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  \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}%
  \thanks{*\,TU Munich, Dept.~of Mathematics, School of CIT; Garching, Germany.
    Email: \texttt{s.sra@tum.de}.}}

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\hypersetup{pdftitle={The Macdonald (q,t)-permanent II: Probability laws and their regions of validity},pdfauthor={Suvrit Sra}}
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\keywords{Macdonald permanent, matrix Laplace transform, complete monotonicity, compound Wishart matrix, infinite divisibility, loop soup, Cox process, Schur measure}

\begin{document}
\title[The Macdonald $(q,t)$-permanent II]{The Macdonald $(q,t)$-permanent II:\\Probability laws and their regions of validity}
\aiauthor
\begin{abstract}
The Macdonald $(q,t)$-permanent replaces the constant cycle weight
of the Vere-Jones $\alpha$-permanent by $(1-t^k)/(1-q^k)$ for a cycle
of length $k$. We determine which of the probability laws built on
the $\alpha$-permanent survive this deformation, and where. The
scalar gamma-type law survives on the whole parameter square, but the
matrix law, a compound Wishart matrix, is more fragile. We prove that
it exists in every dimension exactly when $t\in\{0,q,q^2,\ldots\}$,
and in dimension two exactly when $t\le q$. We conjecture that in
each fixed dimension the matrix law exists exactly on the region that
we show to be necessary, a $q$-analogue of Gindikin's theorem.
\end{abstract}
\preprintslink{macperm2}{29 October 2026}
\maketitle
\addtocontents{toc}{\protect\setcounter{tocdepth}{-1}}
\raggedbottom

\section{Introduction}

The determinant power $\det(I-M)^{-\alpha}$ underlies several
probability laws. Its coefficients are the $\alpha$-permanents of
Vere-Jones, which govern permanental point processes and loop soups
\citep{verejones1988generalization,verejones1997alpha,
shirai2003random,mccullagh2006permanental,lejan2011markov}. As a
Laplace transform it gives a gamma law in one variable for every
$\alpha>0$, but in dimension $d$ far fewer. By Gindikin's theorem,
$\det(I+S)^{-\alpha}$ is the Laplace transform of a random positive
semidefinite $d\times d$ matrix exactly when $\alpha$ lies in the
Wallach set $\{0,1,\ldots,d-1\}\cup(d-1,\infty)$
\citep[Theorems~1.3(b) and~2.2]{scott2014complete}. Only the integers
survive in every dimension, and there the law is the Wishart law, a
sum of $\alpha$ independent Gaussian rank-one matrices.

Replacing the constant cycle weight $\alpha$ by a weight that
depends on the cycle length preserves the generating-function
identities, as the companion paper shows \citep{sra2026macpermI}. It
need not preserve the parameter domains on which those identities
define probability laws, and determining those domains is the
subject of this paper.

For $0\le q,t<1$ and $A=(a_{ij})\in M_n(\C)$, define
\begin{equation}\label{eq:def}
 \Pqt(A)=\nlsum_{\sigma\in S_n}
 w_\sigma\nlprod_{i=1}^n a_{i,\sigma(i)},\qquad
 w_\sigma=\nlprod_{C\in\Cyc(\sigma)}w_{|C|},\qquad
 w_k:=\frac{1-t^k}{1-q^k}.
\end{equation}
Here $\Cyc(\sigma)$ includes the fixed points. At $q=t$ every
cycle weight is one, and as $q\uparrow1$ with $t=q^\alpha$ it tends
to $\alpha$, so \eqref{eq:def} contains the permanent and deforms the
Vere-Jones $\alpha$-permanent. The determinant power becomes a
product of determinant ratios,
\[
 \Zqt(M)=\exp\left(\nlsum_{k\ge1}\frac{w_k}{k}\tr M^k\right)
 =\nlprod_{r\ge0}\frac{\det(I-tq^rM)}{\det(I-q^rM)},
\]
which also makes sense at $M=-S$ for every $S\succeq0$.

In one dimension the deformation costs nothing: $\Zqt(-s)$ is a
Laplace transform on the whole parameter square, and a
Bernoulli--exponential series constructs the law
(Section~\ref{sec:scalar}). In dimension $d$ we ask for a compound
Wishart matrix, that is, a positive semidefinite random matrix $W$
with $\E e^{-\tr SW}=\Zqt(-S)$ for every $S\succeq0$. Its diagonal
moments would be the values of $\Pqt$ on Gram matrices, so such a
matrix would explain at once the positivity studied in Part~I.
Positivity alone falls short, since it tests derivatives only at
the origin, whereas a matrix Laplace transform must be completely
monotone on the whole cone. Nor does the classical theory carry
over. There the Wallach set comes from the symmetry of the cone,
while $\Zqt(-S)$ is invariant only under unitary conjugation.
Even so, the answer has the shape of Gindikin's: the region shrinks
as the dimension grows, and a discrete set survives in every
dimension. We collect our answers for the matrix law in the
following informal theorem.

\begin{dashedbox}
\begin{recalled}{Theorem}[informal; Theorems~\ref{thm:cm-classification} and~\ref{thm:rank-two-Laplace}, Corollary~\ref{cor:matrix-Laplace-necessary}]
Fix $0<q<1$ and $0\le t<1$. The function $\Zqt(-S)$ is the
Laplace transform of a positive semidefinite random matrix in
every dimension if and only if
\[
 t\in\mathcal G(q):=\{0,q,q^2,q^3,\ldots\}.
\]
In dimension two it is one if and only if $0\le t\le q$. In
dimension $d$ it is necessary that
\[
 t\in\mathcal R_d(q):=
 \bigl([0,q^{d-1}]\cup\{q,q^2,\ldots,q^{d-2}\}\bigr)\cap[0,1),
\]
where the finite set is empty for $d\le2$.
\end{recalled}
\end{dashedbox}

Along $t=q^\alpha$ the theorem reads as Gindikin's, since
$\mathcal G(q)$ is the image of the integers, with $t=0$ for
$\alpha=\infty$, and $\mathcal R_d(q)$ is the image of the positive
part of the Wallach set. On $\mathcal G(q)$ the Wishart construction
survives. At $t=q^m$ the cycle weights are the power sums of
$(1,q,\ldots,q^{m-1})$, and a sum of $m$ independent Gaussian
rank-one matrices with these weights has exactly the required
moments (Section~\ref{sec:wishart}). Off $\mathcal G(q)$ a law might
still exist without such a construction. What rules it out is the
moment of $(\det W)^2$ that the law would have to possess. This
moment factors exactly, and it is negative in every sufficiently
large dimension (Section~\ref{subsec:cm}).

In a fixed dimension more survives. Since the transform depends only
on the eigenvalues of $S$, the Harish-Chandra--Itzykson--Zuber
integral turns existence into a positivity condition on derivatives
of the scalar density. In dimension two that condition is
log-concavity, which the density has throughout $0<t<q$, so the
necessary region $\mathcal R_2(q)=[0,q]$ is exact. Whether $\mathcal R_d(q)$
is also sufficient for $d\ge3$ remains open; we state it as a
$q$-analogue of Gindikin's theorem (Conjecture~\ref{conj:q-Gindikin}).

Discrete laws need only nonnegative coefficients, not a mixing
measure, so they survive on larger regions. The closed-walk
expansion of $\Zqt$ gives infinitely divisible occupation laws for
entrywise nonnegative kernels throughout the square
(Theorem~\ref{thm:loops}). The coefficient law of a rank-$d$
positive semidefinite contraction exists on $\mathcal R_d(q)$, and
whether it is infinitely divisible does not depend on $(q,t)$
(Theorem~\ref{thm:coefficient-ID}). Read as correlation functions,
$\Pqt$ gives Cox processes on $\mathcal G(q)$, and on the whole
square for rank-one kernels (Section~\ref{sec:cox}); read as
probabilities, the Schur coefficients give a determinantal measure on
partitions with at most $d$ rows on $\mathcal R_d(q)$
(Theorem~\ref{thm:finite-rank-Schur}). Coefficient positivity alone
does not give a matrix mixing law (Appendix~\ref{app:count-mixing}).

The Macdonald permanent is introduced, and its algebra and
positivity are developed, in Part~I \citep{sra2026macpermI}. The facts needed here are proved
locally, so this paper can be read on its own.
Supplement~II \citep{sra2026macpermSuppII} collects further count
laws, matrix moments, and partition models.

\subsection{Relation to earlier work}

The scalar law used here is the remainder in the
self-decomposition of a Poissonian exponential functional studied
by \citet[Proposition~3.1]{bertoin2004poissonian}; what is new is
the Bernoulli--exponential series, which identifies its moments and
L\'evy measure with the Macdonald cycle weights. Shirai's Wishart
representation expresses repeated $\alpha$-determinants as diagonal
moments and gives their Poisson mixtures
\citep[Theorem~3.8 and Corollary~3.11]{shirai2007remarks}, and
geometric column variances give the corresponding construction here,
with entry formulas that specialize the Wishart permutation calculus
of \citet{graczyk2003complex} and \citet{bryc2008asymptotic}.

The Cox construction uses the permanental processes of
\citet{shirai2003random} and \citet{mccullagh2006permanental}. The loop expansion is a
length-weighted version of the classical permanental construction
\citep{lejan2011markov,li2023note}. The coefficient-law criterion
is related to the multivariate negative-binomial theory of
\citet{griffiths1987class}.
The random-partition application uses the Schur correlation
identities of \citet{okounkov2001infinite}, with positivity
supplied only through the finite number of allowed rows.

\section{The generating function}\label{sec:prelim}

Every law in this paper is read off $\Zqt$, so this section collects
what the later sections need about it: its Taylor coefficients at
matrices with repeated indices, which give the discrete laws; its
invariant polynomial moments, which pose the matrix Laplace problem;
and the signs of its Schur coefficients, which decide where both are
positive. The cycle expansion and the Schur specialization give all
three, with proofs short enough to include here.

We write $[n]=\{1,\ldots,n\}$, $\N=\{0,1,\ldots\}$, and
$M_{d,p}(\C)=\C^{d\times p}$; when $d=p$ we write $M_d(\C)$.
For $A\in M_n(\C)$ and $S\subseteq[n]$, the matrix $A[S]$ is the
principal submatrix indexed by $S$. Empty determinants and permanents,
including $\Pqt(\varnothing)$, are $1$. The notation $A\succeq0$ means that $A$ is Hermitian positive
semidefinite. We use the operator norm $\|\cdot\|_{\mathrm{op}}$,
the spectral radius $\rho(\cdot)$, and inner products that are
conjugate-linear in the first argument.

Repeated indices are how the coefficients of $\Zqt$ become values of
$\Pqt$. For $A\in M_N(\C)$ and $\mathbf m\in\N^N$, let $A[\mathbf m]$
have rows and columns indexed by pairs $(i,a)$ with
$1\le a\le m_i$, and entry $a_{ij}$ at $((i,a),(j,b))$.
Write $\mathbf m!=\nlprod_i m_i!$ and
$x^{\mathbf m}=\nlprod_i x_i^{m_i}$.

\begin{prop}[The repeated-index master identity]\label{prop:master-local}
For $A\in M_N(\C)$ and $X=\diag(x_1,\ldots,x_N)$,
\begin{equation}\label{eq:introZ}
\begin{aligned}
 \Zqt(M)&=
 \exp\left(\nlsum_{k\ge1}\frac{w_k}{k}\tr M^k\right)
 =\nlprod_{r\ge0}\frac{\det(I-tq^rM)}{\det(I-q^rM)},\\
 \Zqt(XA)&=\nlsum_{\mathbf m\in\N^N}
 \frac{\Pqt(A[\mathbf m])}{\mathbf m!}x^{\mathbf m}.
\end{aligned}
\end{equation}
The coefficient identity is formal at zero. For
$0\le q,t<1$ and $\rho(M)<1$, the trace series converges
absolutely and equals the determinant product.
\end{prop}
\begin{proof}
The monomials of $\tr(XA)^k$ are rooted closed words of length
$k$. The factor $1/k$ removes the choice of a cyclic starting
position, and exponentiating collects disjoint cycles into
permutations. With $m_i$ distinguishable copies of label $i$,
the resulting weight is $\Pqt(A[\mathbf m])$; forgetting the
labels within each type divides it by $\mathbf m!$, which proves
the formal identity for arbitrary cycle weights.

For $\rho(M)<1$, the bound
$|\tr M^k|\le N\rho(M)^k$ gives absolute convergence.
Expand $w_k=\nlsum_{r\ge0}(q^{rk}-t^kq^{rk})$ and use
$-\log\det(I-aM)=\nlsum_{k\ge1}a^k\tr M^k/k$.
Absolute convergence permits interchanging the sums and proves
the product formula. At $q=0$ only the term $r=0$ is retained.
\end{proof}

The loop soups and Cox processes below come in families indexed by
an intensity $\beta>0$, and raising $\Zqt$ to the power $\beta$
multiplies every cycle weight by $\beta$. We therefore write
\[
\begin{aligned}
 \per_{q,t}^{(\beta)}(A)
 &:=\nlsum_{\sigma\in S_n}\beta^{c(\sigma)}w_\sigma
 \nlprod_i a_{i,\sigma(i)},\\
 \nlsum_{\mathbf m}\frac{\per_{q,t}^{(\beta)}(A[\mathbf m])}
 {\mathbf m!}x^{\mathbf m}&=\Zqt(XA)^\beta.
\end{aligned}
\]
where $c(\sigma)$ counts cycles and the power is defined near zero
by the logarithm that vanishes there. At $q=t$ this is the classical
$\alpha$-permanent with $\alpha=\beta$, and our convention for it is
$\per_\alpha(A)=\nlsum_\sigma\alpha^{c(\sigma)}\nlprod_i a_{i,\sigma(i)}$,
so that $\per_1=\per$ and $\per_{-1}=(-1)^n\det$. The convention of
Shirai and Takahashi is related by
$\det_\eta(A)=\eta^n\per_{1/\eta}(A)$ for $\eta\ne0$.

The one-variable case of \eqref{eq:introZ} gives the constants $H_n$
that recur below: they are the moments of the scalar law of
Section~\ref{sec:scalar}, the weights of its negative-binomial
companion, and, through the bound \eqref{eq:upper}, the size of the
Cox correlations of Section~\ref{sec:cox}.

\begin{prop}[Scalar coefficients and normalizations]\label{prop:normalizations-local}
Put
\begin{equation}\label{eq:Hn}
 H_n(q,t):=\frac{(t;q)_n}{(q;q)_n},\qquad
 (a;q)_n:=\nlprod_{j=0}^{n-1}(1-aq^j),\qquad
 (a;q)_\infty:=\nlprod_{j\ge0}(1-aq^j).
\end{equation}
Then
\begin{equation}\label{eq:H}
 H(z):=\nlsum_{n\ge0}H_nz^n
 =\exp\left(\nlsum_{k\ge1}\frac{w_kz^k}{k}\right)
 =\frac{(tz;q)_\infty}{(z;q)_\infty},\qquad |z|<1.
\end{equation}
Moreover,
\[
 \Pqt(I_n)=w_1^n,\qquad
 \Pqt(uv^{\mathsf T})=n!H_n(q,t)\nlprod_i u_iv_i,
\]
\[
 \Pqt(A\oplus B)=\Pqt(A)\Pqt(B),\qquad
 \Pqt(DAE)=\left(\nlprod_iD_{ii}E_{ii}\right)\Pqt(A)
\]
for diagonal $D,E$, and
\begin{equation}\label{eq:upper}
 |\Pqt(A)|\le n!H_n(q,t)\nlprod_{i=1}^n a_{ii}
 \qquad(A\succeq0,\ 0\le q,t<1).
\end{equation}
\end{prop}
\begin{proof}
Removing the first factor from each scalar product in
\eqref{eq:introZ} gives $(1-z)H(z)=(1-tz)H(qz)$.
Its constant coefficient is one, and its coefficients satisfy
$(1-q^n)[z^n]H=(1-tq^{n-1})[z^{n-1}]H$ for $n\ge1$.
This recurrence gives \eqref{eq:Hn} and \eqref{eq:H}.
The exponential formula gives
$\nlsum_{\sigma\in S_n}w_\sigma=n!H_n$.
For a rank-one matrix every permutation monomial equals
$\nlprod_i u_iv_i$, which proves its evaluation.
The identity, block-diagonal, and diagonal-scaling formulas follow
directly from \eqref{eq:def}.
$|a_{ij}|\le\sqrt{a_{ii}a_{jj}}$ for $A\succeq0$ bounds every
monomial by $\nlprod_i a_{ii}$; summing the positive
weights gives \eqref{eq:upper}.
\end{proof}

The determinant product also shows how the parameter $t$ composes,
and the scalar law of Section~\ref{sec:scalar} splits accordingly
into independent pieces. For $0\le q,u,v<1$,
\begin{equation}\label{eq:addition}
\begin{aligned}
 \per_{q,uv}(A)
 &=\nlsum_{S\subseteq[n]}u^{|S|}
 \per_{q,v}(A[S])\per_{q,u}(A[S^c]),\\
 \mathcal Z_{q,uv}(M)
 &=\mathcal Z_{q,u}(M)\mathcal Z_{q,v}(uM).
\end{aligned}
\end{equation}
Indeed, the determinant products cancel in the second identity;
the square-free coefficient identity in the first then follows
from \eqref{eq:introZ}.

\subsection{Signs of the Schur coefficients}

The signs of the Schur coefficients of $\Pqt$ decide which of the
laws below exist, so we derive them together with their sign pattern.

A partition $\lambda=(\lambda_1,\ldots,\lambda_\ell)$ is a weakly
decreasing sequence of positive integers. Its size is
$|\lambda|=\nlsum_i\lambda_i$, its length is $\ell(\lambda)=\ell$,
and $\lambda\vdash n$ means $|\lambda|=n$. We identify $\lambda$ with its diagram
$\{(i,j):1\le i\le\ell(\lambda),\,1\le j\le\lambda_i\}$ and write
$\lambda'_j=\#\{i:\lambda_i\ge j\}$ for the conjugate partition.
The hook length of its cell $(i,j)$ is
\[
 h_{ij}=\lambda_i-j+\lambda'_j-i+1.
\]
If $\mu=1^{m_1}2^{m_2}\cdots$ records the multiplicities of its
parts, then
\[
 z_\mu=\nlprod_{k\ge1}k^{m_k}m_k!,\qquad
 w_\mu=\nlprod_iw_{\mu_i}.
\]
There are $n!/z_\mu$ permutations of cycle type $\mu\vdash n$.

For an alphabet $Z=(z_1,z_2,\ldots)$, let
$p_k(Z)=\nlsum_i z_i^k$ and $p_\mu=\nlprod_i p_{\mu_i}$.
We work degree by degree in the symmetric-function algebra
$\Lambda_{\Q(q,t)}=\Q(q,t)[p_1,p_2,\ldots]$, where $\deg p_k=k$.
A specialization is an algebra homomorphism from this algebra,
specified by its values on the $p_k$; it need not be evaluation
at a sequence of numbers.
The complete symmetric functions $h_n$ are defined by
\[
 \nlsum_{n\ge0}h_n(Z)u^n
 =\exp\left(\nlsum_{k\ge1}\frac{p_k(Z)u^k}{k}\right).
\]
The Schur function $s_\lambda(Z)$ is the sum of the monomials
$\nlprod_{(i,j)\in\lambda}z_{T(i,j)}$ over semistandard tableaux $T$
of shape $\lambda$. Thus $s_\lambda(Z)\ge0$ on a nonnegative alphabet, and
$s_\lambda(z_1,\ldots,z_d)=0$ if $\ell(\lambda)>d$.
These conventions follow \citep[Chapter I, Sections 2--3]{macdonald1995}.

Let $\chi^\lambda$ denote the irreducible character of $S_n$
indexed by $\lambda\vdash n$, and put $f^\lambda=\chi^\lambda(e)$.
We use the unnormalized irreducible immanant
\[
 \Imm_\lambda(A)=\nlsum_{\sigma\in S_n}
 \chi^\lambda(\sigma)\nlprod_{i=1}^n a_{i,\sigma(i)}.
\]
Thus $\Imm_{(n)}=\per$ and $\Imm_{(1^n)}=\det$.
The Frobenius identity
\[
 s_\lambda=\nlsum_{\mu\vdash n}
 \frac{\chi^\lambda(\mu)}{z_\mu}p_\mu
\]
relates this normalization to symmetric functions
\citep[Chapter I, Section 7]{macdonald1995}.

The specialization $\Bqt$ is defined by $p_k[\Bqt]=w_k$. It need
not come from a positive alphabet, so its Schur values can be
negative; the next proposition finds exactly where they are not.

\begin{prop}[Hook-content coefficients and their signs]\label{prop:hook-local}
For $0<q<1$ and real $t$,
\begin{equation}\label{eq:schur}
\begin{aligned}
 \Pqt(A)&=\nlsum_{\lambda\vdash n}c_\lambda(q,t)\Imm_\lambda(A),\\
 c_\lambda(q,t)&=\nlprod_{(i,j)\in\lambda}
 \frac{q^{i-1}-tq^{j-1}}{1-q^{h_{ij}}}
 =s_\lambda[\Bqt].
\end{aligned}
\end{equation}
For $d\ge1$, define
\begin{equation}\label{eq:finite-positivity-region}
 \mathcal R_d(q):=
 \bigl([0,q^{d-1}]\cup\{q,q^2,\ldots,q^{d-2}\}\bigr)\cap[0,1),
\end{equation}
where the finite set is empty for $d\le2$.
For $0\le t<1$, the coefficients $c_\lambda(q,t)$ are
nonnegative for every partition with at most $d$ rows if and only
if $t\in\mathcal R_d(q)$. They are nonnegative for every
partition if and only if $t\in\mathcal G(q)=\{0,q,q^2,\ldots\}$.
\end{prop}
\begin{proof}
By the Frobenius identity and character orthogonality, the
coefficient of $\chi^\lambda$ in the class function
$\sigma\mapsto w_\sigma$ is
$\nlsum_{\mu\vdash n}\chi^\lambda(\mu)w_\mu/z_\mu
=s_\lambda[\Bqt]$, which proves the immanant expansion.
At $t=q^m$, the power sums are those of
$(1,q,\ldots,q^{m-1})$, whose Schur principal specialization is
\[
 s_\lambda(1,q,\ldots,q^{m-1})
 =\nlprod_{(i,j)\in\lambda}
 \frac{q^{i-1}-q^mq^{j-1}}{1-q^{h_{ij}}};
\]
see \citet[Chapter I, Section 3]{macdonald1995}.
For fixed $q$, both sides of the proposed hook formula are
polynomials in $t$ of degree at most $n$. Agreement at every
$t=q^m$ proves the formula identically.

If $t\le q^{d-1}$ and $i\le d$, then
$tq^{j-1}\le q^{d+j-2}\le q^{i-1}$, so every numerator
is nonnegative. At $t=q^m$, positivity follows from the positive
alphabet. Conversely, if $t\notin\mathcal R_d(q)$, then
$q^r<t<q^{r-1}$ for some $1\le r\le d-1$; the column
$(1^{r+1})$ has
\[
 c_{(1^{r+1})}(q,t)
 =\frac{\nlprod_{i=0}^r(q^i-t)}{\nlprod_{k=1}^{r+1}(1-q^k)}<0,
\]
because only its last numerator factor is negative.
Intersecting over $d$ leaves precisely
$\mathcal G(q)$.
\end{proof}

The coefficient sign test also gives the Gram-matrix positivity
needed for discrete coefficient laws.
\begin{lemma}[The central operator and Gram matrices]\label{lem:Gram-positivity}
On $(\C^d)^{\otimes n}$ let
$U_\sigma(v_1\otimes\cdots\otimes v_n)
=v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)}$,
and put $T=\nlsum_{\sigma\in S_n}w_\sigma U_\sigma$.
On its $\lambda$-isotypic component, $T$ acts as the scalar
\begin{equation}\label{eq:central-eigen}
 \frac1{f^\lambda}\nlsum_{\sigma\in S_n}
 w_\sigma\chi^\lambda(\sigma)
 =\frac{n!}{f^\lambda}c_\lambda(q,t).
\end{equation}
If $0<q<1$, $t\in\mathcal R_d(q)$, and $A\succeq0$ has
rank at most $d$, then $\Pqt(A)\ge0$, in every size.
\end{lemma}
\begin{proof}
The operator is central because its coefficients depend only on
cycle type, and self-adjoint because $w_{\sigma^{-1}}=w_\sigma$.
Schur's lemma gives the scalar in \eqref{eq:central-eigen} by
taking the trace on the irreducible $S_n$-module of dimension
$f^\lambda$. A conjugacy class of type $\mu$ has $n!/z_\mu$
elements, so the Frobenius formula gives the stated value.
Schur--Weyl decomposition contains only partitions with
$\ell(\lambda)\le d$. Proposition~\ref{prop:hook-local} therefore
makes $T\succeq0$ in the indicated region. If
$A=[\langle u_i,u_j\rangle]$ with $u_i\in\C^d$ and
$u=u_1\otimes\cdots\otimes u_n$, then
$\langle u,Tu\rangle=\Pqt(A)$, after replacing $\sigma$ by
$\sigma^{-1}$ in the sum, so $\Pqt(A)\ge0$.
\end{proof}

\section{Laws in one dimension}\label{sec:scalar}

We first show that the scalar law survives the deformation on the
whole parameter square. It is the
paper's building block: its
moments are the rank-one values of $\Pqt$, its density decides the
matrix problem in dimension two (Section~\ref{subsec:dim-two}), and it
gives Cox intensities throughout the square (Section~\ref{sec:cox}). The transform
$\Zqt(-s)=\nlprod_{r\ge0}(1+tq^rs)/(1+q^rs)$ deforms the gamma
transform $(1+s)^{-\alpha}$, and a Bernoulli--exponential series
realizes it with explicit moments and L\'evy measure.

\begin{prop}[A scalar infinitely divisible law]\label{thm:scalar}
Let $0\le q,t<1$. Take independent random variables
$B_r\sim\Bernoulli(1-t)$ and $E_r\sim\Exp(1)$, for $r\ge0$, and set
\begin{equation*}
 S_{q,t}:=\nlsum_{r\ge0}q^rB_rE_r.
\end{equation*}
When $q=0$, retain only the $r=0$ term. The series converges almost
surely and in $L^1$, and
\begin{align}
 \E e^{-sS_{q,t}}
 &=\frac{(-ts;q)_\infty}{(-s;q)_\infty},
 \qquad s\ge0,
 \label{eq:scalar-lt}\\
 \E S_{q,t}^n&=n!H_n(q,t),\qquad n\ge0,
 \label{eq:scalar-mom}\\
 \cum_k(S_{q,t})&=(k-1)!w_k,\qquad k\ge1.
 \label{eq:scalar-cum}
\end{align}
The law is infinitely divisible. For $0<q<1$, it has zero drift
and L\'evy measure
\begin{equation*}
 \nu_{q,t}(dx)=\frac1x\nlsum_{r\ge0}
 \left(e^{-x/q^r}-e^{-x/(tq^r)}\right)\,dx,
 \qquad x>0,
\end{equation*}
where the second exponential is zero when $t=0$; for $q=0$ only
the $r=0$ term occurs.
\end{prop}
\begin{proof}
The partial sums increase, with expectations converging to
$(1-t)/(1-q)$, so by monotone convergence the series converges
almost surely and in $L^1$. For $s\ge0$, a single summand has
transform
\[
 \E e^{-sq^rB_rE_r}
   =t+\frac{1-t}{1+sq^r}
   =\frac{1+tsq^r}{1+sq^r},
\]
and independence with bounded convergence proves
\eqref{eq:scalar-lt}.

For $0\le u<1$, monotone convergence instead gives
\[
 \E e^{uS_{q,t}}
 =\nlprod_{r\ge0}\frac{1-tuq^r}{1-uq^r}=H(u)<\infty.
\]
Both sides are analytic for $|u|<1$: the exponential moment
dominates every derivative on compact subsets. Differentiation at
zero gives \eqref{eq:scalar-mom}, and differentiation of
$\log H(u)=\nlsum_{k\ge1}w_ku^k/k$ gives \eqref{eq:scalar-cum}.

For $a>0$ and $0\le t<1$, Frullani's integral gives
\[
 \int_0^\infty(1-e^{-sx})
             \frac{e^{-x/a}-e^{-x/(ta)}}x\,dx
 =\log(1+as)-\log(1+tas),
\]
with the second exponential absent at $t=0$. By Tonelli,
$\int_0^\infty x\,\nu_{q,t}(dx)=\nlsum_{r\ge0}(q^r-tq^r)=(1-t)/(1-q)$,
so $\int(1\wedge x)\nu_{q,t}(dx)<\infty$, and summing the
identities over $a=q^r$ yields
\[
 -\log\E e^{-sS_{q,t}}
   =\int_0^\infty(1-e^{-sx})\,\nu_{q,t}(dx).
\]
Infinite divisibility follows: a Poisson random measure on
$(0,\infty)$ with intensity $\nu_{q,t}$ has total mass with this
Laplace transform and finite mean, and intensity
$\nu_{q,t}/h$ gives an $h$th convolution root.
\end{proof}

For $0<q<1$, \citet[Proposition~3.1(i)]{bertoin2004poissonian}
obtains the law of $S_{q,t}$ as the remainder $I_t^{(q)}$ in the
self-decomposition of a Poisson exponential functional: for a
rate-one Poisson process $N_u$,
\[
 I^{(q)}=\int_0^\infty q^{N_u}\,du
 \stackrel d=\nlsum_{r\ge0}q^rE_r,\qquad
 I^{(q)}\stackrel d=t\,I^{(q)\prime}+I_t^{(q)}
\]
with independent terms on the right, and the ratio of the two
transforms is \eqref{eq:scalar-lt}.

The parameter multiplication identity
\eqref{eq:addition} becomes an additive decomposition of
these scalar variables, and at $t=q^m$ cancellation leaves finitely
many exponentials. The decomposition is what reduces the
dimension-two theorem to $q^2\le t<q$
(Theorem~\ref{thm:rank-two-Laplace}).

\begin{corr}[Convolution, boundaries, and the gamma limit]
\label{cor:scalar-limits}
For $0\le q,u,v<1$,
\begin{equation}\label{eq:scalar-add}
 S_{q,uv}\ \stackrel{d}=\ S_{q,u}+uS'_{q,v},
\end{equation}
where the variables on the right are independent and the prime
denotes an independent copy. For every positive integer $m$,
\[
 S_{q,q^m}\stackrel d=\nlsum_{r=0}^{m-1}q^rE_r.
\]
At $q=0$, the distribution is $t\delta_0+(1-t)\Exp(1)$.
For every fixed $\alpha>0$, as $q\uparrow1$ with $t=q^\alpha$,
$S_{q,t}$ converges in distribution and in moments to the gamma law
of shape $\alpha$ and rate $1$.
\end{corr}
\begin{proof}
The product of the Laplace transforms on the right side of
\eqref{eq:scalar-add} is
\[
 \nlprod_{r\ge0}
 \frac{1+usq^r}{1+sq^r}
 \frac{1+uvsq^r}{1+usq^r}
 =\nlprod_{r\ge0}\frac{1+uvsq^r}{1+sq^r}.
\]
Laplace-transform uniqueness proves the decomposition.
For $t=q^m$, the same product cancels to
$\nlprod_{r=0}^{m-1}(1+sq^r)^{-1}$.
The $q=0$ statement follows directly from the defining series.

Since $0<(1-x^\alpha)/(1-x)\le\max(1,\alpha)$ for $0<x<1$ (by
$x^\alpha\ge x$ if $\alpha\le1$, and the mean value theorem
otherwise), the weights
$w_k(q,q^\alpha)$ tend to $\alpha$ and are bounded uniformly in
$q$ and $k$. Hence
\[
 \log\E e^{uS_{q,q^\alpha}}
 =\nlsum_{k\ge1}\frac{w_k(q,q^\alpha)u^k}{k}
 \longrightarrow-\alpha\log(1-u)
\]
uniformly on compact subsets of $|u|<1$; the limit is the gamma
moment generating function, which gives weak convergence and the
moments $\alpha(\alpha+1)\cdots(\alpha+n-1)$.
\end{proof}

The intensity $\beta$ of Section~\ref{sec:prelim} has a scalar
meaning too, which the rank-one Cox processes of
Section~\ref{sec:cox} use. For $\beta>0$, the same Poisson
construction with intensity $\beta\nu_{q,t}$ defines a nonnegative variable
$S_{q,t}^{(\beta)}$ with
$\E e^{uS_{q,t}^{(\beta)}}=H(u)^\beta$ for $u<1$.
The exponential formula gives
\[
 \E\bigl(S_{q,t}^{(\beta)}\bigr)^n
       =\per_{q,t}^{(\beta)}(J_n),
\]
where $J_n$ is the all-ones matrix. Independent addition of the
Poisson random measures adds $\beta$, so the cycle fugacity is the
time parameter of a scalar subordinator.

\subsection{The \texorpdfstring{$(q,t)$}{(q,t)}-negative-binomial law}
The count laws of Section~\ref{sec:loops} have a one-dimensional
prototype, and it already shows their two constructions. The
coefficients of $H$, suitably tilted, are probabilities on $\N$ for
every $(q,t)$: for $0\le z<1$, define $N_{q,t}(z)$ by
\begin{equation}\label{eq:qnb}
 \Prob(N_{q,t}(z)=n)=\frac{H_nz^n}{H(z)},\qquad n\ge0.
\end{equation}
At $z=0$ this means the point mass at zero. We call
\eqref{eq:qnb} the \emph{$(q,t)$-negative-binomial law}.
At $t=0$ it is the Euler distribution, with mass
$(z;q)_\infty z^n/(q;q)_n$.
At $t=q^m$ it is the negative $q$-binomial distribution of the
second kind, with mass
$(z;q)_m(q^m;q)_n z^n/(q;q)_n$; these conventions agree with
\citet[Section~1 and Example~2.5]{charalambides2018class}.

Its generating function is $H(zs)/H(z)$, with
\[
 \log\frac{H(zs)}{H(z)}
 =\nlsum_{k\ge1}\frac{w_kz^k}{k}(s^k-1).
\]
The logarithm gives the first construction: $N_{q,t}(z)$ is the
compound Poisson sum $\nlsum_{k\ge1}kC_k$ with independent
$C_k\sim\Poisson(w_kz^k/k)$, the one-point loop soup of
Theorem~\ref{thm:loops}. The scalar law gives the second: if the law
of $S_{q,t}$ is tilted by $e^{zs}/H(z)$, a conditionally Poisson
variable of mean $zs$ has the same generating function, the one-point
Gaussian tilt of Theorem~\ref{thm:tilt}. Further
representations, factorial cumulants, and the conditioned permutation
law are in Supplement~II \citep{sra2026macpermSuppII}.

\section{The matrix law}\label{sec:matrix}

This section proves the theorem of the introduction on the matrix
law, the compound Wishart matrix that is the matrix analogue of the
scalar law of Section~\ref{sec:scalar}. Gaussian columns build this
law wherever they can produce the cycle weights, which happens
exactly on $\mathcal G(q)$. Elsewhere no such construction exists,
and to rule a law out we use a moment that every representing law
must have.

\subsection{The Wishart construction}\label{sec:wishart}

We first build the law wherever Gaussian columns can produce the
cycle weights. For a Wishart matrix with $m$ independent Gaussian
columns, the Wick formula assigns one column to each cycle of a
pairing, so each cycle contributes the number of columns and the
diagonal moments are $\per_m$ of the covariance submatrix
\citep[Corollary~3.11]{shirai2007remarks}. Weighting the columns
geometrically replaces that count by the power sums of
$(1,q,\ldots,q^{m-1})$, which are the cycle weights at $t=q^m$, and
at $t=0$ in the limit $m=\infty$. We call these weighted columns the
\emph{colors} of the matrix. Proposition~\ref{prop:color-classification}
shows that colors exist only on $\mathcal G(q)$, which is why
Section~\ref{subsec:cm} has to look for laws without them.

We use a standard circular complex Gaussian scalar with density
$\pi^{-1}e^{-|z|^2}$ on $\C$, so its squared modulus has law
$\Exp(1)$. A centered circular complex Gaussian vector with
covariance $\Sigma\succeq0$ means a vector $\Sigma^{1/2}g$, where
$g$ has independent such coordinates. In particular,
$\E zz^*=\Sigma$ and $\E zz^{\mathsf T}=0$.

Let $z^{(r)}$, $r\ge0$, be independent vectors with this law in
$\C^d$. For a positive integer $m$, set
\begin{equation}\label{eq:Wishart}
 W_{q,m}(\Sigma):=\nlsum_{r=0}^{m-1}q^r
                         z^{(r)}z^{(r)*}.
\end{equation}
For $m=\infty$ and $0\le q<1$ the partial sums converge almost
surely in trace norm, their traces having expectation
$(1-q)^{-1}\tr\Sigma$. For finite $m$,
$W_{q,m}(\Sigma)=\Sigma^{1/2}GD_mG^*\Sigma^{1/2}$ with
$D_m=\diag(1,q,\ldots,q^{m-1})$, a classical compound-Wishart
matrix; here $q$ sets a column covariance and does not weight
pairing crossings as in the $q$-Wishart model of
\citet[Definitions~1.2--1.3]{bryc2008compound}.
The Wick formula computes every entry moment, and on the diagonal
they are values of $\Pqt$.

\begin{theorem}[Entry moments and Laplace transform]\label{thm:wishart}
Let $0\le q<1$, let $m\in\{1,2,\ldots\}\cup\{\infty\}$, and let
$\Sigma$ be a positive semidefinite $d\times d$ complex matrix.
Write $W=W_{q,m}(\Sigma)$, with $t=q^m$ for finite $m$ and $t=0$
for $m=\infty$. For every $n\ge1$ and all index lists
$i_1,\ldots,i_n,j_1,\ldots,j_n$ in $[d]$,
\begin{equation}\label{eq:entry-moments}
 \E\nlprod_{a=1}^n W_{i_a j_a}
 =\nlsum_{\sigma\in S_n}w_\sigma
              \nlprod_{a=1}^n\Sigma_{i_a,j_{\sigma(a)}}.
\end{equation}
For every Hermitian $S\succeq0$,
\begin{equation}\label{eq:wishart-LT}
 \E e^{-\tr SW}
 =\nlprod_{r=0}^{m-1}
   \det(I+q^r\Sigma^{1/2}S\Sigma^{1/2})^{-1},
\end{equation}
with an infinite product for $m=\infty$. In particular,
\begin{equation}\label{eq:diag-moment}
 \E\nlprod_{a=1}^nW_{i_ai_a}
 =\Pqt([\Sigma_{i_a i_b}]_{a,b=1}^n).
\end{equation}
\end{theorem}
\begin{proof}
For finite $m$, expand the entries in \eqref{eq:Wishart}:
\[
 \E\nlprod_{a=1}^nW_{i_a j_a}
 =\nlsum_{r_1,\ldots,r_n=0}^{m-1}
 q^{r_1+\cdots+r_n}
 \E\nlprod_{a=1}^n
 z^{(r_a)}_{i_a}\overline{z^{(r_a)}_{j_a}}.
\]
The complex Wick formula gives
\[
 \E\nlprod_{a=1}^n
 z^{(r_a)}_{i_a}\overline{z^{(r_a)}_{j_a}}
 =\nlsum_{\sigma\in S_n}
 \nlprod_{a=1}^n
       \Sigma_{i_a,j_{\sigma(a)}}
       \one_{\{r_a=r_{\sigma(a)}\}}.
\]
A surviving assignment of column indices is constant on
each cycle of $\sigma$. Summing that common index over a cycle of
length $k$ gives $\nlsum_{r=0}^{m-1}q^{rk}=w_k(q,q^m)$.
Different cycles have independent summation indices, which proves
\eqref{eq:entry-moments}.

For a single column, diagonalize the positive semidefinite matrix
$A=\Sigma^{1/2}S\Sigma^{1/2}$. Unitary invariance of the standard
complex Gaussian vector gives
\[
 \E e^{-z^*Sz}
   =\nlprod_{\ell=1}^d\E e^{-\lambda_\ell(A)|g_\ell|^2}
   =\nlprod_{\ell=1}^d(1+\lambda_\ell(A))^{-1}
   =\det(I+A)^{-1}.
\]
Independence proves \eqref{eq:wishart-LT} for finite $m$.
For infinite $m$, bounded convergence passes to the limit on the
left. The product on the right converges because
$0\le\log\det(I+q^rA)\le q^r\tr A$.

To pass entry moments to the infinite sum, choose $a>0$ with
$a\|\Sigma\|<1$ if $\Sigma\ne0$. For every finite partial sum $W_M$,
the same Gaussian integration with the opposite sign gives
\[
 \E e^{a\tr W_M}
   =\nlprod_{r=0}^{M-1}\det(I-aq^r\Sigma)^{-1}.
\]
The logarithms are bounded uniformly in $M$ by
$a\tr\Sigma/((1-a\|\Sigma\|)(1-q))$.
Monotone convergence proves $\E e^{a\tr W}<\infty$.
For $A\succeq0$,
$|A_{ij}|\le\sqrt{A_{ii}A_{jj}}\le\tr A$.
Thus products of entries of $W_M$ are dominated by
$(\tr W)^n$, which is integrable. Dominated convergence proves
\eqref{eq:entry-moments} for $m=\infty$; its cycle factors tend
to $\nlsum_{r\ge0}q^{rk}=(1-q^k)^{-1}$. Taking $j_a=i_a$ gives
\eqref{eq:diag-moment}.
\end{proof}

The permutation expansion is the classical Wishart calculus
(compare \citet[Theorem~2.1]{bryc2008asymptotic} and
\citet{graczyk2003complex}), with the geometric column covariance
supplying the power sums $w_k$; cumulants and invariant moments are
in Supplement~II \citep{sra2026macpermSuppII}. With $\Sigma=I_d$,
\eqref{eq:wishart-LT} is $\Zqt(-S)$, so the theorem already answers
the matrix question on $\mathcal G(q)$.

\subsection{Where Gaussian colors exist}
Can the Gaussian construction be pushed off $\mathcal G(q)$? It
requires a single operator for all cycle lengths: a fixed positive
color covariance $D$ must realize every cycle weight at once,
$w_k=\tr D^k$ for all $k$.

\begin{dashedbox}
\begin{prop}\label{prop:color-classification}
Fix $0<q<1$ and $0\le t<1$. A positive trace-class operator $D$
satisfying $\tr D^k=w_k(q,t)$ for every $k\ge1$ exists if and only if
$t=0$ or $t=q^m$ for a positive integer $m$.
\end{prop}
\end{dashedbox}
\begin{proof}
For $t=q^m$, take $D=\diag(1,q,\ldots,q^{m-1})$.
For $t=0$, take the trace-class operator
$D=\diag(1,q,q^2,\ldots)$ on $\ell^2(\N)$.
Their power sums have the required values.

Conversely, let $d_1,d_2,\ldots\ge0$ be the eigenvalues of a
positive trace-class operator with the asserted traces.
The Schur polynomial $s_\lambda(d_1,\ldots,d_M)$ is nonnegative
for every $M$, by its semistandard-tableau expansion.
As $M\to\infty$, each power sum converges to $w_k$.
Since a fixed Schur function is a polynomial in finitely many
power sums, these nonnegative values converge to
$c_\lambda(q,t)$. Thus $c_\lambda(q,t)\ge0$ for every partition.
Proposition~\ref{prop:hook-local} then requires
$t\in\mathcal G(q)$.
\end{proof}

\subsection{Complete monotonicity in every dimension}\label{subsec:cm}

Off $\mathcal G(q)$ there are no Gaussian colors, and existence means that $\Zqt(-S)$ is completely monotone on the
positive cone; by the
Bernstein--Hausdorff--Widder--Choquet theorem on cones
\citep[Theorem~2.2]{scott2014complete}, this is equivalent to a
positive semidefinite matrix Laplace representation. To exclude a law
we need a property that every representing law would have and that
can be computed without it. The moment of $(\det W)^2$ is such a
property: its sign classifies existence in every dimension at once,
and in dimension two the eigenvalue density decides the rest.

For $0<q<1$, $0\le t<1$, and $S$ in the cone $\pp^r$ of positive
semidefinite $r\times r$ matrices, put
\begin{equation}\label{eq:Lqt}
 L^{(r)}_{q,t}(S):=\Zqt(-S)
 =\nlprod_{j\ge0}\frac{\det(I+tq^jS)}{\det(I+q^jS)},
\end{equation}
a positive function, analytic on a neighborhood of $\pp^r$. For
a Hermitian $B$ write $D_BL(S)=\frac{d}{d\varepsilon}
L(S+\varepsilon B)|_{\varepsilon=0}$. The function $L$ is
\emph{completely monotone} on $\pp^r$ if
\[
 (-1)^kD_{B_1}\cdots D_{B_k}L(S)\ge0
 \qquad(S\succ0,\ B_i\succeq0,\ k\ge0),
\]
the definition of \citet{scott2014complete}. The condition tests
every positive basepoint, not only the origin. At the origin,
with $B_i=u_iu_i^*$, the master theorem \eqref{eq:introZ} identifies the
alternating derivative with $\Pqt([u_i^*u_j])$, so complete
monotonicity of $L^{(r)}_{q,t}$ implies the nonnegativity of
$\Pqt$ on every Gram matrix of vectors in $\C^r$.

The moments that a representing measure would have can be defined
without the measure. Let $\ell^{(r)}_{q,t}$ be the linear
functional on real polynomials on the Hermitian $r\times r$
matrices determined by
\begin{equation}\label{eq:formal}
 \ell^{(r)}_{q,t}\bigl(e^{\tr(XW)}\bigr)=\Zqt(X),
\end{equation}
an identity of Taylor coefficients at $X=0$; the trace pairing is
nondegenerate, so \eqref{eq:formal} fixes every polynomial
moment, and no representing measure is assumed. By
\eqref{eq:introZ},
$\ell^{(r)}_{q,t}(\nlprod_iu_i^*Wu_i)=\Pqt([u_i^*u_j])$. Determinant
powers have rectangular Schur coefficients as moments, and the
square $(\det W)^2$ is the moment that decides existence.

\begin{prop}[Rectangular moments]\label{prop:rectangle}
For positive integers $r,k$,
\begin{equation}\label{eq:rectangle}
 \ell^{(r)}_{q,t}\bigl((\det W)^k\bigr)
 =\left[\nlprod_{a=0}^{k-1}\frac{(r+a)!}{a!}\right]c_{(k^r)}(q,t).
\end{equation}
In particular, with $e_r:=c_{(1^r)}$ and $e_0:=1$,
\begin{align}
 e_r(q,t)&=\nlprod_{j=0}^{r-1}\frac{q^j-t}{1-q^{j+1}},
 \label{eq:column-coeff}\\
 \ell^{(r)}_{q,t}\bigl((\det W)^2\bigr)
 &=r!\,(r+1)!\,\bigl(e_r^2-e_{r-1}e_{r+1}\bigr).
 \label{eq:square}
\end{align}
The identities \eqref{eq:rectangle} and \eqref{eq:square} hold
for arbitrary cycle weights $w$, with $c_\lambda=s_\lambda[\rho_w]$
for the specialization $\rho_w\colon p_k\mapsto w_k$ and
$\ell^{(r)}_w$ defined by \eqref{eq:formal} with
$\mathcal Z_w(X)=\exp(\nlsum_{j\ge1}w_j\tr X^j/j)$.
\end{prop}
\begin{proof}
Apply the Cayley operator $\det(\partial_X)^k$ to the Cauchy
expansion $\Zqt(X)=\nlsum_\lambda c_\lambda(q,t)s_\lambda(X)$, the
Schur Cauchy identity for the specialization $\Bqt$ of
\eqref{eq:schur}, and evaluate at zero. Its action on Schur
polynomials is
\[
 \det(\partial_X)s_\lambda(X)
 =\left[\nlprod_{i=1}^r(\lambda_i+r-i)\right]
          s_{\lambda-(1^r)}(X),
\]
with value zero when $\lambda_r=0$. In degree $kr$, only
$\lambda=(k^r)$ survives $k$ applications. Since
$s_{(k^r)}(X)=\det(X)^k$, the Cayley identity gives the factor in
\eqref{eq:rectangle}; see \citet{caracciolo2013cayley} for that
identity. The column formula \eqref{eq:column-coeff} follows from
\eqref{eq:schur}, or by comparing coefficients in
$(1+tz)E(z)=(1+z)E(qz)$ for
$E(z)=\nlsum_{r\ge0}e_rz^r=\nlprod_{j\ge0}(1+q^jz)/(1+tq^jz)$.
Finally, the dual Jacobi--Trudi
identity gives $c_{(2^r)}=e_r^2-e_{r-1}e_{r+1}$. Nothing in the
argument uses the Macdonald form of the coefficients.
\end{proof}

For the Macdonald weights, the Tur\'an expression \eqref{eq:square}
factors, which makes its sign explicit.

\begin{dashedbox}[supporting]
\begin{theorem}[Exact determinant-square sign]\label{thm:sign}
If $0<t<1$ and $t\ne q^M$ for every $M\ge1$, then
\begin{equation}\label{eq:Turan}
 c_{(2^r)}(q,t)=e_r(q,t)^2\,
       \frac{(1-q)q^{r-1}(1-tq)}
       {(1-q^{r+1})(q^{r-1}-t)}.
\end{equation}
Consequently, if $q^{m+1}<t<q^m$ with $m\ge0$, then
\begin{equation}\label{eq:tailsign}
 \ell^{(r)}_{q,t}\bigl((\det W)^2\bigr)
 \begin{cases}
 >0,&1\le r\le m+1,\\
 <0,&r\ge m+2.
 \end{cases}
\end{equation}
\end{theorem}
\end{dashedbox}
\begin{proof}
No $e_r$ vanishes at these parameters, and
\eqref{eq:column-coeff} gives
\[
 \frac{e_{r-1}e_{r+1}}{e_r^2}
 =\frac{(1-q^r)(q^r-t)}{(q^{r-1}-t)(1-q^{r+1})}.
\]
Subtracting from one yields \eqref{eq:Turan}, because the
numerator simplifies to $(1-q)q^{r-1}(1-tq)$. Every factor in
\eqref{eq:Turan} except $q^{r-1}-t$ is positive, which proves
\eqref{eq:tailsign}.
\end{proof}
At $t=q^M$ the square moments are positive for $r\le M$ and zero
for $r>M$, and at $t=0$ they are positive; these values are read
from \eqref{eq:rectangle}, not by dividing in \eqref{eq:Turan}.
Together with \eqref{eq:tailsign} they say that
$\ell^{(r)}_{q,t}((\det W)^2)\ge0$ exactly when
$t\in\mathcal R_r(q)$, the coefficient region of
Proposition~\ref{prop:hook-local}.

Positivity of the square moment in every dimension excludes every
parameter off the locus, while on the locus the Gaussian
representation gives complete monotonicity.

\begin{dashedbox}
\begin{theorem}[Complete monotonicity in every dimension]
\label{thm:cm-classification}
Let $0<q<1$ and $0\le t<1$, and put
$\mathcal G(q)=\{0,q,q^2,q^3,\ldots\}$. The following are
equivalent.
\begin{enumerate}[label=(\roman*),leftmargin=2.2em]
\item $t\in\mathcal G(q)$.
\item $c_\lambda(q,t)\ge0$ for every partition $\lambda$.
\item $L^{(r)}_{q,t}$ is completely monotone on $\pp^r$ for every
$r\ge1$.
\item $\ell^{(r)}_{q,t}((\det W)^2)\ge0$ for every $r\ge1$.
\item $\ell^{(r)}_{q,t}((\det W)^2)\ge0$ for an unbounded set of
integers $r$.
\end{enumerate}
Each of these conditions implies that $\Pqt(A)\ge0$ for every
complex Hermitian positive semidefinite $A$, of every size and
rank.
\end{theorem}
\end{dashedbox}
\begin{proof}
The equivalence of (i) and (ii) is
Proposition~\ref{prop:hook-local}. If $t=q^M$ or $t=0$,
Theorem~\ref{thm:wishart} with $\Sigma=I_r$ gives
\begin{equation}\label{eq:Gaussian-LT}
 L^{(r)}_{q,t}(S)=\E e^{-\tr(SW)},\qquad
 W=\nlsum_{j=0}^{M-1}q^jg_jg_j^*,
\end{equation}
where $g_j$ are independent standard complex Gaussian vectors in
$\C^r$ and the sum is infinite at $t=0$, converging almost surely
in trace norm. A Laplace transform of a probability measure on
$\pp^r$ is completely monotone, and its polynomial moments are
those of \eqref{eq:formal}, so (i) implies (iii) and (iv).

If (iii) holds, the Bernstein--Hausdorff--Widder--Choquet theorem
on cones \citep[Theorem~2.2]{scott2014complete} gives a
probability measure on $\pp^r$ with Laplace transform
$L^{(r)}_{q,t}$. Analyticity of $L^{(r)}_{q,t}(sI)$ at $s=0$ and
monotone convergence give finite trace moments of every order,
and since $(\det W)^2\le(\tr W)^{2r}$ on $\pp^r$, the square
moment is finite and agrees with \eqref{eq:formal}. Thus (iii)
implies (iv), and (iv) implies (v). Theorem~\ref{thm:sign}
excludes every $t\in(0,1)\setminus\mathcal G(q)$ from (v), which
proves (v) implies (i). The final assertion follows from the diagonal Gaussian
moment identity \eqref{eq:diag-moment}.
\end{proof}

The same moment obstruction restricts existence in a fixed
dimension. A representing law has the formal polynomial moments
of \eqref{eq:formal}; the boundary-moment justification is given
in Appendix~\ref{app:boundary-moments}.

\begin{corr}[A necessary cone-valued Laplace condition]
\label{cor:matrix-Laplace-necessary}
Fix $0<q<1$, $0\le t<1$, and a positive integer $d$.
If a positive semidefinite random matrix $W\in M_d(\C)$ satisfies
\begin{equation*}
 \E e^{-\tr SW}=\mathcal Z_{q,t}(-S),
 \qquad S\succeq0,
\end{equation*}
then $t\in\mathcal R_d(q)$.
\end{corr}

\begin{proof}
For a representing law, $(\det W)^2$ has finite expectation
equal to \eqref{eq:square}. Its nonnegativity and
Theorem~\ref{thm:sign}, including the geometric boundary values,
give $t\in\mathcal R_d(q)$.
\end{proof}

\subsection{The eigenvalue density and dimension two}\label{subsec:dim-two}

Corollary~\ref{cor:matrix-Laplace-necessary} allows a matrix law
only for $t\in\mathcal R_d(q)$; does one exist there? This
subsection reduces the question to the scalar density and answers it
in dimension two. The transform depends only on
the eigenvalues of $S$, so any such $W$ is unitarily invariant, and
the Harish-Chandra--Itzykson--Zuber integral turns its eigenvalue law
into a signed combination of derivatives of the scalar density.
Existence then becomes a positivity question. Let
$\psi_{q,t}$ be the density of $S_{q,t}$. Appendix~\ref{app:density}
writes it as an exponential series $\nlsum_{r\ge0}c_re^{-q^{-r}x}$
(Lemma~\ref{lem:scalar-density-series}), whose first $d-2$ derivatives
vanish at zero when $t<q^{d-1}$, and proves
(Proposition~\ref{prop:matrix-density-criterion}) that for
$0\le t<q^{d-1}$ a positive semidefinite random
$W\in M_d(\C)$ with $\E e^{-\tr SW}=\mathcal Z_{q,t}(-S)$ exists if
and only if
\[
 \det\bigl[(-D)^{\,j-1}\psi_{q,t}(x_i)\bigr]_{i,j=1}^d\ge0
 \qquad(0<x_1<\cdots<x_d),\qquad D=\frac{d}{dx},
\]
and then its ordered eigenvalues have density
$\Delta(x)\det[(-D)^{j-1}\psi_{q,t}(x_i)]/\nlprod_{j=0}^{d-1}j!$ with
$\Delta(x)=\nlprod_{i<j}(x_j-x_i)$; uniqueness of Laplace transforms
also rules out mass on the boundary of the cone.

In dimension two the determinant is $\psi_{q,t}(x)\psi_{q,t}(y)$
times the decrease of $(\log\psi_{q,t})'$ from $x$ to $y$, so the
criterion is log-concavity, which holds throughout the necessary
range.

\begin{theorem}[The rank-two Laplace region]
\label{thm:rank-two-Laplace}
Fix $0<q<1$ and $0\le t<1$. A positive semidefinite random
$W\in M_2(\C)$ with $\E e^{-\tr SW}=\mathcal Z_{q,t}(-S)$ for
$S\succeq0$ exists if and only if $0\le t\le q$. For $0<t<q$, the
density $\psi_{q,t}$ is log-concave, and the matrix law has
ordered-eigenvalue density
\[
 (y-x)\bigl(\psi_{q,t}'(x)\psi_{q,t}(y)
           -\psi_{q,t}(x)\psi_{q,t}'(y)\bigr),
 \qquad 0<x<y.
\]
\end{theorem}
\begin{proof}
First suppose $q^2\le t<q$. The coefficients of
\eqref{eq:scalar-density-series} satisfy
\[
 c_0>0,\qquad
 c_r=\frac{t-q^r}{q(1-q^r)}c_{r-1}.
\]
It gives $c_r\le0$ for every $r\ge1$: at $t=q^2$ the
series terminates after $r=1$, and for $q^2<t<q$ all these
coefficients are negative. Hence
\[
 B(x):=e^x\psi_{q,t}(x)
   =c_0-\nlsum_{r\ge1}(-c_r)e^{-(q^{-r}-1)x}
\]
is positive, increasing, and concave. Thus $\log B$ is concave,
and so is $\log\psi_{q,t}(x)=-x+\log B(x)$.

Convolution of log-concave densities is log-concave, by the
Pr\'ekopa--Leindler inequality, and so is convolution with an
exponential density of any rate.

For $0<t<q^2$, choose an integer $k\ge1$ such that
$q^2\le t/q^k<q$. Repeatedly applying
\eqref{eq:scalar-add} with first parameter $q$ gives
\[
 S_{q,t}\stackrel d=
 \nlsum_{r=0}^{k-1}q^r E_r+q^k S'_{q,t/q^k},
\]
with independent summands. The first part of the proof
handles the last density, and successive exponential convolutions
prove log-concavity of $\psi_{q,t}$.
For $0<x<y$, we now have
\[
 \det\begin{bmatrix}
 \psi_{q,t}(x)&-\psi_{q,t}'(x)\\
 \psi_{q,t}(y)&-\psi_{q,t}'(y)
 \end{bmatrix}
 =\psi_{q,t}(x)\psi_{q,t}(y)
   \bigl((\log\psi_{q,t})'(x)-(\log\psi_{q,t})'(y)\bigr)
 \ge0.
\]
Proposition~\ref{prop:matrix-density-criterion} proves existence
and the stated eigenvalue density.

The endpoints $t=q$ and $t=0$ are the Gaussian constructions
$zz^*$ and $\nlsum_{r\ge0}q^rz^{(r)}z^{(r)*}$ of
Theorem~\ref{thm:wishart}, and
Corollary~\ref{cor:matrix-Laplace-necessary} requires
$t\in\mathcal R_2(q)=[0,q]$.
\end{proof}

\section{Discrete laws}\label{sec:discrete}

This section turns from the matrix law to the discrete laws, which
read $\Zqt$ and $\Pqt$ as probabilities or as correlation functions
rather than as moments. They need only nonnegative weights, not a
matrix mixing law, so they survive on larger regions, which this
section determines.

\subsection{Closed walks and occupation laws}\label{sec:loops}

The count laws are given by the coefficients of $\Zqt$, and two
mechanisms make these coefficients nonnegative. If the kernel is
entrywise nonnegative, the trace expansion is a sum of positive
closed-walk weights, realized by a Poisson collection of weighted
walks, a length-weighted loop soup \citep{li2023note}, in which $w_k$
enters through the intensity of walks of length $k$. If the kernel is
positive semidefinite of rank $d$, the Schur signs of
Section~\ref{sec:prelim} give nonnegativity on $\mathcal R_d(q)$,
whatever the signs of its entries. We begin with the first.

For the loop construction, let $K\in M_N(\R)$ be entrywise nonnegative
with $\rho(K)<1$. Symmetry and positive semidefiniteness are not
required. A closed word of length $k\ge1$ is an ordered tuple
$\ell=(i_1,\ldots,i_k)\in[N]^k$, with $i_{k+1}=i_1$, and its
weight is
\[
 K(\ell):=\nlprod_{a=1}^kK_{i_ai_{a+1}}.
\]
Cyclic rotations are separate words, and the factor $1/k$ goes
into the intensity, so that
$\nlsum_{\ell\in[N]^k}K(\ell)/k=\tr K^k/k$ with no separate
convention for periodic words. With these conventions, a Poisson
collection of words realizes the coefficients of $\Zqt^\beta$ as
probabilities.

\begin{theorem}[A positive length-weighted loop law]\label{thm:loops}
Let $0\le q,t<1$ and $\beta>0$. For each closed word $\ell$ of
length $k$, take an independent Poisson variable $M_\ell$ of mean
\begin{equation*}
 \frac{\beta w_k}{k}K(\ell).
\end{equation*}
Let $N_i$ count all occurrences of vertex $i$ in these words, with
each word counted $M_\ell$ times. Almost surely only finitely many
words are sampled, and, for $0\le s_i\le1$,
\begin{equation}\label{eq:loop-pgf}
 \E\nlprod_i s_i^{N_i}
 =\frac{\Zqt(D_sK)^\beta}{\Zqt(K)^\beta},
 \qquad D_s:=\diag(s_1,\ldots,s_N).
\end{equation}
In particular, the vector $\mathbf N=(N_i)$ has the law
\begin{equation}\label{eq:count-law}
 \Prob(\mathbf N=\mathbf m)
 =\frac{\per_{q,t}^{(\beta)}(K[\mathbf m])}
 {\mathbf m!\,\Zqt(K)^\beta},
 \qquad \mathbf m\in\N^N.
\end{equation}
These occupation laws are infinitely divisible in $\beta$.
\end{theorem}
\begin{proof}
Since $0<w_k\le(1-q)^{-1}$ and $0\le\tr K^k\le N\rho(K)^k$,
the total intensity
\[
 \nlsum_\ell\E M_\ell
 =\beta\nlsum_{k\ge1}\frac{w_k}{k}\tr K^k
 =\beta\log\Zqt(K)
\]
is finite, so almost surely finitely many words are sampled and
every $N_i$ is finite.

For a word $\ell$, write
$s(\ell)=\nlprod_{a=1}^{|\ell|}s_{i_a}$. The Poisson generating
function and independence give
\[
 \log\E\nlprod_i s_i^{N_i}
 =\nlsum_\ell \E M_\ell\bigl(s(\ell)-1\bigr)
 =\beta\nlsum_{k\ge1}\frac{w_k}{k}
 \left(\tr(D_sK)^k-\tr K^k\right).
\]
The series is absolutely convergent because $|s(\ell)-1|\le1$,
and exponentiating proves \eqref{eq:loop-pgf}. Extracting coefficients in the
master theorem with cycle weights $\beta w_k$ proves
\eqref{eq:count-law}. Finally, superposing independent word
families adds intensities, so the law with parameter $\beta$ is
the $r$-fold convolution of the law with parameter $\beta/r$.
\end{proof}

Theorem~\ref{thm:loops} needs nonnegative entries, but a positive
semidefinite $2\times2$ kernel becomes entrywise nonnegative after a
diagonal change of phase, which leaves every closed-walk product
unchanged. Two distinct Gram vectors, repeated arbitrarily often,
therefore give a count law throughout the square, the two-type
counterpart of \citet[Proposition~3.10]{shirai2007remarks}.

\begin{corr}[A two-type infinitely divisible count law]\label{cor:two-type-counts}
Let
$K=\left(\begin{smallmatrix}a&b\\\overline b&d\end{smallmatrix}\right)\succeq0$
with $\norm K<1$, and let $0\le q,t<1$ and $\beta>0$. Then
\begin{equation}\label{eq:two-type-count-law}
 \Prob(N_1=p,N_2=m)
 =\frac{\per_{q,t}^{(\beta)}(K[p,m])}
        {p!\,m!\,\Zqt(K)^\beta},\qquad p,m\in\N,
\end{equation}
defines a compound-Poisson distribution, where $K[p,m]$ repeats
the first index $p$ times and the second $m$ times. Its generating
function is
\[
 \E[s_1^{N_1}s_2^{N_2}]
 =\frac{\Zqt(\diag(s_1,s_2)K)^\beta}{\Zqt(K)^\beta}.
\]
In particular, this distribution is infinitely divisible.
\end{corr}
\begin{proof}
Choose a diagonal unitary matrix $U$ such that $UKU^*$ has
off-diagonal entries $|b|$; if $b=0$, take $U=I$.
Its diagonal entries are $a,d\ge0$, so $UKU^*$ is entrywise
nonnegative. Diagonal unitary conjugation preserves every
closed-walk product and the spectral radius.
Since $\rho(K)=\norm K<1$, Theorem~\ref{thm:loops} applies to
$UKU^*$ and gives the law and its compound-Poisson construction.
\end{proof}

With three or more types, the phases of a positive semidefinite
kernel cannot in general be removed, so the corollary says nothing
about general Gram matrices of rank two, and their coefficients need
another source of positivity.

\subsection{Positive semidefinite kernels}

For $\beta=1$, the Schur coefficient signs give a second route to
nonnegative coefficients, one that allows negative or complex
entries.
\begin{corr}\label{cor:PSD-counts}
Let $d\ge1$ be an integer, and suppose $K\in M_N(\C)$ satisfies
$K\succeq0$, $\norm K<1$, and $\rank K\le d$.
If $0<q<1$ and $t\in\mathcal R_d(q)$, then
\[
 \frac{\Pqt(K[\mathbf m])}{\mathbf m!\,\Zqt(K)},
 \qquad \mathbf m\in\N^N,
\]
is a probability distribution with generating function
$\Zqt(D_sK)/\Zqt(K)$.
\end{corr}
\begin{proof}
Write $K_{ij}=\langle v_i,v_j\rangle$ with $v_i\in\C^d$.
Repeating the vectors according to $\mathbf m$ represents
$K[\mathbf m]$ as a Gram matrix in the same space. Lemma~\ref{lem:Gram-positivity}
therefore makes every coefficient nonnegative. In degree
$n=|\mathbf m|$, only partitions with $\ell(\lambda)\le\min(n,d)$
occur, and $\mathcal R_d(q)\subseteq\mathcal R_{\min(n,d)}(q)$.
The one-variable series $\Zqt(uK)$ has radius of convergence
greater than $1$, since $\norm K<1$. Applying the master theorem
degree by degree and setting $u=1$ gives
\[
 \nlsum_{\mathbf m}\frac{\Pqt(K[\mathbf m])}{\mathbf m!}
 =\Zqt(K).
\]
For $0\le s_i\le1$, nonnegativity lets us evaluate the series at
$s$, proving the last claim.
\end{proof}

Corollary~\ref{cor:PSD-counts} produces the distribution but no
mechanism behind it, such as a mixing law, that is, a random positive
semidefinite matrix given which the counts are Poisson. On
$\mathcal G(q)$ there is one. For $t=q^m$, or $t=0$ with $m=\infty$,
let $W=W_{q,m}(K)$; tilting this Gaussian matrix by
$e^{\tr W}/\Zqt(K)$ and sampling independent Poisson counts with
means $W_{ii}$ gives the coefficient law, and the weighted colors then
have covariances
\[
 C_r=q^rK(I-q^rK)^{-1}.
\]
Theorem~\ref{thm:tilt} proves this and gives the factorial moments.
Infinite divisibility of the count law is a separate question from
that of this mixing law.

\subsection{Infinite divisibility of the coefficient law}
Which coefficient laws are infinitely divisible? The answer is decided
by the logarithm of the generating function, and there the cycle weights enter only as positive factors
$w_{|\mathbf m|}$ in front of closed-walk sums of $K$. The matrix
condition therefore separates from $(q,t)$. In the classical case
this is the multivariate negative-binomial problem of
\citet{griffiths1987class}, and general matrix criteria require care
with zero entries \citep[Section~3]{eisenbaum2017existence}; for
Hermitian matrices the trace expansion gives a direct criterion.
For $K\in M_N(\C)$ with $K\succeq0$ and $\norm K<1$, and
for $\mathbf m\in\N^N\setminus\{\mathbf0\}$, define
\begin{equation*}
 \Gamma_{\mathbf m}(K)
 =\frac1{|\mathbf m|}
   [s^{\mathbf m}]\tr((D_sK)^{|\mathbf m|}),
 \qquad D_s=\diag(s_1,\ldots,s_N).
\end{equation*}
Reversing an ordered closed walk conjugates its weight, so
$\Gamma_{\mathbf m}(K)$ is real. It sums the weights of all
closed walks with occupation vector $\mathbf m$, including the
factor $1/|\mathbf m|$ from the trace logarithm.
\begin{theorem}[Infinite divisibility of coefficient laws]
\label{thm:coefficient-ID}
Let $K\in M_N(\C)$ satisfy $K\succeq0$ and $\norm K<1$,
and let $0\le q,t<1$. Set
\[
 F_{q,t;K}(s)=\frac{\Zqt(D_sK)}{\Zqt(K)}.
\]
The following conditions are equivalent:
\begin{enumerate}[label=\textup{(\roman*)}]
 \item $F_{q,t;K}$ is the probability generating function of
 an infinitely divisible distribution on $\N^N$.
 \item $F_{q,t;K}^{\,\beta}$ is a probability generating
 function for every $\beta>0$.
 \item $\Gamma_{\mathbf m}(K)\ge0$ for every
 $\mathbf m\in\N^N\setminus\{\mathbf0\}$.
\end{enumerate}
Here infinite divisibility is with respect to addition on
$\N^N$. When these conditions hold, the distribution in
\textup{(ii)} is compound Poisson, with jump $\mathbf m$
having rate
\begin{equation}\label{eq:coefficient-Levy-rates}
 \nu_\beta(\mathbf m)
 =\beta\,\frac{1-t^{|\mathbf m|}}{1-q^{|\mathbf m|}}\,
   \Gamma_{\mathbf m}(K).
\end{equation}
In particular, the class of matrices satisfying the conditions
is independent of $(q,t)\in[0,1)^2$.
\end{theorem}
\begin{proof}
Choose $R>1$ with $R\norm K<1$.
On the polydisk $|s_i|<R$ the trace logarithm is analytic and
\[
 -\log\det(I-D_sK)
 =\nlsum_{k\ge1}\frac1k\tr((D_sK)^k)
 =\nlsum_{\mathbf m\ne\mathbf0}
   \Gamma_{\mathbf m}(K)s^{\mathbf m}.
\]
The last series is absolutely summable on $|s_i|\le1$:
Cauchy estimates on any smaller polydisk of radius greater
than one bound the coefficients by a summable geometric
sequence. Since $0<w_k\le(1-q)^{-1}$, the weighted series
is absolutely summable there as well. Consequently
\begin{equation*}
 \log F_{q,t;K}(s)
 =\nlsum_{\mathbf m\ne\mathbf0}
   w_{|\mathbf m|}\Gamma_{\mathbf m}(K)
   (s^{\mathbf m}-1).
\end{equation*}
If \textup{(iii)} holds, this is the exponent of a compound
Poisson generating function with finite total jump rate.
Multiplication by $\beta>0$ gives
\eqref{eq:coefficient-Levy-rates} and proves \textup{(ii)}.
Condition \textup{(ii)} implies \textup{(i)} by taking
$\beta=1/r$ for every positive integer $r$.
Conversely, under \textup{(i)}, let $G_r$ generate an $r$th
convolution root. Since
$G_r(0)^r=F_{q,t;K}(0)>0$, the root with positive constant
term is unique near zero:
\[
 G_r(s)=\exp\left(\frac1r\log F_{q,t;K}(s)\right).
\]
For each fixed $\mathbf m\ne\mathbf0$, coefficient extraction
at zero therefore gives
\[
 [s^{\mathbf m}]G_r(s)
 =\frac1r w_{|\mathbf m|}\Gamma_{\mathbf m}(K)
   +O(r^{-2}).
\]
The left side is nonnegative. Multiplying by $r$ and letting
$r\to\infty$ proves
$w_{|\mathbf m|}\Gamma_{\mathbf m}(K)\ge0$.
Every $w_k$ is strictly positive on $[0,1)^2$, so
\textup{(iii)} follows, and with it parameter independence.
\end{proof}
For real symmetric matrices, the coefficient criterion has a
finite graph-theoretic interpretation. Its classical form is
contained in the Griffiths--Milne theory and in
\citet[Theorem~3.2]{eisenbaum2017existence}. The following proof
also covers zero entries and reducible matrices.
\begin{corr}[Real symmetric matrices]\label{cor:real-coefficient-ID}
Let $K\in M_N(\R)$ be symmetric, $K\succeq0$, and $\norm K<1$.
The equivalent conditions of Theorem~\ref{thm:coefficient-ID}
hold if and only if there is a diagonal matrix
$S=\diag(\varepsilon_1,\ldots,\varepsilon_N)$, with
$\varepsilon_i\in\{-1,1\}$, such that $SKS$ is entrywise
nonnegative.
\end{corr}
\begin{proof}
Signature conjugation preserves every closed-walk product,
because each vertex sign occurs an even number of times.
Thus an entrywise nonnegative $SKS$ makes all
$\Gamma_{\mathbf m}(K)$ nonnegative.
For the converse, form the undirected graph with an edge
$\{i,j\}$ when $i\ne j$ and $K_{ij}\ne0$. Suppose that some
simple cycle has a negative product of edge entries, and choose
one of minimum length $\ell\ge3$. This cycle has no chord.
Indeed, a chord splits it into two shorter simple cycles, and
the product of their signs equals the sign of the original
cycle, because the chord appears twice. One of the shorter
cycles would therefore have negative sign.

Let $V$ be the vertex set of this chordless cycle and let
$\mathbf1_V$ be its indicator vector. A term contributing to
$[s^{\mathbf1_V}]\tr((D_sK)^\ell)$ visits every vertex in $V$
exactly once. The induced graph on $V$ is the cycle itself,
so these terms consist of its two orientations and $\ell$
choices of a starting vertex. No diagonal step or backtracking
walk can contribute to this squarefree monomial. Hence
\[
 \Gamma_{\mathbf1_V}(K)
 =2\nlprod_{\{i,j\}\text{ in the cycle}}K_{ij}<0,
\]
contrary to Theorem~\ref{thm:coefficient-ID}.
All simple cycles therefore have positive sign. In each
connected component, choose a spanning tree and assign vertex
signs successively from a root so that every tree edge becomes
positive under signature conjugation. Each remaining edge
closes a simple cycle with its tree path. Positivity of that
cycle makes the remaining edge positive as well. The diagonal
entries are nonnegative because $K\succeq0$. Combining the
signatures of the components gives $S$.
\end{proof}

The criterion concerns the count kernel $K$, not the covariance of
a mixing law, and it differs from the inverse-$M$-matrix criterion
for continuous multivariate gamma laws: Appendix~\ref{app:count-mixing}
gives a kernel whose counts are infinitely divisible while their
Gaussian mixing law is not.

\subsection{Correlation functions and Cox processes}\label{sec:cox}

We next ask where $\Pqt$ is the correlation function of a point
process, its third probabilistic reading after moments and
probabilities. A Cox process is a
Poisson point process whose intensity is itself random, and when the
intensity is the squared modulus of a Gaussian field one obtains the
permanental (boson) processes of \citet{shirai2003random} and
\citet{mccullagh2006permanental}, whose correlation functions are
$\alpha$-permanents of the covariance kernel. The Gaussian colors of
Section~\ref{sec:wishart} give such processes with $\Pqt$ in place of
$\per_\alpha$ on $\mathcal G(q)$, and the scalar law gives them for rank-one kernels throughout
the square.

The construction runs on a general state space. Let $E$ be a
locally compact, second-countable Hausdorff space with a diffuse Radon measure
$\mu$. Let $g$ be a jointly measurable centered circular complex
Gaussian field, with covariance
$K(x,y)=\E[g(x)\overline{g(y)}]$. Assume that the integral operator
of $K$ is positive and locally trace class and that
$\int_BK(x,x)\,d\mu(x)<\infty$ for every relatively compact
Borel set $B$.
Its restriction to $B$ is an $L^2(B,\mu)$-valued Gaussian random
variable, and its covariance operator $K_B=1_BK1_B$ satisfies
\[
 \tr K_B=\E\int_B|g(x)|^2\,d\mu(x)
        =\int_BK(x,x)\,d\mu(x).
\]
We use this covariance version of the kernel throughout.
A correlation function $\rho_n$ is the density of the $n$th
factorial moment measure:
\[
 \E\nlsum_{(x_1,\ldots,x_n)\in\Xi_{\ne}^n}F(x_1,\ldots,x_n)
 =\int_{E^n}F(x_1,\ldots,x_n)\rho_n(x_1,\ldots,x_n)\,d\mu^n,
\]
for nonnegative measurable $F$, where the sum runs over ordered
tuples of distinct points. Density formulas hold almost
everywhere. The Gaussian colors then give the following process.

\begin{theorem}[Cox realization on the geometric curves]\label{thm:cox}
Take independent copies $g_r$ of $g$ and define
\begin{equation*}
 \Lambda(x)=\nlsum_{r=0}^{m-1}q^r|g_r(x)|^2,
\end{equation*}
where $0\le q<1$, $m$ is a positive integer or $\infty$, and
$t=q^m$, or $t=0$ if $m=\infty$. Given $\Lambda$, let $\Xi$ be
Poisson with intensity $\Lambda(x)\,d\mu(x)$. Then $\Xi$ is locally finite and simple, with
correlations
\begin{equation}\label{eq:cox-rho}
 \rho_n(x_1,\ldots,x_n)
 =\Pqt([K(x_i,x_j)]_{i,j=1}^n).
\end{equation}
For a nonnegative measurable compactly supported function $f$,
put $h=1-e^{-f}$ and $K_h=M_{\sqrt h}KM_{\sqrt h}$, where
$M_v$ denotes multiplication by $v$. The
Laplace functional is
\begin{equation}\label{eq:cox-L}
 \E e^{-\langle f,\Xi\rangle}
 =\nlprod_{r=0}^{m-1}\det(I+q^rK_h)^{-1}
 =\nlprod_{r\ge0}\frac{\det(I+tq^rK_h)}
                         {\det(I+q^rK_h)}.
\end{equation}
For $m=\infty$, the middle product is infinite.
\end{theorem}
\begin{proof}
The expected integrated intensity on $B$ is
\[
 \E\int_B\Lambda\,d\mu
 =\left(\nlsum_{r=0}^{m-1}q^r\right)\tr K_B
 =w_1\tr K_B<\infty.
\]
Thus $\Lambda\,d\mu$ is almost surely locally finite,
simultaneously on a countable exhaustion by relatively compact
sets, and diffuse, so the conditional Poisson process is simple.

Conditional Poisson factorial moments give
$\rho_n(x_1,\ldots,x_n)=\E\nlprod_i\Lambda(x_i)$.
For a finite collection of locations, the vectors
$(g_r(x_1),\ldots,g_r(x_n))$ have covariance
$[K(x_i,x_j)]$. Theorem~\ref{thm:wishart} therefore proves
\eqref{eq:cox-rho}. In the infinite-color case this application
is justified by monotone convergence of products of the
nonnegative partial intensities.
The bound \eqref{eq:upper} gives
\[
 \int_{B^n}\rho_n\,d\mu^n
 \le n!H_n(q,t)(\tr K_B)^n<\infty,
\]
so all local factorial moment measures are finite.

For the Laplace functional, conditional Poisson expectation
gives $\E\exp(-\int h\Lambda\,d\mu)$. The random function
$\sqrt h\,g_r$ is an $L^2$ Gaussian variable with trace-class
covariance $K_h$. If $\lambda_j$ are its nonzero covariance
eigenvalues, its squared norm has the law
$\nlsum_j\lambda_j|\xi_j|^2$, where $\xi_j$ are independent
standard circular complex Gaussians. Finite spectral truncation
and bounded convergence give
\[
 \E\exp\left(-q^r\int h|g_r|^2\,d\mu\right)
 =\nlprod_j(1+q^r\lambda_j)^{-1}
 =\det(I+q^rK_h)^{-1}.
\]
Independence proves the color product. Since
$0\le\log\det(I+q^rK_h)\le q^r\tr K_h$, its logarithm
converges absolutely when there are infinitely many colors.
For finite $m$, cancellation of the numerator factors at indices
$r+m$ gives the last expression in \eqref{eq:cox-L}; the
remaining tail tends to $1$ by the same trace bound.
\end{proof}

Each color contributes an independent permanental process with
kernel $q^rK$, so $\Xi$ is their superposition. For fixed $m$,
letting $q\uparrow1$ in the finite-degree correlations recovers
$\per_m([K(x_i,x_j)])$, which in the Shirai--Takahashi convention is
$\det_{1/m}([mK(x_i,x_j)])$.

The scalar law $S_{q,t}$ of Section~\ref{sec:scalar} exists for
every $(q,t)$ in the square, not only on the curves, and for a
rank-one covariance kernel it supplies the random intensity
directly.
\begin{prop}[Rank-one Cox representation]\label{prop:rankone-Cox}
Suppose $K(x,y)=a(x)\overline{a(y)}$, with $a$ measurable and
$|a|^2$ locally integrable. For $0\le q,t<1$, the Cox
process with random intensity $S_{q,t}|a(x)|^2\,d\mu(x)$ has
correlations $\Pqt([K(x_i,x_j)])$. Replacing $S_{q,t}$ by
$S_{q,t}^{(\beta)}$ gives
$\per_{q,t}^{(\beta)}([K(x_i,x_j)])$ for $\beta>0$.
\end{prop}
\begin{proof}
The scalar variable has finite mean $w_1$, or $\beta w_1$ for
its time-$\beta$ extension. Its intensity is therefore almost
surely locally finite. Conditional Poisson moments give the
$n$th correlation as
$\E S_{q,t}^n\nlprod_i|a(x_i)|^2$. By
Proposition~\ref{thm:scalar} this is
$n!H_n(q,t)\nlprod_i|a(x_i)|^2$, which equals the matrix
function by Proposition~\ref{prop:normalizations-local}. The
moments $\E(S_{q,t}^{(\beta)})^n=\per_{q,t}^{(\beta)}(J_n)$
give the final assertion.
\end{proof}

Further process formulas and boson--fermion superpositions, with
the required determinantal contraction hypotheses, are given in
the supplement \citep{sra2026macpermSuppII}.

\subsection{Schur coefficients as probabilities}\label{sec:processes}

A last probability model takes the Schur coefficients themselves as
the probabilities and survives on all of $\mathcal R_d(q)$, where
the matrix law is only conjectured to exist. On all partitions the
coefficients are nonnegative only on $\mathcal G(q)$, but a Schur
measure supported on partitions with at most $d$ rows needs only the
coefficients visible there, nonnegative on $\mathcal R_d(q)$. Its
determinantal structure survives the restriction, since the
correlation identity of
\citet{okounkov2001infinite} is an identity of symmetric functions,
and positivity enters only when it is evaluated.

Let $a=(a_1,\ldots,a_d)$ with $0\le a_i<1$, $0<q<1$, and
$t\in\mathcal R_d(q)$, and define
\begin{equation}\label{eq:Schur-measure}
 \Prob_{\mathrm{Schur}}(\lambda)
 =\frac{s_\lambda(a)c_\lambda(q,t)}
 {\mathcal Z_{q,t}(\diag a)}.
\end{equation}
Since $s_\lambda(a)=0$ for $\ell(\lambda)>d$, only
nonnegative coefficients from Proposition~\ref{prop:hook-local}
contribute.
The Schur Cauchy identity, applied degree by degree to the
specialization $p_k[\Bqt]=w_k$, gives
\[
 \nlsum_\lambda s_\lambda(a)c_\lambda(q,t)
 =\exp\left(\nlsum_{k\ge1}\frac{w_k}{k}p_k(a)\right)
 =\mathcal Z_{q,t}(\diag a)<\infty.
\]
To justify evaluation rather than only formal equality, first
replace $a$ by $ua$ with $0\le u<1$; the trace series is
analytic beyond $u=1$, and the nonnegative degree coefficients
permit passage to $u=1$. Thus \eqref{eq:Schur-measure} is a
probability law, and it is determinantal, by the following
specialization of \citet[Theorems~1 and~2]{okounkov2001infinite}.
\begin{theorem}[Determinantal correlations in the finite-rank region]
\label{thm:finite-rank-Schur}
Let $d\ge1$, $a=(a_1,\ldots,a_d)$ with $0\le a_i<1$,
$0<q<1$, and $t\in\mathcal R_d(q)$. Under
\eqref{eq:Schur-measure}, the random subset
\[
 X(\lambda)=\{\lambda_i-i+\tfrac12:i\ge1\}
 \subset\mathbb Z+\tfrac12
\]
has determinantal correlations, with the following kernel. Put
$a_*=\max_i a_i$ and
\begin{equation*}
 J(z)=\nlprod_{i=1}^d(1-a_i z)^{-1}
      \frac{(z^{-1};q)_\infty}{(t z^{-1};q)_\infty}.
\end{equation*}
Choose counterclockwise circles of radii $1<r_w<r_z<a_*^{-1}$,
with $a_*^{-1}=\infty$ if $a_*=0$. Define
\begin{equation}\label{eq:Schur-contour-kernel}
 \mathsf K(x,y)=\frac{1}{(2\pi\mathrm i)^2}
 \oint_{|z|=r_z}\oint_{|w|=r_w}
 \frac{J(z)}{J(w)}
 \frac{z^{-x-1/2}w^{y-1/2}}{z-w}\,dw\,dz.
\end{equation}
For every finite set of distinct half-integers $x_1,\ldots,x_k$,
\[
 \Prob\{x_1,\ldots,x_k\in X(\lambda)\}
 =\det[\mathsf K(x_i,x_j)]_{i,j=1}^k.
\]
Equivalently, the kernel is the Laurent coefficient
\begin{equation}\label{eq:Schur-coefficient-kernel}
 \mathsf K(x,y)
 =[z^{x-1/2}w^{-y-1/2}]
   \frac{J(z)}{J(w)(z-w)},
 \qquad
 \frac1{z-w}=\nlsum_{r\ge0}\frac{w^r}{z^{r+1}}.
\end{equation}
The Laurent expansions in this formula are taken in the annulus
$1<|w|<|z|<a_*^{-1}$.
\end{theorem}

Appendix~\ref{app:Schur-proof} proves the theorem, including the
analytic continuation behind the contour formula.

For $0<t<q^{d-1}$ away from the points $q^m$, the specialization
$\Bqt$ is not Schur-positive in all lengths, yet
Theorem~\ref{thm:finite-rank-Schur} still defines a determinantal
process; its points are partition coordinates, not the spatial points
of the Cox process.

The size of the random partition connects back to
Section~\ref{sec:scalar}. Its generating function is
\[
 \E_{\mathrm{Schur}}u^{|\lambda|}
 =\nlprod_{i=1}^d\frac{H(ua_i)}{H(a_i)},
\]
so $|\lambda|$ is a sum of $d$ independent
$(q,t)$-negative-binomial variables $N_{q,t}(a_i)$. The Macdonald
measure with alphabets $a$ and $(1)$ has the same size distribution
but is supported on one-row partitions; the comparison is in
Supplement~II \citep{sra2026macpermSuppII}.

\section{Open questions}\label{sec:scope}

The results leave one question open for the matrix laws and one for
the count laws. The matrix Laplace problem that remains is the
$q$-analogue of Gindikin's theorem from the introduction, with
$\mathcal R_d(q)$ in place of the Wallach set.

\begin{dashedbox}
\begin{conj}[$q$-Gindikin]\label{conj:q-Gindikin}
For $d\ge1$, $0<q<1$, and $0\le t<1$, a random matrix
$W\succeq0$ in $M_d(\C)$ with
$\E e^{-\tr SW}=\mathcal Z_{q,t}(-S)$ for every $S\succeq0$
exists if and only if $t\in\mathcal R_d(q)$.
\end{conj}
\end{dashedbox}

Necessity is Corollary~\ref{cor:matrix-Laplace-necessary} and
dimension two is Theorem~\ref{thm:rank-two-Laplace}, so only
sufficiency in dimensions $d\ge3$, away from $\mathcal G(q)$,
remains. Two reformulations are available. In the continuous range
$t<q^{d-1}$, Proposition~\ref{prop:matrix-density-criterion} reduces
it to a determinant inequality for derivatives of the scalar density,
and Theorem~\ref{thm:basepoint} turns it into a family of positivity
conditions for tilted cycle weights.

For the count laws, the open question is a finite test for infinite
divisibility.

\begin{prob}[Infinitely divisible coefficient laws]
\label{prob:coefficient-ID}
Characterize the complex Hermitian positive semidefinite matrices
$K$ whose closed-walk sums $\Gamma_{\mathbf m}(K)$ are all
nonnegative.
\end{prob}

For real symmetric matrices,
Corollary~\ref{cor:real-coefficient-ID} gives the finite criterion
of signature conjugation to an entrywise nonnegative matrix.
An intrinsic complex-matrix criterion would replace the infinite
family of inequalities in Theorem~\ref{thm:coefficient-ID}.

\begin{aiusage}
The results in this paper were developed through extensive use of an AI
assistant (OpenAI's GPT~Pro). None of the claims were obtained zero-shot:
they emerged through sustained back-and-forth discussion, including failed
attempts, counterexamples, audits, and repairs. The expansion of the
author's notes into the present manuscript, together with its finite
symbolic verification program, also used OpenAI's models, and Anthropic's
Claude (Fable~5.1 and Opus~5.5) prepared this version of the manuscript. Assistant
checks and reruns are not external peer review or formal
verification.
\end{aiusage}

\aiacknowledgements

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\appendix
\addtocontents{toc}{\protect\setcounter{tocdepth}{1}}
\begingroup
\renewcommand{\contentsname}{Appendix contents}
\bigskip
\noindent\begin{minipage}{\linewidth}
\tableofcontents
\end{minipage}
\endgroup

\section{Analytic details for matrix Laplace transforms}\label{app:matrix-Laplace}

The classification in Section~\ref{subsec:cm} uses a representing
law only through its moments and its eigenvalue density, and both
need justification at the boundary of the positive cone. A
representing law must have the formal moments \eqref{eq:formal}, so
that the determinant-square obstruction applies to it, and the scalar
density must vanish to the right order at zero, so that inverting the
Harish-Chandra--Itzykson--Zuber integral leaves no mass on the
boundary. We also record two equivalent forms of complete
monotonicity, finite witnesses near the origin
(Corollary~\ref{cor:local-witness}) and a criterion at scalar
basepoints, which reformulate the open problem of
Section~\ref{sec:scope}.

\subsection{Trace bounds and boundary moments}\label{app:conventions}

For finite matrices, we use the formal and analytic domains of
Proposition~\ref{prop:master-local}.
For operators on an infinite-dimensional Hilbert space,
the determinants are Fredholm determinants of trace-class
perturbations of the identity.

If $T\succeq0$ is trace class and $a\ge0$, its eigenvalues
$\lambda_j\ge0$ satisfy
\[
 0\le\log\det(I+aT)
 =\nlsum_j\log(1+a\lambda_j)
 \le a\nlsum_j\lambda_j=a\tr T.
\]
Thus $\nlsum_rq^r<\infty$ controls the products over Gaussian
colors on every relatively compact set.
All point-process densities are understood
almost everywhere with respect to the reference measure.

\subsubsection{Polynomial moments of a representing law}\label{app:boundary-moments}
A representing law has the formal polynomial moments used in
Theorem~\ref{thm:cm-classification} and
Corollary~\ref{cor:matrix-Laplace-necessary}.

\begin{proof}[Justification of polynomial moments]
Set $T_0=\tr W\ge0$ and
$f(s)=\E e^{-sT_0}=\mathcal Z_{q,t}(-sI)$ for $s\ge0$.
The right side extends analytically to $|s|<1$. For $s>0$,
differentiation under the expectation gives
\[
 (-1)^kf^{(k)}(s)=\E T_0^ke^{-sT_0}.
\]
The derivative is justified because $x^ke^{-sx}$ is bounded on
$[0,\infty)$, locally uniformly for $s>0$. Monotone convergence as
$s\downarrow0$ then proves
$\E T_0^k=(-1)^kf^{(k)}(0)<\infty$.
For any fixed $0<R<1$, Cauchy's estimate gives
$|f^{(k)}(0)|\le M_R k!R^{-k}$, where
$M_R=\max_{|s|=R}|f(s)|$. Hence, for $0<a<R$,
\[
 \E e^{aT_0}
   =\nlsum_{k\ge0}\frac{a^k\E T_0^k}{k!}
   \le\frac{M_R}{1-a/R}<\infty.
\]

For a complex matrix $A$ with $\|A\|<a$, the estimate
$|\tr AW|\le\|A\|T_0$ makes
$F(A)=\E e^{\tr AW}$ analytic in all matrix entries.
The determinant expression $\mathcal Z_{q,t}(A)$ is analytic on
the same sufficiently small neighborhood of zero.
They agree when $A=-S$ is negative definite and small, by
hypothesis, and the identity theorem, applied coordinatewise in a
Hermitian basis of $M_d(\C)$, extends the equality to a complex
neighborhood of zero.

Consequently, every polynomial moment is finite and is obtained by
differentiating $\mathcal Z_{q,t}$ at zero. These are exactly the
formal moments in \eqref{eq:formal}; in particular, this identifies
the determinant-square moment used in the main proof.
\end{proof}

\subsection{The scalar density and the derivative criterion}
\label{app:density}

The rank-two argument requires the two analytic inputs of
Theorem~\ref{thm:rank-two-Laplace}: the exponential series for the
density of $S_{q,t}$, with the boundary values of its derivatives,
and the derivative criterion for the matrix law. Write
\[
 \varphi_{q,t}(s)=\frac{(-ts;q)_\infty}{(-s;q)_\infty},
 \qquad s\ge0,
\]
for the Laplace transform of $S_{q,t}$.

\begin{lemma}[The scalar density]\label{lem:scalar-density-series}
Let $0<q<1$ and $0\le t<1$. The law of $S_{q,t}$ has a strictly
positive smooth density on $(0,\infty)$ given by
\begin{equation}\label{eq:scalar-density-series}
 \psi_{q,t}(x)=\nlsum_{r\ge0}c_r(q,t)e^{-q^{-r}x}.
\end{equation}
For $t>0$, the coefficients are
\begin{equation}\label{eq:scalar-density-coefficients}
 c_r(q,t)=\frac{(t;q)_\infty}{(q;q)_\infty}
       \left(\frac tq\right)^r\frac{(q/t;q)_r}{(q;q)_r};
\end{equation}
at $t=0$, they are
\[
 c_r(q,0)=\frac{(-1)^r q^{r(r-1)/2}}
                  {(q;q)_\infty(q;q)_r}.
\]
The series converges in $L^1(0,\infty)$, and it can be
differentiated term by term on every compact subinterval of
$(0,\infty)$. If $k\ge0$ and $t<q^{k+1}$, its $k$th derivative
extends continuously to zero and
\begin{equation}\label{eq:scalar-density-boundary}
 \psi_{q,t}^{(k)}(0)=0.
\end{equation}
If $d\ge2$ and $t<q^{d-1}$, then
\begin{equation}\label{eq:scalar-density-integrability}
 \int_0^\infty x^\ell
       \bigl|\psi_{q,t}^{(j)}(x)\bigr|\,dx<\infty,
 \qquad \ell\ge0,\quad 0\le j\le d-1,
\end{equation}
for every integer $\ell$.
\end{lemma}
\begin{proof}
For $t>0$, the finite products in
\eqref{eq:scalar-density-coefficients} are bounded in absolute
value, while $(q;q)_r\ge(q;q)_\infty>0$. Consequently
$|c_r(q,t)|\le C(t/q)^r$ for a constant depending on $q,t$.
At $t=0$, the displayed formula decays faster than any geometric
sequence. These bounds give
\[
 \nlsum_{r\ge0}|c_r(q,t)|q^r<\infty
\]
and prove $L^1$ convergence of \eqref{eq:scalar-density-series}.
The additional factor $e^{-q^{-r}x}$ proves local uniform
convergence of every differentiated series for $x>0$.

To identify this initially signed density, let $n\ge0$. The
$q$-binomial theorem gives, for $t>0$,
\begin{align*}
 \int_0^\infty x^n\psi_{q,t}(x)\,dx
 &=n!\frac{(t;q)_\infty}{(q;q)_\infty}
       \nlsum_{r\ge0}\frac{(q/t;q)_r}{(q;q)_r}(tq^n)^r\\
 &=n!\frac{(t;q)_\infty}{(q;q)_\infty}
       \frac{(q^{n+1};q)_\infty}{(tq^n;q)_\infty}
 =n!\frac{(t;q)_n}{(q;q)_n}.
\end{align*}
For $t=0$, Euler's identity
$\nlsum_{r\ge0}(-1)^rq^{r(r-1)/2}z^r/(q;q)_r=(z;q)_\infty$
gives the same result. All termwise integrations are absolute.
Moreover, the absolute value of the signed density has a finite
exponential moment of every order $a<1$, since
$\nlsum_r|c_r|/(q^{-r}-a)<\infty$. Its moment generating
function therefore agrees near zero with that of $S_{q,t}$.
Laplace-transform uniqueness identifies the two measures.

The defining independent series for $S_{q,t}$ shows that
$\Prob(S_{q,t}<\varepsilon)>0$ for every $\varepsilon>0$:
choose a sufficiently long tail with small expectation and
require each of the finitely many remaining terms to be small.
Conditioning on $B_0=1$ then gives, for $x>0$,
\[
 \psi_{q,t}(x)\ge
 (1-t)e^{-x}\Prob(qS_{q,t}<x)>0
\]
(almost everywhere by convolution, then everywhere by
continuity).

If $t<q^{k+1}$, the series for $\psi_{q,t}^{(k)}(0)$ converges
absolutely; for $t>0$ the $q$-binomial theorem gives
\[
 \psi_{q,t}^{(k)}(0)
 =(-1)^k\frac{(t;q)_\infty}{(q;q)_\infty}
       \frac{(q^{-k};q)_\infty}{(t/q^{k+1};q)_\infty}=0.
\]
Euler's identity proves the same assertion at $t=0$.
Finally, termwise integration bounds the integral in
\eqref{eq:scalar-density-integrability} by
\[
 \ell!\nlsum_{r\ge0}|c_r(q,t)|q^{r(\ell+1-j)}.
\]
For $t>0$, this sum is bounded by a geometric series with ratio
$tq^{\ell-j}<1$; for $t=0$, it converges for every $\ell,j$.
The differentiated series also decay exponentially at infinity.
These facts justify integration by parts with the boundary
values in \eqref{eq:scalar-density-boundary}.
\end{proof}

The derivative principle for invariant Hermitian ensembles relates
eigenvalues to diagonal entries
\citep[Proposition~4]{kieburg2023derivative}.
For the present transform, it is useful to construct a finite
signed measure first. Uniqueness then determines whether the
candidate is positive, including whether an additional boundary
component could occur.

\begin{prop}[Derivative criterion]
\label{prop:matrix-density-criterion}
Fix $d\ge2$, $0<q<1$, and $0\le t<q^{d-1}$. Set
\[
 \Delta(x)=\nlprod_{1\le i<j\le d}(x_j-x_i),
 \qquad J_d=\nlprod_{j=0}^{d-1}j!,
 \qquad D=\frac{d}{dx}.
\]
There exists a positive semidefinite random matrix $W\in M_d(\C)$
with Laplace transform
\[
 \E e^{-\tr SW}=\mathcal Z_{q,t}(-S),
 \qquad S\succeq0,
\]
if and only if
\begin{equation}\label{eq:matrix-density-criterion}
 \det\bigl[(-D)^{j-1}\psi_{q,t}(x_i)\bigr]_{i,j=1}^d
 \ge0,\qquad 0<x_1<\cdots<x_d.
\end{equation}
When it exists, the law is unique and unitarily invariant.
Its ordered eigenvalues have density
\begin{equation}\label{eq:matrix-eigenvalue-density}
 \rho_{q,t}^{(d)}(x)=\frac{\Delta(x)}{J_d}
       \det\bigl[(-D)^{j-1}\psi_{q,t}(x_i)\bigr]_{i,j=1}^d,
 \qquad 0<x_1<\cdots<x_d.
\end{equation}
In particular, it assigns no mass to matrices of deficient rank
or with repeated eigenvalues.
\end{prop}
\begin{proof}
The integrability bounds in
\eqref{eq:scalar-density-integrability} make the right side of
\eqref{eq:matrix-eigenvalue-density} absolutely integrable:
expand both the determinant and the Vandermonde into finitely
many products of one-variable functions. It therefore defines
a finite signed eigenvalue measure on the ordered chamber.
Average this measure over normalized Haar measure on $\mathrm U(d)$
through the map $(x,U)\mapsto U\diag(x)U^*$, and call the resulting
finite signed matrix measure $\mu$.

For pairwise distinct $s_i>0$, the normalized
Harish-Chandra--Itzykson--Zuber formula is
\begin{equation}\label{eq:hciz-normalization}
 \int_{\mathrm U(d)}
 e^{-\tr(\diag(s)U\diag(x)U^*)}\,dU
 =J_d\,\frac{\det[e^{-s_i x_j}]_{i,j=1}^d}
                  {\Delta(-s)\Delta(x)}.
\end{equation}
The formula is the analytic continuation of
\citet[equation~(2.29)]{kieburg2023derivative}.
Multiplying \eqref{eq:hciz-normalization} by
\eqref{eq:matrix-eigenvalue-density} cancels $J_d$ and
$\Delta(x)$. The product of the two remaining determinants is
symmetric in the $x_i$. Expanding both determinants and applying
Fubini's theorem (Andr\'eief's identity) gives
\begin{align*}
 \int e^{-\tr(\diag(s)W)}\,d\mu(W)
 &=\frac{\det\left[
       \int_0^\infty e^{-s_i x}
             (-D)^{j-1}\psi_{q,t}(x)\,dx
                    \right]_{i,j=1}^d}{\Delta(-s)}\\
 &=\frac{\det[(-s_i)^{j-1}\varphi_{q,t}(s_i)]_{i,j=1}^d}
              {\Delta(-s)}
 =\nlprod_{i=1}^d\varphi_{q,t}(s_i).
\end{align*}
The factor $d!$ in Andr\'eief's identity is canceled by restricting
the symmetric integral to the ordered chamber. Repeated
integration by parts uses
$\psi_{q,t}^{(k)}(0)=0$ for $0\le k\le d-2$.
Continuity extends the formula to all $s_i\ge0$.
At $s=0$, it shows that $\mu$ has total mass one; conjugation
invariance proves the required transform for every $S\succeq0$.

Laplace transforms on the positive cone determine finite signed
measures there, since the algebra spanned by
$W\mapsto e^{-\tr SW}$, $S\succ0$, is dense in $C_0$ of the cone
by Stone--Weierstrass.
Any probability law with the proposed transform must consequently
equal $\mu$. Since the ordered-eigenvalue map pushes $\mu$
back to the signed density
\eqref{eq:matrix-eigenvalue-density}, positivity is equivalent to
nonnegativity of that density almost everywhere.
Continuity on the open chamber makes this equivalent to
\eqref{eq:matrix-density-criterion}.
This argument also excludes any additional mass on the boundary.
\end{proof}

\subsection{Finite witnesses for the additive kernel}

The negative square moment is visible on finitely many points
near the origin: it is the leading term of a quadratic form in
the additive kernel $(S,S')\mapsto L(S+S')$, whose positive
definiteness in every dimension is another equivalent condition.
\begin{corr}[Finite local kernel witnesses]\label{cor:local-witness}
If $q^{m+1}<t<q^m$ and $r=m+2$, then arbitrarily close to zero
there exist positive definite $S_1,\ldots,S_N\in\pp^r$ with
\[
 \bigl[L^{(r)}_{q,t}(S_i+S_j)\bigr]_{i,j=1}^N\not\succeq0,
 \qquad N\le\binom{r^2+r}{r}.
\]
Positive definiteness of this kernel for every $r$ is equivalent to
the conditions of Theorem~\ref{thm:cm-classification}.
\end{corr}
\begin{proof}
Choose a real basis $B_1,\ldots,B_d$ of the Hermitian $r\times r$
matrices consisting of rank-one positive semidefinite matrices,
so $d=r^2$. There is a real homogeneous polynomial $p$ of degree
$r$ with $p(\tr(B_1W),\ldots,\tr(B_dW))=\det W$. Write
$p(z-\mathbf1)=\nlsum_\beta a_\beta z^\beta$ and set
$S_\beta=\varepsilon(I+\nlsum_j\beta_jB_j)$. The formal exponential
sum $\nlsum_\beta a_\beta e^{-\tr(S_\beta W)}$ equals
$e^{-\varepsilon\tr W}p(e^{-\varepsilon\tr(B_1W)}-1,\ldots,
e^{-\varepsilon\tr(B_dW)}-1)$, whose first nonzero homogeneous
term is $(-\varepsilon)^r\det W$; squaring and applying
\eqref{eq:formal} gives
\[
 \nlsum_{\beta,\gamma}a_\beta a_\gamma
       L^{(r)}_{q,t}(S_\beta+S_\gamma)
 =\varepsilon^{2r}\ell^{(r)}_{q,t}\bigl((\det W)^2\bigr)
       +O(\varepsilon^{2r+1}).
\]
The leading coefficient is negative by Theorem~\ref{thm:sign}, and
there are at most $\binom{d+r}{r}$ multi-indices $\beta$ of degree
at most $r$. On $\mathcal G(q)$ the representation
\eqref{eq:Gaussian-LT} makes the additive kernel positive
definite.
\end{proof}

\subsection{A scalar-basepoint criterion}

Complete monotonicity requires positivity of the derivatives at
every positive basepoint. At a scalar basepoint, these derivatives
are permanents with a new sequence of cycle weights.
For $s\ge0$, the tilted weights are

\begin{equation}\label{eq:tilt}
 v_k(s)=\nlsum_{j\ge0}\left[
 \left(\frac{q^j}{1+sq^j}\right)^k
 -\left(\frac{tq^j}{1+stq^j}\right)^k\right],
 \qquad k\ge1.
\end{equation}
Since $x\mapsto x/(1+sx)$ is increasing, every $v_k(s)$ is
positive, and $v_k(0)=w_k(q,t)$. Normalizing
\eqref{eq:Lqt} at a scalar basepoint gives
\[
 \frac{L^{(r)}_{q,t}(sI-X)}{L^{(r)}_{q,t}(sI)}
 =\exp\left(\nlsum_{k\ge1}\frac{v_k(s)}k\tr X^k\right),
 \qquad\|X\|_{\mathrm{op}}<s+1,
\]
The proof of the master theorem \eqref{eq:introZ} uses only the
exponential form of $\Zqt$, so for arbitrary cycle weights
$v=(v_k)_{k\ge1}$, with
$\per_v(A)=\nlsum_\sigma\nlprod_{C\in\Cyc(\sigma)}v_{|C|}\nlprod_ia_{i,\sigma(i)}$
and $A\in M_n(\C)$,
\begin{equation}\label{eq:general-master}
 [x_1\cdots x_n]
 \exp\left(\nlsum_{k\ge1}\frac{v_k}k\tr\bigl((XA)^k\bigr)\right)
 =\per_v(A),\qquad X=\diag(x_1,\ldots,x_n).
\end{equation}
Applied to the weights $v(s)$ and $A=[u_i^*u_j]$ with
$u_i\in\C^r$, it gives
\begin{equation}\label{eq:tiltderiv}
 (-1)^nD_{u_1u_1^*}\cdots D_{u_nu_n^*}L^{(r)}_{q,t}(sI)
 =L^{(r)}_{q,t}(sI)\per_{v(s)}(A).
\end{equation}
Since every positive definite $S$ lies below a scalar matrix,
scalar basepoints suffice.

\begin{theorem}\label{thm:basepoint}
For $r\ge1$, $0<q<1$, and $0\le t<1$, the following are
equivalent.
\begin{enumerate}[label=(\roman*),leftmargin=2.2em]
\item $L^{(r)}_{q,t}$ is completely monotone on $\pp^r$.
\item $\per_{v(s)}(A)\ge0$ for every $s\ge0$ and every
$A\succeq0$ of rank at most $r$, in every size.
\item The positivity in (ii) holds for $s$ in some unbounded
subset of $[0,\infty)$.
\end{enumerate}
More precisely, positivity at $s=s_0$ implies the alternating
derivative inequalities at every $S$ with $0\prec S\preceq s_0I$.
\end{theorem}
\begin{proof}
By \eqref{eq:tiltderiv}, spectral decomposition of the
directions, and multilinearity, positivity of $\per_{v(s)}$ on
rank at most $r$ is the set of alternating derivative
inequalities at $sI$. Given $0\prec S\preceq sI$, set
$H=sI-S\succeq0$. Since $\|H\|_{\mathrm{op}}<s+1$, Taylor expansion
at $sI$ gives
\[
 (-1)^nD_{B_1}\cdots D_{B_n}L(S)
 =\nlsum_{a\ge0}\frac{(-1)^{n+a}}{a!}
        D_{B_1}\cdots D_{B_n}D_H^aL(sI)\ge0,
\]
each summand being nonnegative. An unbounded set of values $s$
covers every positive definite $S$, so (iii) implies (i).
Clearly (i) implies (ii) and (ii) implies (iii), the case $s=0$
by continuity.
\end{proof}

\section{Gaussian tilting and continuous mixing}\label{app:matrix-moments}

Section~\ref{sec:loops} uses two facts about the Poisson mixture
behind the count laws on $\mathcal G(q)$, proved here: the tilted
Gaussian matrix is the mixing law, and infinite divisibility of the
counts is not inherited by it. The second fact is why
Theorem~\ref{thm:coefficient-ID} is a statement about counts only.

\subsection{Exponential tilting and coefficient laws}
On $\mathcal G(q)$, tilting the Gaussian law identifies its
Poisson mixture and the resolvent covariance of each color.

\begin{theorem}[Exponential tilting and the resolvent change of kernel]\label{thm:tilt}
Suppose $K\succeq0$ and $\norm K<1$. Let $0\le q<1$ and let
$W=W_{q,m}(K)$, where either $m$ is a positive integer and
$t=q^m$, or $m=\infty$ and $t=0$. Tilt the distribution of $W$ by
\begin{equation}\label{eq:Wishart-tilt}
 \frac{e^{\tr W}}{\Zqt(K)}.
\end{equation}
Conditional on the tilted matrix, take independent Poisson
variables of means $W_{ii}$. Their joint law is
\eqref{eq:count-law} at $\beta=1$. Under this tilt the weighted
Gaussian colors have covariance matrices
\begin{equation}\label{eq:resolvent-colors}
 C_r=q^rK(I-q^rK)^{-1},\qquad 0\le r<m.
\end{equation}
More precisely, if $i_1,\ldots,i_n$ is a list in which index $i$
occurs $m_i$ times, then
\[
 \E\nlprod_i(N_i)_{m_i}
 =\nlsum_{\phi:[n]\to\{0,\ldots,m-1\}}
   \nlprod_{r\in\operatorname{im}\phi}
   \per\bigl([(C_r)_{i_a i_b}]_{a,b\in\phi^{-1}(r)}\bigr).
\]
Here $(x)_j=x(x-1)\cdots(x-j+1)$, and for $m=\infty$ the
sum runs over maps into $\N$.
\end{theorem}
\begin{proof}
By the Gaussian transform in Theorem~\ref{thm:wishart},
$\E e^{\tr W}=\Zqt(K)<\infty$, so \eqref{eq:Wishart-tilt}
is normalized. Conditional Poisson expectation gives
\[
 \E\nlprod_i s_i^{N_i}
 =\frac{\E e^{\tr W}
          \exp(\nlsum_i(s_i-1)W_{ii})}{\Zqt(K)}
 =\frac{\Zqt(D_sK)}{\Zqt(K)}.
\]
The master theorem identifies the law from this generating function.
To identify the colors, diagonalize $K$ on its range. A Gaussian
coordinate of variance $\lambda>0$ has density
$(\pi\lambda)^{-1}\exp(-|z|^2/\lambda)$ on $\C$.
Multiplication by $\exp(q^r|z|^2)$ changes its variance to
$\lambda/(1-q^r\lambda)$; the inequality $\norm K<1$ ensures
that this variance is finite. Coordinates of variance zero remain zero.
Thus color $r$ has tilted covariance
$K(I-q^rK)^{-1}$, and multiplying the color itself by $q^{r/2}$
gives \eqref{eq:resolvent-colors}. The tilt factors over colors,
so their independence is preserved.

For infinitely many colors, let $W^{(R)}$ be the sum through
color $R$. The densities
$e^{\tr W^{(R)}}/\E e^{\tr W^{(R)}}$ converge in $L^1$ to
\eqref{eq:Wishart-tilt} by monotone convergence, so every finite
color marginal is the product of the tilts just computed. Moreover,
\[
 \nlsum_{r\ge0}\tr C_r
 \le\frac{\tr K}{1-\norm K}\nlsum_{r\ge0}q^r<\infty,
\]
so the tilted intensity remains finite. This bound and
$\sup_r\norm C_r<\infty$ also give a finite exponential moment
of its trace near zero, by the Gaussian determinant formula.
Thus all the factorial moments below are finite. Conditional Poisson
factorial moments equal products of the means. Expanding those
products over color assignments and applying the complex Gaussian
moment formula proves the stated permanent sum. For infinitely
many colors, the products of diagonal intensities increase to
their limits, so monotone convergence justifies the expansion.
\end{proof}

For one color, the probability coefficients involve
$\per(K[\mathbf m])$, whereas factorial moments involve
$\per(C[\mathbf m])$ with $C=K(I-K)^{-1}$. For several colors
the $C_r$ are not geometric multiples of one matrix, so the
resolvent change applies to each color separately.

\subsection{Counts and their continuous mixing laws}\label{app:count-mixing}

The criterion concerns the count kernel $K$. It is different
from the inverse-$M$-matrix criterion for a continuous
multivariate gamma law. For a real positive definite covariance
$C$, \citet{bapat1989infinite} characterizes infinite
divisibility of $\det(I+CD_u)^{-1/2}$ by the existence of a
signature $S$ such that $SC^{-1}S$ is an $M$-matrix.
In the symmetric positive definite case this means that its
off-diagonal entries are nonpositive. The same criterion
characterizes existence of $\det(I+CD_u)^{-\beta}$ for every
$\beta>0$; see also
\citet[Theorem~5.1]{eisenbaum2017existence}.
For example, take
\[
 K=\frac14
 \begin{pmatrix}2&1&1\\1&2&0\\1&0&2\end{pmatrix}.
\]
Its eigenvalues are $1/2$ and $(2\pm\sqrt2)/4$, so it is a
positive definite contraction. Its coefficient laws are
infinitely divisible by Corollary~\ref{cor:real-coefficient-ID}.
At the classical point $(q,t)=(0,0)$, the corresponding
Gaussian mixing covariance is $C=K(I-K)^{-1}$, and
\[
 C^{-1}=K^{-1}-I
 =\begin{pmatrix}3&-2&-2\\-2&2&1\\-2&1&2\end{pmatrix}.
\]
The product of its three off-diagonal entries is positive.
Signature conjugation preserves this product, whereas three
nonzero nonpositive off-diagonal entries have negative
product. Thus no signature makes $C^{-1}$ an $M$-matrix.
The counts are infinitely divisible, but their mixing law is
not.

\section{The finite-rank determinantal formula}\label{app:Schur-proof}

Theorem~\ref{thm:finite-rank-Schur} applies Okounkov's correlation
formula at a specialization that is not Schur-positive in all
lengths. The formula is an identity of symmetric functions, so it
extends degree by degree to formal specializations; what needs proof
is that it may be evaluated at the given parameters, and this is where
the restriction to $d$ rows enters.

\begin{proof}[Proof of Theorem~\ref{thm:finite-rank-Schur}]
For a specialization $B$ of the symmetric functions, write
\[
 H_B(v)=\nlsum_{j\ge0}h_j[B]v^j
       =\exp\left(\nlsum_{j\ge1}\frac{p_j[B]}jv^j\right).
\]
For ordinary finite alphabets $a,b$,
\citet[Theorems~1 and~2]{okounkov2001infinite} give the
correlation formula with
$J(z)=H_a(z)/H_b(z^{-1})$. His generating function is
$\sqrt{zw}\,J(z)/(J(w)(z-w))$, with coefficient
$z^xw^{-y}$; extracting this coefficient gives
\eqref{eq:Schur-contour-kernel}. In particular, the powers
in the integrand are integers, so the contour formula requires
no choice of a square root.
The identity extends to a formal specialization $B$ degree by
degree. To see this, replace $a$ by $ua$ and expand at $u=0$.
For a fixed set of coordinates, the coefficient of $u^n$ on the
probability side is a symmetric function of $B$ of degree $n$:
only partitions of size at most $n$ enter after division by the
normalizing series. The same statement holds on the determinant
side. Indeed, expand
\[
 H_{ua}(z)=\nlsum_{p\ge0}u^ph_p(a)z^p,
 \qquad
 H_{ua}(w)^{-1}=\nlsum_{s=0}^d(-u)^se_s(a)w^s,
\]
and expand $H_B(w^{-1})$, $H_B(z^{-1})^{-1}$, and $(z-w)^{-1}$
in their indicated powers. At a fixed power $u^n$, coefficient
extraction for $\mathsf K(x,y)$ leaves finitely many terms,
each of degree $n+y-x$ in $B$. In every determinant term the
coordinate differences sum to zero. Thus its coefficient of
$u^n$ has degree $n$ in $B$. Evaluation on finite alphabets
with at least $n$ variables is injective on symmetric functions
of degree $n$. Since the identity holds on those alphabets,
it holds before evaluation and hence after the substitution
$p_j[B]=w_j$.
It remains to justify evaluating the resulting formal identity
at $u=1$. For $B=\Bqt$,
\[
 H_B(v)=H(v)=\frac{(tv;q)_\infty}{(v;q)_\infty}.
\]
Both $J$ and $J^{-1}$ are analytic in $1<|z|<a_*^{-1}$.
Keep the two contours fixed and replace $J$ by
$J_u(z)=H_{ua}(z)/H(z^{-1})$. The contour integrals are
analytic for $|u|<(a_*r_z)^{-1}$, a disk containing $u=1$.
On the probability side, the unnormalized correlation series
converges absolutely for $|u|a_*<1$, because its absolute
value is bounded by
\[
 \nlsum_\lambda |u|^{|\lambda|}
 s_\lambda(a)c_\lambda(q,t)
 =\nlprod_{i=1}^d H(|u|a_i).
\]
Here the nonnegativity of $c_\lambda(q,t)$ is needed only for
$\ell(\lambda)\le d$. The normalizing function
$\nlprod_iH(ua_i)$ is analytic and nonzero on this disk.
The formal identity therefore gives equality of analytic
functions near $u=0$, and analytic continuation gives the
claimed equality at $u=1$. Formula
\eqref{eq:Schur-coefficient-kernel} follows from Cauchy's
coefficient formula and the nesting $r_w<r_z$.
\end{proof}

\end{document}