# Asymptotic theory of Schatten classes

Michael L. Juhos, Zakhar Kabluchko, and Joscha Prochno

###### Abstract

The study of Schatten classes has a long tradition in geometric functional
analysis and related fields. In this paper we study a variety of geometric and
probabilistic aspects of finite-dimensional Schatten classes of not
necessarily square matrices. Among the main results are the exact and
asymptotic volume of the Schatten/$\infty$ unit ball, the boundedness of its
isotropy constant, a Poincaré-Maxwell-Borel lemma for the uniform distribution
on the Schatten/$\infty$ ball, and Sanov/type large deviations principles for
the singular values of matrices sampled uniformly from the Schatten/$p$ unit
ball for any $0<p\leq\infty$.

Keywords. Isotropic constant, large deviations principle, Poincaré-Maxwell-
Borel lemma, Schatten class, volume
MSC. Primary: 46B06, 52A23, 60F05, 60F10 Secondary: 52A22, 60D05

###### Contents

  1. 1 Introduction
    1. 1.1 General overview
    2. 1.2 Organization of the paper
  2. 2 Mathematical setup
    1. 2.1 Norms and spaces
    2. 2.2 Auxiliary notions and results
  3. 3 Geometry of Schatten/̄$p$ balls
  4. 4 Probabilistic results and weak limit theorems for $p=\infty$
  5. 5 Theory of $K$/symmetric measures
  6. 6 Sanov/type LDPs for singular values
    1. 6.1 Prerequisites
    2. 6.2 Results
    3. 6.3 Proof of Theorem G
    4. 6.4 Proof of Theorem H

##  1 Introduction

###  1.1 General overview

The Schatten classes (or Schatten-von Neumann classes) of compact operators
are a classical object of study in functional analysis, and their
finite/dimensional counterparts, that is, spaces of matrices endowed with the
$\ell_{p}$/̄norm of their singular values, have attracted much attention
within the last few decades and many results have been produced in a number of
disciplines, among them local theory of Banach spaces, random matrix theory,
and (asymptotic) convex geometry.

The Schatten/̄$p$ class $S_{p}$, where $p\in(0,\infty]$, consists of all
compact linear operators between two given Hilbert spaces whose sequence of
singular values lies in the sequence space $\ell_{p}$, and thus it subsumes
the special cases of nuclear or trace/class operators ($p=1$) and of Hilbert-
Schmidt operators ($p=2$). It was introduced by Schatten in [48, Chapter 6,
pp. 71] (though he said “completely continuous” for “compact”), having its
roots in earlier works by himself [47] and with von Neumann [49, 50]
concerning nuclear operators on Hilbert spaces; those works were generalized
by Ruston [44] to Banach spaces and by Grothendieck [16] to locally convex
spaces.

From the point of view of the geometry of Banach spaces, Gordon and Lewis [15]
obtained that the space $S_{p}$ for $p\neq 2$ does not have local
unconditional structure and therefore does not admit an unconditional basis,
Tomczak/Jaegerman [55] showed that this space has Rademacher cotype $2$, and
Szarek and Tomczak/Jaegerman [54] provided bounds for the volume ratio of
$S_{1}^{n}$. Saint Raymond [46] computed the asymptotic volumes of real and
complex Schatten balls up to a yet undetermined factor, and König, Meyer, and
Pajor [33] established boundedness of the isotropic constant for all
$p\in[1,\infty)$. Guédon and Paouris [17] proved concentration of mass
properties, Barthe and Cordero/Erausquin [5] studied variance estimates,
Chávez/Domínguez and Kuzarova [8] determined the Gelfand widths of certain
identity mappings between finite/dimensional Schatten classes, Radtke and
Vritsiou [42] proved the thin/shell conjecture and Vritsiou [57] the variance
conjecture for the spectral norm, i.e. $p=\infty$; Hinrichs, Prochno, and
Vybíral [20] computed the entropy numbers for natural embeddings
$S_{p}^{n}\hookrightarrow S_{q}^{n}$, and in [21] the same authors determined
the corresponding Gelfand numbers; Prochno and Strzelecki [40] improved the
last stated results and complemented them by the approximation and Kolmogorov
numbers. In a series of papers, Kabluchko, Prochno, and Thäle [27, 26, 28]
refined some of the previous works by giving the precise asymptotic volume of
Schatten balls, they computed the volume ratio, considered the volume of
intersections in the spirit of Schechtman and Schmuckenschläger [51], and
derived Sanov/type large deviations principles for the empirical measures of
singular values of random matrices sampled from Schatten balls. This last
results stands in line with the study of large deviations for spectral
measures of random matrices initiated by Ben Arous and Guionnet [7], and
Kaufmann and Thäle [29] have generalized the underlying distribution to a
certain mixture of uniform distribution and cone measure on the sphere in the
spirit of Barthe, Guédon, Mendelson, and Naor [6]. The question of volume of
intersections has also been taken up by Sonnleitner and Thäle [53]. Very
recently Dadoun, Fradelizi, Guédon, and Zitt [11] determined the asymptotics
of the moments of inertia of Schatten balls, thereby also providing the limit
of the isotropic constant, and confirmed the variance conjecture for $p>3$.

One of the main deficiencies of most results mentioned above is that they have
only been obtained for Schatten classes of _square_ matrices, some even more
specially for Hermitian matrices, which is in a sense quite restrictive. The
aim of the present article is to extend the scope of some of the results to
Schatten classes of $m$/by/$n$ matrices for arbitrary $m,n\in\mathbb{N}$, in
as far as that is possible or sensible, thereby complementing and generalizing
the existing literature. As we shall see, this is far from being trivial in
several situations.

Let us be more specific about the results obtained in this paper. First we
provide the _exact_ and asymptotic volume of the Schatten/$\infty$ unit ball
of $m$/by/$n$ matrices (Theorem A); the former is done via a Weyl/type
integral transformation formula (similar to the one already used by Saint
Raymond [46]) and evaluating the Selberg integral that appears. Next we
provide bounds for its isotropic constant (Theorem B), thereby adding to the
work of König, Meyer, and Pajor [33]; notably, we are also able to give the
precise limit as $n\to\infty$ in the case when $\lim_{n\to\infty}\frac{m}{n}$
exists. As the last major purely geometrical result, we clarify the relation
between the Hausdorff and cone measures on the Schatten/$p$ unit sphere for
arbitrary $p$ (Theorem C); in the case of the classical $\ell_{p}$/spheres
this was done by Naor and Romik [38, Lemma 2], and, interestingly, the
formulas for the Radon-Nikodým derivatives turn out to be formally equal.

We then continue to explore asymptotic properties of the Schatten/̄$\infty$
balls via a probabilistic representation of the uniform distribution on the
ball (Proposition 4.1) in the spirit of Schechtman and Schmuckenschläger [51]
and of Rachev and Rüschendorf [41], who introduced those for the classical
$\ell_{p}$/balls, leading us to a Poincaré-Maxwell-Borel result (Theorem D)
stating that, fixing the number $m$ of rows and letting the number $n$ of
columns tend to infinity, the projection onto the first $k$ columns converges
to an $m$/by/$k$ matrix with i.i.d. Gaussian components. Under the same
premises we also prove a central limit theorem for the inner product of two
points sampled independently from the unit ball or sphere (Theorem E).

Then we introduce an apparently new notion of symmetry of (probability)
measures w.r.t. a given convex body, and apply this notion, together with
two/sided unitary invariance, to Schatten classes. As the main result (Theorem
F), we produce a probabilistic representation of those random matrices whose
distribution follows this new symmetry w.r.t. Schatten/̄$p$ balls, thereby
subsuming the representations for the singular values applied by Kabluchko,
Prochno, and Thäle [27, 26, 28] in the non/Hermitian case.

Finally, we present Sanov/type large deviations principles for random matrices
sampled from Schatten/$p$ balls for any $p\in(0,\infty]$ (Theorems G and H).
This extends the result for non/Hermitian random matrices given in [28]. It is
important to note here that for non/square matrices the joint density of
singular values contains an additional dimension/dependent nonconstant factor,
which renders the adaption of the proof in places highly nontrivial.

###  1.2 Organization of the paper

Since the present paper contains a wide range of results it is structured
according to the general themes of those results, and the (mostly short)
proofs directly follow their propositions; the exception is Section 6 where
the central results are collected at the beginning and the corresponding
(extensive) proofs are deferred to subsections.

Section 2 lays out the mathematical setup including the norms and spaces, as
well as frequently used auxiliary results.

Section 3 treats the elementary geometric properties of Schatten balls,
including Theorems A, B, and C.

Section 4 contains the results pertaining to the weak limit theorems for the
Schatten/̄$\infty$ ball, that is Proposition 4.1 and Theorems D and E.

Section 5 deals with $K$/̄symmetric measures and concludes with Theorem F.

Finally Section 6 is dedicated to the Sanov/type large deviations principles
for the empirical spectral measures stated in Theorems G and H.

##  2 Mathematical setup

###  2.1 Norms and spaces

Let $\mathbb{K}\in\\{\mathbb{R},\mathbb{C}\\}$, and by $\beta\in\\{1,2\\}$
denote the $\mathbb{R}$/dimension of $\mathbb{K}$. (Most of the present
results remain valid even for $\mathbb{K}=\mathbb{H}$, the skew field of
quaternions; then $\beta=4$.) The nonnegative and positive real numbers are
written $\mathbb{R}_{\geq 0}$ and $\mathbb{R}_{>0}$, resp.; $\mathsf{i}$ is
the imaginary unit, and $\Re(x)$, $\Im(x)$, $\overline{x}$ denote the real
part, imaginary part, and complex conjugate of $x$ (applied entrywise for
vectors and matrices). For the $d$/dimensional Lebesgue measure we write
$\operatorname{\mathit{v}}_{d}$. For $m,n\in\mathbb{N}$ with $m\leq n$
consider $\mathbb{K}^{m\times n}$, the $\beta mn$/dimensional real vector
space of $(m\times n)$/matrices. This is endowed with a Euclidean structure
via the inner product (the _Hilbert-Schmidt_ or _Frobenius inner product_)

$\langle
x,y\rangle:=\Re\operatorname{tr}(xy^{*})=\Re\biggl{(}\sum_{i=1}^{m}\sum_{j=1}^{n}x_{i,j}\overline{y_{i,j}}\biggr{)}\\!,$

where $\operatorname{tr}$ means the trace of a square matrix. $I_{m}$ denotes
the $(m\times m)$/identity matrix. The collection of self/adjoint $(m\times
m)$/matrices is denoted by $\text{Sym}_{m,\beta}$, and that of self/adjoint
and positive/definite $(m\times m)$/matrices by $\text{Pos}_{m,\beta}$. On
$\text{Sym}_{m,\beta}$ define the usual partial order (also called
_Loewner/order_): for $a,b\in\text{Sym}_{m,\beta}$, $a\leq b$ iff $b-a$ is
positive/semidefinite, or equivalently $\xi^{*}a\xi\leq\xi^{*}b\xi$ for all
$\xi\in\mathbb{K}^{m}$. (Elements of $\mathbb{K}^{m}$ are always considered
column/vectors here.)

Recall that for any $x\in\mathbb{K}^{m\times n}$ the $(m\times m)$/matrix
$xx^{*}$ is symmetric and positive/semidefinite, and hence all the latter’s
eigenvalues are nonnegative real numbers, whose nonnegative square/roots are
called the _singular values_ of $x$. Let $s(x):=(s_{i}(x))_{i\leq
m}\in\mathbb{R}^{m}$ be the vector of the singular values of $x$, arranged
non/increasingly.

For $p\in(0,\infty]$ call $\lVert\cdot\rVert_{p}$ the $\ell_{p}$/norm on
$\mathbb{K}^{m}$ (which is a proper quasinorm iff $p<1$), i.e., for
$\xi=(\xi_{i})_{i\leq m}$,

$\lVert\xi\rVert_{p}:=\begin{cases}\bigl{(}\sum_{i=1}^{m}\lvert\xi_{i}\rvert^{p}\bigr{)}^{1/p}&\text{if
}p<\infty,\\\ \max\\{\lvert\xi_{i}\rvert\,:\,i\leq m\\}&\text{if
}p=\infty;\end{cases}$

the corresponding space is written
$\ell_{p,\beta}^{m}:=(\mathbb{K}^{m},\lVert\cdot\rVert_{p})$, the unit ball is
denoted by $\mathbb{B}_{p,\beta}^{m}$, the unit sphere by
$\mathbb{S}_{p,\beta}^{m}$,111Our notation differs from the convention of
writing $\mathbb{S}_{p}^{m-1}$, where in the real case $m-1$ is the dimension
indeed. But already in the comlex case this is contradictory because
$\partial\mathbb{B}_{p,2}^{m}$ has real dimension $2m-1$. Therefore we have
opted to only notate the number of components. and
$\omega_{p,\beta}^{m}:=\operatorname{\mathit{v}}_{\beta
m}(\mathbb{B}_{p,\beta}^{m})$. On $\mathbb{K}^{m\times n}$ define the
_Schatten/̄ $p$/norm_ $\lVert\cdot\rVert_{S_{p}}$ by

$\lVert x\rVert_{S_{p}}:=\lVert
s(x)\rVert_{p}=\operatorname{tr}((xx^{*})^{p/2})^{1/p}\quad\text{for
$x\in\mathbb{K}^{m\times n}$.}$

In particular, $\lVert x\rVert_{S_{\infty}}=s_{1}(x)$ is the _spectral norm_
of $x$ and $\lVert x\rVert_{S_{2}}=\lVert s(x)\rVert_{2}=\langle
x,x\rangle^{1/2}$ is the _Hilbert-Schmidt_ or _Frobenius norm._ The space
$S_{p}:=S_{p,\beta}^{m\times n}:=(\mathbb{K}^{m\times
n},\lVert\cdot\rVert_{S_{p}})$ is called a _(finite/dimensional) Schatten/̄
$p$ class;_ we denote its unit ball by $\mathbb{B}_{S_{p},\beta}^{m\times n}$
and its unit sphere by $\mathbb{S}_{S_{p},\beta}^{m,n}$; additionally define
$\omega_{S_{p},\beta}^{m\times n}:=\operatorname{\mathit{v}}_{\beta
mn}(\mathbb{B}_{S_{p},\beta}^{m\times n})$.

######  Remark 2.1.

The assumption $m\leq n$ is no severe restriction: given arbitrary
$m,n\in\mathbb{N}$ and $x\in\mathbb{K}^{m\times n}$, the Hermitian
positive/semidefinite matrices $x^{*}x$ and $xx^{*}$ have the same positive
eigenvalues with matching multiplicities and the rest are zero, hence singular
values are defined also for “tall” matrices, and we have
$(s_{i}(x^{*}))_{i\leq r}=(s_{i}(x))_{i\leq r}$ and consequently $\lVert
x^{*}\rVert_{S_{p}}=\lVert x\rVert_{S_{p}}$, where $r=\min\\{m,n\\}$. There
are results in the literature that also take the zero singular values into
account, e.g. [19, Theorem 5.5.10].

###  2.2 Auxiliary notions and results

Recall that a matrix $u\in\mathbb{K}^{n\times m}$ is called _semiunitary_
(_semiorthogonal_ if $\mathbb{K}=\mathbb{R}$) iff $u^{*}u=I_{m}$. The
collection of semiunitary (semiorthogonal) $(n\times m)$/matrices is called
the _Stiefel/manifold_ and notated by $\mathrm{U}_{n,m;\beta}$ (it is indeed a
submanifold of $\mathbb{K}^{n\times m}$); in particular
$\mathrm{U}_{m;\beta}:=\mathrm{U}_{m,m;\beta}$ denotes the set of unitary
(orthogonal) matrices, they satisfy $u^{*}u=uu^{*}=I_{m}$.

For any $x\in\mathbb{K}^{m\times n}$ there exist $u\in\mathrm{U}_{n;\beta}$
and $v\in\mathrm{U}_{m;\beta}$ such that $x=v\operatorname{diag}(s(x))u^{*}$,
the so/called _singular/value/decomposition_ of $x$; by slight abuse of
notation, here we mean $\operatorname{diag}(\xi):=(\xi_{i}\delta_{i,j})_{i\leq
m,j\leq n}\in\mathbb{K}^{m\times n}$ for a vector $\mathbb{K}^{m}$, i.e., the
matrix whose entries on the main diagonal are the components of $\xi$ and are
zero else.

The conventions for integration (which can be justified by exploiting that all
domains of integration concerned are Riemannian submanifolds of some full
matrix space) are as follows, where for $\beta=1$ ignore all imaginary parts:
on $\mathbb{K}^{m\times n}$ the volume form is (up to sign, as are the
subsequent)
$\bigwedge_{i=1}^{m}\bigwedge_{j=1}^{n}(\mathrm{d}\Re(x_{i,j})\wedge\mathrm{d}\Im(x_{i,j}))$
(which corresponds to $\beta mn$/̄dimensional Lebesgue measure); on
$\text{Sym}_{m,\beta}$ and $\text{Pos}_{m,\beta}$ it is
$\bigwedge_{i=1}^{m}\bigl{(}\bigwedge_{j=1}^{i}\mathrm{d}\Re(x_{i,j})\wedge\bigwedge_{j=1}^{i-1}\mathrm{d}\Im(x_{i,j})\bigr{)}$;
and on $\mathrm{U}_{n,m;\beta}$,
$\bigwedge_{j=1}^{m}\bigl{(}\bigwedge_{i=j+1}^{n}\Re\bigl{(}\sum_{k=1}^{n}\overline{x_{k,i}}\,\mathrm{d}x_{k,j}\bigr{)}\wedge\bigwedge_{i=j}^{n}\Im\bigl{(}\sum_{k=1}^{n}\overline{x_{k,i}}\,\mathrm{d}x_{k,j}\bigr{)}\bigr{)}$
(where $x_{i,j}$ for $j\geq m+1$ are such that for the augmented matrix,
$(x_{i,j})_{i,j\leq n}\in\mathrm{U}_{n;\beta}$); in all cases we write shortly
$\mathrm{d}x$. Note that the induced measure on $\mathrm{U}_{n,m;\beta}$ is
invariant under left and right multiplication with unitary matrices; in
particular, on $\mathrm{U}_{m;\beta}$ this gives a Haar measure.

For $z\in\mathbb{C}$ with $\Re(z)>\frac{\beta(m-1)}{2}$ define the
multivariate gamma function $\Gamma_{m,\beta}$ by

$\Gamma_{m,\beta}(z)=\int_{\text{Pos}_{m,\beta}}\det(x)^{z-1-\beta(m-1)/2}\,\mathsf{e}^{-\operatorname{tr}(x)}\,\mathrm{d}x=\pi^{\beta
m(m-1)/4}\prod_{k=0}^{m-1}\Gamma\Bigl{(}z-\frac{\beta k}{2}\Bigr{)}.$

The Riemannian volume of $\mathrm{U}_{n;\beta}$ is known at least since
Hurwitz [25] (note that actually Hurwitz considers
$\\{u\in\mathrm{U}_{n;\beta}\,:\,\det(u)=1\\}$, the special unitary group) and
is a special case of the Riemannian volume of $\mathrm{U}_{n,m;\beta}$ which
equals

$\begin{split}\Omega_{n,m;\beta}&:=2^{\beta
m(m-1)/4}\int_{\mathrm{U}_{n,m;\beta}}\mathrm{d}u\\\ &=\frac{2^{m+\beta
m(m-1)/4}\,\pi^{\beta(2mn-m(m-1))/4}}{\prod_{k=0}^{m-1}\Gamma\bigl{(}\frac{\beta(n-k)}{2}\bigr{)}}=\frac{2^{m+\beta
m(m-1)/4}\,\pi^{\beta mn/2}}{\Gamma_{m,\beta}\bigl{(}\frac{\beta
n}{2}\bigr{)}}.\end{split}$ (1)

In particular we set $\Omega_{n;\beta}:=\Omega_{n,n;\beta}$, the volume of
$\mathrm{U}_{n;\beta}$; then $\Omega_{n,m;\beta}=2^{-\beta
m(n-m)/2}\frac{\Omega_{n;\beta}}{\Omega_{n-m;\beta}}$.

The _Selberg integral_ (see [52]) will be of some importance to our results:
let $m\in\mathbb{N}$ and let $\alpha,\beta,\gamma\in\mathbb{C}$ with
$\Re(\alpha),\Re(\beta)>0$ and
$\Re(\gamma)>-\min\bigl{\\{}\frac{1}{m},\frac{\Re(\alpha)}{m-1},\frac{\Re(\beta)}{m-1}\bigr{\\}}$,
then

$\int_{(0,1)^{m}}\prod_{i=1}^{m}\bigl{(}x_{i}^{\alpha-1}\,(1-x_{i})^{\beta-1}\bigr{)}\prod_{1\leq
i<j\leq m}\lvert
x_{i}-x_{j}\rvert^{2\gamma}\,\mathrm{d}x=\prod_{k=0}^{m-1}\frac{\Gamma(\alpha+k\gamma)\Gamma(\beta+k\gamma)\Gamma(\gamma+1+k\gamma)}{\Gamma(\alpha+\beta+(m-1)\gamma+k\gamma)\Gamma(\gamma+1)}.$

Often we are going to need the asymptotics of the product of gamma/functions
as they appear in $\Omega_{n,m;\beta}$ and the Selberg integral. Here we also
remind the reader of the _Landau symbols:_ for sequences
$(a_{n})_{n\in\mathbb{N}}\in\mathbb{K}^{\mathbb{N}}$ and
$(b_{n})_{n\in\mathbb{N}}\in\mathbb{R}_{>0}^{\mathbb{N}}$, we write
$a_{n}=\operatorname{O}(b_{n})$ to mean, $\lvert a_{n}\rvert\leq Cb_{n}$ for
eventually all $n\in\mathbb{N}$, with some $C\in\mathbb{R}_{>0}$; and
$a_{n}=\operatorname{o}(b_{n})$ to mean
$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=0$.

######  Lemma 2.2.

For any $\beta\in\mathbb{R}_{>0}$ we have

$\begin{split}\log\biggl{(}\prod_{k=1}^{n}\Gamma\Bigl{(}\frac{\beta
k}{2}\Bigr{)}\biggr{)}&=\frac{\beta n^{2}}{4}\log\Bigl{(}\frac{\beta
n}{2}\Bigr{)}-\frac{3\beta
n^{2}}{8}+\frac{(\beta-2)n}{4}\log\Bigl{(}\frac{\beta n}{2}\Bigr{)}\\\
&\quad+\frac{n}{4}\log(4\pi^{2}\,\mathsf{e}^{2-\beta})+\frac{\beta^{2}-6\beta+4}{24\beta}\log\Bigl{(}\frac{\beta
n}{2}\Bigr{)}+A_{\beta}+\operatorname{O}\Bigl{(}\frac{1}{n}\Bigr{)},\end{split}$

where $A_{\beta}\in\mathbb{R}$ is some constant.

###### Proof.

Let $\beta\in\mathbb{R}_{>0}$. We notice
$\log\bigl{(}\prod_{k=1}^{n}\Gamma(\frac{\beta
n}{2})\bigr{)}=\sum_{k=1}^{n}\log(\Gamma(\frac{\beta k}{2}))$, hence we apply
the Euler-Maclaurin summation formula to the function
$f\colon\mathbb{R}_{>0}\to\mathbb{R}$, $x\mapsto\log(\Gamma(\frac{\beta
x}{2}))$, that is,

$\sum_{k=1}^{n}f(k)=\int_{1}^{n}f(x)\,\mathrm{d}x+\frac{f(n)+f(1)}{2}+\frac{f^{\prime}(n)-f^{\prime}(1)}{12}+\int_{1}^{n}H_{3}(x)f^{\prime\prime\prime}(x)\,\mathrm{d}x,$

where $H_{3}(x):=\frac{1}{6}B_{3}(x-\lfloor x\rfloor)$ with the
Bernoulli/polynomial $B_{3}(x)=x^{3}-\frac{3}{2}x^{2}+\frac{1}{2}x$.

We already know the asymptotics of the gamma/function through Stirling’s
formula,

$\log(\Gamma(x))=x\log(x)-x-\frac{1}{2}\log(x)+\frac{1}{2}\log(2\pi)+\frac{1}{12x}+R(x),$

where $R$ satisfies $R(x)=\operatorname{O}(x^{-3})$ as $x\to\infty$. Therewith
we get

$\begin{split}\int_{1}^{n}f(x)\,\mathrm{d}x&=\frac{\beta
n^{2}}{4}\log\Bigl{(}\frac{\beta n}{2}\Bigr{)}-\frac{\beta
n^{2}}{8}-\frac{\beta n^{2}}{4}-\frac{n}{2}\log\Bigl{(}\frac{\beta
n}{2}\Bigr{)}+\frac{n}{2}\\\
&\quad+\frac{n}{2}\log(2\pi)+\frac{1}{6\beta}\log\Bigl{(}\frac{\beta
n}{2}\Bigr{)}+c+\int_{1}^{n}R(x)\,\mathrm{d}x,\end{split}$

where we already have gathered all constant terms. Now since $\lvert
R(x)\rvert\leq Cx^{-3}$, $R$ is integrable over $[1,\infty)$, hence the limit
$c_{1}:=\lim_{n\to\infty}\bigl{(}c+\int_{1}^{n}R(x)\,\mathrm{d}x\bigr{)}$
exists in $\mathbb{R}$, and

$\biggl{\lvert}c_{1}-c-\int_{1}^{n}R(x)\,\mathrm{d}x\biggr{\rvert}=\biggl{\lvert}\int_{n}^{\infty}R(x)\,\mathrm{d}x\biggr{\rvert}\leq\frac{C}{2n^{2}};$

this results in

$\begin{split}\int_{1}^{n}f(x)\,\mathrm{d}x&=\frac{\beta
n^{2}}{2}\log\Bigl{(}\frac{\beta n}{2}\Bigr{)}-\frac{3\beta
n^{2}}{8}-\frac{n}{2}\log\Bigl{(}\frac{\beta n}{2}\Bigr{)}\\\
&\quad+\frac{n}{2}\log(2\pi\mathsf{e})+\frac{1}{6\beta}\log\Bigl{(}\frac{\beta
n}{2}\Bigr{)}+c_{1}+\operatorname{O}\Bigl{(}\frac{1}{n^{2}}\Bigr{)}.\end{split}$

Obviously

$\frac{f(n)}{2}=\frac{\beta n}{4}\log\Bigl{(}\frac{\beta
n}{2}\Bigr{)}-\frac{\beta n}{4}-\frac{1}{4}\log\Bigl{(}\frac{\beta
n}{2}\Bigr{)}+c_{2}+\operatorname{O}\Bigl{(}\frac{1}{n}\Bigr{)}.$

We also have, via the polygamma/functions,

$\frac{f^{\prime}(n)}{12}=\frac{\beta}{24}\,\psi_{0}\Bigl{(}\frac{\beta
n}{2}\Bigr{)}=\frac{\beta}{24}\log\Bigl{(}\frac{\beta
n}{2}\Bigr{)}+\operatorname{O}\Bigl{(}\frac{1}{n}\Bigr{)},$

because of $\psi_{0}(x)=\log(x)-\frac{1}{2x}+\operatorname{O}(x^{-2})$. Lastly
we have $f^{\prime\prime\prime}(x)=(\frac{\beta}{2})^{3}\psi_{2}(\frac{\beta
x}{2})$; as $\psi_{2}(x)=\operatorname{O}(x^{-2})$ and $H_{3}$ is bounded, the
function $H_{3}f^{\prime\prime\prime}$ is integrable over $[1,\infty)$, thus
the limit
$c_{3}:=\lim_{n\to\infty}\int_{1}^{n}H_{3}(x)f^{\prime\prime\prime}(x)\,\mathrm{d}x$
exists in $\mathbb{R}$, and the error satisfies

$\biggl{\lvert}c_{3}-\int_{1}^{n}H_{3}(x)f^{\prime\prime\prime}(x)\,\mathrm{d}x\biggr{\rvert}=\biggl{\lvert}\int_{n}^{\infty}H_{3}(x)f^{\prime\prime\prime}(x)\,\mathrm{d}x\biggr{\rvert}\leq\frac{C}{n}$

with some constant $C$ (different from the previous occurrence). The statement
of the lemma now follows by gathering terms and defining
$A_{\beta}:=c_{1}+c_{3}+c_{3}$. ∎

######  Remark 2.3.

For $\beta\in\\{1,2\\}$ the statement of Lemma 2.2 could also be obtained by
expressing the product in terms of the _Barnes/ $G$/̄function_ whose
asymptotics are well/known. But this approach fails for general $\beta$, and
even for $\beta=1$ the calculations are tedious.

For the present investigations the following transformation formulae for
integrals with respect to two important matrix factorizations turn out to be
useful. (Others exist that we do not need.)

######  Proposition 2.4.

Let $f\colon\mathbb{K}^{m\times n}\to\mathbb{K}$ be measurable and either
nonnegative or integrable.

  1. 1.

(Polar decomposition [9, Lemma 2.1], also [58, Proposition 2.27])222Note that
[9] works with “tall” matrices, i.e. $m\geq n$, so the representation needs to
be transposed; also the measure on $\mathrm{U}_{n,m;\beta}$ is normalized
differently.

$\int_{\mathbb{K}^{m\times
n}}f(x)\,\mathrm{d}x=2^{-m}\int_{\text{Pos}_{m,\beta}}\int_{\mathrm{U}_{n,m;\beta}}f(r^{1/2}\,u^{*})\,\mathrm{d}u\,\det(r)^{\beta(n-m+1)/2-1}\,\mathrm{d}r.$

  2. 2.

(Singular value decomposition [10, Lemma 1.5.3, (i)])

$\displaystyle\int_{\mathbb{K}^{m\times n}}f(x)\,\mathrm{d}x$
$\displaystyle=\frac{2^{\beta n(n-1)/4+\beta m(m-1)/4+\beta m/2}}{2^{\beta
mn/2}\,m!\Omega_{1;\beta}^{m}\Omega_{n-m;\beta}}\int_{\mathbb{R}_{>0}^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(s)u^{*})\,\mathrm{d}u\,\mathrm{d}v$
$\displaystyle\mspace{230.0mu}\cdot\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)-1}\prod_{1\leq
i<j\leq m}\lvert s_{i}^{2}-s_{j}^{2}\rvert^{\beta}\,\mathrm{d}s.$

In particular, if $f=F\circ s$, where $F\colon\mathbb{R}^{m}\to\mathbb{K}$
(“$f$ depends only on the singular values of $x$”), then [3, Proposition
4.1.3]

$\int_{\mathbb{K}^{m\times n}}F(s(x))\,\mathrm{d}x=\frac{2^{\beta
m/2}\,\Omega_{m;\beta}\Omega_{n;\beta}}{2^{\beta
mn/2}\,m!\Omega_{1;\beta}^{m}\Omega_{n-m;\beta}}\int_{\mathbb{R}_{>0}^{m}}F(s)\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)-1}\prod_{1\leq
i<j\leq m}\lvert s_{i}^{2}-s_{j}^{2}\rvert^{\beta}\,\mathrm{d}s.$

######  Remark 2.5.

Some comments on Proposition 2.4, part 2, seem in place. The form given in the
present work does not completely match the one given in [10]. Apart from the
same notational differences as mentioned in Footnote 2, the major discrepancy
is the domain of integration for the singular values: Chikuse assumes them to
be ordered decreasingly, whereas we impose no such restriction. But this poses
no serious obstacle, for the following argument: define
$W^{m}:=\\{x\in\mathbb{R}_{>0}^{m}\,:\,x_{1}>\dotsb>x_{m}\\}$, then almost
every $s\in\mathbb{R}_{>0}^{m}$ can be uniquely written $s=\tau s^{\prime}$,
where $\tau\in\mathrm{U}_{m;\beta}$ is a permutation matrix and $s^{\prime}\in
W^{m}$; therewith
$v\operatorname{diag}(s)u^{*}=(v\tau)\operatorname{diag}(s^{\prime})(u\operatorname{diag}(\tau,I_{n-m}))^{*}$;
also write $h(s):=\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)-1}\prod_{i<j}\lvert
s_{i}^{2}-s_{j}^{2}\rvert^{\beta}$, then clearly $h(\tau s)=h(s)$. Now
decompose $\mathbb{R}_{>0}^{m}=\bigcup_{\tau\text{ perm.\ m.}}\tau W^{m}$,
neglecting the set of vectors for which at least two coordinates are equal.
Then for any permutation matrix $\tau$,

$\int_{\tau
W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(s)u^{*})h(s)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}s\\\
\begin{aligned}
&=\int_{W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(\tau^{\mathsf{T}}s)u^{*})h(\tau^{\mathsf{T}}s)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}s\\\
&=\int_{W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\tau^{\mathsf{T}}\operatorname{diag}(s)(u\operatorname{diag}(\tau^{\mathsf{T}},I_{n-m}))^{*})h(s)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}s\\\
&=\int_{W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(s)u^{*})h(s)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}s,\end{aligned}$

where we also have used that on $\mathrm{U}_{n;\beta}$ and
$\mathrm{U}_{m;\beta}$ we are integrating with respect to the (scaled) Haar
measures. This leads to

$\int_{\mathbb{R}_{>0}^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(s)u^{*})h(s)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}s\\\
\begin{aligned} &=\sum_{\tau\text{ perm.\ m.}}\int_{\tau
W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(s)u^{*})h(s)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}s\\\
&=m!\int_{W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(s)u^{*})h(s)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}s.\end{aligned}$

Summing up, an additional factor $m!$ has to be taken into account.

Lastly, for any set $A$ we denote the (set/theoretic) indicator function by
$\operatorname{1}_{A}$, that is, $\operatorname{1}_{A}(x)=1$ if $x\in A$, and
$\operatorname{1}_{A}(x)=0$ else. Further notions and notations shall be
introduced on the fly as needed.

##  3 Geometry of Schatten/̄$p$ balls

The first result is an explicit expression for the exact and asymptotic $\beta
mn$/̄dimensional Lebesgue volume of $\mathbb{B}_{S_{\infty},\beta}^{m\times
n}$. In the case $m=n$ this has already been accomplished in principle by
Saint Raymond [46, Théorème 1], though he does not evaluate the Selberg
integral encountered. A statement concerning
$\mathbb{B}_{S_{p},\beta}^{m\times n}$ for $p<\infty$ is found in Proposition
6.10.

######  Theorem A.

The exact volume of $\mathbb{B}_{S_{\infty},\beta}^{m\times n}$ is given by

$\begin{split}\omega_{S_{\infty},\beta}^{m\times
n}&=\frac{\prod_{k=0}^{m-1}\Gamma(1+\frac{\beta
k}{2})\prod_{k=0}^{n-1}\Gamma(1+\frac{\beta
k}{2})}{\prod_{k=0}^{m+n-1}\Gamma(1+\frac{\beta k}{2})}\,\pi^{\beta mn/2}\\\
&=\frac{\Gamma_{m,\beta}(1+\frac{\beta(m-1)}{2})\,\Gamma_{n,\beta}(1+\frac{\beta(n-1)}{2})}{\Gamma_{m+n,\beta}(1+\frac{\beta(m+n-1)}{2})}\,\pi^{\beta
mn}.\end{split}$

Hence follow the asymptotics of the volume radius,

$(\omega_{S_{\infty},\beta}^{m\times n})^{1/(\beta
mn)}=\Bigl{(}\frac{2\pi\mathsf{e}^{3/2}}{\beta(m+n)}\Bigr{)}^{1/2}\Bigl{(}\frac{m}{m+n}\Bigr{)}^{(m+2/\beta-1)/(4n)}\Bigl{(}\frac{n}{m+n}\Bigr{)}^{(n+2/\beta-1)/(4m)}\bigl{(}1+r_{\beta}(m,n)\bigr{)},$

where $r_{\beta}\colon\mathbb{N}^{2}\to\mathbb{R}$ satisfies $C_{1}\leq
mn\,r_{\beta}(m,n)+\frac{\beta^{2}-6\beta+4}{24\beta^{2}}\log\bigl{(}\frac{1}{m}+\frac{1}{n}\bigr{)}\leq
C_{2}$ for all $m,n\in\mathbb{N}$ and $\beta\in\\{1,2\\}$, with global
constants $C_{1},C_{2}\in\mathbb{R}$.

In particular, consider $m$ as depending on $n$, then if
$c:=\lim_{n\to\infty}\frac{m}{n}\in[0,1]$ exists, then

$\lim_{n\to\infty}(\beta n)^{1/2}(\omega_{S_{\infty},\beta}^{m\times
n})^{1/(\beta
mn)}=\Bigl{(}\frac{2\pi\mathsf{e}^{3/2}}{1+c}\Bigr{)}^{1/2}(1+c)^{-1/(4c)}\Bigl{(}1+\frac{1}{c}\Bigr{)}^{-c/4},$

where in the case $c=0$ we interpret $(1+c)^{1/c}=\mathsf{e}$ and
$(1+\frac{1}{c})^{c}=1$.

###### Proof.

We apply Proposition 2.4, part 2, to
$f:=\operatorname{1}_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}=\operatorname{1}_{\mathbb{B}_{\infty,1}^{m}}{}\circ s$ and get

$\omega_{S_{\infty},\beta}^{m\times n}=\frac{2^{\beta
m/2}\,\Omega_{m;\beta}\Omega_{n;\beta}}{2^{\beta
mn/2}\,m!\Omega_{1;\beta}^{m}\Omega_{n-m;\beta}}\int_{[0,1]^{m}}\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)-1}\prod_{1\leq
i<j\leq m}\lvert s_{i}^{2}-s_{j}^{2}\rvert^{\beta}\,\mathrm{d}s.$

Via the transformation $s_{i}\mapsto s_{i}^{1/2}$ the last integral is
identified as Selberg’s integral whose exact value equals

$\displaystyle\int_{[0,1]^{m}}\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)-1}\prod_{1\leq
i<j\leq m}\lvert s_{i}^{2}-s_{j}^{2}\rvert^{\beta}\,\mathrm{d}s$
$\displaystyle=2^{-m}\prod_{k=0}^{m-1}\frac{\Gamma(\frac{\beta(n-m+1)}{2}+\frac{\beta
k}{2})\Gamma(1+\frac{\beta k}{2})\Gamma(\frac{\beta}{2}+1+\frac{\beta
k}{2})}{\Gamma(\frac{\beta n}{2}+1+\frac{\beta
k}{2})\Gamma(\frac{\beta}{2}+1)}.$

Plugging in and rearranging terms yields the claimed result.

The asymptotic formulas are an application of Lemma 2.2; in order to apply it,
it helps to rearrange terms a bit via $\Gamma(x+1)=x\Gamma(x)$, like this,

$\frac{\prod_{k=0}^{m-1}\Gamma(1+\frac{\beta
k}{2})\prod_{k=0}^{n-1}\Gamma(1+\frac{\beta
k}{2})}{\prod_{k=0}^{m+n-1}\Gamma(1+\frac{\beta
k}{2})}=\frac{2}{\beta}\,\frac{\Gamma(m)\Gamma(n)\Gamma(\frac{\beta(m+n)}{2})}{\Gamma(\frac{\beta
m}{2})\Gamma(\frac{\beta
n}{2})\Gamma(m+n)}\,\frac{\prod_{k=1}^{m}\Gamma(\frac{\beta
k}{2})\prod_{k=1}^{n}\Gamma(\frac{\beta
k}{2})}{\prod_{k=1}^{m+n}\Gamma(\frac{\beta k}{2})}.$

Now thence follows, after gathering terms,

$\begin{split}(\omega_{S_{\infty},\beta}^{m\times n})^{1/(\beta
mn)}&=\Bigl{(}\frac{2\pi\mathsf{e}^{3/2}}{\beta(m+n)}\Bigr{)}^{1/2}\Bigl{(}\frac{m}{m+n}\Bigr{)}^{(m+2/\beta-1)/(4n)}\Bigl{(}\frac{n}{m+n}\Bigr{)}^{(n+2/\beta-1)/(4m)}\\\
&\quad\cdot\Bigl{(}\frac{mn}{m+n}\Bigr{)}^{(\beta^{2}-6\beta+4)/(24\beta^{2}mn)}\mathsf{e}^{(c_{\beta}+\operatorname{O}(1/m+1/n))/(mn)},\end{split}$

where
$c_{\beta}:=\frac{A_{\beta}}{\beta}+\frac{\beta^{2}-18\beta+4}{24\beta^{2}}\log\bigl{(}\frac{\beta}{2}\bigr{)}$.
Note $\frac{m+n}{mn}=\frac{1}{m}+\frac{1}{n}\leq 2$ because of $m,n\geq 1$;
furthermore by the inequality of arithmetic and geometric means
$\frac{1}{m}+\frac{1}{n}\geq 2(mn)^{-1/2}$ and hence
$(\frac{1}{n}+\frac{1}{m})^{1/(mn)}\to 1$ as $n\to\infty$. Therefore

$\Bigl{(}\frac{mn}{m+n}\Bigr{)}^{(\beta^{2}-6\beta+4)/(24\beta^{2}mn)}\mathsf{e}^{(c_{\beta}+\operatorname{O}(1/m+1/n))/(mn)}\\\
=1+\frac{1}{mn}\Bigl{\\{}-\frac{\beta^{2}-6\beta+4}{24\beta^{2}}\log\Bigl{(}\frac{1}{m}+\frac{1}{n}\Bigr{)}+c_{\beta}+\operatorname{O}\Bigl{(}\frac{1}{m}+\frac{1}{n}\Bigr{)}\Bigr{\\}}+R_{\beta}(m,n),$

where $R_{\beta}(m,n)$ is of the order of the square of the first/order term;
from this it also follows that $mn\,R_{\beta}(m,n)$ is bounded. Naming

$r_{\beta}(m,n):=\frac{1}{mn}\Bigl{\\{}-\frac{\beta^{2}-6\beta+4}{24\beta^{2}}\log\Bigl{(}\frac{1}{m}+\frac{1}{n}\Bigr{)}+c_{\beta}+\operatorname{O}\Bigl{(}\frac{1}{m}+\frac{1}{n}\Bigr{)}\Bigr{\\}}+R_{\beta}(m,n),$

we arrive at the claimed result.

Assume now that $c=\lim_{n\to\infty}\frac{m}{n}\in[0,1]$ exists, then we
define $c_{n}:=\frac{m}{n}\in(0,1)$ and we can write $m=c_{n}n$. Using this
yields

$\Bigl{(}\frac{m}{m+n}\Bigr{)}^{(m+2/\beta-1)/n}\Bigl{(}\frac{n}{m+n}\Bigr{)}^{(n+2/\beta-1)/m}=\Bigl{(}1+\frac{1}{c_{n}}\Bigr{)}^{-c_{n}+(1-2/\beta)/n}(1+c_{n})^{-1/c_{n}+(1-2/\beta)/m}.$

Now $\lim_{n\to\infty}(1+c_{n})^{-1/c_{n}}=(1+c)^{-1/c}$ ($=\mathsf{e}^{-1}$
if $c=0$); furthermore $\lim_{n\to\infty}(1+c_{n})^{(1-2/\beta)/m}=1$, which
is clear if $c=0$, irrespective of whether $m$ is bounded or not, and in the
case $c>0$ we have $\lim_{n\to\infty}m=\infty$ and the result follows again;
the limit $\lim_{n\to\infty}(1+\frac{1}{c_{n}})^{-c_{n}}=(1+\frac{1}{c})^{-c}$
($=1$ if $c=0$) is unremarkable; and concerning the remaining term observe
$c_{n}n=m\geq 1$, thus $1\leq 1+\frac{1}{c_{n}}\leq 1+n$; hence irrespective
of the value of $\beta$, $(1+\frac{1}{c_{n}})^{(1-2/\beta)/n}$ is sandwiched
between $1$ and $(1+n)^{(1-2/\beta)/n}$, either of which converges to $1$ as
$n\to\infty$, and therefore also

$\Bigl{(}1+\frac{1}{c_{n}}\Bigr{)}^{(1-2/\beta)/n}\xrightarrow[n\to\infty]{}1.\qed$

######  Remark 3.1.

In the quadratic case (i.e. $m=n$) Saint Raymond has derived the following
asymptotics [46, Corollaire 4]:

$(\omega_{S_{\infty},\beta}^{n\times n})^{1/(\beta
n^{2})}\sim\frac{1}{2}\sqrt{\frac{2\pi\mathsf{e}^{3/2}}{\beta n}},$

where $a_{n}\sim b_{n}$ means $\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=1$. It is
easily seen that Theorem A reproduces his result when taking $c=1$.

Recall that a _convex body_ is a subset $K\subset\mathbb{R}^{d}$
($d\in\mathbb{N}$) that is compact, convex, and has nonempty interior. A
convex body $K$ is said to be _isotropic_ iff
$\operatorname{\mathit{v}}_{d}(K)=1$, it is centred, i.e.,

$\int_{K}x\,\mathrm{d}x=0,$

and

$\exists L\in[0,\infty)\forall\theta\in\mathbb{R}^{d}\colon\int_{K}\langle
x,\theta\rangle^{2}\,\mathrm{d}x=L^{2}\lVert\theta\rVert_{2}^{2}.$

The last condition can be expressed equivalently as

$\exists L\in[0,\infty)\forall
i\in[1,d]\colon\int_{K}x_{i}^{2}\,\mathrm{d}x=L^{2}\quad\wedge\quad\forall
i,j\in[1,d]\colon\biggl{(}i\neq
j\Longrightarrow\int_{K}x_{i}x_{j}\,\mathrm{d}x=0\biggr{)}.$

For convex bodies in $\mathbb{C}^{d}$ one would have to check the above
conditions for the $2d$ real components
$\Re(x_{1}),\Im(x_{1}),\dotsc,\Re(x_{d}),\Im(x_{d})$. We provide below an
equivalent characterization using directly the complex components.

######  Lemma 3.2.

A centred convex body $K\subset\mathbb{C}^{d}$ with unit volume is isotropic
iff both of the following conditions are satisfied:

$\exists L\in\mathbb{R}\forall i\in[1,d]\colon\int_{K}\lvert
x_{i}\rvert^{2}\,\mathrm{d}x=L^{2}$ (2)

and

$\forall i,j\in[1,d],i\neq
j\colon\int_{K}x_{i}^{2}\,\mathrm{d}x=\int_{K}x_{i}x_{j}\,\mathrm{d}x=\int_{K}x_{i}\overline{x_{j}}\,\mathrm{d}x=0.$
(3)

###### Proof.

Let $i,j\in[1,d]$ with $i\neq j$, then using $\Re(x)=\frac{x+\overline{x}}{2}$
and $\Im(x)=\frac{x-\overline{x}}{2\mathsf{i}}$ we can rewrite,

$\displaystyle\int_{K}\Re(x_{i})^{2}\,\mathrm{d}x$
$\displaystyle=\int_{K}\frac{1}{4}(x_{i}^{2}+2x_{i}\overline{x_{i}}+\overline{x_{i}^{2}})\,\mathrm{d}x=\int_{K}\frac{\Re(x_{i}^{2})}{2}\,\mathrm{d}x+\int_{K}\frac{\lvert
x_{i}\rvert^{2}}{2}\,\mathrm{d}x,$ (4) and similarly,
$\displaystyle\int_{K}\Re(x_{i})\Re(x_{j})\,\mathrm{d}x$
$\displaystyle=\int_{K}\frac{\Re(x_{i}x_{j})}{2}\,\mathrm{d}x+\int_{K}\frac{\Re(x_{i}\overline{x_{j}})}{2}\,\mathrm{d}x,$
(5) $\displaystyle\int_{K}\Im(x_{i})^{2}\,\mathrm{d}x$
$\displaystyle=\int_{K}\frac{1}{-4}(x_{i}^{2}-2x_{i}\overline{x_{i}}+\overline{x_{i}^{2}})\,\mathrm{d}x=-\int_{K}\frac{\Re(x_{i}^{2})}{2}\,\mathrm{d}x+\int_{K}\frac{\lvert
x_{i}\rvert^{2}}{2}\,\mathrm{d}x,$ (6)
$\displaystyle\int_{K}\Im(x_{i})\Im(x_{j})\,\mathrm{d}x$
$\displaystyle=-\int_{K}\frac{\Re(x_{i}x_{j})}{2}\,\mathrm{d}x+\int_{K}\frac{\Re(x_{i}\overline{x_{j}})}{2}\,\mathrm{d}x,$
(7) and without the restriction $i\neq j$,
$\displaystyle\int_{K}\Re(x_{i})\Im(x_{j})\,\mathrm{d}x$
$\displaystyle=\int_{K}\frac{\Im(x_{i}x_{j})}{2}\,\mathrm{d}x-\int_{K}\frac{\Im(x_{i}\overline{x_{j}})}{2}\,\mathrm{d}x.$
(8)

$\Rightarrow$: Let $K$ be isotropic, then we know from (4) and (6), for some
$L\in\mathbb{R}$,

$\int_{K}\frac{\Re(x_{i}^{2})}{2}\,\mathrm{d}x+\int_{K}\frac{\lvert
x_{i}\rvert^{2}}{2}\,\mathrm{d}x=L^{2}=-\int_{K}\frac{\Re(x_{i}^{2})}{2}\,\mathrm{d}x+\int_{K}\frac{\lvert
x_{i}\rvert^{2}}{2}\,\mathrm{d}x;$

substract the two equations to get

$\int_{K}\Re(x_{i}^{2})\,\mathrm{d}x=0,$

and add them to get

$\int_{K}\lvert x_{i}\rvert^{2}\,\mathrm{d}x=2L^{2}.$

In like manner we obtain from (5) and (7) that

$\int_{K}\frac{\Re(x_{i}x_{j})}{2}\,\mathrm{d}x+\int_{K}\frac{\Re(x_{i}\overline{x_{j}})}{2}\,\mathrm{d}x=0=-\int_{K}\frac{\Re(x_{i}x_{j})}{2}\,\mathrm{d}x+\int_{K}\frac{\Re(x_{i}\overline{x_{j}})}{2}\,\mathrm{d}x,$

hence

$\int_{K}\Re(x_{i}x_{j})\,\mathrm{d}x=0=\int_{K}\Re(x_{i}\overline{x_{j}})\,\mathrm{d}x.$

Swap $i$ and $j$ in Equation (8), then
$\Im(x_{j}\overline{x_{i}})=\Im(\overline{x_{i}\overline{x_{j}}})=-\Im(x_{i}\overline{x_{j}})$,
and therefore

$\int_{K}\frac{\Im(x_{i}x_{j})}{2}\,\mathrm{d}x-\int_{K}\frac{\Im(x_{i}\overline{x_{j}})}{2}\,\mathrm{d}x=0=\int_{K}\frac{\Im(x_{i}x_{j})}{2}\,\mathrm{d}x+\int_{K}\frac{\Im(x_{i}\overline{x_{j}})}{2}\,\mathrm{d}x,$

and again from both subtracting and adding there follow

$\int_{K}\Im(x_{i}x_{j})\,\mathrm{d}x=0=\int_{K}\Im(x_{i}\overline{x_{j}})\,\mathrm{d}x.$

Putting all things together yields the claim.

$\Leftarrow$: Separate the real and imaginary parts in Condition (3) (all are
zero), and together with (2) substitute them in Equations (4)-(8) to get the
desired isotropy. ∎

######  Remark 3.3.

In relatively recent years interest in the study of complex convex bodies has
grown, in the sense that the ambient space $\mathbb{C}^{d}$ is considered over
$\mathbb{C}$, not over $\mathbb{R}$; examples are the works [32, 30, 1, 24,
31, 23, 14]. This also leads to a different notion of isotropy; more
specifically, a compact convex set $K\subset\mathbb{C}^{d}$ is called
_(complex) isotropic_ iff it has unit volume, it is centred, and it possesses
a constant $L\in[0,\infty)$ such that,

$\forall\theta\in\mathbb{C}^{d}\colon\int_{K}\lvert
x^{*}\theta\rvert^{2}\,\mathrm{d}x=L^{2}\lVert\theta\rVert_{2}^{2};$

contrast this to _real isotropy_ where the last condition reads,

$\forall\theta\in\mathbb{C}^{d}\colon\int_{K}\Re(x^{*}\theta)^{2}\,\mathrm{d}x=L^{2}\lVert\theta\rVert_{2}^{2}.$

These two notions are not equivalent, the class of complex isotropic bodies is
strictly larger; see [34, Theorem 2.1].

The authors stress that the present article considers _real_ isotropy only.

######  Proposition 3.4.

The volume/normalized ball, $(\omega_{S_{p},\beta}^{m\times n})^{-1/(\beta
mn)}\,\mathbb{B}_{S_{p},\beta}^{m\times n}$, is isotropic.

###### Proof.

Since $(\omega_{S_{p},\beta}^{m\times n})^{-1/(\beta
mn)}\,\mathbb{B}_{S_{p},\beta}^{m\times n}$ differs from
$\mathbb{B}_{S_{p},\beta}^{m\times n}$ only by a dilation, it suffices to
consider the latter. Being a norm unit ball,
$\mathbb{B}_{S_{p},\beta}^{m\times n}$ is centred; thence it remains to check
the second moments, and we will exploit the unitary invariance of
$\mathbb{B}_{S_{p},\beta}^{m\times n}$ throughout (inherited from the singular
values). Let $(i,j)\in m\times n$, then let $v\in\mathrm{U}_{m;\beta}$ swap
the first and $i$th row and $u\in\mathrm{U}_{n;\beta}$ swap the first and
$j$th column, effecting

$\int_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}x_{i,j}^{2}\,\mathrm{d}x=\int_{v\mathbb{B}_{S_{p},\beta}^{m\times
n}u}x_{1,1}^{2}\,\mathrm{d}x=\int_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}x_{1,1}^{2}\,\mathrm{d}x,$

so the diagonal moments are equal to a common value.

Now let $(i,j),(i^{\prime},j^{\prime})\in m\times n$ with
$(i,j)\neq(i^{\prime},j^{\prime})$. If $i\neq i^{\prime}$, define
$v\in\mathrm{U}_{m;\beta}$ to multiply the $i^{\prime}$th row by $-1$;
therewith

$\int_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}x_{i,j}x_{i^{\prime},j^{\prime}}\,\mathrm{d}x=\int_{v\mathbb{B}_{S_{p},\beta}^{m\times
n}}x_{i,j}(-x_{i^{\prime},j^{\prime}})\,\mathrm{d}x=-\int_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}x_{i,j}x_{i^{\prime},j^{\prime}}\,\mathrm{d}x,$

which results in the integral being zero. If $i=i^{\prime}$, then $j\neq
j^{\prime}$ by assumption; and then multiply the $j^{\prime}$th column by
$-1$. Eventually all off/diagonal moments are zero, and in the case $\beta=1$
this finishes the proof.

In the case $\beta=2$ we apply Lemma 3.2; performing the same steps as before
with the integral $\int_{\mathbb{B}_{S_{p},2}^{m\times
n}}x_{i,j}\overline{x_{i^{\prime},j^{\prime}}}\,\mathrm{d}x$ we see, for
$(i,j)=(i^{\prime},j^{\prime})$,

$\int_{\mathbb{B}_{S_{p},2}^{m\times n}}\lvert
x_{i,j}\rvert^{2}\,\mathrm{d}x=\int_{\mathbb{B}_{S_{p},2}^{m\times n}}\lvert
x_{1,1}\rvert^{2}\,\mathrm{d}x,$

and for $(i,j)\neq(i^{\prime},j^{\prime})$,

$\int_{\mathbb{B}_{S_{p},2}^{m\times
n}}x_{i,j}\overline{x_{i^{\prime},j^{\prime}}}\,\mathrm{d}x=0.$

It remains to show $\int_{\mathbb{B}_{S_{p},2}^{m\times
n}}x_{1,1}^{2}\,\mathrm{d}x=0$, but in order to do so let
$v\in\mathrm{U}_{m;2}$ multiply the first row by $\mathsf{i}$, therewith

$\int_{\mathbb{B}_{S_{p},2}^{m\times
n}}x_{1,1}^{2}\,\mathrm{d}x=\int_{v^{*}\mathbb{B}_{S_{p},2}^{m\times
n}}(\mathsf{i}x_{1,1})^{2}\,\mathrm{d}x=-\int_{\mathbb{B}_{S_{p},2}^{m\times
n}}x_{1,1}^{2}\,\mathrm{d}x,$

which leads to the desired statement. ∎

For a convex body $K\subset\mathbb{R}^{d}$ which is isotropic up to dilation,
the isotropy constant $L_{K}\in(0,\infty)$ is defined via

$\displaystyle L_{K}^{2}$
$\displaystyle=\frac{1}{d\operatorname{\mathit{v}}_{d}(K)^{1+2/d}}\int_{K}\lVert
x\rVert_{2}^{2}\,\mathrm{d}x$
$\displaystyle=\frac{1}{d\operatorname{\mathit{v}}_{d}(K)^{2/d}}\,\frac{\int_{K}\lVert
x\rVert_{2}^{2}\,\mathrm{d}x}{\int_{K}\mathrm{d}x}.$

######  Theorem B.

The isotropy constant $L_{\mathbb{B}_{S_{\infty},\beta}^{m\times n}}$ remains
bounded in $m$ and $n$; to be precise,

$L_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}^{2}=\frac{1}{2\pi\mathsf{e}^{3/2}}\,c_{\beta}(m,n),$

where we know

$\mathsf{e}^{1/2}\Bigl{(}1+\frac{C_{1}}{mn}\Bigr{)}\leq
c_{\beta}(m,n)\leq\mathsf{e}^{2}\Bigl{(}1+C_{2}\Bigl{(}\frac{1}{m}+\frac{1}{n}\Bigr{)}\Bigr{)}$

for all $m,n\in\mathbb{N}$, with global constants $C_{1},C_{2}\in\mathbb{R}$.

In particular, consider $m$ as depending on $n$, then if
$c:=\lim_{n\to\infty}\frac{m}{n}\in[0,1]$ exists, then

$\lim_{n\to\infty}L_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}^{2}=\frac{1}{2\pi\mathsf{e}^{3/2}}(1+c)^{1/(2c)}\Bigl{(}1+\frac{1}{c}\Bigr{)}^{c/2},$

with the same interpretations for the case $c=0$ as in Theorem A.

###### Proof.

Note that in our case $\langle x,x\rangle=\lVert s(x)\rVert_{2}^{2}$ and
$\operatorname{1}_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}=\operatorname{1}_{\mathbb{B}_{\infty,1}^{m}}\circ s$, therefore

$\frac{\int_{\mathbb{B}_{S_{\infty},\beta}^{m\times n}}\langle
x,x\rangle\,\mathrm{d}x}{\int_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}\mathrm{d}x}=\frac{\int_{\mathbb{R}_{>0}^{m}}\operatorname{1}_{\mathbb{B}_{\infty,1}^{m}}(s)\lVert
s\rVert_{2}^{2}\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)-1}\prod_{i<j}\lvert
s_{i}^{2}-s_{j}^{2}\rvert^{\beta}\,\mathrm{d}s}{\int_{\mathbb{R}_{>0}^{m}}\operatorname{1}_{\mathbb{B}_{\infty,1}^{m}}(s)\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)-1}\prod_{i<j}\lvert
s_{i}^{2}-s_{j}^{2}\rvert^{\beta}\,\mathrm{d}s};$

but via the transformation $s_{i}\mapsto s_{i}^{1/2}$ this quotient equals

$\frac{\int_{[0,1]^{m}}\lVert
s\rVert_{1}\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)/2-1}\prod_{i<j}\lvert
s_{i}-s_{j}\rvert^{\beta}\,\mathrm{d}s}{\int_{[0,1]^{m}}\prod_{i=1}^{m}s_{i}^{\beta(n-m+1)/2-1}\prod_{i<j}\lvert
s_{i}-s_{j}\rvert^{\beta}\,\mathrm{d}s},$

and after observing that the integrand of the enumerator is symmetric, we can
apply a theorem of Aomoto’s [4] and obtain the exact value

$\frac{\int_{\mathbb{B}_{S_{\infty},\beta}^{m\times n}}\langle
x,x\rangle\,\mathrm{d}x}{\int_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}\mathrm{d}x}=\frac{mn}{m+n+\frac{2}{\beta}-1}.$

Using this together with the asymptotic volume from Theorem A, we have

$L_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}^{2}=\frac{m+n}{2\mathsf{e}^{3/2}\,\pi(m+n+\frac{2}{\beta}-1)}\Bigl{(}\frac{m+n}{m}\Bigr{)}^{(m+2/\beta-1)/(2n)}\Bigl{(}\frac{m+n}{n}\Bigr{)}^{(n+2/\beta-1)/(2m)}\bigl{(}1+r_{\beta}(m,n)\bigr{)}^{-2},$

with $r_{\beta}(m,n)$ as in Theorem A. In order to find suitable bounds, we
consider

$\begin{split}\log(c_{\beta}(m,n))&:=\log\Bigl{(}\frac{m+n}{m+n+\frac{2}{\beta}-1}\Bigr{)}+\frac{m+\frac{2}{\beta}-1}{2n}\log\Bigl{(}\frac{m+n}{m}\Bigr{)}\\\
&\quad+\frac{n+\frac{2}{\beta}-1}{2m}\log\Bigl{(}\frac{m+n}{n}\Bigr{)}-2\log\bigl{(}1+r_{\beta}(m,n)\bigr{)}.\end{split}$

Recall the general estimates for the logarithm: for any $x>0$ we have

$1-\frac{1}{x}\leq\log(x)\leq x-1.$

On the one hand, together with
$\frac{c_{1}}{mn}\log(\frac{1}{m}+\frac{1}{n})\leq\log(1+r_{\beta}(m,n))\leq\frac{c_{2}}{mn}$
with suitable global constants $c_{1},c_{2}\in\mathbb{R}$, this yields the
lower bound

$\displaystyle\log(c_{\beta}(m,n))$ $\displaystyle\geq
1-\frac{m+n+\frac{2}{\beta}-1}{m+n}+\frac{m+\frac{2}{\beta}-1}{2n}\Bigl{(}1-\frac{m}{m+n}\Bigr{)}+\frac{n+\frac{2}{\beta}-1}{2m}\Bigl{(}1-\frac{n}{m+n}\Bigr{)}-\frac{2c_{2}}{mn}$
$\displaystyle=\frac{1}{2}-\frac{2c_{2}}{mn}.$

On the other hand for an upper bound we get

$\displaystyle\log(c_{\beta}(m,n))$
$\displaystyle\leq\frac{m+n}{m+n+\frac{2}{\beta}-1}-1+\frac{m+\frac{2}{\beta}-1}{2n}\Bigl{(}\frac{m+n}{m}-1\Bigr{)}$
$\displaystyle\quad+\frac{n+\frac{2}{\beta}-1}{2m}\Bigl{(}\frac{m+n}{n}-1\Bigr{)}-\frac{2c_{1}}{mn}\log\Bigl{(}\frac{1}{m}+\frac{1}{n}\Bigr{)}$
$\displaystyle=\frac{m+n}{m+n+\frac{2}{\beta}-1}+\frac{2-\beta}{2\beta
m}+\frac{2-\beta}{2\beta
n}-\frac{2c_{1}}{mn}\log\Bigl{(}\frac{1}{m}+\frac{1}{n}\Bigr{)};$

obviously $\frac{m+n}{m+n+\frac{2}{\beta}-1}\leq\frac{m+n}{m+n-1}\leq 2$, and
it remains to show that $\frac{1}{m}+\frac{1}{n}$ dominates
$\frac{1}{mn}\log(\frac{1}{m}+\frac{1}{n})$ when at least $n$ grows
indefinitely. But we have

$\frac{\frac{1}{mn}\log(\frac{1}{m}+\frac{1}{n})}{\frac{1}{m}+\frac{1}{n}}=\frac{\log(m+n)}{m+n}-\frac{\log(mn)}{m+n};$

clearly the first summand is bounded, and via the inequality of arithmetic and
geometric means we know $0\leq\log(mn)\leq 2\log(m+n)-2\log(2)$ and therefore
also the second summand is bounded.

The value of $\lim_{n\to\infty}L_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}^{2}$ in the case $\lim_{n\to\infty}\frac{m}{n}=c$ is a direct consequence
of the corresponding values of $\lim_{n\to\infty}(\beta
n)^{1/2}(\omega_{S_{\infty},\beta}^{m\times n})^{1/(\beta mn)}$ given in
Theorem A. ∎

######  Remark 3.5.

Although König, Meyer, Pajor [33] claim that their theorem includes the case
$p=\infty$ (that is, the spectral norm), their method of proof actually does
not cover that, because the integral in question contains the exponential
$\mathsf{e}^{-\lVert x\rVert_{p}^{p}}$ which in the limit case equals just
$\operatorname{1}_{\mathbb{B}_{\infty,1}^{d}}(x)$, being an exponential no
longer and rendering their integral transfrom moot.

Define a _star body_ to be a compact set $K\subset\mathbb{R}^{d}$ with zero in
its interior which is star/shaped w.r.t. zero, i.e., $tK\subset K$ for all
$t\in[0,1]$, and whose Minkowski/functional $\lvert\cdot\rvert_{K}$ is
continuous. Any convex body with zero in its interior is a star body. For such
a star body $K$ define the associated _(normalized) cone measure_
$\kappa=\kappa_{K}$ on its boundary $\partial K$ by

$\kappa(A):=\frac{\operatorname{\mathit{v}}_{d}\bigl{(}\\{tx\,:\,t\in[0,1],x\in
A\\}\bigr{)}}{\operatorname{\mathit{v}}_{d}(K)}$

for any measurable $A\subset\partial K$; it is the unique measure on $\partial
K$ which satisfies the following polar integration formula (see [38,
Proposition 1]): for any $f\colon\mathbb{R}^{d}\to\mathbb{R}$ measurable and
nonnegative,

$\int_{\mathbb{R}^{d}}f(x)\,\mathrm{d}x=d\operatorname{\mathit{v}}_{d}(K)\int_{\mathbb{R}_{\geq
0}}\int_{\partial
K}f(r\theta)\,\mathrm{d}\kappa(\theta)\,r^{d-1}\,\mathrm{d}r.$ (9)

A useful identity for evaluating integrals w.r.t. the cone measure can be
obtained by integrating $x\mapsto f(\frac{x}{\lvert x\rvert_{K}})$ over $K$,
to wit,

$\int_{\partial
K}f(\theta)\,\mathrm{d}\kappa(\theta)=\frac{1}{\operatorname{\mathit{v}}_{d}(K)}\int_{K}f\Bigl{(}\frac{x}{\lvert
x\rvert_{K}}\Bigr{)}\,\mathrm{d}x.$ (10)

The surface measure on $\partial K$ is defined to be the $(d-1)$/̄dimensional
Hausdorff measure restricted to $\partial K$. The two measures are related
according to the following statement.

######  Lemma 3.6.

Let $\nu(x)$ be the outer unit normal vector in $x\in\partial K$, then for
$\operatorname{\mathcal{H}}^{d-1}$/̄almost every $x\in\partial K$,

$\mathrm{d}\\!\operatorname{\mathcal{H}}^{d-1}(x)=\frac{d\operatorname{\mathit{v}}_{d}(K)}{\langle\nu(x),x\rangle}\,\mathrm{d}\kappa(x).$

###### Proof.

This is a rewriting of Naor and Romik [38, Lemma 1]. Adapting their notation,
note that $\operatorname{\mathcal{H}}^{d-1}=\operatorname{area}(\partial
K)\sigma_{K}$; it remains to treat
$\lVert\nabla\lvert\cdot\rvert_{K}(x)\rVert_{2}$. But in their proof they
derive $\langle\nabla\lvert\cdot\rvert_{K}(x),x\rangle=1$, and therewith

$\lVert\nabla\lvert\cdot\rvert_{K}(x)\rVert_{2}=\frac{\lVert\nabla\lvert\cdot\rvert_{K}(x)\rVert_{2}}{\langle\nabla\lvert\cdot\rvert_{K}(x),x\rangle}=\frac{1}{\bigl{\langle}\frac{\nabla\lvert\cdot\rvert_{K}(x)}{\lVert\nabla\lvert\cdot\rvert_{K}(x)\rVert_{2}},x\bigr{\rangle}},$

from which the claim follows because of
$\nu(x)=\frac{\nabla\lvert\cdot\rvert_{K}(x)}{\lVert\nabla\lvert\cdot\rvert_{K}(x)\rVert_{2}}$.
∎

######  Remark 3.7.

Rewritten as
$\operatorname{\mathit{v}}_{d}(K)\,\mathrm{d}\kappa(x)=\frac{1}{d}\langle\nu(x),x\rangle\,\mathrm{d}\\!\operatorname{\mathcal{H}}^{d-1}(x)$,
the formula has a nice elementary interpretation: fix a point $x\in\partial
K$, then this determines an “infinitesimal cone” with apex $o$ and an
“infinitesimal base” tangent to $\partial K$ at $x$; the height of this cone
equals $\lVert\langle\nu(x),x\rangle\nu(x)\rVert_{2}=\langle\nu(x),x\rangle$
(the length of the orthogonal projection of $x$ onto $\mathbb{R}\nu(x)$), the
base measures $\mathrm{d}\\!\operatorname{\mathcal{H}}^{d-1}(x)$, and
therefore the cone has volume
$\frac{1}{d}\langle\nu(x),x\rangle\,\mathrm{d}\\!\operatorname{\mathcal{H}}^{d-1}(x)$.
But by the very definition of the cone measure this also equals
$\operatorname{\mathit{v}}_{d}(K)\,\mathrm{d}\kappa(x)$.

As a consequence of Lemma 3.6, if $x\mapsto\langle\nu(x),x\rangle$ is constant
on $\partial K$, then the normalized measures coincide. This is known to be
the case for $\mathbb{S}_{p,\beta}^{m}$ iff $p\in\\{1,2,\infty\\}$, going back
to [41, p. 1314]. Let us fix the notation
$\kappa_{p,\beta}^{m}:=\kappa_{\mathbb{B}_{p,\beta}^{m}}$ and
$\kappa_{S_{p},\beta}^{m\times n}:=\kappa_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}$.

######  Lemma 3.8.

Let $p\in\mathbb{R}$, then the trace power map

$\tau_{p}\colon\left\\{\begin{aligned} \text{Pos}_{m,\beta}&\to\mathbb{R}\\\
x&\mapsto\operatorname{tr}(x^{p})\end{aligned}\right.$

is continuously differentiable and for any $x\in\text{Pos}_{m,\beta}$,

$\nabla\tau_{p}(x)=px^{p-1}.$

###### Proof.

Case $p\in\mathbb{N}_{0}$: Continuous differentiability is obvious, as is the
gradient, for $p\in\\{0,1\\}$. So let $p\geq 2$, then the map
$f\colon\text{Pos}_{m,\beta}\to\text{Pos}_{m,\beta}$, $x\mapsto x^{p}$, is
continuously differentiable with derivative
$\mathrm{d}f(x)(h)=\sum_{k=1}^{p}x^{k-1}hx^{p-k}$ for any
$x\in\text{Pos}_{m,\beta}$ and $h\in\text{Sym}_{m,\beta}$. This implies,
because of $\tau_{p}=\operatorname{tr}\circ f$,

$\displaystyle\mathrm{d}\tau_{p}(x)(h)$
$\displaystyle=\operatorname{tr}(\mathrm{d}f(x)(h))=\operatorname{tr}\biggl{(}\sum_{k=1}^{p}x^{k-1}hx^{p-k}\biggr{)}$
$\displaystyle=\sum_{k=1}^{p}\operatorname{tr}(x^{k-1}hx^{p-k})=\sum_{k=1}^{p}\operatorname{tr}(x^{p-1}h)$
$\displaystyle=p\operatorname{tr}(x^{p-1}h)=\langle px^{p-1},h\rangle.$

(Note $\operatorname{tr}(ab)\in\mathbb{R}$ for any
$a,b\in\text{Sym}_{m,\beta}$.)

Case $p$ arbitrary: Write $x^{p}=\mathsf{e}^{p\log(x)}$, then this already
suggests continuous differentiability, and expanding the exponential we have

$\tau_{p}(x)=\sum_{k=0}^{\infty}\frac{p^{k}}{k!}\operatorname{tr}(\log(x)^{k}).$

Call $f_{k}(x):=\operatorname{tr}(\log(x)^{k})$, then we have to determine
$\nabla f_{k}(x)$ and show that $\sum_{k\geq 0}\frac{p^{k}}{k!}\nabla f_{k}$
converges locally uniformly on $\text{Pos}_{m,\beta}$; this will yield
$\nabla\tau_{p}=\sum_{k=0}^{\infty}\frac{p^{k}}{k!}\nabla f_{k}$.

From the case of integer powers and the chain rule we know
$\mathrm{d}f_{k}(x)(h)=\langle k\log(x)^{k-1},\mathrm{d}\log(x)(h)\rangle$ for
any $x\in\text{Pos}_{m,\beta}$ and $h\in\text{Sym}_{m,\beta}$. In order to
determine $\mathrm{d}\log(x)(h)$ consider the exponential map again, that is,
$\mathsf{e}^{x}=\sum_{k=0}^{\infty}\frac{1}{k!}x^{k}$. Call $g_{k}(x):=x^{k}$,
then each $g_{k}$ is differentiable with derivative
$\mathrm{d}g_{k}(x)(h)=\sum_{l=1}^{k}x^{l-1}hx^{k-l}$; therefore
$\lVert\mathrm{d}g_{k}(x)(h)\rVert_{S_{2}}\leq k\lVert
x\rVert_{S_{2}}^{k-1}\lVert h\rVert_{S_{2}}$ (any submultiplicative norm would
do) and for the operator norm we get $\lVert\mathrm{d}g_{k}(x)\rVert\leq
k\lVert x\rVert_{S_{2}}^{k-1}$. Now given any $R\in(0,\infty)$ we find

$\sum_{k=0}^{\infty}\frac{1}{k!}\lVert\mathrm{d}g_{k}(x)\rVert\leq\sum_{k=0}^{\infty}\frac{1}{k!}\,k\lVert
x\rVert_{S_{2}}^{k-1}\leq\sum_{k=0}^{\infty}\frac{1}{k!}R^{k}=\mathsf{e}^{R}<\infty$

for all $x\in\text{Sym}_{m,\beta}$ with $\lVert x\rVert_{S_{2}}\leq R$; thus
$\sum_{k\geq 0}\frac{1}{k!}\mathrm{d}g_{k}$ is locally uniformly convergent on
$\text{Sym}_{m,\beta}$ and hence
$\mathrm{d}\exp=\sum_{k=0}^{\infty}\frac{1}{k!}\mathrm{d}g_{k}$. From that and
from the identity $x=\mathsf{e}^{\log(x)}$ follows, for any
$x\in\text{Pos}_{m,\beta}$ and $h\in\text{Sym}_{m,\beta}$,

$h=\mathrm{d}\exp(\log(x))(\mathrm{d}\log(x)(h))=\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{l=1}^{k}\log(x)^{l-1}\mathrm{d}\log(x)(h)\log(x)^{k-l}.$

Fix a spectral decomposition $x=u\operatorname{diag}(\lambda)u^{*}$ with
$\lambda\in\mathbb{R}_{>0}^{m}$ and $u\in\mathrm{U}_{m;\beta}$ and define
$\xi:=u^{*}hu$ and $\eta:=u^{*}\mathrm{d}\log(x)(h)u$, then the above equation
is equivalent to

$\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{l=1}^{k}\log(\operatorname{diag}(\lambda))^{l-1}\eta\log(\operatorname{diag}(\lambda))^{k-l}=\xi.$

Let $i\in[1,n]$, then the $(i,i)$/component of the equation reads

$\xi_{i,i}=\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{l=1}^{k}\log(\lambda_{i})^{l-1}\eta_{i,i}\log(\lambda_{i})^{k-l}=\eta_{i,i}\sum_{k=0}^{\infty}\frac{1}{k!}\,k\log(\lambda_{i})^{k-1}=\eta_{i,i}\mathsf{e}^{\log(\lambda_{i})}=\lambda_{i}\eta_{i,i}$

which yields the solution

$\eta_{i,i}=\frac{\xi_{i,i}}{\lambda_{i}}.$

Therewith we can continue,

$\displaystyle\mathrm{d}f_{k}(x)(h)$ $\displaystyle=\langle
k\log(x)^{k-1},\mathrm{d}\log(x)(h)\rangle$ $\displaystyle=\langle
k\log(\operatorname{diag}(\lambda))^{k-1},\eta\rangle=\sum_{i=1}^{m}k\log(\lambda_{i})^{k-1}\,\eta_{i,i}$
$\displaystyle=\sum_{i=1}^{m}k\log(\lambda_{i})^{k-1}\lambda_{i}^{-1}\,\xi_{i,i}=\langle
k\log(\operatorname{diag}(\lambda))^{k-1}\operatorname{diag}(\lambda)^{-1},\xi\rangle$
$\displaystyle=\langle k\log(x)^{k-1}x^{-1},h\rangle,$

and so $\nabla f_{k}(x)=k\log(x)^{k-1}x^{-1}$. Finally let $R\in(1,\infty)$,
then we have, for all $x\in\text{Pos}_{m,\beta}$ such that
$\frac{1}{R}\leq\lambda_{m}(x)\leq\lambda_{1}(x)\leq R$,

$\lVert\nabla f_{k}(x)\rVert_{S_{2}}\leq
k\lVert\log(x)\rVert_{S_{2}}^{k-1}\lVert x^{-1}\rVert_{S_{2}}\leq
k(m^{1/2}\log(R))^{k-1}(m^{1/2}R),$

and since

$\sum_{k=0}^{\infty}\frac{p^{k}}{k!}\,k(m^{1/2}\log(R))^{k-1}(m^{1/2}R)=pm^{1/2}R\mathsf{e}^{pm^{1/2}\log(R)}=pm^{1/2}\,R^{pm^{1/2}+1}<\infty,$

the series $\sum_{k\geq 0}\frac{p^{k}}{k!}\nabla f_{k}$ converges locally
uniformly on $\text{Pos}_{m,\beta}$ as desired, so we obtain, for any
$x\in\text{Pos}_{m,\beta}$,

$\nabla\tau_{p}(x)=\sum_{k=0}^{\infty}\frac{p^{k}}{k!}k\log(x)^{k-1}x^{-1}=p\mathsf{e}^{p\log(x)}x^{-1}=px^{p-1}.\qed$

######  Theorem C.

For $\operatorname{\mathcal{H}}^{\beta mn-1}$/almost all
$x\in\mathbb{S}_{S_{p},\beta}^{m,n}$,

$\langle\nu(x),x\rangle=\begin{cases}\lVert
x\rVert_{S_{2p-2}}^{1-p}&\textit{if }p<\infty,\\\ 1&\textit{if
}p=\infty;\end{cases}$

in particular the cone measure $\kappa_{S_{p},\beta}^{m\times n}$ coincides
with the normalized Hausdorff/measure on $\mathbb{S}_{S_{p},\beta}^{m,n}$ iff
$p\in\\{1,2,\infty\\}$. Thereby also,

$\operatorname{\mathcal{H}}^{\beta
mn-1}(\mathbb{S}_{S_{p},\beta}^{m,n})=\begin{cases}\beta
m^{3/2}n\omega_{S_{1},\beta}^{m\times n}&\text{if }p=1,\\\ \beta
mn\omega_{S_{p},\beta}^{m\times n}&\text{if }p\in\\{2,\infty\\}.\end{cases}$

###### Proof.

Case $p<\infty$: Define the map

$g\colon\left\\{\begin{aligned} \mathbb{K}^{m\times n}&\to\mathbb{R}\\\
x&\mapsto\operatorname{tr}((xx^{*})^{p/2})\end{aligned}\right.,$

then $\mathbb{S}_{S_{p},\beta}^{m,n}=g^{-1}\\{1\\}$, and via Lemma 3.8 $g$ is
differentiable at any $x\in\mathbb{K}^{m\times n}$ such that
$xx^{*}\in\text{Pos}_{m,\beta}$, which is almost everywhere, and then, for any
$h\in\mathbb{K}^{m\times n}$,

$\displaystyle\mathrm{d}g(x)(h)$
$\displaystyle=\Bigl{\langle}\frac{p}{2}(xx^{*})^{p/2-1},hx^{*}+xh^{*}\Bigr{\rangle}$
$\displaystyle=\langle p(xx^{*})^{p/2-1}x,h\rangle,$

which means

$\nabla g(x)=p(xx^{*})^{p/2-1}x.$

From this follow, for almost any $x\in\mathbb{S}_{S_{p},\beta}^{m,n}$, both

$\displaystyle\langle\nabla g(x),x\rangle$
$\displaystyle=p\langle(xx^{*})^{p/2-1}x,x\rangle=p\langle(xx^{*})^{p/2-1}xx^{*},I_{m}\rangle$
$\displaystyle=p\langle(xx^{*})^{p/2},I_{m}\rangle=p\operatorname{tr}((xx^{*})^{p/2})=p,$

and

$\displaystyle\lVert\nabla g(x)\rVert_{S_{2}}^{2}$
$\displaystyle=p^{2}\langle(xx^{*})^{p/2-1}x,(xx^{*})^{p/2-1}x\rangle=p^{2}\langle(xx^{*})^{p-2}xx^{*},I_{m}\rangle$
$\displaystyle=p^{2}\operatorname{tr}((xx^{*})^{p-1})=p^{2}\lVert
x\rVert_{S_{2p-2}}^{2p-2},$

which imply

$\langle\nu(x),x\rangle=\frac{\langle\nabla g(x),x\rangle}{\lVert\nabla
g(x)\rVert_{S_{2}}}=\frac{1}{\lVert x\rVert_{S_{2p-2}}^{p-1}},$

as claimed.

Case $p=\infty$: Define $D:=\\{x\in\mathbb{K}^{m\times n}\,:\,s_{2}(x)<1\\}$,
then $D$ is an open subset of $\mathbb{K}^{m\times n}$, and we can write
$\mathbb{S}_{S_{\infty},\beta}^{m,n}=D\cap\\{x\in\mathbb{K}^{m\times
n}\,:\,\det(I_{m}-xx^{*})=0\\}$ up to the null set $\\{x\in\mathbb{K}^{m\times
n}\,:\,s_{1}(x)=s_{2}(x)=1\\}$. Define the map

$g\colon\left\\{\begin{aligned} D&\to\mathbb{R}\\\
x&\mapsto\det(I_{m}-xx^{*})\end{aligned}\right.,$

then $g$ is continuously differentiable,
$\mathbb{S}_{S_{\infty},\beta}^{m,n}=g^{-1}\\{0\\}$, and for any
$x,h\in\mathbb{K}^{m\times n}$ the derivative is

$\displaystyle\mathrm{d}g(x)(h)$
$\displaystyle=\operatorname{tr}\bigl{(}\operatorname{adj}(I_{d}-xx^{*})(-hx^{*}-xh^{*})\bigr{)}$
$\displaystyle=-2\langle\operatorname{adj}(I_{m}-xx^{*})x,h\rangle,$

where $\operatorname{adj}$ denotes the adjugate of a matrix, and hence,

$\nabla g(x)=-2\operatorname{adj}(I_{m}-xx^{*})x.$

For any $x\in\mathbb{K}^{m\times n}$ with singular value decomposition
$x=v\operatorname{diag}(s(x))u^{*}$ we have

$\operatorname{adj}(I_{m}-xx^{*})=\sum_{i=1}^{m}\biggl{(}\prod_{\begin{subarray}{c}j=1\\\
j\neq
i\end{subarray}}^{m}\bigl{(}1-s_{j}(x)^{2}\bigr{)}\biggr{)}v_{i}v_{i}^{*},$

where $v_{1},\dotsc,v_{m}\in\mathbb{K}^{m}$ are the columns of $v$; in
particular for all $x\in\mathbb{S}_{S_{\infty},\beta}^{m,n}$ we get

$\operatorname{adj}(I_{m}-xx^{*})x=\biggl{(}\prod_{j=2}^{m}\bigl{(}1-s_{j}(x)^{2}\bigr{)}\biggr{)}v_{1}(x^{*}v_{1})^{*}=\alpha
v_{1}u_{1}^{*},$

with $\alpha:=\prod_{j=2}^{m}(1-s_{j}(x)^{2})$ and $u_{1}\in\mathbb{K}^{n}$
the first column of $u$. Thus $\nabla g(x)\neq 0$ for all
$x\in\mathbb{S}_{S_{\infty},\beta}^{m,n}$ which have simple maximal singular
value $1$, and this means that $\mathbb{S}_{S_{\infty},\beta}^{m,n}$ has a
continuous outer unit normal vector field almost everywhere and up to sign it
is given by $\frac{\nabla g(x)}{\lVert\nabla g(x)\rVert_{S_{2}}}$ at
$x\in\mathbb{S}_{S_{\infty},\beta}^{m,n}$.

Let $x\in\mathbb{S}_{S_{\infty},\beta}^{m,n}$ with simple singular value $1$,
then

$\displaystyle\langle\nabla g(x),x\rangle$ $\displaystyle=-2\alpha\langle
v_{1}u_{1}^{*},x\rangle$ $\displaystyle=-2\alpha\langle v_{1},xu_{1}\rangle$
$\displaystyle=-2\alpha\langle v_{1},v_{1}\rangle=-2\alpha,$

and similarly,

$\displaystyle\lVert\nabla g(x)\rVert_{S_{2}}^{2}$
$\displaystyle=4\alpha^{2}\langle v_{1}u_{1}^{*},v_{1}u_{1}^{*}\rangle$
$\displaystyle=4\alpha^{2}\langle v_{1}^{*}v_{1},u_{1}^{*}u_{1}\rangle$
$\displaystyle=4\alpha^{2}\langle 1,1\rangle=4\alpha^{2},$

and this yields

$\frac{\lvert\langle\nabla g(x),x\rangle\rvert}{\lVert\nabla
g(x)\rVert_{S_{2}}}=1.$

This also implies that the cone and normalized Hausdorff/measures conincide.

In the case $p=1$ we have

$\operatorname{tr}((xx^{*})^{p-1})=\operatorname{tr}(I_{m})=m$

which leads to $\langle\nu(x),x\rangle=m^{-1/2}$, which is constant; and in
the case $p=2$ we see

$\lVert x\rVert_{S_{2p-2}}^{p-1}=\lVert x\rVert_{S_{2}}=1$

for all $x\in\mathbb{S}_{S_{2},\beta}^{m,n}$, which is constant too. (Also
notice that $S_{2,\beta}^{m\times n}$ and $\ell_{2,1}^{\beta mn}$ are
isomorphic Euclidean spaces.) For $p\notin\\{1,2,\infty\\}$ it suffices to
show that $\lVert\cdot\rVert_{S_{2p-2}}^{2p-2}$ attains at least two different
values on $\mathbb{S}_{S_{p},\beta}^{m,n}$ at points with full rank, since
then by continuity it cannot be constant outside of a null set. But this is
achieved with, say, $m^{-1/p}\operatorname{diag}(1,\dotsc,1)$ ($m$ ones) and
$(m-\frac{1}{2})^{-1/p}\operatorname{diag}(1,\dotsc,1,2^{-1/p})$ ($m-1$ ones).

The volume of $\mathbb{S}_{S_{p},\beta}^{m,n}$ follows from Lemma 3.6 and the
concrete values of $\langle\nu(x),x\rangle$. ∎

The ball $\mathbb{B}_{S_{\infty},\beta}^{m\times n}$ can be described without
direct reference to singular values, as the following proposition states.

######  Proposition 3.9.

The following identity holds true,

$\mathbb{B}_{S_{\infty},\beta}^{m\times n}=\\{x\in\mathbb{K}^{m\times
n}\,:\,xx^{*}\leq I_{m}\\}.$

###### Proof.

Call $B:=\\{x\in\mathbb{K}^{m\times n}\,:\,xx^{*}\leq I_{m}\\}$. Let
$x\in\mathbb{K}^{m\times n}$, then $x\in B$ iff for all
$\xi\in\mathbb{K}^{m}$, $\xi^{*}xx^{*}\xi\leq\xi^{*}\xi$, equivalently $\lVert
x^{*}\xi\rVert_{2}^{2}\leq\lVert\xi\rVert_{2}^{2}$, this itself means
$s_{1}(x^{*})=s_{1}(x)\leq 1$, which is equivalent to
$x\in\mathbb{B}_{S_{\infty},\beta}^{m\times n}$. ∎

The “boundary” $S_{\beta}^{m,n}:=\\{x\in\mathbb{K}^{m\times
n}\,:\,xx^{*}=I_{m}\\}$ is easy to identify: a close glance at the definition
of the Stiefel manifold (see beginning of Section 2.2) reveals
$S_{\beta}^{m,n}=\\{x^{*}\,:\,x\in\mathrm{U}_{n,m;\beta}\\}=\\{x^{\mathsf{T}}\,:\,x\in\mathrm{U}_{n,m;\beta}\\}$.
Also note the chain of inclusions
$S_{\beta}^{m,n}\subset\mathbb{S}_{S_{\infty},\beta}^{m,n}\subset\mathbb{B}_{S_{\infty},\beta}^{m\times
n}$ (concerning the first inclusion observe $s(x)=(1)_{i\leq m}$ for any $x\in
S_{\beta}^{m,n}$).

######  Proposition 3.10.

Let $r\in\mathbb{K}^{m\times m}$ be regular, then the following hold true.

  1. 1.

$\\{x\in\mathbb{K}^{m\times n}\,:\,xx^{*}\leq
rr^{*}\\}=r\mathbb{B}_{S_{\infty},\beta}^{m\times n}$, and
$\operatorname{\mathit{v}}_{\beta mn}(r\mathbb{B}_{S_{\infty},\beta}^{m\times
n})=\lvert\det(r)\rvert^{\beta n}\omega_{S_{\infty},\beta}^{m\times n}$.

  2. 2.

$\partial(r\mathbb{B}_{S_{\infty},\beta}^{m\times
n})=r\mathbb{S}_{S_{\infty},\beta}^{m\times n}$, and

$\operatorname{\mathcal{H}}^{\beta
mn-1}(r\mathbb{S}_{S_{\infty},\beta}^{m,n})=\frac{\lvert\det(r)\rvert^{\beta(n-1)}\operatorname{\mathcal{H}}^{\beta
m-1}(r\mathbb{S}_{2,\beta}^{m})}{\operatorname{\mathcal{H}}^{\beta
m-1}(\mathbb{S}_{2,\beta}^{m})}\operatorname{\mathcal{H}}^{\beta
mn-1}(\mathbb{S}_{S_{\infty},\beta}^{m,n}).$

  3. 3.

$\\{x\in\mathbb{K}^{m\times n}\,:\,xx^{*}=rr^{*}\\}=rS_{\beta}^{m,n}$, and

$\operatorname{\mathcal{H}}^{d}(rS_{\beta}^{m,n})=2^{-\beta
m(m-1)/4}\lvert\det(r)\rvert^{\beta(n-m+1)-1}\prod_{1\leq j<i\leq
m}(s_{j}(r)^{2}+s_{i}(r)^{2})^{\beta/2}\cdot\Omega_{n,m;\beta},$

where $d=\frac{\beta m(2n-m+1)}{2}-m$.

###### Proof.

Before proving the volumes, we argue the claimed set identities, and we fix
the map $f\colon\mathbb{K}^{m\times n}\to\mathbb{K}^{m\times n}$, $x\mapsto
rx$. $f$ is linear and bijective, hence a homeomorphism, and this immediately
gives $\partial(r\mathbb{B}_{S_{\infty},\beta}^{m\times
n})=r\partial\mathbb{B}_{S_{\infty},\beta}^{m\times
n}=r\mathbb{S}_{S_{\infty},\beta}^{m\times n}$. Concerning
$r\mathbb{B}_{S_{\infty},\beta}^{m\times n}$ note that $xx^{*}\leq rr^{*}$ iff
$\xi^{*}xx^{*}\xi\leq\xi^{*}rr^{*}\xi$ for all $\xi\in\mathbb{K}^{m}$; now
substitute $\eta:=r^{*}\xi$, which is bijective, and then
$\eta^{*}r^{-1}xx^{*}(r^{*})^{-1}\eta\leq\eta^{*}\eta$ for all
$\eta\in\mathbb{K}^{m}$, which means $(r^{-1}x)(r^{-1}x)^{*}\leq I_{m}$ and
so, by Proposition 3.9, $r^{-1}x\in\mathbb{B}_{S_{\infty},\beta}^{m\times n}$,
hence the claim. Concerning $rS_{\beta}^{m,n}$, note $xx^{*}=rr^{*}$ iff
$r^{-1}xx^{*}(r^{*})^{-1}=I_{m}$, and the claim follows.

We now turn to the volumes. The general tool is to calculate the determinant
of $f$ restricted to the tangent spaces of the respective manifolds; to that
end we are going to choose orthonormal bases as fits our need.

  1. 1.

Fix a singular value decomposition $r=v\operatorname{diag}(s(r))u^{*}$. Here
$f$ is acting on the whole space $\mathbb{K}^{m\times n}$ which is its own
tangent space, and we choose as basis elements the matrices
$ue_{i}e_{j}^{\mathsf{T}}$, additionally $\mathsf{i}ue_{i}e_{j}^{\mathsf{T}}$
in the complex case, then

$f(ue_{i}e_{j}^{\mathsf{T}})=rue_{i}e_{j}^{\mathsf{T}}=s_{i}(r)ve_{i}e_{j}^{\mathsf{T}},\quad
f(\mathsf{i}ue_{i}e_{j}^{\mathsf{T}})=\mathsf{i}rue_{i}e_{j}^{\mathsf{T}}=s_{i}(r)\mathsf{i}ve_{i}e_{j}^{\mathsf{T}}$

for all $i\in[1,m],j\in[1,n]$; and because also the
$ve_{i}e_{j}^{\mathsf{T}}$–and $\mathsf{i}ve_{i}e_{j}^{\mathsf{T}}$–together
constitute an orthonormal basis of $\mathbb{K}^{m\times n}$, we get

$\det(f)=\biggl{(}\prod_{i=1}^{m}\prod_{j=1}^{n}s_{i}(r)\biggr{)}^{\beta}=\biggl{(}\prod_{i=1}^{m}s_{i}(r)\biggr{)}^{\beta
n}=\lvert\det(r)\rvert^{\beta n},$

concluding this part.

  2. 2.

From the proof of Theorem C we know
$\nu(x)=v_{1}u_{1}^{*}=ve_{1}e_{1}^{\mathsf{T}}u^{*}$ for
$x\in\mathbb{S}_{S_{\infty},\beta}^{m,n}$ with singular value decomposition
$x=v\operatorname{diag}(s(x))u^{*}$ such that $s_{1}(x)=1$ is isolated.
Therefore an orthonormal base of the tangent space at $x$ is given by
$v_{i}u_{j}^{*}$ for all $(i,j)\in[1,m]\times[1,n]$ except $(i,j)=(1,1)$–in
the complex case additionally $\mathsf{i}v_{i}u_{j}^{*}$ for all
$(i,j)\in[1,m]\times[1,n]$. Because $u$, $v$ depend on $x$ it makes no sense
to fix a decomposition of $r$ beforehand. Instead, given $x$, define
$t:=v^{*}rv\in\mathbb{K}^{m\times m}$; then
$rv_{i}=\sum_{k=1}^{m}t_{k,i}v_{k}$ for any $i\in[1,m]$. Let $j\in[1,n]$ and
$i\in[1,m]$, then

$f(v_{i}u_{j}^{*})=rv_{i}u_{j}^{*}=\sum_{k=1}^{m}t_{k,i}v_{k}u_{j}^{*}=\sum_{k=1}^{m}\Re(t_{k,i})v_{k}u_{j}^{*}+\sum_{k=1}^{m}\Im(t_{k,i})\mathsf{i}v_{k}u_{j}^{*},$

and specially for the complex case,

$f(\mathsf{i}v_{i}u_{j}^{*})=rv_{i}u_{j}^{*}=-\sum_{k=1}^{m}\Im(t_{k,i})v_{k}u_{j}^{*}+\sum_{k=1}^{m}\Re(t_{k,i})\mathsf{i}v_{k}u_{j}^{*}.$

This means that for any $j\in[1,n]$, the subspace spanned by $v_{i}u_{j}^{*}$
with $i\in[1,m]$–additionally $\mathsf{i}v_{i}u_{j}^{*}$ in the complex
case–is $f$/̄invariant and with respect to that orthonormal basis the matrix
of $f$ is given by $t$–by $\bigl{(}\begin{smallmatrix}\Re(t)&-\Im(t)\\\
\Im(t)&\Re(t)\end{smallmatrix}\bigr{)}$ in the complex case. Also note that
for distinct $j$ those subspaces are mutually orthogonal. Thus for any
$j\in[2,n]$ the associated subspace makes the contribution
$\lvert\det(t)\rvert^{\beta}=\lvert\det(r)\rvert^{\beta}$ to the functional
determinant, yielding a total of $\lvert\det(r)\rvert^{\beta(n-1)}$.

For $j=1$ though we lack $f(v_{1}u_{1}^{*})$, therefore the matrix of $f$ is
the ($\beta m\times(\beta m-1)$)/matrix
$\mathbf{t}\bigl{(}\begin{smallmatrix}0\\\ I_{\beta
m-1}\end{smallmatrix}\bigr{)}$, that is $\mathbf{t}$ sans the first column;
here we mean $\mathbf{t}:=t$ in the real case and
$\mathbf{t}:=\bigl{(}\begin{smallmatrix}\Re(t)&-\Im(t)\\\
\Im(t)&\Re(t)\end{smallmatrix}\bigr{)}$ in the complex case. The contribution
to the functional determinant in this case is the Gramian

$G(x):=\det\left((0,I_{\beta
m-1})\mathbf{t}^{\mathsf{T}}\mathbf{t}\begin{pmatrix}0\\\ I_{\beta
m-1}\end{pmatrix}\right)^{1/2}=\det\left((0,I_{\beta
m-1})\mathbf{v}^{\mathsf{T}}\mathbf{r}^{\mathsf{T}}\mathbf{r}\mathbf{v}\begin{pmatrix}0\\\
I_{\beta m-1}\end{pmatrix}\right)^{1/2},$

where we have used that $ab$ corresponds to $\mathbf{a}\mathbf{b}$, that
$a^{*}$ corresponds to $\mathbf{a}^{\mathsf{T}}$, and that
$v\in\mathrm{U}_{m;\beta}$ corresponds to $\mathbf{v}\in\mathrm{U}_{\beta
m;1}$. Summing up, the functional determinant of $f$ at $x$ is given by

$D(x):=\lvert\det(r)\rvert^{\beta(n-1)}G(x);$

note that it depends on $x$ only through $v$, not $u$ nor $s(x)$. This allows
us to compute the volume of $r\mathbb{S}_{S_{\infty},\beta}^{m,n}$ as follows,
where we use Theorem C to transform from Hausdorff- to cone measure,

$\displaystyle\operatorname{\mathcal{H}}^{\beta
mn-1}(r\mathbb{S}_{S_{\infty},\beta}^{m,n})$
$\displaystyle=\int_{\mathbb{S}_{S_{\infty},\beta}^{m,n}}D(x)\,\mathrm{d}\\!\operatorname{\mathcal{H}}^{\beta
mn-1}(x)$ $\displaystyle=\operatorname{\mathcal{H}}^{\beta
mn-1}(\mathbb{S}_{S_{\infty},\beta}^{m,n})\int_{\mathbb{S}_{S_{\infty},\beta}^{m,n}}D(x)\,\mathrm{d}\kappa_{S_{\infty},\beta}^{m\times
n}(x)$ $\displaystyle=\frac{\operatorname{\mathcal{H}}^{\beta
mn-1}(\mathbb{S}_{S_{\infty},\beta}^{m,n})}{\omega_{S_{\infty},\beta}^{m\times
n}}\int_{\mathbb{B}_{S_{\infty},\beta}^{m\times n}}D\Bigl{(}\frac{x}{\lVert
x\rVert_{S_{\infty}}}\Bigr{)}\,\mathrm{d}x$
$\displaystyle=\frac{\operatorname{\mathcal{H}}^{\beta
mn-1}(\mathbb{S}_{S_{\infty},\beta}^{m,n})}{\omega_{S_{\infty},\beta}^{m\times
n}}\,Cm!\int_{W^{m}}\int_{\mathrm{U}_{n;\beta}}\int_{\mathrm{U}_{m;\beta}}\operatorname{1}_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}(v\operatorname{diag}(\sigma)u^{*})$ $\displaystyle\qquad\cdot
D\Bigl{(}\frac{v\operatorname{diag}(\sigma)u^{*}}{\lVert
v\operatorname{diag}(\sigma)u^{*}\rVert_{S_{\infty}}}\Bigr{)}h(\sigma)\,\mathrm{d}v\,\mathrm{d}u\,\mathrm{d}\sigma$
$\displaystyle=\operatorname{\mathcal{H}}^{\beta
mn-1}(\mathbb{S}_{S_{\infty},\beta}^{m,n})\,\frac{2^{\beta
m(m-1)/4}}{\Omega_{m;\beta}}\,\lvert\det(r)\rvert^{\beta(n-1)}\int_{\mathrm{U}_{m;\beta}}G(v\operatorname{diag}(e_{1}))\,\mathrm{d}v;$

we note a few points: in the fourth line we have abbreviated
$C:=\frac{2^{\beta m/2+\beta m(m-1)/4+\beta n(n-1)/4}}{2^{\beta
mn/2}m!\Omega_{1;\beta}^{m}\Omega_{n-m;\beta}}$ and
$h(\sigma):=\prod_{j=1}^{m}\sigma_{j}^{\beta(n-m+1)-1}\prod_{1\leq j<i\leq
m}\lvert\sigma_{i}^{2}-\sigma_{j}^{2}\rvert^{\beta}$, and as we need the
singular values to be in decreasing order for $G$ we have integrated over
$W^{m}$, cf. Remark 2.5; for the last line we have invested the structure of
$D$ and the knowledge that $G$ only depends on $v$, therefore we have inserted
some constant values for $\sigma$ and $u$, namely $e_{1}$ and $I_{n}$, resp.
In order to identify the last integral we rely on the following transformation
formula, which can be derived from the coarea formula in considering the map
$\mathrm{U}_{m;\beta}\to\mathbb{S}_{2,\beta}^{m}$ given by $v\mapsto ve_{1}$:

$\int_{\mathrm{U}_{m;\beta}}f(v)\,\mathrm{d}v=\int_{\mathbb{S}_{2,\beta}^{m}}\int_{\mathrm{U}_{m-1;\beta}}f\left(\hat{\theta}\begin{pmatrix}1&0\\\
0&w\end{pmatrix}\right)\,\mathrm{d}w\,\mathrm{d}\\!\operatorname{\mathcal{H}}^{\beta
m-1}(\theta),$ (11)

where $\hat{\theta}\in\mathrm{U}_{m;\beta}$ is such that
$\hat{\theta}e_{1}=\theta$. In our situation we need to consider, in the real
case,

$\hat{\theta}\begin{pmatrix}1&0\\\ 0&w\end{pmatrix}\begin{pmatrix}0\\\
I_{m-1}\end{pmatrix}=\hat{\theta}\begin{pmatrix}0\\\ I_{m-1}\end{pmatrix}w,$

and similarly in the complex case,

$\hat{\bm{\theta}}\begin{pmatrix}1&0&0&0\\\ 0&\Re(w)&0&-\Im(w)\\\ 0&0&1&0\\\
0&\Im(w)&0&\Re(w)\end{pmatrix}\begin{pmatrix}0\\\
I_{2m-1}\end{pmatrix}=\hat{\bm{\theta}}\begin{pmatrix}0\\\
I_{2m-1}\end{pmatrix}\begin{pmatrix}\Re(w)&0&-\Im(w)\\\ 0&1&0\\\
\Im(w)&0&\Re(w)\end{pmatrix}\\!;$

now we recall that a Gramian is orthogonally invariant, hence

$\displaystyle\int_{\mathrm{U}_{m;\beta}}G(v\operatorname{diag}(e_{1}))\,\mathrm{d}v$
$\displaystyle=\int_{\mathbb{S}_{2,\beta}^{m}}\int_{\mathrm{U}_{m-1;\beta}}\det\left((0,I_{\beta
m-1})\hat{\bm{\theta}}^{\mathsf{T}}\mathbf{r}^{\mathsf{T}}\mathbf{r}\hat{\bm{\theta}}\begin{pmatrix}0\\\
I_{\beta
m-1}\end{pmatrix}\right)^{1/2}\,\mathrm{d}w\,\mathrm{d}\\!\operatorname{\mathcal{H}}^{\beta
m-1}(\theta)$
$\displaystyle=\frac{\Omega_{m-1;\beta}}{2^{\beta(m-1)(m-2)/4}}\int_{\mathbb{S}_{2,\beta}^{m}}\det\left((0,I_{\beta
m-1})\hat{\bm{\theta}}^{\mathsf{T}}\mathbf{r}^{\mathsf{T}}\mathbf{r}\hat{\bm{\theta}}\begin{pmatrix}0\\\
I_{\beta
m-1}\end{pmatrix}\right)^{1/2}\mathrm{d}\\!\operatorname{\mathcal{H}}^{\beta
m-1}(\theta).$

The conclusion of this part is reached through the identities

$\int_{\mathbb{S}_{2,\beta}^{m}}\det\left(0,I_{\beta
m-1})\hat{\bm{\theta}}^{\mathsf{T}}\mathbf{r}^{\mathsf{T}}\mathbf{r}\hat{\bm{\theta}}\begin{pmatrix}0\\\
I_{\beta
m-1}\end{pmatrix}\right)^{1/2}\mathrm{d}\\!\operatorname{\mathcal{H}}^{\beta
m-1}(\theta)=\operatorname{\mathcal{H}}^{\beta
m-1}(r\mathbb{S}_{2,\beta}^{m})$

and

$\Omega_{m;\beta}=2^{\beta(m-1)/2}\,\Omega_{m-1;\beta}\operatorname{\mathcal{H}}^{\beta
m-1}(\mathbb{S}_{2,\beta}^{m}),$

the first of which can be proved by considering the map
$\mathbb{S}_{2,\beta}^{m}\to\mathbb{K}^{m}$ given by $\theta\mapsto r\theta$,
leading to calculations which essentially are the same as those above for
$\mathbb{S}_{S_{\infty},\beta}^{m,n}$ without writing the $u_{j}$ and only
considering $j=1$; the latter can be proved either from the volume formula (1)
or from the transformation formula (11).

  3. 3.

Fix a singular value decomposition $r=v\operatorname{diag}(s(r))u^{*}$. We
already know $S_{\beta}^{m,n}=\\{w^{*}\,:\,w\in\mathrm{U}_{n,m;\beta}\\}$ and
hence $rS_{\beta}^{m,n}=\\{(wr^{*})^{*}\,:\,w\in\mathrm{U}_{n,m;\beta}\\}$,
and for any $w\in\mathrm{U}_{n,m;\beta}$ the tangent space is spanned by the
matrices $E_{i,j}$–additionally $F_{i,j}$ in the complex case–given by

$\displaystyle F_{j,j}$
$\displaystyle:=\mathsf{i}wue_{j}e_{j}^{\mathsf{T}}u^{*}$ $\displaystyle\text{
for }j\in[1,m],$ $\displaystyle E_{i,j}$
$\displaystyle:=\frac{w}{\sqrt{2}}\,u(e_{i}e_{j}^{\mathsf{T}}-e_{j}e_{i}^{\mathsf{T}})u^{*}$
$\displaystyle\text{ for }1\leq j<i\leq m,$ $\displaystyle F_{i,j}$
$\displaystyle:=\frac{\mathsf{i}w}{\sqrt{2}}\,u(e_{i}e_{j}^{\mathsf{T}}+e_{j}e_{i}^{\mathsf{T}})u^{*}$
$\displaystyle\text{ for }1\leq j<i\leq m,$ $\displaystyle E_{i,j}$
$\displaystyle:=\hat{w}e_{i}e_{j}^{\mathsf{T}}u^{*}$ $\displaystyle\text{ for
}i\in[m+1,n],j\in[1,m],$ $\displaystyle F_{i,j}$
$\displaystyle:=\mathsf{i}\hat{w}e_{i}e_{j}^{\mathsf{T}}u^{*}$
$\displaystyle\text{ for }i\in[m+1,n],j\in[1,m],$

where $\hat{w}\in\mathrm{U}_{n,\beta}$ is an extension of $w$ with suitable
$n-m$ columns. Their respective images are, in the same order,

$\displaystyle f(F_{j,j})$
$\displaystyle=s_{j}(r)\mathsf{i}wue_{j}e_{j}^{\mathsf{T}}v^{*},$
$\displaystyle f(E_{i,j})$
$\displaystyle=\frac{w}{\sqrt{2}}\,u(s_{j}(r)e_{i}e_{j}^{\mathsf{T}}-s_{i}(r)e_{j}e_{i}^{\mathsf{T}})v^{*},$
$\displaystyle f(F_{i,j})$
$\displaystyle=\frac{\mathsf{i}w}{\sqrt{2}}\,u(s_{j}(r)e_{i}e_{j}^{\mathsf{T}}+s_{i}(r)e_{j}e_{i}^{\mathsf{T}})v^{*},$
$\displaystyle f(E_{i,j})$
$\displaystyle=s_{j}(r)\hat{w}e_{i}e_{j}^{\mathsf{T}}v^{*},$ $\displaystyle
f(F_{i,j})$
$\displaystyle=s_{j}(r)\mathsf{i}\hat{w}e_{i}e_{j}^{\mathsf{T}}v^{*}.$

Clearly the elements of each line are orthogonal to the other lines, and the
elements of the first, fourth and fifth lines also are orthogonal among
themselves and their norms are the indicated singular values. Orthogonality is
even true for the second and third lines, although less obvious: let $1\leq
j<i\leq m$ and $1\leq j^{\prime}<i^{\prime}\leq m$, then

$\langle
f(E_{i,j}),f(E_{i^{\prime},j^{\prime}})\rangle=\frac{s_{i}(r)s_{i^{\prime}}(r)+s_{j}(r)s_{j^{\prime}}(r)}{2}\delta_{i,i^{\prime}}\delta_{j,j^{\prime}}-\frac{s_{i}(r)s_{j^{\prime}}(r)+s_{i^{\prime}}(r)s_{j}(r)}{2}\delta_{i,j^{\prime}}\delta_{i^{\prime},j};$

but for our choice of indices always
$\delta_{i,j^{\prime}}\delta_{i^{\prime},j}=0$, and so orthgonality follows.
The analogous calculation goes through for $f(F_{i,j})$. Therefore the
functional determinant equals

$\biggl{(}\prod_{j=1}^{m}s_{j}(r)\biggr{)}^{\beta-1}\biggl{(}\prod_{1\leq
j<i\leq
m}\frac{(s_{j}(r)^{2}+s_{i}(r)^{2})^{1/2}}{\sqrt{2}}\biggr{)}^{\beta}\biggl{(}\prod_{1\leq
j\leq m<i\leq n}s_{j}(r)\biggr{)}^{\beta}\\\ =2^{-\beta
m(m-1)/4}\lvert\det(r)\rvert^{\beta(n-m+1)-1}\prod_{1\leq j<i\leq
m}(s_{j}(r)^{2}+s_{i}(r))^{\beta/2}.$

Since this is indpendent of $w$, this implies directly

$\operatorname{\mathcal{H}}^{d}(\mathrm{U}_{n,m;\beta}r^{*})=2^{-\beta
m(m-1)/4}\lvert\det(r)\rvert^{\beta(n-m+1)-1}\prod_{1\leq j<i\leq
m}(s_{j}(r)^{2}+s_{i}(r))^{\beta/2}\cdot\Omega_{n,m;\beta},$

and because adjoining is an isometry the result for $S_{\beta}^{m,n}$ follows.
∎

######  Remark 3.11.

The factor $\operatorname{\mathcal{H}}^{\beta m-1}(r\mathbb{S}_{2,\beta}^{m})$
in Proposition 3.10, part 2, is the surface area of an ellipsoid, as such it
cannot be expressed in terms of elementary functions of $s(r)$ (the semiaxes
of the ellipsoid).

##  4 Probabilistic results and weak limit theorems for $p=\infty$

For all subsequent probabilistic results we suppose a sufficiently rich
probability space $(\Omega,\mathcal{A},\operatorname{\mathbb{P}})$ on which
all random variables that appear in the present paper are defined. The
expected value of a random variable $X$ w.r.t. $\operatorname{\mathbb{P}}$ is
denoted $\operatorname{\mathbb{E}}[X]$ and is to be understood component/wise
for vector/valued $X$. $X\sim\mu$ means the law, or distribution, of $X$ is
$\mu$, and $\stackrel{{\scriptstyle\text{d}}}{{=}}$ denotes equality in law.
For $A\subset\mathbb{R}^{d}$ with
$\operatorname{\mathit{v}}_{d}(A)\in\mathbb{R}_{>0}$ the uniform distribution
on $A$ (w.r.t. Lebesgue measure) is $\operatorname{\mathcal{U}}(A)$, and
$\operatorname{\mathcal{U}}(\mathrm{U}_{n,m;\beta})$ means the normalized
invariant measure on $\mathrm{U}_{n,m;\beta}$ (see also Section 2.2).
Convergence in law, in probability, and almost surely are denoted by
$\xrightarrow[]{\text{d}}$, $\xrightarrow[]{\mathbb{P}}$, and
$\xrightarrow[]{\text{a.s.}}$ respectively.

The following can be considered a variant of the Schechtman-Zinn
representation of the uniform distribution on the $\ell_{p}$/balls. It uses
the _matrix/variate type/̄I beta/distribution,_ written
$B_{m,\beta}^{\text{I}}(a,b)$ and defined as follows (see [18, Definition
5.2.1]): suppose $a,b>\frac{\beta(m-1)}{2}$, then
$B_{m,\beta}^{\text{I}}(a,b)$ is the probability distribution on
$\text{Sym}_{m,\beta}$ with density

$x\mapsto\frac{\Gamma_{m,\beta}(a+b)}{\Gamma_{m,\beta}(a)\Gamma_{m,\beta}(b)}\det(x)^{a-1-\beta(m-1)/2}\det(I_{m}-x)^{b-1-\beta(m-1)/2}\operatorname{1}_{(0,I_{m})}(x),$

where of course $(0,I_{m}):=\\{x\in\text{Sym}_{m,\beta}\,:\,0<x<I_{m}\\}$.

######  Proposition 4.1.

A random variable $X\in\mathbb{K}^{m\times n}$ is distributed uniformly on
$\mathbb{B}_{S_{\infty},\beta}^{m\times n}$ iff there exist independent random
variables $R\in\mathbb{K}^{m\times m}$ and $U\in\mathbb{K}^{n\times m}$ such
that $R\sim B_{m,\beta}^{\text{I}}(\frac{\beta n}{2},\frac{\beta(m-1)}{2}+1)$,
$U\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n,m;\beta})$, and

$X\stackrel{{\scriptstyle\text{d}}}{{=}}R^{1/2}\,U^{*}.$

######  Remark 4.2.

Note that from $X\stackrel{{\scriptstyle\text{d}}}{{=}}R^{1/2}\,U^{*}$ follow
$R\stackrel{{\scriptstyle\text{d}}}{{=}}XX^{*}$ and
$U\stackrel{{\scriptstyle\text{d}}}{{=}}X^{*}(XX^{*})^{-1/2}$.

###### Proof.

The statement is a simple consequence of Proposition 2.4, part 1. We have
$X\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{\infty},\beta}^{m\times n})$
iff, for any $f\colon\mathbb{K}^{m\times n}\to\mathbb{R}$ nonnegative and
measurable,

$\displaystyle\operatorname{\mathbb{E}}[f(X)]$
$\displaystyle=\frac{1}{\omega_{S_{\infty},\beta}^{m\times
n}}\int_{\mathbb{K}^{m\times
n}}\operatorname{1}_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}(x)f(x)\,\mathrm{d}x$
$\displaystyle=\frac{2^{-m}}{\omega_{S_{\infty},\beta}^{m\times
n}}\int_{\text{Pos}_{m,\beta}}\int_{\mathrm{U}_{n,m;\beta}}\operatorname{1}_{\mathbb{B}_{S_{\infty},\beta}^{m\times
n}}(r^{1/2}\,u^{*})f(r^{1/2}\,u^{*})\,\mathrm{d}u\det(r)^{\beta(n-m+1)/2-1}\,\mathrm{d}r$
$\displaystyle=\frac{2^{-m}}{\omega_{S_{\infty},\beta}^{m\times
n}}\int_{\text{Pos}_{m,\beta}}\int_{\mathrm{U}_{n,m;\beta}}f(r^{1/2}\,u^{*})\,\mathrm{d}u\operatorname{1}_{(0,I_{m}]}(r)\det(r)^{\beta(n-m+1)/2-1}\,\mathrm{d}r$
$\displaystyle=\operatorname{\mathbb{E}}[f(R^{1/2}\,U^{*})],$

where $R$ and $U$ have the claimed joint distribution. (This is obvious for
$U$; concerning $R$ one may compare our result with the known density.) In the
last integral, $r^{1/2}\,u^{*}\in\mathbb{B}_{S_{\infty},\beta}^{m\times n}$
iff $r\leq I_{m}$, and note that $(0,I_{m}]\setminus(0,I_{m})$ is negligible.
∎

Another representation, related to the first one, can be obtained in the
special case $n=dm$ for some $d\in\mathbb{N}$. Here we may identify
$\mathbb{K}^{m\times dm}\cong(\mathbb{K}^{m\times m})^{d}$, constisting of
$d$/̄dimensional row vectors whose entries are from $\mathbb{K}^{m\times m}$.
Then, for $x=(x_{1},\dotsc,x_{d}),y=(y_{1},\dotsc,y_{d})\in\mathbb{K}^{m\times
dm}$, $xy^{*}=\sum_{i=1}^{d}x_{i}y_{i}^{*}$, and hence also $\langle
x,y\rangle=\sum_{i=1}^{d}\langle x_{i},y_{i}\rangle$. It uses the
_matrix/variate type/̄I Dirichlet/distribution_
$D_{m,\beta}^{\text{I}}(a_{1},\dotsc,a_{d};a_{d+1})$, with
$a_{1},\dotsc,a_{d+1}>\frac{\beta(m-1)}{2}$, defined to be the probability
distribution on $\text{Sym}_{m,\beta}^{d}$ with density (see [18, Definition
6.2.1])

$\begin{split}(x_{1},\dotsc,x_{d})&\mapsto\frac{\Gamma_{m,\beta}\bigl{(}\sum_{i=1}^{d+1}a_{i}\bigr{)}}{\prod_{i=1}^{d+1}\Gamma_{m,\beta}(a_{i})}\prod_{i=1}^{d}\det(x_{i})^{a_{i}-1-\beta(m-1)/2}\\\
&\qquad\cdot\det\biggl{(}I_{m}-\sum_{i=1}^{d}x_{i}\biggr{)}^{a_{d+1}-1-\beta(m-1)/2}\operatorname{1}_{\Delta_{m,\beta}^{d}}(x_{1},\dotsc,x_{d}),\end{split}$

where
$\Delta_{m,\beta}^{d}:=\bigl{\\{}(x_{1},\dotsc,x_{d})\in\text{Sym}_{m,\beta}^{d}\,:\,x_{1},\dotsc,x_{d}>0\wedge\sum_{i=1}^{d}x_{i}<I_{m}\bigr{\\}}$.

######  Remark 4.3.

With this componentwise interpretation the spectral norm unit ball can be
written $\mathbb{B}_{S_{\infty},\beta}^{m\times dm}=\bigl{\\{}(x_{i})_{i\leq
d}\in(\mathbb{K}^{m\times m})^{d}\,:\,\sum_{i=1}^{d}x_{i}x_{i}^{*}\leq
I_{m}\bigr{\\}}$, which looks like a natural generalization of the usual
$\ell_{2}$/̄ball $\mathbb{B}_{2,\beta}^{d}=\bigl{\\{}(x_{i})_{i\leq
d}\in\mathbb{K}^{d}\,:\,\sum_{i=1}^{d}\lvert x_{i}\rvert^{2}\leq 1\bigr{\\}}$,
replacing the scalar components with matrix/valued ones. In truth this was the
starting point of the present investigations.

######  Proposition 4.4.

A random vector $X\in\mathbb{K}^{m\times dm}$ is distributed uniformly on
$\mathbb{B}_{S_{\infty},\beta}^{m\times dm}$ iff there exist independent
random variables $R=(R_{1},\dotsc,R_{d})\in\mathbb{K}^{m\times dm}$ and
$U_{1},\dotsc,U_{d}\in\mathbb{K}^{m\times m}$ such that $R\sim
D_{m,\beta}^{\text{I}}(\frac{\beta m}{2}^{[d]};\frac{\beta(m-1)}{2}+1)$ (i.e.,
the parameter $\frac{m}{2}$ occurs $d$ times),
$U_{i}\sim\operatorname{\mathcal{U}}(\mathrm{U}_{m;\beta})$ for each
$i\in\\{1,\dotsc,d\\}$, and

$X\stackrel{{\scriptstyle\text{d}}}{{=}}(R_{1}^{1/2}\,U_{1},\dotsc,R_{d}^{1/2}\,U_{d}).$

###### Proof.

Again we apply Proposition 2.4, part 1, this time separately to each of the
$d$ components of $\mathbb{K}^{m\times dm}=\mathbb{K}^{m\times
m}\times\dotsm\times\mathbb{K}^{m\times m}$. We have
$X\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{\infty},\beta}^{m\times dm})$
iff, for any $f\colon\mathbb{K}^{m\times dm}\to\mathbb{R}$ measurable and
nonnegative,

$\displaystyle\operatorname{\mathbb{E}}[f(X)]$
$\displaystyle=\frac{1}{\omega_{S_{\infty},\beta}^{m\times
dm}}\int_{\mathbb{R}^{m\times
dm}}\operatorname{1}_{\mathbb{B}_{S_{\infty},\beta}^{m\times
dm}}(x)f(x)\,\mathrm{d}x$
$\displaystyle=\frac{(2^{-m})^{d}}{\omega_{S_{\infty},\beta}^{m\times
dm}}\int_{\text{Pos}_{m,\beta}}\int_{\mathrm{U}_{m;\beta}}\dotsi\int_{\text{Pos}_{m,\beta}}\int_{\mathrm{U}_{m;\beta}}\operatorname{1}_{\mathbb{B}_{S_{\infty},\beta}^{m\times
dm}}(r_{1}^{1/2}\,u_{1}^{*},\dotsc,r_{d}^{1/2}\,u_{d}^{*})$
$\displaystyle\mspace{100.0mu}\cdot
f(r_{1}^{1/2}\,u_{1}^{*},\dotsc,r_{d}^{1/2}\,u_{d}^{*})\prod_{i=1}^{d}\det(r_{i})^{\beta/2-1}\prod_{i=1}^{d}(\mathrm{d}u_{i}\,\mathrm{d}r_{i})$
$\displaystyle=\frac{2^{-dm}}{\omega_{S_{\infty},\beta}^{m\times
dm}}\int_{\Delta_{m,\beta}^{d}}\int_{\mathrm{U}_{m;\beta}^{d}}f(r_{1}^{1/2}\,u_{1},\dotsc,r_{d}^{1/2}\,u_{d})\prod_{i=1}^{d}\mathrm{d}u_{i}\prod_{i=1}^{d}\det(r_{i})^{\beta/2-1}\,\mathrm{d}r$
$\displaystyle=\operatorname{\mathbb{E}}[f(R_{1}^{1/2}\,U_{1},\dotsc,R_{d}^{1/2}\,U_{d})],$

where the desired conclusion is reached by noticing a few points: firstly,
$(r_{1}^{1/2}\,u_{1},\dotsc,r_{d}^{1/2}\,u_{d})\in\mathbb{B}_{S_{\infty},\beta}^{m\times
dm}$ iff $\sum_{i=1}^{d}r_{i}\leq I_{m}$; secondly, we have transformed
$u_{i}\mapsto u_{i}^{*}$, which has no import since we are integrating with
respect to the (scaled) Haar measure; thirdly and lastly, the distribution of
the $U_{i}$ is obvious again, and that of $(R_{1},\dotsc,R_{d})$ can be read
off by comparing with the known density. ∎

Next we present a Poincaré-Maxwell-Borel/type result. It is useful to keep the
following relationships in mind. The standard Gaussian distribution on
$\mathbb{R}$ is denoted by $\mathcal{N}_{1}$, defined by its Lebesgue/density
$x\mapsto(2\pi)^{-1/2}\mathsf{e}^{-x^{2}/2}$. The standard Gaussian
distribution on $\mathbb{C}$ is denoted by $\mathcal{N}_{2}$ and defined to be
the law of $2^{-1/2}(x+\mathsf{i}y)$, where
$(x,y)\sim\mathcal{N}_{1}\otimes\mathcal{N}_{1}$. Note
$\operatorname{\mathbb{E}}[\lvert X\rvert^{2}]=1$ for
$X\sim\mathcal{N}_{\beta}$. The content of the following proposition is
standard from random matrix theory, especially as used in statistics; textbook
references are [18, 35].

######  Proposition 4.5.

  1. 1.

Let $X\sim\mathcal{N}_{\beta}^{\otimes(m\times n)}$, then
$XX^{*}\sim\Gamma_{m,\beta}(\frac{\beta n}{2},\frac{2}{\beta}I_{m})$, where
the matrix/variate gamma/distribution $\Gamma_{m,\beta}(\alpha,b)$ with
$\alpha>\frac{\beta(m-1)}{2}$ and $b\in\text{Pos}_{m,\beta}$ is defined on
$\text{Pos}_{m,\beta}$ by its density

$x\mapsto\frac{\det(x)^{\alpha-1-\beta(m-1)/2}}{\det(b)^{\alpha}\Gamma_{m,\beta}(\alpha)}\,\mathsf{e}^{-\operatorname{tr}(b^{-1}x)}.$

  2. 2.

$B_{m,\beta}^{\text{I}}(\alpha_{1},\alpha_{2})=D_{m,\beta}^{\text{I}}(\alpha_{1};\alpha_{2})$.

  3. 3.

Let $d\in\mathbb{N}$, $\alpha_{1},\dotsc,\alpha_{d+1}>\frac{\beta(m-1)}{2}$,
$b\in\text{Pos}_{m,\beta}$, and $(X_{i})_{i\leq
d+1}\sim\bigotimes_{i=1}^{d+1}\Gamma_{m,\beta}(\alpha_{i},b)$. Define
$Y_{i}:=\bigl{(}\sum_{j=1}^{d+1}X_{j}\bigr{)}^{-1/2}X_{i}\bigl{(}\sum_{j=1}^{d+1}X_{j}\bigr{)}^{-1/2}$
for $i\in[1,d]$ and $Y_{d+1}:=\sum_{i=1}^{d+1}X_{i}$, then $(Y_{i})_{i\leq d}$
and $Y_{d+1}$ are independent, and $(Y_{i})_{i\leq d}\sim
D_{m,\beta}^{\text{I}}(\alpha_{1},\dotsc,\alpha_{d};\alpha_{d+1})$ and
$Y_{d+1}\sim\Gamma_{m,\beta}\bigl{(}\sum_{i=1}^{d+1}\alpha_{i},b\bigr{)}$.

  4. 4.

Consequently, given $n_{1},\dotsc,n_{d+1}\in\mathbb{N}_{\geq m}$, then
$(X_{i})_{i\leq d}\sim D_{m,\beta}^{\text{I}}(\frac{\beta
n_{1}}{2},\dotsc,\frac{\beta n_{d}}{2};\frac{\beta n_{d+1}}{2})$ iff there
exists $(G_{i})_{i\leq
d+1}\sim\bigotimes_{i=1}^{d+1}\mathcal{N}_{\beta}^{\otimes(m\times n_{i})}$
such that
$X_{i}\stackrel{{\scriptstyle\text{d}}}{{=}}\bigl{(}\sum_{j=1}^{d+1}G_{j}G_{j}^{*}\bigr{)}^{-1/2}G_{i}G_{i}^{*}\bigl{(}\sum_{j=1}^{d+1}G_{j}G_{j}^{*}\bigr{)}^{-1/2}$
for each $i\in[1,d]$.

######  Theorem D.

Consider $m$ fixed, let $k\in\mathbb{N}$, and let $(X_{n})_{n\geq m}$ be a
sequence of random variables such that either

  * •

$X_{n}\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{\infty},\beta}^{m\times
n})$ for each $n\geq m$, or

  * •

$X_{n}\sim\operatorname{\mathcal{U}}(\mathbb{S}_{S_{\infty},\beta}^{m,n})$ for
each $n\geq m$, or

  * •

$X_{n}\sim\operatorname{\mathcal{U}}(S_{\beta}^{m,n})$ for each $n\geq m$.

Denote the projection of $X_{n}$ onto the first $k$ columns by $X_{n}^{(k)}$.
Then in any case

$(n^{1/2}X_{n}^{(k)})_{n\geq\max\\{m,k\\}}\xrightarrow[]{\text{d}}\mathcal{N}_{\beta}^{\otimes(m\times
k)}.$

###### Proof.

We begin with the case $X_{n}\sim\operatorname{\mathcal{U}}(S_{\beta}^{m,n})$,
then $X_{n}^{*}\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n,m;\beta})$. Let
$(g_{n})_{n\geq 1}$ be an i.i.d./sequence of $\mathcal{N}_{\beta}^{\otimes
m}$/distributed vectors and set $G_{n}:=(g_{1},\dotsc,g_{n})$ for $n\geq 1$,
then $G_{n}\sim\mathcal{N}_{\beta}^{\otimes(m\times n)}$, and by Proposition
5.5 we have
$X_{n}^{*}\stackrel{{\scriptstyle\text{d}}}{{=}}G_{n}^{*}(G_{n}G_{n}^{*})^{-1/2}$.
It follows that

$X_{n}^{(k)}\stackrel{{\scriptstyle\text{d}}}{{=}}(G_{n}G_{n}^{*})^{-1/2}G_{k}.$

By the strong law of large numbers,

$\frac{1}{n}\,G_{n}G_{n}^{*}=\frac{1}{n}\sum_{i=1}^{n}g_{i}g_{i}^{*}\xrightarrow[n\to\infty]{\text{a.s.}}\operatorname{\mathbb{E}}[g_{1}g_{1}^{*}]=I_{m},$
(12)

and therefore,

$n^{1/2}X_{n}^{(k)}\stackrel{{\scriptstyle\text{d}}}{{=}}\Bigl{(}\frac{1}{n}\,G_{n}G_{n}^{*}\Bigr{)}^{-1/2}G_{k}\xrightarrow[n\to\infty]{\text{a.s.}}G_{k},$

which proves the claim.

We proceed with the case
$X_{n}\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{\infty},\beta}^{m\times
n})$, then by Proposition 4.1

$X_{n}\stackrel{{\scriptstyle\text{d}}}{{=}}R_{n}^{1/2}\,U_{n}^{*}$

with independent random variables $R_{n}\sim
B_{m,\beta}^{\text{I}}(\frac{\beta n}{2},\frac{\beta(m-1)}{2}+1)$ and
$U_{n}\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n,m;\beta})$. The first case
implies

$(n^{1/2}{U_{n}^{*}}^{(k)})_{n\geq\max\\{m,k\\}}\xrightarrow[]{\text{d}}G_{k}.$

Using the probabilistic representation of $B_{m,\beta}^{\text{I}}(\frac{\beta
n}{2},\frac{\beta(m-1)}{2}+1)$ from Proposition 4.5 we can write

$R_{n}\stackrel{{\scriptstyle\text{d}}}{{=}}(G_{n}G_{n}^{*}+H)^{-1/2}G_{n}G_{n}^{*}(G_{n}G_{n}^{*}+H)^{-1/2},$

where $H\sim\Gamma_{m,\beta}(\frac{\beta(m-1)}{2}+1,\frac{2}{\beta}I_{m})$ is
independent from $(g_{j})_{j\in\mathbb{N}}$. Since $(n^{-1}H)_{n\geq
1}\xrightarrow[]{\text{a.s.}}0$, together with Equation (12) we infer

$(R_{n})_{n\geq m}\xrightarrow[]{\mathbb{P}}I_{m},$

and therewith we conclude

$n^{1/2}X_{n}^{(k)}\stackrel{{\scriptstyle\text{d}}}{{=}}R_{n}^{1/2}\,n^{1/2}{U_{n}^{*}}^{(k)}\xrightarrow[n\to\infty]{\text{d}}G_{k},$

as desired.

Lastly we consider the case
$X_{n}\sim\operatorname{\mathcal{U}}(\mathbb{S}_{S_{\infty},\beta}^{m,n})=\kappa_{S_{\infty},\beta}^{m\times
n}$. Let
$Y_{n}\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{\infty},\beta}^{m\times
n})$, then by Proposition 5.4,

$X_{n}\stackrel{{\scriptstyle\text{d}}}{{=}}\frac{Y_{n}}{\lVert
Y_{n}\rVert_{S_{\infty}}}$

and by the same proposition and by Remark 5.2, part 1, the distribution of
$\lVert Y_{n}\rVert_{S_{\infty}}$ has density $r\mapsto\beta mnr^{\beta
mn-1}\operatorname{1}_{[0,1]}(r)$. The latter fact implies $\lVert
Y_{n}\rVert_{S_{\infty}}^{\beta mn}\sim\operatorname{\mathcal{U}}([0,1])$ and
therefore

$(\lVert Y_{n}\rVert_{S_{\infty}})_{n\geq m}\xrightarrow[]{\text{a.s.}}1.$

(First use $\lVert
Y_{n}\rVert_{S_{\infty}}\stackrel{{\scriptstyle\text{d}}}{{=}}U^{1/(\beta
mn)}$ with $U\sim\operatorname{\mathcal{U}}([0,1])$ to get convergence in
distribution; because the limit is constant, promote to convergence in
probability; use Borel-Cantelli to obtain almost sure convergence.) This
finally leads to

$n^{1/2}X_{n}^{(k)}\stackrel{{\scriptstyle\text{d}}}{{=}}\frac{n^{1/2}Y_{n}^{(k)}}{\lVert
Y_{n}\rVert_{S_{\infty}}}\xrightarrow[n\to\infty]{\text{d}}G_{k},$

and the proof is complete. ∎

######  Remark 4.6.

Essentially the same result, but using a very different method, for the case
of $\operatorname{\mathcal{U}}(S_{\beta}^{m,n})$ has been obtained recently by
Petrov and Vershik [56, Theorem 3.1].

The next proposition has a nice geometric interpretation: pick two points
independently from either $\mathbb{B}_{S_{\infty},\beta}^{m\times n}$,
$\mathbb{S}_{S_{\infty},\beta}^{m,n}$, or $S_{\beta}^{m,n}$, then they will be
nearly orthogonal, when the dimension is high. This will be formulated as a
distributional limit theorem.

######  Theorem E.

Let $(X_{n})_{n\geq m}$ and $(Y_{n})_{n\geq m}$ be sequences of random
variables such that individually their distributions satisfy the same
conditions as in Theorem D, and additionally let $\\{X_{n},Y_{n}\\}$ be
independent for each $n\geq m$. Then

$(n^{1/2}X_{n}Y_{n}^{*})_{n\geq
m}\xrightarrow[]{\text{d}}\mathcal{N}_{\beta}^{\otimes(m\times m)},$

and

$((\beta m^{-1}\,n)^{1/2}\langle X_{n},Y_{n}\rangle)_{n\geq
m}\xrightarrow[]{\text{d}}\mathcal{N}_{1}.$

###### Proof.

We use the same notation as in the proof of Theorem D. Note that the second
convergence is an almost immediate consequence of the first one: apply the
real part of the trace to get

$\frac{(\beta n)^{1/2}}{m^{1/2}}\langle
X_{n},Y_{n}\rangle=\frac{\beta^{1/2}}{m^{1/2}}\Re\operatorname{tr}(n^{1/2}X_{n}Y_{n}^{*})\xrightarrow[n\to\infty]{\text{d}}\frac{\beta^{1/2}}{m^{1/2}}\Re\operatorname{tr}(G_{m})=\frac{\beta^{1/2}}{m^{1/2}}\sum_{i=1}^{m}\Re(g_{i,i})\sim\mathcal{N}_{1},$

since the $\Re(g_{i,i})$ are i.i.d. Gaussian with zero mean and variance
$\frac{1}{\beta}$.

Concerning the first statement, it suffices to consider the case
$X_{n},Y_{n}\sim\operatorname{\mathcal{U}}(S_{\beta}^{m,n})$; then use the
same arguments as in the proof of Theorem D to get rid of any ‘radial’
components of the random variables. Write

$X_{n}\stackrel{{\scriptstyle\text{d}}}{{=}}(G_{n}G_{n}^{*})^{-1/2}G_{n}\quad\text{and}\quad
Y_{n}\stackrel{{\scriptstyle\text{d}}}{{=}}(G_{n}^{\prime}(G_{n}^{\prime})^{*})^{-1/2}G_{n}^{\prime},$

where $G_{n}^{\prime}=(g_{1}^{\prime},\dotsc,g_{n}^{\prime})$ is an
independent copy of $G_{n}$, to obtain

$\displaystyle n^{1/2}X_{n}Y_{n}^{*}$
$\displaystyle\stackrel{{\scriptstyle\text{d}}}{{=}}n^{1/2}(G_{n}G_{n}^{*})^{-1/2}G_{n}(G_{n}^{\prime})^{*}(G_{n}^{\prime}(G_{n}^{\prime})^{*})^{-1/2}$
$\displaystyle=\Bigl{(}\frac{1}{n}G_{n}G_{n}^{*}\Bigr{)}^{-1/2}\Bigl{(}\frac{1}{\sqrt{n}}G_{n}(G_{n}^{\prime})^{*}\Bigr{)}\Bigl{(}\frac{1}{n}G_{n}^{\prime}(G_{n}^{\prime})^{*}\Bigr{)}^{-1/2}.$

Because of Equation (12) the square root terms are accounted for and they do
not contribute to the limit. The middle term can be written out,

$G_{n}(G_{n}^{\prime})^{*}=\sum_{j=1}^{n}g_{j}(g_{j}^{\prime})^{*},$

and this is a sum of i.i.d. random variables; the summands are centred because
of

$\operatorname{\mathbb{E}}[g_{1}g_{1}^{\prime*}]=\operatorname{\mathbb{E}}[g_{1}]\operatorname{\mathbb{E}}[g_{j}^{\prime}]^{*}=oo^{\mathsf{T}}=0,$

their covariance matrix equals

$\displaystyle\operatorname{Cov}[\operatorname{vec}(g_{1}(g_{1}^{\prime})^{*})]=\operatorname{\mathbb{E}}[\operatorname{vec}(g_{1}(g_{1}^{\prime})^{*})\operatorname{vec}(g_{1}(g_{1}^{\prime})^{*})^{*}]=\overline{\operatorname{Cov}[g_{1}^{\prime}]}\otimes\operatorname{Cov}[g_{1}]=I_{m}\otimes
I_{m}=I_{m^{2}},$

and in the complex case their pseudo/covariance matrix (also called relation
matrix) equals

$\operatorname{Rel}[\operatorname{vec}(g_{1}(g_{1}^{\prime})^{*})]=\operatorname{\mathbb{E}}[\operatorname{vec}(g_{1}(g_{1}^{\prime})^{*})\operatorname{vec}(g_{1}(g_{1}^{\prime})^{*})^{\mathsf{T}}]=\overline{\operatorname{Rel}[g_{1}^{\prime}]}\otimes\operatorname{Rel}[g_{1}]=0\otimes
0=0.$

Therefore by the multivariate central limit theorem

$\frac{1}{\sqrt{n}}G_{n}(G_{n}^{\prime})^{*}\xrightarrow[n\to\infty]{\text{d}}G_{m}.$

This finally yields

$n^{1/2}X_{n}Y_{n}^{*}\xrightarrow[n\to\infty]{\text{d}}G_{m}.\qed$

##  5 Theory of $K$/symmetric measures

Referring back to Remark 4.2, we see that if
$X\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{\infty},\beta}^{m\times n})$,
then
$X^{*}(XX^{*})^{-1/2}\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n,m;\beta})$.
But it is also well known that if $X\sim\mathcal{N}_{1}^{\otimes(m\times n)}$,
then
$X^{\mathsf{T}}(XX^{\mathsf{T}})^{-1/2}\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n,m;1})$
too (see [9, Theorem 2.2. and top of p. 270], e.g.). This can be generalized.
But first we need to introduce a notion of symmetry of a measure.

######  Definition 5.1.

Let $d\in\mathbb{N}$ and let $K\subset\mathbb{R}^{d}$ be a star body. A
Borel/measure $\mu$ on $\mathbb{R}^{d}$ is called _$K$ /̄symmetric_ iff there
exists a Borel/measure $\rho$ on $\mathbb{R}_{\geq 0}$ such that for any
measurable, nonnegative map $f\colon\mathbb{R}^{d}\to\mathbb{R}$,

$\int_{\mathbb{R}^{d}}f(x)\,\mathrm{d}\mu(x)=\int_{\mathbb{R}_{\geq
0}}\int_{\partial
K}f(r\theta)\,\mathrm{d}\kappa_{K}(\theta)\,\mathrm{d}\rho(r),$

where recall that $\kappa_{K}$ denotes the normalized cone measure on
$\partial K$.

######  Remark 5.2.

  1. 1.

By the polar integration formula (9) the uniform distribution on $K$ is
$K$/̄symmetric with
$\frac{\mathrm{d}\rho}{\mathrm{d}r}(r)=dr^{d-1}\operatorname{1}_{[0,1]}(r)$;
and $\kappa_{K}$ itself is $K$/̄symmetric with $\rho=\delta_{1}$.

  2. 2.

Given a $K$/̄symmetric measure $\mu$, the ‘radial measure’ $\rho$ as
introduced above is unique. (This can be proved by inserting
$f=\operatorname{1}_{A}$ with $A=\\{r\theta\,:\,r\in A_{1},\theta\in\partial
K\\}$ for some Borel/set $A_{1}\subset\mathbb{R}_{\geq 0}$.)

That Definition 5.1 captures the symmetry of $K$ in a natural way is argued by
the following statement which tells us that the density of a $K$/̄symmetric
measure is constant on the dilates of $\partial K$.

######  Proposition 5.3.

Let $d$, $K$ be as in Definition 5.1, and let $\mu$ be absolutely continuous
w.r.t. $\operatorname{\mathit{v}}_{d}$. Then $\mu$ is $K$/̄symmetric iff its
$\operatorname{\mathit{v}}_{d}$/̄density is of the form
$f\circ\lvert\cdot\rvert_{K}$ with some measurable, nonnegative
$f\colon\mathbb{R}_{\geq 0}\to\mathbb{R}$; and then
$\frac{\mathrm{d}\rho(r)}{\mathrm{d}r}=d\operatorname{\mathit{v}}_{d}(K)f(r)r^{d-1}$.

###### Proof.

$\Rightarrow$: Call
$F:=\frac{\mathrm{d}\mu}{\mathrm{d}\\!\operatorname{\mathit{v}}_{d}}$. Let
$h\colon\mathbb{R}^{d}\to\mathbb{R}$ be measurable and nonnegative. Then, with
a suitable measure $\rho$,

$\displaystyle\int_{\mathbb{R}_{\geq 0}}\int_{\partial
K}h(r\theta)\,\mathrm{d}\kappa_{K}(\theta)\,\mathrm{d}\rho(r)$
$\displaystyle=\int_{\mathbb{R}^{d}}h(x)\,\mathrm{d}\mu(x)=\int_{\mathbb{R}^{d}}h(x)F(x)\,\mathrm{d}x$
(13) $\displaystyle=d\operatorname{\mathit{v}}_{d}(K)\int_{\mathbb{R}_{\geq
0}}\int_{\partial
K}h(r\theta)F(r\theta)\,\mathrm{d}\kappa_{K}(\theta)\,r^{d-1}\,\mathrm{d}r.$

Choose $h=h_{1}\circ\lvert\cdot\rvert_{K}$ with some measurable, nonnegative
$h_{1}\colon\mathbb{R}_{\geq 0}\to\mathbb{R}$, then $h(r\theta)=h_{1}(r)$ for
any $r>0$ and $\theta\in\partial K$, and therewith

$\int_{\mathbb{R}_{\geq
0}}h_{1}(r)\,\mathrm{d}\rho(r)=d\operatorname{\mathit{v}}_{d}(K)\int_{\mathbb{R}_{\geq
0}}h_{1}(r)\int_{\partial
K}F(r\theta)\,\mathrm{d}\kappa_{K}(\theta)\,r^{d-1}\,\mathrm{d}r.$ (14)

Define $f(r):=\int_{\partial K}F(r\theta)\,\mathrm{d}\kappa_{K}(\theta)$ for
$r\geq 0$, then Equation 14 tells us
$\frac{\mathrm{d}\rho}{\mathrm{d}r}(r)=d\operatorname{\mathit{v}}_{d}(K)f(r)r^{d-1}$,
and going back to Equation 13, again with $h$ arbitrary, we have

$\displaystyle\int_{\mathbb{R}^{d}}h(x)F(x)\,\mathrm{d}x$
$\displaystyle=\int_{\mathbb{R}_{\geq 0}}\int_{\partial
K}h(r\theta)\,\mathrm{d}\kappa_{K}(\theta)\,d\operatorname{\mathit{v}}_{d}(K)f(r)r^{d-1}\,\mathrm{d}r$
$\displaystyle=d\operatorname{\mathit{v}}_{d}(K)\int_{\mathbb{R}_{\geq
0}}\int_{\partial K}h(r\theta)f(\lvert
r\theta\rvert_{K})\,\mathrm{d}\kappa_{K}(\theta)\,r^{d-1}\,\mathrm{d}r$
$\displaystyle=\int_{\mathbb{R}^{d}}h(x)f(\lvert x\rvert_{K})\,\mathrm{d}x,$

and from this we obtain $F=f\circ\lvert\cdot\rvert_{K}$, as claimed.

$\Leftarrow$: Let $h\colon\mathbb{R}^{d}\to\mathbb{R}$ be measurable and
nonnegative. Then,

$\displaystyle\int_{\mathbb{R}^{d}}h(x)\,\mathrm{d}\mu(x)$
$\displaystyle=\int_{\mathbb{R}^{d}}h(x)f(\lvert x\rvert_{K})\,\mathrm{d}x$
$\displaystyle=d\operatorname{\mathit{v}}_{d}(K)\int_{\mathbb{R}_{\geq
0}}\int_{\partial K}h(r\theta)f(\lvert
r\theta\rvert_{K})r^{d-1}\,\mathrm{d}\kappa_{K}(\theta)\,\mathrm{d}r$
$\displaystyle=\int_{\mathbb{R}_{\geq 0}}\int_{\partial
K}h(r\theta)\,\mathrm{d}\kappa_{K}(\theta)\,d\operatorname{\mathit{v}}_{d}(K)f(r)r^{d-1}\,\mathrm{d}r,$

hence the claim follows with
$\frac{\mathrm{d}\rho}{\mathrm{d}r}(r):=d\operatorname{\mathit{v}}_{d}(K)f(r)r^{d-1}$.
∎

######  Proposition 5.4.

Let $d$, $K$, $\mu$, $\rho$ be as in Definition 5.1, with $\mu$ a probabiliy
measure, and let $X$ be a $\mathbb{R}^{d}$/̄valued random variable. Then
$X\sim\mu$ iff there exist independent random variables $R$ and $\Theta$ such
that $R\sim\rho$, $\Theta\sim\kappa_{K}$, and
$X\stackrel{{\scriptstyle\text{d}}}{{=}}R\Theta$. In that case $\lvert
X\rvert_{K}\stackrel{{\scriptstyle\text{d}}}{{=}}R$.

If in addition $\mu\\{o\\}=0$, then $\frac{X}{\lvert
X\rvert_{K}}\stackrel{{\scriptstyle\text{d}}}{{=}}\Theta$, and $\lvert
X\rvert_{K}$ and $\frac{X}{\lvert X\rvert_{K}}$ are stochastically
independent.

###### Proof.

The representation $X\stackrel{{\scriptstyle\text{d}}}{{=}}R\Theta$ together
with the properties of $R$ and $\Theta$ are a mere rewriting of Definition
5.1. Because of $\Theta\sim\kappa_{K}$ we have $\Theta\in\partial K$ almost
surely, hence $\lvert\Theta\rvert_{K}=1$ almost surely, also $R\geq 0$ almost
surely, and thus $\lvert
X\rvert_{K}\stackrel{{\scriptstyle\text{d}}}{{=}}\lvert
R\Theta\rvert_{K}=R\lvert\Theta\rvert_{K}=R$.

Note $\operatorname{\mathbb{P}}[X=o]=\mu\\{o\\}$ and
$X=o\Longleftrightarrow\lvert X\rvert_{K}=0$, hence if $\mu\\{o\\}=0$, then
$\frac{X}{\lvert X\rvert_{K}}$ is almost surely defined. In the latter case,
let $f\colon\mathbb{R}_{\geq 0}\to\mathbb{R}$ and $g\colon\partial
K\to\mathbb{R}$ be measurable and nonnegative, then,

$\displaystyle\operatorname{\mathbb{E}}\Bigl{[}f(\lvert
X\rvert_{K})g\Bigl{(}\frac{X}{\lvert X\rvert_{K}}\Bigr{)}\Bigr{]}$
$\displaystyle=\int_{\mathbb{R}_{\geq 0}}\int_{\partial K}f(\lvert
r\theta\rvert_{K})g\Bigl{(}\frac{r\theta}{\lvert
r\theta\rvert_{K}}\Bigr{)}\,\mathrm{d}\kappa_{K}(\theta)\,\mathrm{d}\rho(r)$
$\displaystyle=\int_{\mathbb{R}_{\geq 0}}\int_{\partial
K}f(r)g(\theta)\,\mathrm{d}\kappa_{K}(\theta)\,\mathrm{d}\rho(r)$
$\displaystyle=\int_{\mathbb{R}_{\geq 0}}f(r)\,\mathrm{d}\rho(r)\int_{\partial
K}g(\theta)\,\mathrm{d}\kappa_{K}(\theta).$

From that follow the distribution of $\frac{X}{\lvert X\rvert_{K}}$ and the
independence of $\lvert X\rvert_{K}$ and $\frac{X}{\lvert X\rvert_{K}}$. ∎

Thus equipped we turn to probability distributions on $\mathbb{K}^{m\times
n}$.

######  Proposition 5.5.

Let $K\subset\mathbb{K}^{m\times n}$ be a star body and let $X$ be an
$\mathbb{K}^{m\times n}$/̄valued random variable whose distribution is
$K$/̄symmetric with radial distribution $\rho$.

  1. 1.

$XX^{*}$ is almost surely invertible, iff $X\neq 0$ almost surely, iff
$\rho\\{0\\}=0$.

  2. 2.

If $\rho\\{0\\}=0$ and additionally $K$ is right/unitarily invariant (that is
$Ku=K$ for all $u\in\mathrm{U}_{n;\beta}$), then $XX^{*}$ and
$X^{*}(XX^{*})^{-1/2}$ are stochastically independent, and
$X^{*}(XX^{*})^{-1/2}\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n,m;\beta})$.

  3. 3.

If $K$ is right/unitarily invariant and the distribution of $X$ has
$\operatorname{\mathit{v}}_{\beta mn}$/̄density $h\circ\lvert\cdot\rvert_{K}$,
then the distribution of $XX^{*}$ has density proportional to $r\mapsto
h(\lvert r^{1/2}\rvert_{\lvert K\rvert})\det(r)^{\beta(n-m+1)/2-1}$, where
$\lvert K\rvert:=\\{(xx^{*})^{1/2}\,:\,x\in K\\}\subset\text{Sym}_{m,\beta}$.

###### Proof.

  1. 1.

Write $X\stackrel{{\scriptstyle\text{d}}}{{=}}R\Theta$ with $R\sim\rho$ and
$\Theta\sim\kappa_{K}$ independent, according to Proposition 5.4; then $X=0$
iff $\lvert X\rvert_{K}\stackrel{{\scriptstyle\text{d}}}{{=}}R=0$, hence
$\operatorname{\mathbb{P}}[X=0]=\operatorname{\mathbb{P}}[R=0]=\rho\\{0\\}$.

Next note that $XX^{*}$ is invertible iff $X$ has full rank, iff $s_{m}(X)>0$;
using the representation of $X$ his yields

$\begin{split}\operatorname{\mathbb{P}}[s_{m}(X)>0]&=\operatorname{\mathbb{P}}[s_{m}(R\Theta)>0]=\operatorname{\mathbb{P}}[Rs_{m}(\Theta)>0]\\\
&=\operatorname{\mathbb{P}}[R>0\wedge
s_{m}(\Theta)>0]=\operatorname{\mathbb{P}}[R>0]\operatorname{\mathbb{P}}[s_{m}(\Theta)>0].\end{split}$

Trivially $\operatorname{\mathbb{P}}[R>0]=1-\rho\\{0\\}$; take
$U\sim\operatorname{\mathcal{U}}([0,1])$ independent of $\Theta$, then
$U^{1/(\beta mn)}\Theta\sim\operatorname{\mathcal{U}}(K)$ (this follows from
Remark 5.2, part 1) and $\operatorname{\mathbb{P}}[U^{1/(\beta mn)}>0]=1$, and
by doing the same calculation as before,

$1=\frac{\operatorname{\mathit{v}}_{\beta mn}(\\{x\in
K\,:\,s_{m}(x)>0\\})}{v_{\beta
mn}(K)}=\operatorname{\mathbb{P}}[s_{m}(U^{1/(\beta
mn)}\Theta)>0]=\operatorname{\mathbb{P}}[s_{m}(\Theta)>0].$

This establishes $\operatorname{\mathbb{P}}[s_{m}(X)>0]=1-\rho\\{0\\}$, thus
$XX^{*}$ is almost surely invertible iff $\rho\\{0\\}=0$.

  2. 2.

Let $f\colon\\!\mathrm{U}_{n,m;\beta}\to\mathbb{R}$ and
$g\colon\\!\text{Pos}_{m,\beta}\to\mathbb{R}$ be measurable and nonnegative,
then

$\displaystyle\operatorname{\mathbb{E}}[f(X^{*}(XX^{*})^{-1/2})g(XX^{*})]$
$\displaystyle=\int_{\mathbb{R}_{\geq 0}}\int_{\partial
K}f(r\theta^{*}(r^{2}\theta\theta^{*})^{-1/2})g(r^{2}\theta\theta^{*})\,\mathrm{d}\kappa_{K}(\theta)\,\mathrm{d}\rho(r)$
$\displaystyle=\int_{\partial
K}f(\theta^{*}(\theta\theta^{*})^{-1/2})\int_{\mathbb{R}_{\geq
0}}g(r^{2}\theta\theta^{*})\,\mathrm{d}\rho(r)\,\mathrm{d}\kappa_{K}(\theta)$
$\displaystyle=\frac{1}{\operatorname{\mathit{v}}_{\beta
mn}(K)}\int_{K}f(x^{*}(xx^{*})^{-1/2})\int_{\mathbb{R}_{\geq
0}}g\Bigl{(}\frac{r^{2}xx^{*}}{\lvert
x\rvert_{K}^{2}}\Bigr{)}\mathrm{d}\rho(r)\,\mathrm{d}x$
$\displaystyle=\frac{2^{-m}}{\operatorname{\mathit{v}}_{\beta
mn}(K)}\int_{\text{Pos}_{m,\beta}}\int_{\mathrm{U}_{n,m;\beta}}\operatorname{1}_{K}(t^{1/2}u^{*})f(ut^{1/2}(t^{1/2}u^{*}ut^{1/2})^{-1/2})$
$\displaystyle\quad\cdot\int_{\mathbb{R}_{\geq
0}}g\Bigl{(}\frac{r^{2}t^{1/2}u^{*}ut^{1/2}}{\lvert
t^{1/2}u^{*}\rvert_{K}^{2}}\Bigr{)}\mathrm{d}\rho(r)\,\mathrm{d}u\det(t)^{\beta(n-m+1)/2-1}\,\mathrm{d}t.$

We have $t^{1/2}u^{*}\in K$ iff $t^{1/2}\in\lvert K\rvert$; indeed, concerning
“$\Rightarrow$” let $t^{1/2}u^{*}\in K$, then
$(t^{1/2}u^{*}ut^{1/2})^{1/2}=t^{1/2}\in\lvert K\rvert$. Concerning
“$\Leftarrow$” let $t^{1/2}\in\lvert K\rvert$, then there is some $x\in K$
such that $t^{1/2}=(xx^{*})^{1/2}$; let $x=\lvert x\rvert v^{*}$ be its polar
decomposition, then $t^{1/2}=\lvert x\rvert$; and as $K$ is right/unitarily
invariant we get $x\hat{v}\hat{u}^{*}\in K$, and $x\hat{v}\hat{u}^{*}=\lvert
x\rvert v^{*}\hat{v}\hat{u}^{*}=\lvert x\rvert u^{*}=t^{1/2}u^{*}$. This also
implies $\operatorname{1}_{K}(t^{1/2}u^{*})=\operatorname{1}_{\lvert
K\rvert}(t^{1/2})$. Hence we may continue,

$\displaystyle\operatorname{\mathbb{E}}[f(X^{*}(XX^{*})^{-1/2})g(XX^{*})]$
$\displaystyle=\frac{2^{-m}}{\operatorname{\mathit{v}}_{\beta
mn}(K)}\int_{\text{Pos}_{m,\beta}}\int_{\mathrm{U}_{n,m;\beta}}\operatorname{1}_{\lvert
K\rvert}(t^{1/2})f(u)$ $\displaystyle\quad\cdot\int_{\mathbb{R}_{\geq
0}}g\Bigl{(}\frac{r^{2}t}{\lvert
t^{1/2}\rvert_{K}^{2}}\Bigr{)}\mathrm{d}\rho(r)\,\mathrm{d}u\det(t)^{\beta(n-m+1)/2-1}\,\mathrm{d}t$
$\displaystyle=\frac{2^{-m}}{\operatorname{\mathit{v}}_{\beta
mn}(K)}\int_{\mathrm{U}_{n,m;\beta}}f(u)\,\mathrm{d}u$
$\displaystyle\quad\cdot\int_{\text{Pos}_{m,\beta}}\operatorname{1}_{\lvert
K\rvert}(t^{1/2})\int_{\mathbb{R}_{\geq 0}}g\Bigl{(}\frac{r^{2}t}{\lvert
t^{1/2}\rvert_{K}^{2}}\Bigr{)}\,\mathrm{d}\rho(r)\det(t)^{\beta(n-m+1)/2-1}\,\mathrm{d}t.$

This reveals the claimed independence of $X^{*}(XX^{*})^{-1/2}$ and $XX^{*}$,
and the distribution of $X^{*}(XX^{*})^{-1/2}$.

  3. 3.

Let the distribution of $X$ have $\operatorname{\mathit{v}}_{\beta
mn}$/̄density $h\circ\lvert\cdot\rvert_{K}$ according to Proposition 5.3 and
let $g$ be as before, then,

$\displaystyle\operatorname{\mathbb{E}}[g(XX^{*})]$
$\displaystyle=\int_{\mathbb{K}^{m\times n}}g(xx^{*})h(\lvert
x\rvert_{K})\,\mathrm{d}x$
$\displaystyle=2^{-m}\int_{\text{Pos}_{m,\beta}}\int_{\mathrm{U}_{n,m;\beta}}g(r^{1/2}u^{*}ur^{1/2})h(\lvert
r^{1/2}u^{*}\rvert_{K})\,\mathrm{d}u\det(r)^{\beta(n-m+1)/2-1}\,\mathrm{d}r$
$\displaystyle=2^{-m}\,\frac{\Omega_{n,m;\beta}}{2^{\beta
m(m-1)/4}}\int_{\text{Pos}_{m,\beta}}g(r)h(\lvert r^{1/2}\rvert_{\lvert
K\rvert})\det(r)^{\beta(n-m+1)/2-1}\,\mathrm{d}r.$

This last line yields the claimed density of the distribution of $XX^{*}$. ∎

######  Remark 5.6.

  1. 1.

Because $\mathbb{B}_{S_{p},\beta}^{m\times n}$ is both left- and
right/unitarily invariant, all of Proposition 5.5 is applicable to it.

  2. 2.

The classical case $X\sim\mathcal{N}_{\beta}^{\otimes(m\times n)}$ is
recovered by noticing its density $(\frac{\beta}{2\pi})^{\beta
mn/2}\mathsf{e}^{-\beta\operatorname{tr}(xx^{*})/2}=(\frac{\beta}{2\pi})^{\beta
mn/2}\mathsf{e}^{-\beta\lVert x\rVert_{S_{2}}^{2}/2}$, that is,
$\mathcal{N}_{\beta}^{\otimes(m\times n)}$ is
$\mathbb{B}_{S_{2},\beta}^{m\times n}$/̄symmetric. As an aside, $XX^{*}$ is
Wishart/distributed, which fact can also be inferred from part 3 given above.

  3. 3.

Part 3 of Proposition 5.5 is always satisfied for
$X\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{p},\beta}^{m\times n})$
(density $(\omega_{S_{p},\beta}^{m\times
n})^{-1}\operatorname{1}_{[0,1]}\circ\lVert\cdot\rVert_{S_{p}}$).

Proposition 5.5 partly overlaps with [18, Theorem 8.2.5], as follows from the
next statement. But in Proposition 5.5 $X$ need not have a
$\operatorname{\mathit{v}}_{\beta mn}$/̄density in order to ensure almost sure
invertibility of $XX^{*}$ and to achieve
$X^{*}(XX^{*})^{-1/2}\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n,m;\beta})$.
(Of course, left or right unitary invariance is not sufficient for
invertibility of $XX^{*}$: just take
$X:=V\operatorname{diag}(1,\dotsc,1,0)U^{*}$ with
$U\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n;\beta})$ and
$V\sim\operatorname{\mathcal{U}}(\mathrm{U}_{m;\beta})$, for instance.)

In the sequel $\mathfrak{S}_{m}$ denotes the symmetric group of degree
$m\in\mathbb{N}$.

######  Proposition 5.7.

A $\mathbb{K}^{m\times n}$/̄valued random variable $X$ has a two/sided
unitarily invariant distribution, that is
$X\stackrel{{\scriptstyle\text{d}}}{{=}}vXu$ for all
$u\in\mathrm{U}_{n;\beta}$ and $v\in\mathrm{U}_{m;\beta}$, if and only if
there exist independent random variables $S$, $U$, and $V$ such that $S$ takes
values in $\mathbb{R}^{m}$,
$U\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n;\beta})$,
$V\sim\operatorname{\mathcal{U}}(\mathrm{U}_{m;\beta})$, and
$X\stackrel{{\scriptstyle\text{d}}}{{=}}V\operatorname{diag}(S)U^{*}$.

Moreover, if $X$ has a two/sided unitarily invariant distribution, then $S$
can be constructed such that $\operatorname{\mathbb{P}}[S\in\mathbb{R}_{\geq
0}^{m}]=1$, it has exchangeable coordinates and then $(s_{\pi(i)}(X))_{i\leq
m}\stackrel{{\scriptstyle\text{d}}}{{=}}S$ with
$\pi\sim\operatorname{\mathcal{U}}(\mathfrak{S}_{m})$ independent of $X$, or
such that $\operatorname{\mathbb{P}}[S\in W^{m}]=1$ (recall
$W^{m}=\\{(x_{i})_{i\leq m}\in\mathbb{R}^{m}\,:\,0\leq x_{m}\leq\dotsb\leq
x_{1}\\}$) and then $s(X)\stackrel{{\scriptstyle\text{d}}}{{=}}S$.

###### Proof.

$\Leftarrow$: This is trivial because
$\operatorname{\mathcal{U}}(\mathrm{U}_{d;\beta})$ is a Haar measure and thus
$wW\stackrel{{\scriptstyle\text{d}}}{{=}}W$ for
$W\sim\operatorname{\mathcal{U}}(\mathrm{U}_{d;\beta})$ and any
$w\in\mathrm{U}_{d;\beta}$.

$\Rightarrow$: Call $\mu:=\operatorname{\mathbb{P}}\circ X^{-1}$; then we
have, for any $f\colon\mathbb{K}^{m\times n}\to\mathbb{R}$ measurable and
nonneagtive and any $u\in\mathrm{U}_{n;\beta}$ and $v\in\mathrm{U}_{m;\beta}$,

$\operatorname{\mathbb{E}}[f(X)]=\operatorname{\mathbb{E}}[f(vXu)]=\int_{\mathbb{K}^{m\times
n}}f(vxu)\,\mathrm{d}\mu(x).$

Integrate on both sides over the unitary groups to obtain

$\displaystyle\frac{\Omega_{m;\beta}\Omega_{n;\beta}}{2^{\beta m(m-1)/4+\beta
n(n-1)/4}}\operatorname{\mathbb{E}}[f(X)]$
$\displaystyle=\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}\int_{\mathbb{K}^{m\times
n}}f(vxu)\,\mathrm{d}\mu(x)\mathrm{d}u\,\mathrm{d}v$
$\displaystyle=\int_{\mathbb{K}^{m\times
n}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(vxu)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}\mu(x).$

For any $x\in\mathbb{K}^{m\times n}$ choose a singular value decomposition
$x=v_{x}\operatorname{diag}(s(x))u_{x}^{*}$ and get

$\displaystyle\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(vxu)\,\mathrm{d}u\,\mathrm{d}v$
$\displaystyle=\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(vv_{x}\operatorname{diag}(s(x))u_{x}^{*}u)\,\mathrm{d}u\,\mathrm{d}v$
$\displaystyle=\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(s(x))u^{*})\,\mathrm{d}u\,\mathrm{d}v,$

where we have used unitary invariance of the respective Haar measures and also
its invariance under adjoining. This leads to (abbreviating the normalization
constant by $c$)

$\displaystyle\operatorname{\mathbb{E}}[f(X)]$
$\displaystyle=c\int_{\mathbb{K}^{m\times
n}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(s(x))u^{*})\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}\mu(x)$
$\displaystyle=c\int_{W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(\sigma)u^{*})\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}(\mu\circ
s^{-1})(\sigma),$

and this yields $X=V\operatorname{diag}(S)U^{*}$ with independent $S$, $U$,
and $V$ such that $U\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n;\beta})$,
$V\sim\operatorname{\mathcal{U}}(\mathrm{U}_{m;\beta})$, and $S\sim\mu\circ
s^{-1}=\operatorname{\mathbb{P}}{}\circ s(X)^{-1}$, so
$s(X)\stackrel{{\scriptstyle\text{d}}}{{=}}S$.

Now let $P\sim\operatorname{\mathcal{U}}(\mathfrak{S}_{m})$ be independent of
all other random variables and set $T:=(S_{P(i)})_{i\leq m}$, then
$T\in\mathbb{R}_{\geq 0}^{m}$ almost surely, $T$ has exchangeable coordinates,
and $T\stackrel{{\scriptstyle\text{d}}}{{=}}(s_{P(i)}(X))_{i\leq m}$, and we
also have

$\displaystyle\operatorname{\mathbb{E}}[f(VTU^{*})]$
$\displaystyle=c\,\frac{1}{m!}\sum_{\pi\in\mathfrak{S}_{m}}\int_{W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}((\sigma_{\pi(i)})_{i\leq
m})u^{*})\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}(\mu\circ s^{-1})(\sigma)$
$\displaystyle=c\,\frac{1}{m!}\sum_{\pi\in\mathfrak{S}_{m}}\int_{W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(\sigma)u^{*})\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}(\mu\circ
s^{-1})(\sigma)$
$\displaystyle=c\int_{W^{m}}\int_{\mathrm{U}_{m;\beta}}\int_{\mathrm{U}_{n;\beta}}f(v\operatorname{diag}(\sigma)u^{*})\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}(\mu\circ
s^{-1})(\sigma)$ $\displaystyle=\operatorname{\mathbb{E}}[f(X)],$

where from the first to second line we have used
$\operatorname{diag}((\sigma_{\pi(i)})_{i\leq
m})=\tau_{\pi}\operatorname{diag}(\sigma)\operatorname{diag}(\tau_{\pi},I_{n-m})^{\mathsf{T}}$
with $\tau_{\pi}=(\delta_{\pi(i),j})_{(i,j)\in n\times n}$ and unitary
invariance, as in Remark 2.5. This proves
$X\stackrel{{\scriptstyle\text{d}}}{{=}}VTU^{*}$. ∎

######  Remark 5.8.

Proposition 5.7 tells us that if the distribution of $X$ is two/sided
unitarily invariant, then it is alredy completely determined by the
distribution of the singular values of $X$, in the sense that
$X\stackrel{{\scriptstyle\text{d}}}{{=}}V\operatorname{diag}(s(X))U^{*}$ with
independent $U\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n;\beta})$ and
$V\sim\operatorname{\mathcal{U}}(\mathrm{U}_{m;\beta})$.

######  Proposition 5.9.

Any $\mathbb{B}_{S_{p},\beta}^{m\times n}$/̄symmetric measure on
$\mathbb{K}^{m\times n}$ is two/sided unitarily invariant.

###### Proof.

Let $\mu$ be a $\mathbb{B}_{S_{p},\beta}^{m\times n}$/̄symmetric measure on
$\mathbb{K}^{m\times n}$ with radial measure $\rho$, and let
$u\in\mathrm{U}_{n;\beta}$ and $v\in\mathrm{U}_{m;\beta}$. Keeping in mind
that singular values, and hence also $\lVert\cdot\rVert_{S_{p}}$ and
$\mathbb{B}_{S_{p},\beta}^{m\times n}$, are two/sided unitarily invariant, let
$f\colon\mathbb{K}^{m\times n}\to\mathbb{R}$ be measurable and nonnegative,
then we have for any $r\in\mathbb{R}_{\geq 0}$

$\displaystyle\int_{\mathbb{S}_{S_{p},\beta}^{m,n}}f(rv\theta
u)\,\mathrm{d}\kappa_{S_{p},\beta}^{m\times n}(\theta)$
$\displaystyle=\frac{1}{\omega_{S_{p},\beta}^{m\times
n}}\int_{\mathbb{K}^{m\times
n}}\operatorname{1}_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}(x)f\Bigl{(}\frac{rvxu}{\lVert x\rVert_{S_{p}}}\Bigr{)}\,\mathrm{d}x$
$\displaystyle=\frac{1}{\omega_{S_{p},\beta}^{m\times
n}}\int_{\mathbb{K}^{m\times
n}}\operatorname{1}_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}(v^{*}xu^{*})f\Bigl{(}\frac{rx}{\lVert
v^{*}xu^{*}\rVert_{S_{p}}}\Bigr{)}\,\mathrm{d}x$
$\displaystyle=\frac{1}{\omega_{S_{p},\beta}^{m\times
n}}\int_{\mathbb{K}^{m\times
n}}\operatorname{1}_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}(x)f\Bigl{(}\frac{rx}{\lVert x\rVert_{S_{p}}}\Bigr{)}\,\mathrm{d}x$
$\displaystyle=\int_{\mathbb{S}_{S_{p},\beta}^{m,n}}f(r\theta)\,\mathrm{d}\kappa_{S_{p},\beta}^{m\times
n}(\theta).$

Now intgerate w.r.t. $\rho$ in order to get

$\displaystyle\int_{\mathbb{K}^{m\times n}}f(vxu)\,\mathrm{d}\mu(x)$
$\displaystyle=\int_{\mathbb{R}_{\geq
0}}\int_{\mathbb{S}_{S_{p},\beta}^{m,n}}f(rv\theta
u)\,\mathrm{d}\kappa_{S_{p},\beta}^{m\times n}(\theta)\,\mathrm{d}\rho(r)$
$\displaystyle=\int_{\mathbb{R}_{\geq
0}}\int_{\mathbb{S}_{S_{p},\beta}^{m,n}}f(r\theta)\,\mathrm{d}\kappa_{S_{p},\beta}^{m\times
n}(\theta)\,\mathrm{d}\rho(r)=\int_{\mathbb{K}^{m\times
n}}f(x)\,\mathrm{d}\mu(x),$

and the claim follows. ∎

The next theorem is a major generalization of the probabilistic representation
of singular values of random matrices as used in [27] and [28].

######  Theorem F.

Let $X$ be a $\mathbb{K}^{m\times n}$/̄valued random variable.

  1. 1.

$X$ has a $\mathbb{B}_{S_{p},\beta}^{m\times n}$/symmetric distribution with
radial distribution $\rho$ if and only if there exist independent random
variables $R$, $U$, $V$, and $Y$ such that $R\sim\rho$,
$U\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n;\beta})$,
$V\sim\operatorname{\mathcal{U}}(\mathrm{U}_{m;\beta})$, and $Y=(Y_{i})_{i\leq
m}$ is $\mathbb{R}^{m}$/valued with Lebesgue/density

$g(y_{1},\dotsc,y_{m}):=\frac{1_{\mathbb{R}_{>0}^{m}}(y)}{Z_{m,n,p,\beta}}\mathsf{e}^{-\beta
n\lVert
y\rVert_{p/2}^{p/2}}\prod_{i=1}^{m}y_{i}^{\beta(n-m+1)/2-1}\prod_{1\leq
i<j\leq m}\lvert y_{i}-y_{j}\rvert^{\beta}$ (15)

(here $Z_{m,n,p,\beta}\in\mathbb{R}_{>0}$ is the appropriate normalization
constant, and we interpret $\mathsf{e}^{-\beta n\lVert
y\rVert_{p/2}^{p/2}}:=\operatorname{1}_{\mathbb{B}_{\infty,1}^{m}}(y)$ for
$p=\infty$), and

$X\stackrel{{\scriptstyle\text{d}}}{{=}}R\,V\,\dfrac{\operatorname{diag}\bigl{(}(Y_{i}^{1/2})_{i\leq
m}\bigr{)}}{\lVert Y\rVert_{p/2}^{1/2}}\,U^{*}.$

  2. 2.

Special cases of 1.:

    1. (a)

$X\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{p},\beta}^{m\times n})$ iff
$R\stackrel{{\scriptstyle\text{d}}}{{=}}W^{1/(\beta mn)}$, where
$W\sim\operatorname{\mathcal{U}}([0,1])$ is independent of all other random
variables.

    2. (b)

$X\sim\kappa_{S_{p},\beta}^{m\times n}$ iff $R=1$ almost surely.

  3. 3.

For the particular value $p=\infty$ we have the following additional
representations:

    1. (a)

$X\sim\operatorname{\mathcal{U}}(\mathbb{B}_{S_{\infty},\beta}^{m\times n})$
if and only if
$X\stackrel{{\scriptstyle\text{d}}}{{=}}V\operatorname{diag}((Y_{i}^{1/2})_{i\leq
m})U^{*}$, where $U$, $V$, and $Y$ have the same meanings as in 1.

    2. (b)

$X\sim\kappa_{S_{\infty},\beta}^{m\times n}$ if and only if
$X\stackrel{{\scriptstyle\text{d}}}{{=}}V\operatorname{diag}\bigl{(}(Y_{i}^{1/2})_{i\leq
m}\bigr{)}U^{*}$, where $U$ und $V$ have the same meanings as in 1., and
$Y=(Y_{1},\dotsc,Y_{m})^{\mathsf{T}}$ is $\mathbb{R}^{m}$/valued with
$Y_{1}=1$ almost surely and $(Y_{2},\dotsc,Y_{m})$ has Lebesgue/density
proportional to

$(y_{2},\dotsc,y_{m})\mapsto g(1,y_{2},\dotsc,y_{m}).$

###### Proof.

For ease of notation we abbreviate normalization constants (whose value also
may change from one occurrence to the other) and set
$h(x):=\operatorname{1}_{\mathbb{R}_{>0}^{m}}(x)\cdot\prod_{i=1}^{m}x_{i}^{\beta(n-m+1)-1}\cdot\prod_{1\leq
i<j\leq m}\lvert x_{i}^{2}-x_{j}^{2}\rvert^{\beta}$.

  1. 1.

$\Rightarrow$: From Proposition 5.4 we know
$X\stackrel{{\scriptstyle\text{d}}}{{=}}R\Theta$ where $R$ and $\Theta$ are
independent, $R\sim\rho$ and $\Theta\sim\kappa_{S_{p},\beta}^{m\times n}$.
Since $\kappa_{S_{p},\beta}^{m\times n}$ is $\mathbb{B}_{S_{p},\beta}^{m\times
n}$/̄symmetric (see Remark 5.2), from Propositions 5.7 and 5.9 we know
$\Theta\stackrel{{\scriptstyle\text{d}}}{{=}}V\operatorname{diag}(S)U^{*}$
with independent variables $S$, $U$, and $V$ such that
$S\stackrel{{\scriptstyle\text{d}}}{{=}}(s_{P(i)}(\Theta))_{i\leq m}$,
$U\sim\operatorname{\mathcal{U}}(\mathrm{U}_{n;\beta})$, and
$V\sim\operatorname{\mathcal{U}}(\mathrm{U}_{m;\beta})$, where
$P\sim\operatorname{\mathcal{U}}(\mathfrak{S}_{m})$ is independent of all
others. Thence it suffices to establish
$S\stackrel{{\scriptstyle\text{d}}}{{=}}\lVert
Y\rVert_{p/2}^{-1/2}(Y_{i}^{1/2})_{i\leq m}$ where $Y$ has the claimed
distribution. We further define $x_{\pi}:=(x_{\pi(i)})_{i\leq m}$ for
$x\in\mathbb{K}^{m}$ and $\pi\in\mathfrak{S}_{m}$. So let
$f\colon\mathbb{R}^{m}\to\mathbb{R}$ be measurable and nonnegative, then

$\operatorname{\mathbb{E}}[f(S)]=\operatorname{\mathbb{E}}[f(s(\Theta)_{P})]=\frac{1}{m!}\sum_{\pi\in\mathfrak{S}_{m}}\int_{\mathbb{S}_{S_{p},\beta}^{m,n}}f(s(\theta)_{\pi})\,\mathrm{d}\kappa_{S_{p},\beta}^{m\times
n}(\theta);$

express the integral over $\mathbb{S}_{S_{p},\beta}^{m,n}$ as one over
$\mathbb{B}_{S_{p},\beta}^{m\times n}$ and use the positive homogeneity of
singular values to get

$\operatorname{\mathbb{E}}[f(S)]=\frac{C}{m!}\sum_{\pi\in\mathfrak{S}_{m}}\int_{\mathbb{B}_{S_{p},\beta}^{m\times
n}}f\Bigl{(}\frac{s(x)_{\pi}}{\lVert x\rVert_{S_{p}}}\Bigr{)}\,\mathrm{d}x;$

now use Proposition 2.4, part 2, noting that the integrand already depends
only on the singular values of $x$,

$\operatorname{\mathbb{E}}[f(S)]=\frac{C_{1}}{m!}\sum_{\pi\in\mathfrak{S}_{m}}\int_{\mathbb{B}_{p,1}^{m}}f\Bigl{(}\frac{s(\operatorname{diag}(\sigma))_{\pi}}{\lVert\sigma\rVert_{p}}\Bigr{)}\,h(\sigma)\,\mathrm{d}\sigma;$

as in Remark 2.5 we split
$\mathbb{R}_{>0}^{m}=\bigcup_{\pi^{\prime}\in\mathfrak{S}_{m}}\tau_{\pi^{\prime}}W^{m}$
(up to null sets), transform $\tau_{\pi^{\prime}}W^{m}$ to $W^{m}$ while
bearing the symmetry of $\lVert\cdot\rVert_{p}$ and $h$ in mind, and use
$s(\operatorname{diag}(\sigma_{\pi^{\prime}}))=\sigma$ for $\sigma\in W^{m}$,
so

$\displaystyle\int_{\mathbb{B}_{p,1}^{m}}f\Bigl{(}\frac{s(\operatorname{diag}(\sigma))_{\pi}}{\lVert\sigma\rVert_{p}}\Bigr{)}\,h(\sigma)\,\mathrm{d}\sigma$
$\displaystyle=\sum_{\pi^{\prime}\in\mathfrak{S}_{m}}\int_{\tau_{\pi^{\prime}}W^{m}}\operatorname{1}_{\mathbb{B}_{p,1}^{m}}(\sigma)f\Bigl{(}\frac{s(\operatorname{diag}(\sigma))_{\pi}}{\lVert\sigma\rVert_{p}}\Bigr{)}\,h(\sigma)\,\mathrm{d}\sigma$
$\displaystyle=\sum_{\pi^{\prime}\in\mathfrak{S}_{m}}\int_{W^{m}}\operatorname{1}_{\mathbb{B}_{p,1}^{m}}(\sigma)f\Bigl{(}\frac{s(\operatorname{diag}(\sigma_{\pi^{\prime}}))_{\pi}}{\lVert\sigma\rVert_{p}}\Bigr{)}\,h(\sigma)\,\mathrm{d}\sigma$
$\displaystyle=\sum_{\pi^{\prime}\in\mathfrak{S}_{m}}\int_{W^{m}}\operatorname{1}_{\mathbb{B}_{p,1}^{m}}(\sigma)f\Bigl{(}\frac{\sigma_{\pi}}{\lVert\sigma\rVert_{p}}\Bigr{)}\,h(\sigma)\,\mathrm{d}\sigma$
$\displaystyle=m!\int_{\tau_{\pi^{-1}}W^{m}}\operatorname{1}_{\mathbb{B}_{p,1}^{m}}(\sigma)f\Bigl{(}\frac{\sigma}{\lVert\sigma\rVert_{p}}\Bigr{)}\,h(\sigma)\,\mathrm{d}\sigma;$

therefore

$\displaystyle\operatorname{\mathbb{E}}[f(S)]$
$\displaystyle=C_{1}\sum_{\pi\in\mathfrak{S}_{m}}\int_{\tau_{\pi^{-1}}W^{m}}\operatorname{1}_{\mathbb{B}_{p,1}^{m}}(\sigma)f\Bigl{(}\frac{\sigma}{\lVert\sigma\rVert_{p}}\Bigr{)}\,h(\sigma)\,\mathrm{d}\sigma$
$\displaystyle=C_{1}\int_{\mathbb{R}^{m}}\operatorname{1}_{\mathbb{B}_{p,1}^{m}}(\sigma)f\Bigl{(}\frac{\sigma}{\lVert\sigma\rVert_{p}}\Bigr{)}\,h(\sigma)\,\mathrm{d}\sigma.$

Introduce polar coordinates in $\mathbb{R}^{m}$ and use the homogeneity of $h$
to arrive at

$\operatorname{\mathbb{E}}[f(S)]=C_{2}\int_{\mathbb{S}_{p,1}^{m}}f(\theta)h(\theta)\,\mathrm{d}\kappa_{p,1}^{m}(\theta),$

which shows that the distribution of $S$ has $\kappa_{p,1}^{m}$/̄density
proportional to $h$. It remains to prove that $\lVert
Y\rVert_{p/2}^{-1/2}(Y_{i}^{1/2})_{i\leq m}$ also has
$\kappa_{p,1}^{m}$/̄density proportional to $h$; so let $f$ be as before, then
via substitution $x:=(y_{i}^{1/2})_{i\leq m}$ and polar integration, and
noting that $g$ and $h$ contain the indicator function
$\operatorname{1}_{\mathbb{R}_{>0}^{m}}$,

$\displaystyle\operatorname{\mathbb{E}}\biggl{[}f\biggl{(}\frac{(Y_{i}^{1/2})_{i\leq
m}}{\lVert Y\rVert_{p/2}^{1/2}}\biggr{)}\biggr{]}$
$\displaystyle=\int_{\mathbb{R}^{m}}f\biggl{(}\frac{(y_{i}^{1/2})_{i\leq
m}}{\lVert y\rVert_{p/2}^{1/2}}\biggr{)}g(y)\,\mathrm{d}y$
$\displaystyle=2^{m}\int_{\mathbb{R}^{m}}f\Bigl{(}\frac{x}{\lVert
x\rVert_{p}}\Bigr{)}g((x_{i}^{2})_{i\leq m})\prod_{i=1}^{m}x_{i}\,\mathrm{d}x$
$\displaystyle=C\int_{\mathbb{R}^{m}}f\Bigl{(}\frac{x}{\lVert
x\rVert_{p}}\Bigr{)}\mathsf{e}^{-\beta n\lVert
x\rVert_{p}^{p}}h(x)\,\mathrm{d}x$ $\displaystyle=C_{1}\int_{\mathbb{R}_{\geq
0}}\int_{\mathbb{S}_{p,1}^{m}}f\Bigl{(}\frac{r\theta}{\lVert
r\theta\rVert_{p}}\Bigr{)}\mathsf{e}^{-\beta n\lVert
r\theta\rVert_{p}^{p}}h(r\theta)r^{m-1}\,\mathrm{d}\kappa_{p,1}^{m}(\theta)\,\mathrm{d}r$
$\displaystyle=C_{1}\int_{\mathbb{R}_{\geq 0}}r^{\beta mn-1}\mathsf{e}^{-\beta
nr^{p}}\,\mathrm{d}r\int_{\mathbb{S}_{p,1}^{m}}f(\theta)h(\theta)\,\mathrm{d}\kappa_{p,1}^{m}(\theta),$

where for the last step we have invested the homogeneity of $h$; and we have
finished.

$\Leftarrow$: Because of Proposition 5.4 it suffices to prove $\Theta:=\lVert
Y\rVert_{p/2}^{-1/2}\,V\operatorname{diag}((Y_{i}^{1/2})_{i\leq
m})U^{*}\sim\kappa_{S_{p},\beta}^{m\times n}$, which makes sense since
$\lVert\Theta\rVert_{S_{p}}=\lVert
Y\rVert_{p/2}^{-1/2}\lVert\operatorname{diag}((Y_{i}^{1/2})_{i\leq
m})\rVert_{S_{p}}=1$. From the proof of $\Rightarrow$ we already know that
$\lVert Y\rVert_{p/2}^{-1/2}(Y_{i}^{1/2})_{i\leq m}$ has
$\kappa_{p,1}^{m}$/̄density proportional to $h$. So let