu@caltech.edu}
\date{\today}
\begin{abstract}
The Gibbs distribution for $n$ particles of the Coulomb gas on the unit
circle at
inverse temperature $\beta$ is given by\\[1mm]
\hbox{}\hfill\hbox{$\displaystyle
\mathbb{E}^{\beta}_{n}(f)=
\frac{1}{Z_{n,\beta}} \int\!\!\cdots\!\!\int\,
f(e^{i\theta_1},\ldots,e^{i\theta_n})|\Delta(e^{i\theta_1},\ldots,e^{i\theta_n}
)|^{\beta} \frac{d\theta_1}{2\pi}\cdots\frac{d\theta_n}{2\pi}
$}\hfill\hbox{}\\[1mm]
for any symmetric function $f$, where $\Delta$ denotes the Vandermonde
determinant and
$Z_{n,\beta}$ the normalization constant.
We will describe an ensemble of (sparse) random matrices whose eigenvalues follow this distribution.
Our approach combines elements from the theory of orthogonal polynomials on the
unit circle with ideas from recent work of Dumitriu and Edelman.
In particular, we resolve a question left open by them: find a tri-diagonal model for the Jacobi ensemble.
\end{abstract}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In 1962, Dyson \cite{Dyson} introduced three ensembles of random unitary matrices with
a view to simplifying the study of energy level behavior in complex quantum systems.
Earlier work in this direction, pioneered by Wigner, focused on ensembles of Hermitian matrices.
The simplest of these three models is the unitary ensemble, which is just the group $\U(n)$
of $n\times n$ unitary matrices together with its Haar measure. The induced probability
measure on the eigenvalues is given by the Weyl integration formula
(cf. \cite[\S VII.4]{Weyl}): for any symmetric function of the eigenvalues,
\begin{equation}\label{E:WeylIF}
\Exp(f) = \frac{1}{n!} \int_0^{2\pi}\!\! \cdots \int_0^{2\pi}
f(e^{i\theta_1},\ldots,e^{i\theta_n})
\bigl|\Delta(e^{i\theta_1},\ldots,e^{i\theta_n})\bigr|^2
\frac{d\theta_1}{2\pi} \cdots \frac{d\theta_n}{2\pi}
\end{equation}
where $\Delta$ denotes the Vandermonde determinant,
\begin{equation}\label{VDefn}
\Delta(z_1,\ldots,z_n) = \prod_{1\leq j < k \leq n} \!\! (z_k - z_j)
=
\begin{vmatrix}
1 & \ldots & 1 \\
\vdots & & \vdots \\
z_1^{n-1} & \cdots & z_n^{n-1}
\end{vmatrix}.
\end{equation}
The orthogonal ensemble consists of symmetric $n\times n$ unitary matrices together with the unique measure that
is invariant under $U\mapsto W^T U W$ for all $W\in \U(n)$. Alternatively, if $U$ is chosen according
to the unitary ensemble, then $U^TU$ is distributed as a random element from the orthogonal ensemble.
The distribution of eigenvalues is given by \eqref{E:WeylIF} but with $|\Delta|^2$ replaced
by $|\Delta|$ and a new normalization constant.
The symplectic ensemble is a little more complicated. Let $Z$ denote the $2n\times 2n$ block diagonal matrix
\begin{equation}
\begin{bmatrix}
0 & 1 & & & \\
-1 & 0 & & & \\
& &\ddots& & \\
& & & 0 & 1 \\
& & & -1 & 0
\end{bmatrix}
\end{equation}
and define the dual of a matrix by $U^R=Z^TU^TZ$. The symplectic ensemble consists of self-dual unitary
$2n\times 2n$ matrices; the measure is that induced from $\U(2n)$ by the map $U\mapsto U^R U$.
(This is the unique measure invariant under $V\mapsto W^R V W$ for all $W\in \U(2n)$.)
The eigenvalues of such matrices are doubly degenerate and the pairs are distributed on the circle
as in \eqref{E:WeylIF} but now with $|\Delta|^2$ replaced by $|\Delta|^4$. Again, the normalization
constant needs to be changed.
Dyson also observed that these eigenvalue distributions correspond to the Gibbs distribution for the classical
Coulomb gas on the circle at three different temperatures. Let us elaborate.
Consider $n$ identically charged particles confined to move on the unit circle in the complex plane.
Each interacts with the others through the usual Coulomb potential, $-\log|z_i-z_j|$, which gives rise to
the Hamiltonian
$$
H(z_1,\ldots,z_n) = \sum_{1\leq j1$) if
\begin{equation}\label{E:ThetaDefn}
\Exp\{f(X)\} = \tfrac{\nu-1}{2\pi} \int\!\!\!\int_\Disk f(z)
(1-|z|^2)^{(\nu-3)/2} \,d^2z.
\end{equation}
For $\nu\geq2$ an integer, this has the following geometric interpretation: If $v$ is chosen from the
unit sphere $S^\nu$ in $\Reals^{\nu+1}$ at random according to the usual surface measure, then $v_1+iv_2$
is $\Theta_\nu$-distributed. (See Corollary~\ref{C:A3}.)
As a continuation of this geometric picture, we shall say that $X$ is $\Theta_1$-distributed if it is
uniformly distributed on the unit circle in $\Cmplx$.
\end{definition}
Let us now describe the family of matrix models.
\begin{theorem}\label{T:1}
Given $\beta>0$, let $\alpha_k\sim\Theta_{\beta(n-k-1)+1}$ be independent random variables for $0\leq k\leq n-1$,
$\rho_k=\sqrt{1-|\alpha_k|^2}$, and define
$$
\Xi_k = \begin{bmatrix} \bar\alpha_k & \rho_k \\ \rho_k & -\alpha_k
\end{bmatrix}
$$
for $0\leq k\leq n-2$, while $\Xi_{-1}=[1]$ and $\Xi_{n-1}=[\bar\alpha_{n-1}]$ are $1 \times1$ matrices.
From these, form the $n\times n$ block-diagonal matrices
$$
L=\diag\bigl(\Xi_0 ,\Xi_2,\Xi_4,\ldots\bigr) \quad\text{and}\quad
M=\diag\bigl(\Xi_{-1},\Xi_1,\Xi_3,\ldots\bigr).
$$
Both $LM$ and $ML$ give {\rm(}sparse{\rm)} matrix models for the Coulomb gas at inverse temperature $\beta$.
That is, their eigenvalues are distributed according to \eqref{CGbeta}.
\end{theorem}
\begin{remark}
As each of the $\Xi_k$ is unitary, so are $L$ and $M$. (In the case of $\Xi_{n-1}$, we should reiterate
that $\alpha_{n-1}\sim\Theta_{1}$ is uniformly distributed on the unit circle.) As a result, the eigenvalues
of $LM$ and $ML$ lie on the unit circle. Note also that, since $M$ conjugates one to the other, $LM$ and
$ML$ have the same eigenvalues.
\end{remark}
In proving this theorem, we will be following the recent paper of Dumitriu
and Edelman \cite{DumE} rather closely, while incorporating the nuances
of the theory of polynomials orthogonal on the unit circle. The matrices $L$ and $M$ that appear in the theorem
have their origin in the work of Cantero, Moral, and Vel\'asquez \cite{CMV}; this is discussed in Section~\ref{S:2}.
Dumitriu and Edelman constructed tri-diagonal matrix models for two of
the three standard examples of the Coulomb gas on the real line.
A model for the third will be constructed below.
The simplest way to obtain a normalizable Gibbs measure on the real line is to add an external harmonic
potential $V(x)=\tfrac{1}{2}x^2$. This gives rise to the probability measure
\begin{equation}\label{CGRL}
\Exp(f) \propto
\int\!\! \cdots \!\! \int f(x_1,\ldots,x_n) \,
\bigl|\Delta(x_1,\ldots,x_n)\bigr|^\beta \prod_j e^{-V(x_j)} \,
dx_1\cdots dx_n
\end{equation}
on $\Reals^n$. This is known as the Hermite ensemble, because of its intimate connection to the
orthogonal polynomials of the same name, and when $\beta=1$, $2$, or $4$, arises as the eigenvalue
distribution in the classical Gaussian ensembles of random matrix theory. Dumitriu and Edelman showed
that \eqref{CGRL} is the distribution of eigenvalues for a symmetric tri-diagonal matrix with independent
entries (modulo symmetry). The diagonal entries have standard Gaussian distribution and the lower diagonal
entries are $2^{-1/2}$ times a $\chi$-distributed random variable with the number of degrees of freedom equal
to $\beta$ times the number of the row.
The second example treated by Dumitriu and Edelman is the Laguerre ensemble. In statistical circles, this
is known as the Wishart ensemble, special cases of which arise in the empirical determination of the covariance
matrix of a multivariate Gaussian distribution. For this ensemble, one needs to modify the distribution given in
\eqref{CGRL} in two ways: each particle $x_j$ is confined to lie in $[0,\infty)$ and is subject
to the external potential $V(x)=-a\log(x)+x$, where $a>-1$ is a parameter.
In \cite{DumE}, it is shown that if $B$ is a certain $n\times n$ matrix with
independent $\chi$-distributed entries on the main and sub-diagonal (the number of degrees of freedom depends
on $a$, $\beta$, and the element in question) and zeros everywhere else, then the eigenvalues of
$L=BB^T$ follow this distribution.
The third canonical form of the Coulomb gas on $\Reals$ is the Jacobi ensemble.
The distribution is as in \eqref{CGRL}, but now the particles are
confined to lie within $[-2,2]$ and are subject to
the external potential $V(x)=-a\log(2-x)-b\log(2+x)$, where $a,b>-1$
are parameters. This corresponds to the
probability measure on $[-2,2]^n$ that is proportional to
\begin{equation}\label{E:JE}
\bigl|\Delta(x_1,\ldots,x_n)\bigr|^\beta \prod_j \,
(2-x_j)^{a}(2+x_j)^{b} \, dx_1\cdots dx_n.
\end{equation}
The partition function (or normalization coefficient) was determined by
Selberg~\cite{Selberg}. This
will be discussed in Section~\ref{SandA}.
Dumitriu and Edelman did not give a matrix model for this ensemble, listing it as an open problem.
We present a tri-diagonal matrix model in Theorem~\ref{T:2} below. The independent parameters follow
a beta distribution:
\begin{definition}
A real-valued random variable $X$ is said to be beta-distributed with parameters $s,t>0$,
which we denote by $X\sim B(s,t)$, if
\begin{equation}\label{E:beta}
\Exp\{f(X)\} = \frac{2^{1-s-t}\Gamma(s+t)}{\Gamma(s)\Gamma(t)}
\int_{-1}^1 f(x) (1-x)^{s-1}(1+x)^{t-1} \, dx.
\end{equation}
Note that $B(\tfrac{\nu}{2},\tfrac{\nu}{2})$ is the distribution of the
first component of a random vector from the $\nu$-sphere. (See Corollary~\ref{C:A3}.)
\end{definition}
\begin{theorem}\label{T:2}
Given $\beta>0$, let $\alpha_k$, $0\leq k\leq 2n-2$, be distributed as follows
\begin{equation}\label{DistAR}
\alpha_k \sim \begin{cases} B(\tfrac{2n-k-2}{4}\beta + a +
1,\tfrac{2n-k-2}{4}\beta + b + 1) &
\text{$k$ even,} \\
B(\tfrac{2n-k-3}{4}\beta + a+b+2,\tfrac{2n-k-1}{4}\beta) & \text{$k$
odd.}
\end{cases}
\end{equation}
Let $\alpha_{2n-1}=\alpha_{-1}=-1$ and define
\begin{align}
b_{k+1} &= (1-\alpha_{2k-1})\alpha_{2k} -
(1+\alpha_{2k-1})\alpha_{2k-2} \label{E:BofG}\\
a_{k+1} &= \big\{ (1-\alpha_{2k-1})(1-\alpha_{2k}^2)(1+\alpha_{2k+1})
\big\}^{1/2} \label{E:AofG}
\end{align}
for $0\leq k \leq n-1$; then the eigenvalues of the tri-diagonal
matrix
$$
J = \begin{bmatrix}
b_1 & a_1 & & \\
a_1 & b_2 &\ddots & \\
&\ddots&\ddots & {\!a_{n-1}\!} \\
& &{\!a_{n-1}\!}& b_n
\end{bmatrix}
$$
are distributed according to the the Jacobi ensemble \eqref{E:JE}.
\end{theorem}
We know of two other papers which discuss the Jacobi ensemble in a manner inspired by
the work of Dumitriu--Edelman: \cite[\S 4.2]{FR} and \cite{Lip}. These results are however of a rather
different character; in particular, we contend that Theorem~\ref{T:2} is the true Jacobi
ensemble analogue of the results of \cite{DumE}.
In Section~\ref{SandA}, we show how the ideas developed in the earlier parts of this paper
lead to new derivations of the classical integrals of Aomoto \cite{Aomoto} and Selberg \cite{Selberg}.
The main novelty of these proofs is their directness: They treat all values of $\beta$ and
$n$ on an equal footing. In particular, we do not prove the result for $\beta$ an integer
and then make recourse to Carlson's Theorem. These remarks are also applicable to the proof
of \eqref{CGpart} given at the end of Section~\ref{S:3}.
\smallskip
\noindent
\textit{Acknowledgements:} We would like to thank Barry Simon for encouragement and for
access to preliminary drafts of his forthcoming book \cite{Simon}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Overview of the Proofs and Background Material}\label{S:2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
We begin by examining the $\beta=2$ case of Theorem~\ref{T:1}, that is, Haar measure on the unitary group.
Rather than study the eigenvalues as the fundamental statistical object, we will consider the spectral
measure associated to $U$ and the vector $e_1=(1,0,\ldots,0)^T$. It will be denoted by $d\mu$.
As Haar measure is invariant under conjugation, any choice of unit vector $e_1$ leads to the same
probability distribution on $d\mu$.
The most natural coordinates for $d\mu$ are the eigenvalues $e^{i\theta_1},\ldots,e^{i\theta_n}$
and the mass that $d\mu$ gives to them: $\mu_1=\mu(\{e^{i\theta_1}\}),\ldots,\mu_{n-1}=\mu(\{e^{i\theta_{n-1}}\})$.
As $\int d\mu =1$, we omit $\mu_n=\mu(\{e^{i\theta_n}\})$.
Note that we have chosen not to order the eigenvalues, which means that the natural parameter space gives
an $n!$-fold cover of the set of measures. We have already used this way of thinking a number of times,
beginning with \eqref{E:WeylIF}.
The above system of coordinates does not cover the possibility that $U$ has degenerate eigenvalues. However,
as the Weyl integration formula shows, the set of such $U$ has zero Haar measure;
in fact, the density vanishes quadratically at these points.
The reason for this is worth repeating (cf.\ \cite[\S VII.4]{Weyl}): The submanifold where two eigenvalues
coincide has co-dimension three in $\U(n)$; one degree of freedom is lost from the reduction of the number
of eigenvalues and two more are lost in the reduction from two orthogonal one-dimensional eigenspaces to
a single two-dimensional eigenspace. One should compare this to spherical polar coordinates in $\Reals^3$
where $r=0$ is a submanifold of co-dimension three and consequently, the density also vanishes to second order.
In Section~\ref{S:dmuUSO}, we will determine the probability distribution on $d\mu$ induced from Haar measure
on $\U(n)$, in the $\theta,\mu$ coordinates. Conjugation invariance of Haar measure implies that
the eigenvalues and masses are statistically independent; it is then easy to see that the
former are distributed as in \eqref{E:WeylIF} and $(\mu_1,\ldots,\mu_n)$ is uniformly distributed on
the simplex $\sum \mu_j =1$. See Proposition~\ref{P:Un}. This implies that
$d\mu$ gives non-zero weight to each of the eigenvalues with probability one. As a consequence, we can always
recover the eigenvalues from $d\mu$.
We will now introduce different coordinates, $(\alpha_0,\ldots,\alpha_{n-1})$, for $d\mu$
that arise in the study of orthogonal polynomials on the unit circle.
The monomials $1,z,\ldots,z^{n-1}$ form a basis for $L^2(d\mu)$ and so, applying the Gram--Schmidt procedure,
we can construct an orthogonal basis of monic polynomials: $\Phi_j$, $0\leq j j+1
\end{cases}
\end{equation}
where $\rho_j=\sqrt{1-|\alpha_j|^2}$ and $\alpha_{-1}=-1$.
\end{lemma}
\begin{proof}
As $d\mu$ is the spectral measure for $(H,e_1)$ and as $L^2(d\mu)$ has the same dimension as the space
on which the operator $H$ acts, there must be an orthonormal basis $f_1,\ldots,f_n$ for $L^2(d\mu)$
with $f_0\equiv 1$ such that $H$ represents $f(z)\mapsto zf(z)$ in this basis. This is just the spectral
theorem combined with the fact that $e_1$ must be cyclic (for otherwise, $L^2(d\mu)$ wouldn't have full
dimension).
From the cyclicity argument we also learn that $H$ must have a strictly positive sub-diagonal.
To finish the proof of the first claim, we need only show that $f_j(z)=\phi_{j-1}(z)$; that is,
that the orthonormal basis in question is precisely that of the orthonormal polynomials. Because $H$
is in Hessenberg form with positive sub-diagonal, the standard basis vectors arise from applying the
Gram--Schmidt procedure to $e_1,He_1,\ldots,H^{n-1}e_1$. Consequently, the vectors $f_j$ must be
the result of applying the same procedure to $1,z,\ldots,z^{n-1}$; that is, $f_j$ must be $\phi_{j-1}$.
The first part of \eqref{GGTentry} merely reexpresses what we have just proved. The second follows from
the recursion relations and \eqref{PhiNorm}; however, the proof is not particularly enlightening and
we refer the reader to \cite{GeronMS} for details.
\end{proof}
We will now apply the Householder algorithm outlined earlier to a matrix chosen at random from $\U(n)$.
By the lemma above, this will allow us to determine the induced distribution on the Verblunsky parameters
associated to the spectral measure for $(U,e_1)$.
\begin{prop}\label{P:UnG}
Let $d\mu$ be the spectral measure for $(U,e_1)$ with $U$ chosen at
random from $\U(n)$ according to Haar measure.
In terms of the Verblunsky parameters, $\alpha_0,\ldots,\alpha_{n-2}$,
and $\alpha_{n-1}=e^{i\phi}$, this probability
distribution is given by
\begin{equation}\label{Uingamma}
\tfrac{1}{2}\tfrac{(n-1)!}{\pi^n}\prod_{k=0}^{n-2}
(1-|\alpha_k|^2)^{n-2-k}
\, d^2 \alpha_0\cdots d^2\alpha_{n-2} \, d\phi.
\end{equation}
That is, the Verblunsky parameters are independent and
$\alpha_j\sim\Theta_{2n-2j-1}$.
\end{prop}
\begin{proof}
The key to applying the Householder algorithm to a random element
$U\in\U(n)$ is the following realization of Haar measure:
Choose the first column at random from the unit sphere; then choose the
second column from the unit sphere of
vectors orthogonal to the first; then the third column and so forth.
In this way, one could say that the columns of $U$
form a random orthonormal basis for $\Cmplx^n$. (That this is indeed
Haar measure is a simple consequence
of invariance under left multiplication by unitary matrices.)
The first column of $U$ is a random vector from the unit sphere. After
left multiplication by the appropriate reflection $R$,
the new first column takes the form $[\bar\alpha_0,b,0,\ldots,0]^T$
where $\bar\alpha_0$ is the the original $(1,1)$ entry
of $U$ and so $\Theta_{2n-1}$-distributed, while $b$ has modulus
$\rho_0$ and arbitrary argument. Subsequent left
multiplication by $D$ converts the first column to
$[\bar\alpha_0,\rho_0,0,\ldots,0]^T$, as it will remain.
The other columns are still orthogonal to the first; indeed,
they form a random orthonormal basis for the orthogonal complement of
the first column.
Right multiplication by $RD^\dagger$ leaves the first column untouched
while orthogonally intermixing the other columns.
Of course, this means that they remain a random orthonormal basis for
the orthogonal complement of the first column. (Remember, Haar measure
is also invariant under right multiplication by a unitary.)
For the subsequent columns, the procedure is similar. Let us skip
ahead to dealing with the $k$th column.
From the unitarity of the matrix $H$ from the previous lemma,
$$
X=
\begin{bmatrix}
\rho_0\rho_1\rho_2\cdots\rho_{k-2} \\
-\alpha_0\rho_1\rho_2\cdots\rho_{k-2} \\
-\alpha_1\rho_2\cdots\rho_{k-2} \\
\vdots \\
-\alpha_{k-3}\rho_{k-2} \\
-\alpha_{k-2} \\
0 \\
\vdots \\
0
\end{bmatrix}
$$
is a unit vector orthogonal to the first $k-1$ columns. As the $k$th
column is a random vector orthogonal
to the first $k-1$ columns, its inner product with $X$ is distributed
as the top entry of a random vector
from the $(2n-2k+1)$-sphere and is independent of
$\alpha_1,\ldots,\alpha_{k-2}$. Let us call this inner
product $\bar\alpha_{k-1}$, noting that this implies $\alpha_{k-1}$ is
$\Theta_{2n-2k+1}$-distributed
as stated in the proposition.
We now multiply the matrix at hand from the left by the appropriate
reflection and rotation to bring the $k$th
column into the desired
form. Neither of these operations alters the top $k$ rows and so the
inner product of the $k$th column
with $X$ is unchanged. But now the $k$th column is uniquely
determined; it must be $\bar\alpha_{k-1}X+\rho_{k-1}e_{k+1}$,
just as in \eqref{GGTentry}.
Lastly, we should multiply on the right by $RD^\dagger$, but this
leaves the first $k$ columns unchanged while
orthogonally intermixing the other columns. In this way, we obtain a
matrix whose first $k$ columns conform to the
structure of $H$, while the remaining columns form a random basis for
the orthogonal complement of the span of
those $k$ columns.
In this way, we can proceed inductively until we reach the last column.
It is obliged to be a random orthonormal basis
for the one-dimensional space orthogonal to the preceding $n-1$ columns
and hence a random unimodular multiple,
say $\bar\alpha_{n-1}$, of $X$. This is why the last Verblunsky
parameter is $\Theta_1$-distributed.
We have now conjugated $U$ to a matrix in the form of \eqref{GGTentry}
with parameters distributed as stated
in the proposition. The vector $e_1$ is unchanged under the action of
each of the conjugating matrices and
consequently, these are precisely the Verblunsky parameters of $d\mu$.
\end{proof}
We now turn to the study of Haar measure on $\SO(2n)$. The proofs
follow those given above pretty closely.
\begin{prop}\label{SO2n}
If $U$ is chosen at random from $\SO(2n)$ according to Haar measure,
then the spectral measure
$d\mu$ associated to $(U,e_1)$ is distributed as
\begin{equation}\label{SOinzmu}
\frac{(n-1)!}{2^{n-1}\ n!}
\bigl|\Delta\bigl(2\cos\theta_1,\ldots,2\cos\theta_n\bigr)\bigr|^2
\frac{d\theta_1}{\pi}\cdots\frac{d\theta_n}{\pi} d\mu_1\cdots
d\mu_{n-1}
\end{equation}
where $\theta_j$ and $\mu_j$ are the coordinates given in
\eqref{Ocoords}.
\end{prop}
\begin{proof}
By the Weyl integration formula for $\SO(2n)$, the marginal distribution of the
eigenvalues is as above. (See \cite[\S VII.9]{Weyl}.)
If the eigenvalues are prescribed, say $e^{\pm i\theta_1},\ldots,e^{\pm i\theta_n}$, then
the conditional distribution of $U$ is given by taking a fixed matrix with this spectrum and
conjugating it by a random element from $\SO(2n)$. The natural choice for this fixed matrix is
block diagonal:
$$
U_0 = \diag\left(
\begin{bmatrix} \cos(\theta_1) & \sin(\theta_1) \\ -\sin(\theta_1) &
\cos(\theta_1)\end{bmatrix},\ldots,
\begin{bmatrix} \cos(\theta_n) & \sin(\theta_n) \\
-\sin(\theta_n)&\cos(\theta_n)\end{bmatrix}\right).
$$
From this we see that the $d\mu$ is the spectral measure for $U_0$ and a random vector from the $(2n-1)$-sphere.
The proposition then follows by diagonalizing $U_0$ and applying Corollary~\ref{C:A3}.
\end{proof}
\begin{prop}\label{SO2nG}
Let $U$ be chosen from $\SO(2n)$ according to Haar measure and let
$d\mu$ denote the spectral measure
for $(U,e_1)$. In terms of the Verblunsky parameters, the probability
distribution on $d\mu$ is
\begin{equation}\label{SOing}
\tfrac{(n-1)!}{\pi^n} \prod_{k=0}^{2n-2}
(1-\alpha_k^2)^{\frac{2n-k-3}{2}}\,d\alpha_0\cdots d\alpha_{2n-2}
\end{equation}
with $\alpha_{2n-1}=-1$.
That is, the Verblunsky parameters are independent and $\alpha_k\sim
B(\frac{2n-k-1}{2},\frac{2n-k-1}{2})$.
\end{prop}
\begin{proof}
While we may use the Householder algorithm as set out above, the fact
that we are now dealing with
real-valued matrices allows the following simplification: the vector
defining the reflection is
as in \eqref{E:vDef}, but now with
$$
\alpha = a_{k+1,k} - \sqrt{a_{k+1,k}^2 + \cdots + a_{2n,k}^2}
$$
instead of \eqref{E:alphaDef}. This permits us to forgo the
conjugation by $D$.
Haar measure on $\SO(2n)$ can be realized by choosing the first column
as a random vector from the
unit sphere in $\Reals^{2n}$, and then the second as a random vector
orthogonal to the first, and so
forth. However, the fact that the matrix has determinant one means that
the first $2n-1$ columns
completely determine the last. One may say that the columns of $U$
form a random positively oriented
basis for $\Reals^{2n}$.
Proceeding as in the proof of Proposition~\ref{P:UnG}, we see that
$\alpha_{k-1}$ is defined as the
inner product of a specific vector $X$ with a random unit vector from
the $(2n-k-2)$-sphere of vectors
orthogonal to the first $k-1$ columns. It follows from
Corollary~\ref{C:A2} that $\alpha_{k-1}\sim
B(\frac{2n-k-2}{2},\frac{2n-k-2}{2})$ as stated above.
The last column of $H$, and hence $\alpha_{2n-1}$, is uniquely determined by the fact that $\det(H)=1$.
It is just a matter of using \eqref{GGTentry} to determine which value of $\alpha_{2n-1}$ makes
this determinant one; moreover, by continuity of the determinant, it suffices to consider the case where
all other $\alpha$s are zero. This gives
$$
1=\det(H)=-\alpha_{-1}\alpha_{2n-1}\sign(\sigma)=-\alpha_{2n-1}
$$
where $\sigma$ is the cyclic permutation $j\mapsto j+1 \mod 2n$, which is odd.
Lastly, we should justify the normalization coefficient given in \eqref{SOing}; what appears there
is very much simpler than one would expect from \eqref{E:beta}. This
simplification is based on the duplication formula for the $\Gamma$ function:
$
\sqrt{\pi}\Gamma(2t) = 2^{2t-1}\Gamma(t)\Gamma(t+\frac12).
$
Specifically, beginning with \eqref{E:beta},
$$
\int_{-1}^1 (1-\alpha^2)^{t-1} \, d\alpha =
\frac{2^{1-2t}\Gamma(2t)}{\Gamma(t)^2}
= \frac{\Gamma(t+\frac12)}{\sqrt{\pi}\,\Gamma(t)},
$$
which causes the product of normalization coefficients to telescope:
$$
\prod_{k=0}^{2n-2} \frac{\Gamma(\frac{2n-k}{2}) }{ \sqrt{\pi}\,\Gamma(\frac{2n-k-1}{2}) }
= \pi^{\frac{1}{2}-n} \frac{\Gamma(n)}{\Gamma(\tfrac{1}{2})}
= \frac{(n-1)!}{\pi^n},
$$
as given in \eqref{SOing}.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Proof of Theorem 1}\label{S:3}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $d\mu$ be the measure on $S^1$ given by
\begin{equation}\label{S3mu}
\int f d\mu = \sum_{j=1}^n \mu_j f(e^{i\theta_j})
\end{equation}
with $\theta_j\in[0,2\pi)$ distinct and $\sum \mu_j=1$.
As discussed in the Introduction, this measure is uniquely determined
by its Verblunsky parameters $\alpha_0,\ldots,\alpha_{n-2}\in\Disk$ and
$\alpha_{n-1}=e^{i\phi}\in S^1$.
It is difficult to find functions of $d\mu$ that admit simple expressions in terms of both
$\theta_j,\mu_j$ and the Verblunsky parameters. One such quantity is the determinant of the
associated Toeplitz matrix; this is the subject of the next lemma.
\begin{lemma}\label{VandUn}
If $d\mu$ is a probability measure of the form given in \eqref{S3mu} and $\alpha_0,\ldots,\alpha_{n-1}$
its Verblunsky coefficients, then
\begin{equation}\label{VUnit}
|\Delta(z_1,\ldots,z_n)|^2 \, \prod_{j=1}^n \mu_j = \prod_{k=0}^{n-2}
(1-|\alpha_k|^2)^{n-k-1}.
\end{equation}
\end{lemma}
\begin{proof}
Let $c_k=\sum_{j=1}^{n} \mu_j z_j^k$ denote the moments of $d\mu$.
We will prove that both sides of \eqref{VUnit} are equal to the
determinant of the
$n\times n$ Toeplitz matrix associated to $d\mu$:
$$
T=\begin{bmatrix}
c_0 & c_{-1} & \cdots & c_{1-n} \\
c_1 & c_0 & \cdots & c_{2-n} \\
\vdots & \vdots & \ddots & \vdots \\
c_{n-1} & c_{n-2} & \cdots & c_0 \\
\end{bmatrix}.
$$
If we define $A$ and $M$ by
$$
A=\begin{bmatrix}
1 & 1 & \cdots & 1 \\
z_1 & z_2 & \cdots & z_n \\
\vdots & \vdots & \ddots & \vdots \\
z_1^{n-1} & z_2^{n-1} & \cdots & z_n^{n-1}
\end{bmatrix}
\quad
M=\begin{bmatrix}
\mu_1 & 0 & \cdots & 0 \\
0 & \mu_2 & \cdots & 0 \\
\vdots &\vdots & \ddots & \vdots \\
0 & 0 & \cdots & \mu_n
\end{bmatrix},
$$
then $T = A M A^\dagger$. Consequently,
$$
\det(T) = \bigl|\det(A)\bigr|^2 \, \det(M) =
|\Delta(z_1,\ldots{},z_n)|^2 \, \prod_{j=1}^n \mu_j.
$$
We will now show that the right-hand side of \eqref{VUnit} is equal to
$\det(T)$.
To this end, let
$$
B=\begin{bmatrix}
\Phi_0(z_1) & \Phi_0(z_2) & \cdots & \Phi_0(z_n) \\
\Phi_1(z_1) & \Phi_1(z_2) & \cdots & \Phi_1(z_n) \\
\vdots & \vdots & \ddots & \vdots \\
\Phi_{n-1}(z_1) & \Phi_{n-1}(z_2) & \cdots & \Phi_{n-1}(z_n)
\end{bmatrix},
$$
which has the same determinant as $A$ because each can be reduced to the other by elementary row
operations. From the orthogonality property of the $\{\Phi_j\}$, it follows that
$B M B^\dagger$ is the diagonal matrix whose entries are the squares of the $L^2(d\mu)$-norms of
$\Phi_0,\Phi_1,\ldots{,}\Phi_{n-1}$. Therefore by \eqref{PhiNorm},
$$
\det(T) = \det(B M B^\dagger) = \prod_{k=0}^{n-1} \bigl\|\Phi_k\bigr\|^2_{L^2(d\mu)}
= \prod_{k=0}^{n-2} (1-|\alpha_k|^2)^{n-k-1},
$$
just as was required.
\end{proof}
Both expressions for the Toeplitz determinant are well known; indeed, the argument presented above and its
Hankel-matrix analog play a central role in random matrix theory.
We are now ready to state and prove the main result of this section.
Please note that neither measure given below is normalized; however, they do have the same
normalization coefficient. It is calculated in Lemma~\ref{Somewhere} where it is used to
give an independent proof of \eqref{CGpart}.
\begin{prop}\label{ItU}
The following formulae express the same measure on the manifold of
probability distributions on $S^1$ supported at $n$ points:
\begin{equation}\label{qqqqq}
\frac{2^{1-n}}{n!} |\Delta(e^{i\theta_1},\ldots,e^{i\theta_n})|^{\beta}
\prod_{j=1}^n
\mu_j^{\frac{\beta}{2}-1} \, d\theta_1 \cdots d\theta_n\,d\mu_1 \cdots
d\mu_{n-1}
\end{equation}
in the $(\theta,\mu)$-coordinates of \eqref{S3mu}, and
\begin{equation}\label{ggggg}
\prod_{k=0}^{n-2} (1-|\alpha_k|^2)^{\frac{\beta}{2}(n-k-1)-1} \, d^2
\alpha_0 \cdots d^2\alpha_{n-2}\,d\phi
\end{equation}
in terms of the Verblunsky parameters.
\end{prop}
\begin{proof}
When $\beta=2$, this follows immediately from Propositions~\ref{P:Un} and~\ref{P:UnG}.
To obtain the general-$\beta$ version of \eqref{qqqqq} from the $\beta=2$ version, one
has to multiply by
\begin{equation}\label{VUnitLHS}
|\Delta(e^{i\theta_1},\ldots,e^{i\theta_n})|^{\beta-2} \prod_{j=1}^n
\mu_j^{\frac{\beta}{2}-1},
\end{equation}
while the same transformation of $\beta$ in \eqref{ggggg} is effected by multiplying by
\begin{equation}\label{VUnitRHS}
\prod_{k=0}^{n-2} (1-|\alpha_k|^2)^{(\frac{\beta}{2}-1)(n-k-1)}.
\end{equation}
But, \eqref{VUnitLHS} and \eqref{VUnitRHS} are equal; they are either side of \eqref{VUnit}
raised to the power $\frac{\beta}{2}-1$.
\end{proof}
\begin{proof}[Proof of Theorem 1]
Theorem~\ref{T:1} is an immediate corollary of Proposition~\ref{ItU} and results from \cite{CMV}:
The Verblunsky parameters of the spectral measure for $(LM,e_1)$ are precisely the $\alpha$s
that appear in the definition of $L$ and $M$. Consequently, if the Verblunsky parameters
are distributed according to \eqref{ggggg}, then the eigenvalues are distributed as in \eqref{CGbeta}.
\end{proof}
It is fair to suggest that studying the unitary group is a rather roundabout way of proving the above proposition.
We simply do not know a better way. The natural suggestion is to first calculate the Jacobian for the map
from the $(\theta,\mu)$-coordinates to the Verblunsky parameters and to proceed from there. While we can
determine this Jacobian \textit{a posteriori} by employing a cunning idea of Dumitriu and
Edelman, we do not have a direct derivation.
The idea of Dumitriu and Edelman can be summarized as follows:
\begin{lemma}\label{L:Jacobian}
Suppose $\phi:\mathcal{O}_1\to\mathcal{O}_2$ is an $N$-fold cover of $\mathcal{O}_2$ by $\mathcal{O}_1$,
both of which are open subsets of $\Reals^n$. If the measure $f(x)\,d^nx$ is the symmetric pull-back
of the measure $g(y)\,d^ny$, with both $f$ and $g$ positive, then the Jacobian of $\phi$ is given by
$$
|\phi'(x)|=\frac{N f(x)}{g\circ\phi\,(x)}
$$
for any $x\in\mathcal{O}_1$.
\end{lemma}
\begin{proof}
For every $x\in\mathcal{O}_1$,
$$
f(x)\,d^nx = \phi^*\bigl(\tfrac{1}{N}g(y)\,d^ny\bigr) = \tfrac{1}{N}[g\circ\phi](x)|\phi'(x)| \,d^nx,
$$
which proves the lemma.
\end{proof}
Proposition~\ref{ItU} allows us to apply this lemma to the current situation.
The map of $\theta,\mu$ to the Verblunsky parameters is an $n!$-fold cover and so we obtain
\begin{align}\label{JU}
\left|\frac{\partial(\alpha,\phi)}{\partial(\theta,\mu)}\right|
&= 2^{1-n} \frac{|\Delta(e^{i\theta_1},\ldots,e^{i\theta_n})|^2 }
{\prod_{k=0}^{n-2} (1-|\alpha_k|^2)^{n-k-2}} \\
&= 2^{1-n} \frac{\prod_{k=0}^{n-2} (1-|\alpha_k|^2) }
{\prod_{j=0}^{n} \mu_j }
\end{align}
where one should regard the Verblunsky parameters as functions of the $\theta$s and $\mu$s.
The formulae above correspond to applying Proposition~\ref{ItU} with $\beta=2$ and $\beta=0$,
respectively. Of course, one can also use other values of $\beta$, but the resulting formulae
are related to one another by Lemma~\ref{VandUn}.
Earlier we promised to determine the (common) normalization coefficient for the measures
\eqref{qqqqq} and \eqref{ggggg}. We also promised to give a new derivation of the partition
function, \eqref{CGpart}, for the Coulomb gas. We will now settle these obligations.
\begin{lemma}\label{Somewhere}
The integral of \eqref{ggggg} is
\begin{equation}\label{SW1}
\int\!\!\cdots\!\!\int \prod_{k=0}^{n-2}
(1-|\alpha_k|^2)^{\frac{\beta}{2}(n-k-1)-1} \,
d^2 \alpha_0 \cdots d^2\alpha_{n-2}\,d\phi =
\frac{(2\pi)^n}{\beta^{n-1}(n-1)!}
\end{equation}
while
\begin{align}\label{SW2}
\int\!\! \cdots \!\! \int
\bigl|\Delta(e^{i\theta_1},\ldots,e^{i\theta_n})\bigr|^\beta \,
\frac{d\theta_1}{2\pi} \cdots \frac{d\theta_n}{2\pi}
= \frac{\Gamma(\tfrac12\beta n + 1)}{\bigl[\Gamma(\tfrac12\beta +
1)\bigr]^n},
\end{align}
in agreement with \eqref{CGbeta} and \eqref{CGpart}.
\end{lemma}
\begin{proof}
Each of the integrals in \eqref{SW1} is rendered trivial by switching to polar coordinates:
$$
\int (1-|z|^2)^{(t/2)-1} \,d^2 z = 2\pi t^{-1}.
$$
It is this integral that gives the normalization coefficient in the definition of the
$\Theta$ distributions; cf. \eqref{E:ThetaDefn}.
The proof of \eqref{SW2} begins with the evaluation of the Dirichlet integral
$$
\int_\triangle \prod_{j=1}^n \mu_j^{\frac{\beta}{2}-1} \, d\mu_1 \cdots
d\mu_{n-1}
= \frac{\Gamma(\tfrac{\beta}{2})^n}{\Gamma(\tfrac{n\beta}{2})},
$$
which is derived in the proof of Lemma~\ref{L:A4}; see
\eqref{DirichletI}.
As the two measures in Proposition~\ref{ItU} have the same total mass,
the integral of
\eqref{qqqqq} is given by \eqref{SW1}. Combining this with the
Dirichlet integral above
leads us to
\begin{align*}
\text{LHS \eqref{SW2}}
=\frac{n!}{2^{1-n}}
\frac{1}{\beta^{n-1}(n-1)!}\frac{\Gamma(\tfrac{n\beta}{2})}{\Gamma(\tfrac{\beta}{2})^n}
=\frac{\Gamma(\tfrac{n\beta}{2}+1)}{\Gamma(\tfrac{\beta}{2}+1)^n},
\end{align*}
which is exactly \eqref{CGpart}.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Proof of Theorem~2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As explained in Section~\ref{S:2}, Theorem~\ref{T:2} is an immediate corollary of Proposition~\ref{OIt}
and the Geronimus relations. As a result, the primary purpose of this section is to prove this proposition.
Throughout this section, $d\mu$ will denote a probability measure of the form given in \eqref{Ocoords}.
In particular, it is symmetric with respect to complex conjugation and the last Verblunsky parameter,
$\alpha_{2n-1}$, is equal to $-1$. We will also use the notation $x_j=2\cos(\theta_j)$ repeatedly.
In addition to Lemma~\ref{VandUn} from the previous section, two further lemmas are required.
They are the following:
\begin{lemma}\label{VandVand}
If $x_j=2\cos(\theta_j)$, $1\leq j \leq n$, then
\begin{equation}
|\Delta(e^{\pm i\theta_1}, e^{\pm i\theta_2},\ldots, e^{\pm i\theta_n})|
= |\Delta(x_1,x_2,\ldots,x_n)|^2 \, \prod_{l=1}^n
|2\sin(\theta_l)|
\end{equation}
where the left-hand side is shorthand for the Vandermonde of the $2n$ quantities
$e^{i\theta_1},e^{-i\theta_1}, \ldots, e^{i\theta_n}, e^{-i\theta_n}$.
\end{lemma}
\begin{proof}
By expanding the Vandermonde as in \eqref{VDefn},
\begin{align*}
&\phantom{{}={}} |\Delta(e^{\pm i\theta_1}, e^{\pm i\theta_2},\ldots, e^{\pm i\theta_n})| \\
&=\prod_{l=1}^n \bigl|e^{i\theta_l}-e^{-i\theta_l}\bigr|
\prod_{j