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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 I: Algebra and positivity},pdfauthor={Suvrit Sra}}
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\keywords{Macdonald permanent, alpha-permanent, immanant, kernel, Wallach set}

\begin{document}
\title[The Macdonald $(q,t)$-permanent I]{The Macdonald $(q,t)$-permanent I:\\Algebra and positivity}
\aiauthor
\begin{abstract}
  We introduce the Macdonald $(q,t)$-permanent, a two-parameter cycle-weighted permanent that contains
  the usual permanent and determinant, as well as the $\alpha$-permanent of Vere-Jones, as degenerate
  cases. We prove a MacMahon-type master theorem for its generating function, and determine when it
  induces a valid kernel on tuples of vectors. Nonnegativity of the $(q,t)$-permanent itself
  on semidefinite matrices turns out to be subtler: for $t=q^\alpha$ nonnegativity
  holds exactly at the positive Wallach exponents; for a fixed $q$ we determine the region in rank two
  and for higher ranks under an additional condition, while for small $q$ positivity can even reach
  beyond the kernel region. Whether positivity forces $t\in\{0,q,q^2,\ldots\}$ remains an intriguing open problem.
\end{abstract}
\preprintslink{macperm1}{29 October 2026}
\maketitle
\addtocontents{toc}{\protect\setcounter{tocdepth}{-1}}
\raggedbottom

\vspace*{-8pt}
\section{Introduction}

The $\alpha$-permanent of \citet{verejones1988generalization} weights
each cycle of a permutation by the same scalar $\alpha$. It
interpolates between the permanent and the determinant, and through
it gamma laws, Wishart moments, loop occupation fields, and
permanental point processes are expressed
\citep{verejones1997alpha,shirai2003random,mccullagh2006permanental,li2023note}.
In each of these uses it is evaluated on a positive semidefinite
matrix, and its value is a moment, so it must be nonnegative. It is
known that, for $\alpha>0$, $\per_\alpha\ge0$ on positive
semidefinite matrices of every size exactly when $\alpha$ is a
positive integer \citep[Theorem~2.3]{branden2012solutions}.

We introduce a deformation in which the weight also depends on the
cycle length, suggested by Macdonald's power-sum scalar product. For
$0\le q,t<1$ and $A=(a_{ij})\in M_n(\C)$ we 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_{k\in\Cyc(\sigma)}w_k(q,t),\qquad
 w_k(q,t):=\frac{1-t^k}{1-q^k}.
\end{equation}
Here $\Cyc(\sigma)$ includes fixed points, and $w_k=w_k(q,t)$. We call this function the
\emph{Macdonald $(q,t)$-permanent}; its dependence on cycle length
distinguishes it from the inversion-weighted $q$-permanent of
\citet{bapat1994inequalities} and from the quantum immanants of
\citet{okounkov1996quantum}. On $q=t$ all cycle weights are one, so
$\per_{q,q}=\per$, and for $t=q^\alpha$ they tend to $\alpha$ as
$q\uparrow1$, recovering $\per_\alpha$.

The key property of $\Pqt$ that we study is this paper is: \emph{for which $(q,t)$ is it nonnegative on
  positive semidefinite matrices?} Two identities make this question accessible. The first is a
MacMahon-type master theorem, where we introduce the generating function 
\begin{equation}\label{eq:introZ}
 \Zqt(M):=\exp\left(\nlsum_{k\ge1}\frac{w_k}{k}\tr M^k\right)
 =\nlprod_{j\ge0}\frac{\det(I-tq^jM)}{\det(I-q^jM)},
\end{equation}
which generates $\Pqt$ on matrices with repeated indices. Here $\Pqt$ is the square-free polarization of
its degree-$n$ component, the one-row Macdonald function $Q_{(n)}$. The second identity comes from
expanding $Q_{(n)}$ in Schur functions
\[
 \Pqt(A)=\nlsum_{\lambda\vdash n}c_\lambda(q,t)\Imm_\lambda(A),
 \qquad
 c_\lambda(q,t)=\nlprod_{(i,j)\in\lambda}
       \frac{q^{i-1}-tq^{j-1}}{1-q^{h_{ij}}},
\]
where $\Imm_\lambda$ is the irreducible immanant and $h_{ij}$
is a hook length (Section~\ref{sec:object}). 

For tuples $\mathbf u=(u_1,\ldots,u_n)$ and
$\mathbf v=(v_1,\ldots,v_n)$ of vectors in $\C^d$ we define the ``kernel''
\begin{equation}\label{eq:kernel}
 \mathscr K_{q,t}^{n,d}(\mathbf u,\mathbf v)
 :=\Pqt\bigl([\langle u_i,v_j\rangle]_{i,j=1}^n\bigr).
\end{equation}
The first problem asks whether~\eqref{eq:kernel} is a true kernel, i.e., whether it is positive
semidefinite; the second one asks if $\Pqt(A)\ge0$ for every positive semidefinite $A$. For $0<q<1$ and
$r\ge1$, put
\begin{equation*}
 \mathcal R_r(q):=
 \bigl([0,q^{r-1}]\cup\{q,q^2,\ldots,q^{r-2}\}\bigr)\cap[0,1).
\end{equation*}
Then, the first main result of our paper is the following.
\begin{dashedbox}
\begin{theorem}[Kernel region]\label{thm:kernel}
Let $n,d\ge1$, $0<q<1$, $0\le t<1$, and $r=\min(n,d)$.
The kernel \eqref{eq:kernel} is positive semidefinite if and only if
$t\in\mathcal R_r(q)$. In particular,
$\mathcal R_1(q)=[0,1)$. For every $t\le0$, the kernel is positive
semidefinite for all $n,d\ge1$.
\end{theorem}
\end{dashedbox}
The proof is Schur--Weyl duality: since $w_\sigma$ is a class
function, the operator $\nlsum_\sigma w_\sigma U_\sigma$ that permutes
the factors of $(\C^d)^{\otimes n}$ acts on each isotypic component as
a positive multiple of $c_\lambda(q,t)$, so positivity amounts to
$c_\lambda(q,t)\ge0$ for partitions with at most $d$ rows
(Section~\ref{sec:kernel}). The same sign test governs a matrix-ball
kernel and the total positivity of a scalar sequence
(Theorem~\ref{thm:ball}). As $r$ grows, the regions shrink to
$\{0,q,q^2,\ldots\}$ (Corollary~\ref{cor:all-rank}).

On $\mathcal R_r(q)$ the Schur expansion also gives scalar
positivity on matrices of rank at most $r$, in every size
(Theorem~\ref{thm:scalar-PSD}). Is this region also necessary? Along
$t=q^\alpha$ it becomes the positive part of the Wallach set for
$r\times r$ matrices \citep{faraut1990function},
$\mathcal W_r:=\{\alpha>0:\alpha\in\{1,2,\ldots\}\text{ or }\alpha\ge r-1\}$,
and across the whole deformation the answer is yes.
\begin{dashedbox}
\begin{theorem}[Positivity throughout the deformation]
\label{thm:deformation-wallach}
Fix $r\ge1$ and $\alpha>0$, and let $A$ range over the complex
positive semidefinite matrices of every size and of rank at most $r$.
The following are equivalent: (i) $\alpha\in\mathcal W_r$;
(ii) $\per_{q,q^\alpha}(A)\ge0$ for every $q\in(0,1)$ and every
such $A$; (iii) $\per_\alpha(A)\ge0$ for every such $A$. If
$\alpha\notin\mathcal W_r$, one such $A$ has
$\per_{q,q^\alpha}(A)<0$ for all $q$ in an interval $(q_0,1)$.
\end{theorem}
\end{dashedbox}
The proof passes to the limit $q\uparrow1$, where the negative
matrices of \citet[Theorem~2.3]{branden2012solutions} have rank at
most $\lfloor\alpha\rfloor+2$ (Section~\ref{subsec:deformation}).
Intersecting over $r$ extends the positive-exponent half of
Br\"and\'en's theorem across the deformation
(Corollary~\ref{cor:deformation-integers}).

At a fixed $q$ the picture is subtler. The region $\mathcal R_r(q)$ is still exact in rank two, and it
remains exact in every rank $r\ge3$ once $q$ exceeds an explicit threshold
(Theorem~\ref{thm:scalar-rank-two}, Corollary~\ref{cor:scalar-Haar-classification}). The negative
matrices come from random binary polynomials in rank two and from random projections in higher rank. For
small $q$, however, scalar positivity reaches beyond $\mathcal R_r(q)$ at every finite rank
(Theorem~\ref{thm:rank-budget}). Whether scalar positivity in all ranks forces $t\in\{0,q,q^2,\ldots\}$
remains open (Conjecture~\ref{conj:scalar}). The exact regions in sizes three and four are in
Appendix~\ref{app:size-three}.

Our tools are classical: Macdonald's symmetric-function theory \citep[Chapters I and VI]{macdonald1995},
the Schur-function and immanant correspondence of \citet{goulden1992immanants}, the Wallach set~\citep{faraut1990function}; and also the notable construction of
\citet{branden2012solutions}. The Jack limit of our master theorem is the $\alpha$-extension of
MacMahon's \citep{foata1988laguerre,verejones1988generalization}, and the block inequalities of
Theorem~\ref{thm:block-inequalities} transfer Lieb's permanent inequality to $\Pqt$ along Frenkel's
partition argument \citep{frenkel2009remarks}.

\section{Cycle weights and the Schur expansion}\label{sec:object}

Both positivity problems of the introduction turn on the signs of
the hook-content coefficients $c_\lambda(q,t)$, so we derive these
first. They appear once $\Pqt$ is read as a symmetric function: the
cycle weights $w_k$ are the power sums of a Macdonald specialization,
square-free polarization turns power sums into permutation sums, and
the coefficients are Schur functions evaluated at that specialization.

\subsection{Notation}

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.

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}.

Our convention for the Vere-Jones parameter is
\[
 \per_\alpha(A)=\nlsum_{\sigma\in S_n}\alpha^{c(\sigma)}
 \nlprod_i a_{i,\sigma(i)},
\]
where $c(\sigma)$ is the number of cycles. In particular,
$\per_1=\per$ and $\per_{-1}=(-1)^n\det$.
The alternative convention
$\det_\eta(A)=\nlsum_\sigma\eta^{n-c(\sigma)}
\nlprod_i a_{i,\sigma(i)}$ satisfies
$\det_\eta(A)=\eta^n\per_{1/\eta}(A)$ for $\eta\ne0$.
The standard Jack parameter is the reciprocal of $\alpha$
in the degeneration considered below.

\subsection{Polarization and the master theorem}\label{sec:master}

The definition \eqref{eq:def} sums over permutations, whereas
Macdonald's theory works with power sums, that is, with traces of
matrix powers. The bridge is polarization: evaluate a symmetric
function at $XA$, with $X=\diag(x_1,\ldots,x_n)$ a diagonal matrix
of variables, and keep the coefficient of $x_1\cdots x_n$.
The trace of a matrix power is a sum over closed walks on the index
set, whereas a permutation is a collection of disjoint cycles, that
is, of closed walks with distinct vertices. Extracting the coefficient
of $x_1\cdots x_n$ keeps exactly the walks whose vertices are
distinct. This observation transfers symmetric-function identities to matrix identities
without assuming that the matrix is diagonalizable.

For a square matrix $M$, we evaluate a symmetric function $f$ by the
algebra homomorphism
$p_k\mapsto\tr M^k$. If the eigenvalues of $M$, counted with algebraic
multiplicity, are $\xi_1,\ldots,\xi_N$, this evaluation agrees with
$f(\xi_1,\ldots,\xi_N)$. For a homogeneous symmetric function of degree
$n$ and a matrix $A\in M_n(\C)$, define its square-free matrix
polarization by
\[
 \mathfrak M_n(f)(A):=[x_1\cdots x_n]f(XA),
 \qquad X=\diag(x_1,\ldots,x_n).
\]

\begin{lemma}[Polarization dictionary]\label{lem:dictionary}
If $f=\nlsum_{\mu\vdash n}a_\mu p_\mu$, then
\begin{equation}\label{eq:dictionary}
 \mathfrak M_n(f)(A)=\nlsum_{\sigma\in S_n}
 z_{\rho(\sigma)}a_{\rho(\sigma)}
 \nlprod_{i=1}^n a_{i,\sigma(i)},
\end{equation}
where $\rho(\sigma)$ is the cycle type of $\sigma$. In particular,
$\mathfrak M_n(s_\lambda)=\Imm_\lambda$.
\end{lemma}
\begin{proof}
Expanding a trace gives
\[
 \tr((XA)^k)=
 \nlsum_{i_1,\ldots,i_k=1}^n
 x_{i_1}\cdots x_{i_k}
 a_{i_1i_2}a_{i_2i_3}\cdots a_{i_ki_1}.
\]
A term contributing to $[x_1\cdots x_n]p_\mu(XA)$ uses every label
exactly once across all trace factors. Each factor therefore specifies
a cycle on distinct labels, and different factors use disjoint labels.
For a fixed permutation of type $\mu=1^{m_1}2^{m_2}\cdots$, each
$k$-cycle has $k$ possible starting points in its trace. Its $m_k$
cycles of length $k$ can be assigned to the corresponding trace factors
in $m_k!$ orders. The resulting multiplicity is
$\nlprod_k k^{m_k}m_k!=z_\mu$, which proves
\eqref{eq:dictionary}.

The Frobenius identity gives
$\mathfrak M_n(s_\lambda)=\Imm_\lambda$, as in
\citet{goulden1992immanants}.
\end{proof}

The one-row Macdonald functions are characterized by the
generating function
\citep[Chapter VI, Sections 2 and 4]{macdonald1995}
\begin{equation}\label{eq:one-row}
 \nlsum_{n\ge0}Q_{(n)}(Z;q,t)u^n
 =\exp\left(\nlsum_{k\ge1}\frac{w_k}{k}p_k(Z)u^k\right).
\end{equation}
The identity holds over $\C(q,t)$ and specializes wherever its
coefficients are defined; the full Macdonald basis and its
polarizations are in Supplement~I \citep{sra2026macpermSuppI}.

The positivity arguments evaluate $\Pqt$ on Gram matrices whose
vectors repeat and on matrices of every size at once, and the
generating function $\Zqt$ does both: allowing repeated labels in
the polarization turns its Taylor coefficients into $\Pqt$ of the
matrices with repeated rows and columns.
\begin{dashedbox}[supporting]
\begin{prop}[Macdonald polarization and repeated indices]\label{thm:master}
For every $A\in M_n(\C)$,
\begin{equation}\label{eq:polarization}
 \Pqt(A)=[x_1\cdots x_n]Q_{(n)}(XA;q,t).
\end{equation}
For $A\in M_N(\C)$ and $\mathbf m=(m_1,\ldots,m_N)\in\N^N$,
let $A[\mathbf m]$ be the matrix whose rows and columns are indexed by
the pairs $(i,a)$ with $1\le i\le N$ and $1\le a\le m_i$, and whose
$((i,a),(j,b))$ entry is $a_{ij}$.
Then, as a formal power series,
\begin{equation*}
 \Zqt(XA)=\nlsum_{\mathbf m\in\N^N}
 \frac{\Pqt(A[\mathbf m])}{\mathbf m!}x^{\mathbf m},
 \qquad
 \mathbf m!:=\nlprod_{i=1}^N m_i!,\quad
 x^{\mathbf m}:=\nlprod_{i=1}^N x_i^{m_i}.
\end{equation*}
For $|q|<1$ and fixed $t\in\C$, the determinant product in
\eqref{eq:introZ} agrees with the trace exponential near $M=0$.
For $0\le q,t<1$, both define the same analytic function on
$\{M:\rho(M)<1\}$.

\end{prop}
\end{dashedbox}
\begin{proof}
Expanding the exponential in \eqref{eq:one-row} by cycle lengths gives
\[
 Q_{(n)}(Z;q,t)
 =\nlsum_{\mu\vdash n}\frac{w_\mu}{z_\mu}p_\mu(Z).
\]
Lemma~\ref{lem:dictionary} proves \eqref{eq:polarization}. Since
$Q_{(j)}(XA)$ is homogeneous of degree $j$ in $x$, the same coefficient
can be extracted from $\Zqt(XA)=\nlsum_{j\ge0}Q_{(j)}(XA;q,t)$.

To account for repeated labels, put $B=A[\mathbf m]$ and introduce
one variable $y_{ia}$ for each of its indices. Write
$Y=\diag(y_{ia})$ and $z_i=\nlsum_{a=1}^{m_i}y_{ia}$, with an empty sum
equal to zero. Expansion of the trace shows that
\[
 \tr((YB)^k)
 =\nlsum_{i_1,\ldots,i_k}
 z_{i_1}\cdots z_{i_k}
 a_{i_1i_2}\cdots a_{i_ki_1}
 =\tr((\diag(z_1,\ldots,z_N)A)^k).
\]
Consequently $\Zqt(YB)=\Zqt(\diag(z_1,\ldots,z_N)A)$.
In a monomial $z^{\mathbf k}$, the coefficient of
$\nlprod_{i,a}y_{ia}$ vanishes unless $\mathbf k=\mathbf m$; in that
case it equals $\mathbf m!$. The square-free identity for $B$ thus
gives
\[
 \Pqt(A[\mathbf m])
 =\mathbf m!\,[x^{\mathbf m}]\Zqt(XA),
\]
including $\mathbf m=\mathbf0$ by the empty-matrix convention.

For the determinant identity, take $M$ sufficiently small that
$\|M\|<1$ and $|t|\|M\|<1$. Absolute convergence permits the expansion
\begin{align*}
 \log\nlprod_{r\ge0}
 \frac{\det(I-tq^rM)}{\det(I-q^rM)}
 &=\nlsum_{r\ge0}\nlsum_{k\ge1}
 \frac{q^{rk}-(tq^r)^k}{k}\tr M^k\\
 &=\nlsum_{k\ge1}\frac{1-t^k}{1-q^k}
 \frac{\tr M^k}{k}.
\end{align*}
with the logarithms that vanish at $M=0$. Exponentiation proves
the product formula near zero.

For $0\le q,t<1$ and $\rho(M)\le a<1$, $0<w_k\le(1-q)^{-1}$ and
$|\tr M^k|\le Na^k$, so the trace series converges locally
uniformly on $\rho(M)<1$ and its exponential is analytic there;
the determinant product converges locally uniformly as well, its
factors differing from $1$ by $O(q^r)$, and the two agree by
analytic continuation from zero. At $q=0$ only the factor $r=0$
remains.
\end{proof}

The rank-one case of the master theorem computes the constant
$H_n(q,t)$ below, which is the one-row Schur coefficient and hence the
constant in the scalar lower bound of Theorem~\ref{thm:scalar-PSD}; it
also gives the elementary symmetries used throughout. We use
$(a;q)_\infty=\nlprod_{j\ge0}(1-aq^j)$ when $|q|<1$.

\begin{corr}[Elementary normalizations]\label{cor:normalizations}
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).
\end{equation}
Then
\[
 \Pqt(I_n)=w_1^n,\qquad
 \Pqt(uv^{\mathsf T})
 =n!H_n(q,t)\nlprod_{i=1}^n u_iv_i.
\]
For square matrices $A,B$, one has
$\Pqt(A\oplus B)=\Pqt(A)\Pqt(B)$.
For $n\times n$ diagonal matrices $D,E$, one has
\[
 \Pqt(DAE)
 =\left(\nlprod_{i=1}^nD_{ii}E_{ii}\right)\Pqt(A).
\]
If $A\succeq0$ and $0\le q,t<1$, then
\begin{equation}\label{eq:upper}
 |\Pqt(A)|\le n!H_n(q,t)\nlprod_{i=1}^n a_{ii}.
\end{equation}
\end{corr}
\begin{proof}
Only the identity permutation contributes on $I_n$. To evaluate the
rank-one case, define
\[
 F(z):=\frac{(tz;q)_\infty}{(z;q)_\infty}
 =\nlsum_{j\ge0}h_jz^j.
\]
Removing the first factors gives
$(1-z)F(z)=(1-tz)F(qz)$. Comparing coefficients yields
\[
 h_0=1,\qquad
 (1-q^j)h_j=(1-tq^{j-1})h_{j-1}\quad(j\ge1),
\]
and hence $h_j=H_j(q,t)$. For $M=Xuv^{\mathsf T}$,
$\tr M^k=(\nlsum_i x_iu_iv_i)^k$. The master theorem reduces to
$F(\nlsum_i x_iu_iv_i)$, whose square-free coefficient is
$n!H_n\nlprod_i u_iv_i$.

The direct-sum and diagonal-scaling identities hold term by term
in \eqref{eq:def}. For \eqref{eq:upper}, $|a_{ij}|\le\sqrt{a_{ii}a_{jj}}$
gives $|\nlprod_ia_{i,\sigma(i)}|\le\nlprod_ia_{ii}$, and the weights
$w_\sigma\ge0$ sum to $\Pqt(\one\one^{\mathsf T})=n!H_n$.
\end{proof}

\subsection{The hook-content coefficients}\label{sec:schur}

Since $w_\sigma$ is a class function on $S_n$, $\Pqt$ is a linear
combination of the irreducible immanants $\Imm_\lambda$. Its
coefficients are explicit products over the cells of $\lambda$.
Their signs determine the kernel region and give a sufficient
condition for scalar positivity. The coefficients are evaluations at the specialization
\begin{equation*}
 p_k[\Bqt]=w_k,\qquad \Bqt=\frac{1-t}{1-q},
\end{equation*}
This is plethystic notation for the specialization
$p_k\mapsto w_k$, rather than a quotient of alphabets. In
particular, $p_k[Z\Bqt]=p_k(Z)w_k$.

\begin{dashedbox}[supporting]
\begin{prop}[Hook-content immanant expansion]\label{thm:schur}
For $0<q<1$, arbitrary real $t$, and $A\in M_n(\C)$,
\begin{equation}\label{eq:schur}
 \Pqt(A)=\nlsum_{\lambda\vdash n}c_\lambda(q,t)\Imm_\lambda(A),
 \qquad
 c_\lambda(q,t)=\nlprod_{(i,j)\in\lambda}
 \frac{q^{i-1}-tq^{j-1}}{1-q^{h_{ij}}}.
\end{equation}
The same identity holds at $q=0$ by taking limits.
\end{prop}
\end{dashedbox}
\begin{proof}
By the generating function of the $h_n$ and \eqref{eq:one-row},
$Q_{(n)}(Z;q,t)=h_n[Z\Bqt]$, and the Schur Cauchy identity
\citep[Chapter I, Section 4]{macdonald1995} gives
\[
 h_n[Z\Bqt]
 =\nlsum_{\lambda\vdash n}s_\lambda[\Bqt]s_\lambda(Z).
\]
Lemma~\ref{lem:dictionary} therefore identifies
$c_\lambda(q,t)$ with $s_\lambda[\Bqt]$.

To evaluate this specialization, first take $t=q^m$ for an integer
$m\ge1$. Its power sums are those of the finite alphabet
$(1,q,\ldots,q^{m-1})$. The principal specialization of a Schur
function \citep[Chapter I, Section 3]{macdonald1995} is
\[
 s_\lambda(1,q,\ldots,q^{m-1})
 =q^{\nlsum_i(i-1)\lambda_i}
 \nlprod_{(i,j)\in\lambda}
 \frac{1-q^{m+j-i}}{1-q^{h_{ij}}}
 =\nlprod_{(i,j)\in\lambda}
 \frac{q^{i-1}-q^mq^{j-1}}{1-q^{h_{ij}}}.
\]
For fixed $q\in(0,1)$, both $s_\lambda[\Bqt]$ and the product in
\eqref{eq:schur} are polynomials in $t$ of degree at most $n$, and
they agree at the infinitely many points $t=q^m$, hence
identically. As $q\downarrow0$ every denominator has a nonzero
limit, which gives the boundary identity.
\end{proof}

The diagonal specialization $q=t$ gives the permanent. For
$t=q^\alpha$, $\alpha>0$, each cycle weight tends to $\alpha$
as $q\uparrow1$, so the finite permutation sum tends to
$\per_\alpha$. At $t=q^m$, $m\ge1$ integral, the weights are
the power sums of the alphabet $(1,q,\ldots,q^{m-1})$:
\[
 w_k(q,q^m)=\nlsum_{a=0}^{m-1}q^{ak}.
\]
The boundary formulas and their proofs are collected in
Proposition~\ref{thm:limits}.

\section{Kernel positivity and total positivity}\label{sec:kernel}

This section proves Theorem~\ref{thm:kernel}. Because $w_\sigma$ is
a class function, the kernel \eqref{eq:kernel} is a matrix coefficient
of a central element of the group algebra of $S_n$ acting on
$(\C^d)^{\otimes n}$, and Schur--Weyl duality diagonalizes such
elements with the coefficients $c_\lambda(q,t)$ as eigenvalues; kernel
positivity is therefore exactly their sign test. The same test
reappears before polarization, for the determinant product on matrix
balls and for total positivity of $H_n$, which shows that
$\mathcal R_r(q)$ belongs to the specialization itself rather than to
the permanent. The necessity half uses linear combinations of simple
tensors, which a single Gram matrix need not detect, so it says
nothing yet about scalar positivity.

In terms of quadratic forms, a kernel $\mathscr K$ on tuples is
positive semidefinite if, for every finite family
$\mathbf u^{(1)},\ldots,\mathbf u^{(L)}$ and every
$z_1,\ldots,z_L\in\C$,
\[
 \nlsum_{a,b=1}^L\overline{z_a}z_b
 \mathscr K(\mathbf u^{(a)},\mathbf u^{(b)})\ge0.
\]
Each immanant term of \eqref{eq:schur} is such a kernel, which is
the sufficiency half of the sign test.

\begin{lemma}[Immanants on Gram matrices]\label{lem:immanant-kernel}
For $\lambda\vdash n$, the kernel
$\Imm_\lambda([\langle u_i,v_j\rangle])$ on $(\C^d)^n$
is positive semidefinite. It vanishes identically when
$\ell(\lambda)>d$.
\end{lemma}
\begin{proof}
Let $P_\lambda$ be the central orthogonal projection onto the
$\lambda$-isotypic component of $(\C^d)^{\otimes n}$.
The central idempotent formula gives
\[
 \Imm_\lambda([\langle u_i,v_j\rangle])
 =\frac{n!}{f^\lambda}
 \langle u_1\otimes\cdots\otimes u_n,
 P_\lambda(v_1\otimes\cdots\otimes v_n)\rangle.
\]
Applying $P_\lambda$ to both tensors makes this a Gram kernel.
Schur--Weyl decomposition has no such component when
$\ell(\lambda)>d$.
\end{proof}

\begin{proof}[\textbf{Proof of Theorem~\ref{thm:kernel}}]
Let $\mathcal T=(\C^d)^{\otimes n}$ with its tensor-product inner
product, and let
\[
 U_\sigma(v_1\otimes\cdots\otimes v_n)
 :=v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)}.
\]
The $U_\sigma$ form a unitary representation of $S_n$, and
$T=\nlsum_{\sigma\in S_n}w_\sigma U_\sigma$ is self-adjoint, since
inverse permutations have the same weight, with
\[
 \langle u_1\otimes\cdots\otimes u_n,\,
 T(v_1\otimes\cdots\otimes v_n)\rangle
 =\mathscr K_{q,t}^{n,d}(\mathbf u,\mathbf v).
\]
Simple tensors span $\mathcal T$, so kernel positivity is
equivalent to $T\succeq0$.

Schur--Weyl decomposition separates the irreducible $S_n$ actions:
\[
 \mathcal T\cong
 \bigoplus_{\substack{\lambda\vdash n\\\ell(\lambda)\le d}}
 S_\lambda(\C^d)\otimes V_\lambda.
\]
Here $V_\lambda$ is the irreducible $S_n$-module with character
$\chi^\lambda$ and $S_\lambda(\C^d)$ its multiplicity space, the
Schur module, which is nonzero exactly when $\ell(\lambda)\le d$.

Since $w_\sigma$ depends only on cycle type, Schur's lemma makes $T$
a scalar on each $V_\lambda$ factor. Taking its trace on that factor
gives 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}
Indeed, the class of cycle type $\mu$ has size $n!/z_\mu$, and the
Frobenius formula evaluates the resulting class sum as
$n!s_\lambda[\Bqt]$. Since every summand with $\ell(\lambda)\le d$
occurs, $T\succeq0$ if and only if $c_\lambda(q,t)\ge0$ for every
$\lambda\vdash n$ with $\ell(\lambda)\le d$.

If $0\le t\le q^{r-1}$, every relevant numerator
$q^{i-1}-tq^{j-1}$ is nonnegative. At $t=q^m$, with
$1\le m\le r-2$, a partition with more than $m$ rows has a
zero factor at $(m+1,1)$, while all numerators of the remaining
partitions are positive. For $t\le0$ all numerators are positive.
The denominators are positive in each case. This proves
sufficiency and the final assertion.

Suppose instead that $t\notin\mathcal R_r(q)$. Then $r\ge2$ and
there is an integer $m\in\{1,\ldots,r-1\}$ such that
$q^m<t<q^{m-1}$. The hook
$\lambda=(n-m,1^m)$ has $m+1\le d$ rows and is therefore present.
Its first-row factors $1-tq^{j-1}$ and the factors $q^{i-1}-t$ for
$2\le i\le m$ are positive, while the bottom factor $q^m-t$ is
negative. Thus this coefficient is negative,
and \eqref{eq:central-eigen} excludes kernel positivity.
\end{proof}

As the rank grows the regions shrink, and their intersection is the
locus on which we conjecture scalar positivity in all ranks
(Conjecture~\ref{conj:scalar}).
\begin{corr}\label{cor:all-rank}
For $0<q<1$ and $0\le t<1$, \eqref{eq:kernel} is positive for all
$n,d\ge1$ exactly when
\[
 t=0\quad\text{or}\quad t=q^m
 \quad\text{for an integer }m\ge1.
\]
In the positive-kernel region,
\[
 |\mathscr K(\mathbf u,\mathbf v)|^2
 \le\mathscr K(\mathbf u,\mathbf u)
 \mathscr K(\mathbf v,\mathbf v).
\]
\end{corr}
\begin{proof}
Each $q^m$ and $0$ belong to every region, while any other $t>0$
lies between two successive powers of $q$ and is excluded for large
$r$. The inequality is the
$2\times2$ case of positivity.
\end{proof}

\begin{rmk}[A necessary distinction]\label{rem:distinction}
The negative eigenspace in the proof of Theorem~\ref{thm:kernel}
contains a linear combination of simple tensors, but need not contain a
simple tensor on which the form is negative. The theorem therefore
classifies kernel positivity, not nonnegativity of $\Pqt(A)$ on
single Gram matrices.
The distinction also arises for classical $\alpha$-permanents
\citep{branden2012solutions}.

For example, every $2\times2$ positive semidefinite matrix satisfies
\[
 \Pqt(A)=w_1^2a_{11}a_{22}+w_2|a_{12}|^2\ge0
 \qquad(0\le q,t<1).
\]
For $n=d=2$ and $0<q<1$, however,
Theorem~\ref{thm:kernel} requires $t\le q$.
The scalar and kernel conditions already differ in this dimension.

\end{rmk}

\subsection{Finite-order total positivity and matrix-ball kernels}

The determinant product itself defines a kernel before
polarization. Its arguments lie in the operator-norm ball
\[
 \mathbb B_{d,p}:=\{U\in\C^{d\times p}:\|U\|_{\mathrm{op}}<1\}.
\]
For $U,V\in\mathbb B_{d,p}$, set
\begin{equation}\label{eq:ball-kernel}
 \mathscr H_{q,t}(U,V):=
 \nlprod_{j\ge0}\frac{\det(I-tq^jUV^*)}{\det(I-q^jUV^*)}.
\end{equation}
By Sylvester's identity $\det(I_d-UV^*)=\det(I_p-V^*U)$ the side
is immaterial; no denominator vanishes since
$\|UV^*\|_{\mathrm{op}}<1$, and for fixed real $t$ the product
converges locally uniformly.

The same coefficients also govern a purely scalar notion of
positivity, the total positivity of the sequence $H_n=\Pqt(J_n)/n!$
of Corollary~\ref{cor:normalizations}. For a real sequence $h=(h_j)_{j\ge0}$ with $h_0=1$, set $h_j=0$
for $j<0$. The sequence is a \emph{P\'olya frequency sequence of
order $r$}, or $\mathrm{PF}_r$, if every minor of size at most $r$
of the matrix $(h_{j-i})_{i,j\ge0}$ is nonnegative, and
$\mathrm{PF}_\infty$ if this holds for every $r$. Total positivity is a
sign condition on Toeplitz minors, and the next lemma turns it into
the Schur sign test of Theorem~\ref{thm:kernel}.

\begin{lemma}[Toeplitz minors and Schur positivity]
\label{lem:pf-schur}
Let $\rho:\Lambda_{\R}\to\R$ be a specialization, and set
$h_j=h_j[\rho]$. Then $(h_j)_{j\ge0}$ is $\mathrm{PF}_r$ if and
only if
\[
 s_\lambda[\rho]\ge0
 \qquad\text{for every partition }\lambda
       \text{ with }\ell(\lambda)\le r.
\]
\end{lemma}

The proof of Lemma~\ref{lem:pf-schur} is in
Appendix~\ref{app:kernel-details}. Combined with the sign test in the
proof of Theorem~\ref{thm:kernel}, the lemma shows that one region
governs three positivity notions at once.
\begin{dashedbox}
\begin{theorem}[Finite-order total positivity and matrix kernels]
\label{thm:ball}
Let $d,p\ge1$, $0<q<1$, $0\le t<1$, and $r=\min(d,p)$.
The following conditions are equivalent:
\begin{enumerate}
\item The sequence $H_n=(t;q)_n/(q;q)_n$, $n\ge0$, is
      $\mathrm{PF}_r$.
\item For every $n\ge1$, the kernel \eqref{eq:kernel} is positive
      semidefinite on $(\C^r)^n$.
\item The kernel \eqref{eq:ball-kernel} is positive semidefinite
      on $\mathbb B_{d,p}$.
\item $t\in\mathcal R_r(q)$.
\end{enumerate}
For $t\le0$, the first three conditions hold for every $d,p\ge1$.
\end{theorem}
\end{dashedbox}

The coefficient identity behind the theorem is
\[
 \mathscr H_{q,t}(U,V)
 =\nlsum_{\lambda}c_\lambda(q,t)s_\lambda(UV^*).
\]
For the map induced by $U$ on Schur modules,
\[
 s_\lambda(UV^*)
 =\tr\bigl(S_\lambda(U)S_\lambda(V)^*\bigr).
\]
This is a positive semidefinite kernel, nonzero precisely when
$\ell(\lambda)\le\min(d,p)$. Distinct partitions give independent
polynomial coefficient spaces, so positivity of the kernel
also forces every contributing $c_\lambda$ to be nonnegative.
Lemma~\ref{lem:pf-schur} identifies the same condition with
total positivity. The coefficient separation and convergence
arguments are proved in Appendix~\ref{app:kernel-details}.

In every degree and rank, Corollary~\ref{cor:all-rank} reduces
the positive-square locus to $t=0,q,q^2,\ldots$. At $t=q^m$
the specialization is the finite alphabet $(1,q,\ldots,q^{m-1})$;
at $t=0$ it is the summable alphabet $(q^j)_{j\ge0}$.
These are the Edrei--Thoma specializations of the present family
in the positive square. Their character interpretation and the extension to $t<0$
are given in Supplement~I \citep{sra2026macpermSuppI}.

\section{Scalar positivity}\label{sec:positivity}

Sufficiency of $\mathcal R_r(q)$ for scalar positivity comes free
from Section~\ref{sec:kernel}, since every immanant is nonnegative on
a Gram matrix; the difficulty is necessity. A negative coefficient
gives a negative direction of the kernel, but that direction is a
linear combination of tensors, whereas a scalar counterexample must
be a single Gram matrix. The necessity results below therefore each
rest on a construction of negative matrices, and which one works
depends on what is held fixed: the deformation, the rank, or the
size. For small $q$ none works at a fixed rank, and we prove
positivity beyond the kernel region instead. Throughout,
$\mathcal W_r$ is the Wallach set of the introduction.

\subsection{Scalar positivity and the Wallach set}
\label{subsec:deformation}

Where all contributing coefficients are nonnegative, the one-row
term of the Schur expansion alone is a lower bound.

\begin{theorem}[Rank-sensitive scalar positivity]\label{thm:scalar-PSD}
Let $A\in M_n(\C)$ be Hermitian positive semidefinite of rank
at most $r$, where $1\le r\le n$. If $0<q<1$ and
$t\in\mathcal R_r(q)$, then
\begin{equation*}
 \Pqt(A)\ge H_n(q,t)\per(A)
          \ge H_n(q,t)\nlprod_{i=1}^n a_{ii}\ge0.
\end{equation*}
The same inequalities hold for every real $t\le0$.
In either range, $\Pqt(A)=0$ if and only if some diagonal
entry of $A$ is zero.
\end{theorem}
\begin{proof}
Write $A$ as a Gram matrix of vectors in $\C^r$.
Lemma~\ref{lem:immanant-kernel} gives
$\Imm_\lambda(A)\ge0$. It also gives
$\Imm_\lambda(A)=0$ when $\ell(\lambda)>r$, since the
corresponding summand is absent from $(\C^r)^{\otimes n}$.

The sign analysis in the proof of Theorem~\ref{thm:kernel}
makes every contributing coefficient nonnegative in the stated
ranges. The one-row term of the Schur expansion is
$c_{(n)}(q,t)\Imm_{(n)}(A)=H_n(q,t)\per(A)$.
Discarding the other terms proves the first inequality.
The permanent lower bound is Lieb's inequality iterated over
singletons \citep[equation (10)]{frenkel2009remarks}.

The factor $H_n(q,t)$ is strictly positive in the stated
ranges. Hence a matrix with all diagonal entries positive
has $\Pqt(A)>0$. If $a_{ii}=0$, positive semidefiniteness
forces the entire $i$th row and column to vanish, so every
permutation monomial is zero.
\end{proof}

The sufficiency in Theorem~\ref{thm:scalar-PSD} holds at every
$q$, and in the Jack limit $q\uparrow1$ it becomes the classical
nonnegativity of $\per_\alpha$ on Gram matrices of rank at most
$r$. Theorem~\ref{thm:deformation-wallach} says that across the
whole deformation nothing more is true. Its negative matrices come
from the classical case: the limit $q\uparrow1$ carries a negative
value of $\per_\alpha$ back to negative values of
$\per_{q,q^\alpha}$, and the rank-tracked witness is
Lemma~\ref{lem:classical-scalar-witness} in
Appendix~\ref{app:scalar-proofs}.

\begin{proof}[\textbf{Proof of Theorem~\ref{thm:deformation-wallach}}]
If $\alpha\in\mathcal W_r$, then $q^\alpha\in\mathcal R_r(q)$ for
every $q\in(0,1)$, and Theorem~\ref{thm:scalar-PSD}, applied with
$\min(n,r)$ in place of $r$, gives (ii), since
$\mathcal R_r(q)\subseteq\mathcal R_h(q)$ for $h\le r$.
Since $(1-q^{\alpha k})/(1-q^k)\to\alpha$ as $q\uparrow1$ for
each positive integer $k$, and the defining permutation sum for a
fixed $A$ is finite, $\per_{q,q^\alpha}(A)\to\per_\alpha(A)$, so
(ii) implies (iii).

If $\alpha\notin\mathcal W_r$, then $\alpha$ is nonintegral and
$s=\lfloor\alpha\rfloor+2\le r$. Lemma~\ref{lem:classical-scalar-witness}
supplies a complex positive semidefinite matrix $A$ of rank at
most $s$ with $\per_\alpha(A)<0$.
For this $A$ the limit above gives
$\per_{q,q^\alpha}(A)\to\per_\alpha(A)<0$, so
$\per_{q,q^\alpha}(A)<0$ for all $q$ close to one. This is the
last assertion, and it shows that (ii) fails, so (ii) implies (i).
\end{proof}
Intersecting the sets $\mathcal W_r$ over $r$ leaves the positive
integers, which gives the counterpart across the deformation of
the positive-exponent half of
\citet[Theorem~2.3]{branden2012solutions}.
\begin{corr}[Unrestricted rank]\label{cor:deformation-integers}
Let $\alpha>0$. Then $\per_{q,q^\alpha}(A)\ge0$ for every
$q\in(0,1)$, every $n\ge1$, and every complex positive
semidefinite $A\in M_n(\C)$ if and only if $\alpha$ is a positive
integer.
\end{corr}
\begin{proof}
The condition is statement (ii) of
Theorem~\ref{thm:deformation-wallach} for every $r$, hence
equivalent to $\alpha\in\bigcap_r\mathcal W_r=\{1,2,\ldots\}$.
\end{proof}

\subsection{Scalar positivity in rank two, uniformly in the size}

In rank two, an unbounded immanant ratio settles necessity at
every $q$.

\begin{theorem}[Scalar positivity in rank two, uniformly in the size]
\label{thm:scalar-rank-two}
Fix $0\le q,t<1$. The inequality $\Pqt(A)\ge0$ holds for
every complex Hermitian positive semidefinite matrix $A$ of rank
at most two, in every size, if and only if $t\le q$.
\end{theorem}

When the rank is fixed and the size grows, more Gram
configurations are tested, and the two-row immanants, whose
coefficients are negative above $t=q$, get room to outweigh the
permanent. The following lemma shows that in rank two the
standard immanant can be arbitrarily large relative to the
permanent, which fixed small sizes do not allow.

\begin{lemma}[A rank-two immanant ratio]\label{lem:scalar-rank-two-ratio}
Let $\mathsf H_n=\nlsum_{j=1}^n j^{-1}$. For every $n\ge2$,
there is an $n\times n$ complex positive semidefinite correlation
matrix $A$ of rank two such that
\begin{equation}\label{eq:scalar-rank-two-ratio}
 \frac{\Imm_{(n-1,1)}(A)}{\per A}
 \ge \frac{n\mathsf H_n-1}{n+1}.
\end{equation}
In particular, these ratios are unbounded as the size varies.
\end{lemma}

\begin{proof}
For a polynomial $f$ on $\C^2$, use the Gaussian norm
\[
 \|f\|_G^2=\E|f(g_1,g_2)|^2,
\]
where $g_1,g_2$ are independent standard circular complex
Gaussians, normalized by $\E|g_i|^2=1$.
Consider the random homogeneous polynomial
\[
 p(z_1,z_2)=\nlsum_{j=0}^n c_j\binom nj^{1/2}
                   z_1^{n-j}z_2^j,
 \qquad c_0,\ldots,c_n\ \text{independent standard complex Gaussians}.
\]
Its covariance is $\E p(z)\overline{p(w)}=\langle w,z\rangle^n$,
so its distribution is invariant under unitary changes of the two
variables. Almost surely it has $n$ distinct projective zeros and
factors as $p=C\nlprod_{i=1}^n\ell_i$, where each linear form
$\ell_i$ has a coefficient vector of norm one. The Gram matrix $A$
of these coefficient vectors is a correlation matrix of rank two.
The complex Gaussian moment formula gives
\[
 \|p\|_G^2=|C|^2\per A,\qquad
 \|p/\ell_i\|_G^2=|C|^2\per A[[n]\setminus\{i\}].
\]
With these norms, define
\[
 Q(p)=\nlsum_{i=1}^n\frac{\|p/\ell_i\|_G^2}{\|p\|_G^2}.
\]
The standard character satisfies
$\chi^{(n-1,1)}(\sigma)=\#\{i:\sigma(i)=i\}-1$.
Since $a_{ii}=1$, it follows that
\begin{equation}\label{eq:scalar-standard-character}
 \frac{\Imm_{(n-1,1)}(A)}{\per A}=Q(p)-1.
\end{equation}

The root-average calculation in Appendix~\ref{app:scalar-proofs}
gives
\[
 \E Q(p)=\frac{n(\mathsf H_n+1)}{n+1}.
\]
Together with \eqref{eq:scalar-standard-character}, this implies
that some realization satisfies \eqref{eq:scalar-rank-two-ratio}.
\end{proof}
Above $t=q$ every two-row coefficient is negative, so the
standard-immanant term alone can outweigh the
permanent, while all other two-row terms have the same sign.

\begin{lemma}[Two-row coefficients]\label{lem:two-row-sign}
Let $n\ge2$ and $0<q<t<1$. Every coefficient
$c_\lambda(q,t)$ with $\ell(\lambda)=2$ is negative, and
\begin{equation*}
 \frac{c_{(n-1,1)}(q,t)}{H_n(q,t)}
 =-\frac{(t-q)(1-q^{n-1})}
         {(1-q)(1-tq^{n-1})}.
\end{equation*}
Hence every positive semidefinite matrix $A$ of rank at most two
with positive diagonal satisfies
\begin{equation}\label{eq:rank-two-scalar-upper}
 \frac{\Pqt(A)}{H_n(q,t)\per A}
 \le 1-
 \frac{(t-q)(1-q^{n-1})}
      {(1-q)(1-tq^{n-1})}
 \frac{\Imm_{(n-1,1)}(A)}{\per A}.
\end{equation}
\end{lemma}

\begin{proof}
For a two-row partition, the numerator at $(2,1)$ is $q-t<0$.
Every first-row numerator is $1-tq^{j-1}>0$, and every other
second-row numerator is
$q-tq^{j-1}=q(1-tq^{j-2})>0$. All hook denominators are
positive, so the coefficient has exactly one negative factor.

The Jacobi--Trudi identity gives
$s_{(n-1,1)}=h_{n-1}h_1-h_n$. Since
$h_j[B_{q,t}]=H_j(q,t)$ and
$H_n/H_{n-1}=(1-tq^{n-1})/(1-q^n)$, it follows that
\begin{align*}
 \frac{c_{(n-1,1)}}{H_n}
 &=\frac{(1-t)(1-q^n)}
         {(1-q)(1-tq^{n-1})}-1\\
 &=\frac{(q-t)(1-q^{n-1})}
         {(1-q)(1-tq^{n-1})}.
\end{align*}
On a Gram matrix of rank at most two the immanants with more than
two rows vanish and the others are nonnegative, as in the proof
of Theorem~\ref{thm:scalar-PSD}; the one-row term of the Schur
expansion is $H_n(q,t)\per A$, and dropping every two-row term
except that of the standard immanant, all of them nonpositive,
gives \eqref{eq:rank-two-scalar-upper}, whose denominator is
positive because $H_n(q,t)>0$ and $\per A\ge\nlprod_i a_{ii}>0$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:scalar-rank-two}]
For $0<q<1$ and $t\le q$, Theorem~\ref{thm:scalar-PSD}
with $r=\min(n,2)$ gives nonnegativity in rank at most two, since
$\mathcal R_2(q)=[0,q]$; at $q=t=0$ the function is the ordinary
permanent, which is nonnegative on positive semidefinite
matrices.

Suppose $0<q<t<1$. The coefficient of the immanant ratio in
\eqref{eq:rank-two-scalar-upper} tends to $(t-q)/(1-q)>0$ as
$n\to\infty$, while $\mathsf H_n\to\infty$, so there is an
$n\ge2$ with
\begin{equation*}
 \frac{(t-q)(1-q^{n-1})}
      {(1-q)(1-tq^{n-1})}
 \frac{n\mathsf H_n-1}{n+1}>1.
\end{equation*}
Lemma~\ref{lem:scalar-rank-two-ratio} supplies a rank-two
correlation matrix for which the last ratio in
\eqref{eq:rank-two-scalar-upper} is at least
$(n\mathsf H_n-1)/(n+1)$. The right side is therefore strictly
negative, and $\Pqt(A)<0$.

At $q=0<t<1$, the hook formula \eqref{eq:HL} and rank-two
vanishing give
\[
 \per_{0,t}(A)
 =(1-t)\bigl(\per A-t\Imm_{(n-1,1)}(A)\bigr),
\]
so an $n$ with $t(n\mathsf H_n-1)/(n+1)>1$ and the matrix of
Lemma~\ref{lem:scalar-rank-two-ratio} again give
$\per_{0,t}(A)<0$.
\end{proof}

The negative matrices of Theorem~\ref{thm:scalar-rank-two} are
complex, and their number of distinct Gram vectors grows with the
size; in size three the region is larger
(Theorem~\ref{thm:scalar-three}), and with only two distinct
vectors, repeated, positivity holds on the whole square
(Corollary~\ref{cor:two-type-positive}).

\subsection{Scalar counterexamples from random projections}

In rank three and higher, generic homogeneous polynomials in
more than two variables do not factor into linear forms. We
obtain negative values by averaging $\Pqt$ over the unitary
orbit of an orthogonal projection of rank $r$. The average
turns the immanant expansion \eqref{eq:schur} into explicit
terms, one per partition. Each term has an exponential rate in
the size, with factors that grow only polynomially; the largest
rates determine the sign of the sum.

\begin{theorem}[Scalar counterexamples at fixed $q$]
\label{thm:scalar-Haar-strip}
Let $r\ge2$, $0<q<1$, and
\begin{equation}\label{eq:haar-strip-threshold}
 (r+1)q^{r-1}>\nlsum_{j=0}^{r-1}q^j.
\end{equation}
For every $t$ with $q^{r-1}<t<q^{r-2}$, there is an integer
$n_0=n_0(q,t,r)\ge r$ such that the random rank-$r$ projection
$P_n$ in Lemma~\ref{lem:haar-immanant-average} satisfies
\[
 \E\Pqt(P_n)<0\qquad\text{for every }n\ge n_0.
\]
Consequently, for every $n\ge n_0$ there is a complex Hermitian
positive semidefinite matrix $A\in M_n(\C)$ of rank $r$ with
$\Pqt(A)<0$.
The integer $n_0$ can be chosen uniformly for $t$ in any compact
subinterval of $(q^{r-1},q^{r-2})$.
\end{theorem}

Through the Schur expansion, the average of $\Pqt$ over the orbit
reduces to averages of single immanants, and these are explicit.

\begin{lemma}[Haar-averaged immanants]
\label{lem:haar-immanant-average}
For $1\le r\le n$ and Haar distributed $U\in\mathrm U(n)$, set
\[
 Q_r=\diag(I_r,0_{n-r}),\qquad P_n=UQ_rU^*.
\]
For every $\lambda\vdash n$,
\begin{equation}\label{eq:haar-immanant-average}
 \E\Imm_\lambda(P_n)
 =f^\lambda\frac{s_\lambda(1^r)}{s_\lambda(1^n)}
 =\frac{n!\,s_\lambda(1^r)}
        {\displaystyle\nlprod_{(i,j)\in\lambda}(n+j-i)}.
\end{equation}
In particular, the expectation vanishes when $\ell(\lambda)>r$.
\end{lemma}

The formula follows by averaging $P_n^{\otimes n}$ on each
Schur--Weyl component (Appendix~\ref{app:scalar-proofs}).

For $q^{r-1}<t<q^{r-2}$, the partitions with $r$ rows have
negative coefficients, and those with fewer rows have positive
coefficients. Write $p_i=\lambda_i/n$, appending zero parts.
Up to factors bounded above and below by powers of $n$, the
absolute value of a term in the expected Schur expansion has
exponential rate $\Phi(p)-1$, where
\[
 \Phi(p)=(\log q)\nlsum_{i=1}^r(i-1)p_i
 -\nlsum_{i=1}^r\bigl((1+p_i)\log(1+p_i)-p_i\bigr).
\]
On the ordered simplex this function is strictly concave, with
unique maximizer under \eqref{eq:haar-strip-threshold}
\[
 p_i^*=\frac{(r+1)q^{i-1}}{\nlsum_{j=0}^{r-1}q^j}-1>0.
\]
Thus the largest rate occurs among negative terms and is
strictly larger than every rate on the boundary $p_r=0$,
where the positive terms lie. This strict gap forces the
expectation to be negative for large $n$. Uniform estimates,
partition rounding, and the compact-$t$ assertion are proved in
Appendix~\ref{app:scalar-proofs}.
For $r=3$ the condition reads $q>(1+\sqrt{13})/6$, and the
theorem then gives rank-three negative matrices throughout
$q^2<t<q$.

With Theorem~\ref{thm:scalar-PSD}, this theorem classifies scalar
positivity at bounded rank in every size.
\begin{corr}[Scalar positivity at fixed rank]
\label{cor:scalar-Haar-classification}
Let $r\ge3$, let $0<q<1$ satisfy
\eqref{eq:haar-strip-threshold}, and let $0\le t<1$.
Then $\Pqt(A)\ge0$ for every complex Hermitian positive
semidefinite matrix $A$, of every order and of rank at most $r$,
if and only if
\[
 t\in\mathcal R_r(q)
 =[0,q^{r-1}]\cup\{q,q^2,\ldots,q^{r-2}\}.
\]
\end{corr}

\begin{proof}
Sufficiency is Theorem~\ref{thm:scalar-PSD}, applied with
$\min(n,r)$ in place of $r$, since
$\mathcal R_r(q)\subseteq\mathcal R_h(q)$ for $h\le r$.
For necessity, the threshold can be rewritten as
\[
 \nlsum_{j=1}^{r-1}(q^{-j}-1)<1.
\]
Every summand is positive, so the same condition holds with
$r$ replaced by any $h$ with $2\le h\le r$.
If $t\notin\mathcal R_r(q)$, then
$q^{h-1}<t<q^{h-2}$ for one such $h$.
Theorem~\ref{thm:scalar-Haar-strip}, applied in rank $h$, supplies
a positive semidefinite matrix of rank at most $r$ with negative
scalar value.
\end{proof}

\subsection{Scalar positivity beyond the kernel region}\label{subsec:rank-budget}

For small $q$, the scalar region can extend beyond
$\mathcal R_R(q)$ in every size at a fixed rank bound $R$.
The comparison is with $t=q^2$, where two Gaussian colors
realize the weights. A bound on the cost of deleting a Gaussian
factor controls the omitted colors uniformly in the size.
\begin{theorem}[Positivity beyond any prescribed finite rank]
\label{thm:rank-budget}
For every integer $R\ge3$ there are $q,t$ with $q^2<t<q$, so
that $t\notin\mathcal R_R(q)$, such that
\begin{equation}\label{eq:robust}
 \Pqt(A)\ge\tfrac12\per_{q,q^2}(A)\ge0
\end{equation}
for every complex Hermitian positive semidefinite $A$ of rank at
most $R$, in every size. One choice is $q=1/(4R)$ and
$t=q^2+\delta_R$ with $\delta_R$ as in \eqref{eq:delta}.
\end{theorem}

The comparison writes both $\per_{q,q^2}$ and $\per_{q,0}$ as
moments of Gaussian quadratic forms, the first with two colors and
the second with infinitely many. For arbitrary cycle weights $w$,
write $\per_w$ for the sum in \eqref{eq:def} with those weights.

\begin{lemma}[Gaussian cycle moments]\label{lem:gaussian-cycle-moments}
Let $A=[u_i^*u_j]_{i,j=1}^n$ with $u_i\in\C^R$, and let
$g_a$ be independent standard circular complex Gaussian vectors
in $\C^R$. If $(c_a)$ is a finite or summable sequence of
nonnegative numbers, put $w_k=\nlsum_a c_a^k$. Then
\[
 \E\nlprod_{i=1}^n\left(\nlsum_a c_a|u_i^*g_a|^2\right)
 =\per_w(A).
\]
In particular, $c_a=q^a$ for $0\le a<m$ gives
$\per_{q,q^m}(A)$, and $c_a=q^a$ for all $a\ge0$ gives
$\per_{q,0}(A)$.
\end{lemma}
\begin{proof}
For finitely many colors, expand the product and apply the
complex Wick formula. A pairing permutation $\sigma$ requires
the color index to be constant on each cycle $C$, whose color
sum is $\nlsum_a c_a^{|C|}$. The covariance factors give the
permutation monomial of $A$ (replacing $\sigma$ by its inverse
does not change the sum), proving the identity.

For a summable sequence, each nonnegative sum in the product is
finite almost surely, since its expectation is
$\|u_i\|^2\nlsum_a c_a<\infty$. Products of the partial sums
increase, so monotone convergence passes the expectation to the
limit. Each $w_k$ is finite and is the limit of its finite-color
sum; the permutation sum has only finitely many terms and
therefore converges to $\per_w(A)$. The geometric choices give
the stated Macdonald weights.
\end{proof}

The colors beyond the first two add a nonnegative term $u_i^*Yu_i$ to
each factor, and the following bound controls the cost of that
addition with a constant independent of the size.
\begin{lemma}[Uniform Gaussian deletion]\label{lem:gaussian-deletion}
Let $k\ge2$, let $C=\diag(c_1,\ldots,c_k)$ with every $c_a>0$,
and let $g_1,\ldots,g_k$ be independent standard complex Gaussian
vectors in $\C^R$. For $u_1,\ldots,u_n\in\C^R$ with Gram matrix
$A=[u_i^*u_j]$, put
\[
 L_i=\nlsum_{a=1}^kc_a|u_i^*g_a|^2,\quad
 F_C(A)=\E\nlprod_{i=1}^nL_i,\quad
 d=Rk,\quad
 \beta=\tr C-\max_ac_a,\quad
 K=\frac{d-1}{\beta}.
\]
The following bounds then hold:
\begin{align}
 \nlsum_{i=1}^na_{ii}F_C(A[[n]\setminus\{i\}])
 &\le\frac{n(d-1)}{\beta(n+d-1)}F_C(A)\le KF_C(A),
 \label{eq:delete}\\
 \E\nlprod_{i=1}^n(L_i+xa_{ii})&\le e^{Kx}F_C(A)\qquad(x\ge0).
 \label{eq:shift}
\end{align}
For deterministic $Y\succeq0$, moreover,
$\E\nlprod_i(L_i+u_i^*Yu_i)\le e^{K\tr Y}F_C(A)$.
\end{lemma}

The proof of Lemma~\ref{lem:gaussian-deletion} is in
Appendix~\ref{app:scalar-proofs}.
\begin{proof}[Proof of Theorem~\ref{thm:rank-budget}]
Take $q=1/(4R)$, write $F(A)=\per_{q,q^2}(A)$, and let
$A=[u_i^*u_j]$ with $u_i\in\C^R$. With $g_0,g_1,\ldots$
independent standard complex Gaussian vectors in $\C^R$,
Lemma~\ref{lem:gaussian-cycle-moments} gives
\[
 F(A)=\E\nlprod_i\bigl(|u_i^*g_0|^2+q|u_i^*g_1|^2\bigr),
 \qquad
 \per_{q,0}(A)=\E\nlprod_i\bigl(|u_i^*g_0|^2+q|u_i^*g_1|^2
                                +u_i^*Yu_i\bigr),
\]
where $Y=\nlsum_{j\ge2}q^jg_jg_j^*$ is independent of $g_0,g_1$.
The last assertion of Lemma~\ref{lem:gaussian-deletion} with
$C=\diag(1,q)$, so that $K=(2R-1)/q$, averaged over $Y$, gives
$\per_{q,0}(A)\le F(A)\,\E e^{K\tr Y}$. Here
$\E e^{K\tr Y}=\nlprod_{j\ge2}(1-Kq^j)^{-R}$ and
$\nlsum_{j\ge2}Kq^j=(2R-1)q/(1-q)<1$, so the inequality
$\nlprod_j(1-b_j)\ge1-\nlsum_jb_j$ yields
\begin{equation}\label{eq:tailcompare}
 \per_{q,0}(A)\le M_RF(A),\qquad
 M_R\le\overline M_R:=\left(\frac{1-q}{1-2Rq}\right)^R.
\end{equation}

For $\ell(\lambda)\le R$, the hook formula \eqref{eq:schur} gives
\[
 D_\lambda(t):=\frac{c_\lambda(q,t)}{c_\lambda(q,0)}
 =\nlprod_{(i,j)\in\lambda}(1-tq^{j-i}),
 \qquad
 B_R:=\frac{\nlsum_{i=1}^Rq^{1-i}}{1-q},
\]
and for $0\le t\le2q^2$ the product rule gives
$|D_\lambda'(t)|\le C_R:=B_Re^{2q^2B_R}$, uniformly in $\lambda$
and its size. On $A$ the immanants in \eqref{eq:schur} are
nonnegative and only partitions with at most $R$ rows contribute,
with positive coefficients $c_\lambda(q,0)$, so
\[
 |\Pqt(A)-\per_{q,s}(A)|
 \le C_R|t-s|\per_{q,0}(A)\qquad(s,t\in[0,2q^2]).
\]
With these constants, choose
\begin{equation}\label{eq:delta}
 \delta_R=\min\{q^2,(2C_R\overline M_R)^{-1}\},
 \qquad t=q^2+\delta_R.
\end{equation}
Then $q^2<t\le2q^2<q$, and \eqref{eq:tailcompare} gives
$\Pqt(A)\ge F(A)-C_R\delta_R\overline M_RF(A)\ge\tfrac12F(A)$,
which is \eqref{eq:robust}.
\end{proof}

Remark~\ref{rem:rank-three-instance} gives the explicit
rank-three example $q=1/10$, $t=21/2000$.

\section{Block and diagonal inequalities}\label{subsec:frenkel}

After nonnegativity, the classical question for the permanent on
positive semidefinite matrices is how it compares with the product of
its values on complementary principal blocks, and Lieb's inequality
says that it dominates that product. The Schur expansion cannot carry
this over to $\Pqt$, since it compares $\Pqt$ with the permanent only
through its one-row term. A second expansion, in products of ordinary
permanents of principal blocks, does: its coefficients are
nonnegative, and Lieb's inequality applies block by block. Because
the coefficients are indexed by set partitions of $[n]$, the
hypothesis here depends on the matrix size, not on the rank.

\begin{theorem}[Block and diagonal inequalities]\label{thm:block-inequalities}
Let $n\ge1$, $0<q<1$, $t\in\mathcal R_n(q)$, and let
$A\in M_n(\C)$ be Hermitian positive semidefinite.
For $I\sqcup J=[n]$, set $D=A[I]\oplus A[J]$, after a
simultaneous reordering of rows and columns. Then
\[
 \Pqt(A)\ge\Pqt(D)=\Pqt(A[I])\Pqt(A[J]),
 \qquad
 \Pqt(A)\ge w_1^n\nlprod_{i=1}^n a_{ii}.
\]
\end{theorem}
\begin{proof}
Let $\Pi_n$ denote the set partitions of $[n]$.
Proposition~\ref{prop:positive-partition-expansion} gives
\[
 \Pqt(A)=\nlsum_{\pi\in\Pi_n}a_\pi(q,t)
                  \nlprod_{B\in\pi}\per A[B],
 \qquad a_\pi(q,t)\ge0.
\]
Lieb's inequality on each block gives
$\per A[B]\ge\per A[B\cap I]\per A[B\cap J]=\per D[B]$
\citep[equation (10)]{frenkel2009remarks}.
Multiplying these nonnegative factors and summing proves the
block inequality; block multiplicativity gives the equality.
Iterating over singleton blocks gives the diagonal bound because
$\Pqt([a_{ii}])=w_1a_{ii}$.
\end{proof}

The coefficient signs follow from a divided-difference identity
in Appendix~\ref{app:frenkel}. Signed Fischer inequalities and a real-Gaussian comparison
are proved in Supplement~I \citep{sra2026macpermSuppI}. The sufficient region
$\mathcal R_n(q)$ is not asserted to be necessary for each
inequality: the block inequality holds throughout the square
in size two.

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

The scalar problem that remains concerns a fixed $q$ and matrices
of unbounded size and rank, where neither the limit $q\uparrow1$ nor
a rank bound is available. The kernel locus suggests its answer.
\begin{dashedbox}
\begin{conj}[Scalar Macdonald positivity]\label{conj:scalar}
For every $0<q<1$ and $0\le t<1$, the inequality $\Pqt(A)\ge0$
holds for every complex Hermitian positive semidefinite matrix
$A$, of every size, if and only if $t\in\{0,q,q^2,q^3,\ldots\}$.
\end{conj}
\end{dashedbox}

Sufficiency is Theorem~\ref{thm:scalar-PSD}. Necessity holds for
$t>q$ by the rank-two theorem, and in the further strips reached
by Theorem~\ref{thm:scalar-Haar-strip}. The small-$q$ comparison
shows that no fixed rank bound suffices to supply witnesses
uniformly over all excluded parameters. In particular, the pair
$q=1/10$, $t=21/2000$ requires a negative witness of rank at
least four, if one exists. No such witness is supplied here.
The endpoint $t=1$, outside the stated range, would add the
identically zero function in every positive size. Even at a fixed
rank, the question is open.

\begin{prob}[The scalar region at a prescribed $q$ and rank]
\label{prob:fixed-q}
For $0<q<1$ and $r\ge3$, determine the set of $t$ for which
$\Pqt(A)\ge0$ for every size and every complex $A\succeq0$ of rank
at most $r$.
\end{prob}

The set contains $\mathcal R_r(q)$, equals it under the Haar
threshold, and can be strictly larger for small $q$.
Determining its boundary requires understanding scalar
evaluations when positive and negative immanant terms coexist.
Even the shape of the region above $q^2$ at small $q$ is open:
the present comparison gives a right neighborhood of $q^2$ but
does not determine how far it extends.

The exact fixed-size regions are likewise unknown for $n\ge5$.
The thresholds for $n=3,4$ and their explicit witnesses are in
Appendix~\ref{app:size-three}. The broader immanant and signed-inequality questions accompany
their corresponding extensions in Supplement~I.

\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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\end{thebibliography}

\appendix
\addtocontents{toc}{\protect\setcounter{tocdepth}{1}}
\begingroup
\renewcommand{\contentsname}{Appendix contents}
\bigskip
\noindent\begin{minipage}{\linewidth}
\tableofcontents
\end{minipage}
\endgroup
\section{Total positivity and matrix-ball proofs}\label{app:kernel-details}

Theorem~\ref{thm:ball} identifies three positivity notions with one
sign test, and the main text gives only the mechanism. Two facts make
the identification rigorous. Toeplitz minors of $(h_{j-i})$ are skew
Schur functions, so total positivity of order $r$ is Schur
positivity in at most $r$ rows; and the Schur components of the
matrix-ball kernel live in independent polynomial spaces, so
positivity of the kernel forces the sign of each coefficient
separately.

\subsection{Toeplitz minors}
\begin{proof}[Proof of Lemma~\ref{lem:pf-schur}]
Consider a $k\times k$ Toeplitz minor, with increasing row
indices $a_1<\cdots<a_k$ and column indices $b_1<\cdots<b_k$.
If $a_i>b_i$ for some $i$, its first $i$ columns are supported in
at most the first $i-1$ rows, so the minor vanishes.
Otherwise set
\[
 \lambda_i=b_{k+1-i}+i-k,\qquad
 \mu_i=a_{k+1-i}+i-k,\qquad 1\le i\le k.
\]
Both $\lambda$ and $\mu$ are partitions, padded with zeros, and
$\mu\subseteq\lambda$. Reversing both orders and
transposing gives the skew Jacobi--Trudi identity
\citep[Chapter I, Section 5]{macdonald1995}
\[
 \det[h_{b_j-a_i}]_{i,j=1}^k
 =\det[h_{\lambda_i-\mu_j-i+j}]_{i,j=1}^k
 =s_{\lambda/\mu}[\rho].
\]
The two reversals have cancelling signs.

The Littlewood--Richardson rule
\citep[Chapter I, Section 9]{macdonald1995} expands this skew
function as
\[
 s_{\lambda/\mu}
 =\nlsum_\nu c^\lambda_{\mu\nu}s_\nu,
 \qquad c^\lambda_{\mu\nu}\in\mathbb N.
\]
If $c^\lambda_{\mu\nu}>0$, the symmetry
$c^\lambda_{\mu\nu}=c^\lambda_{\nu\mu}$ implies
$\nu\subseteq\lambda$. Thus every contributing $\nu$ has at most
$k$ rows. Schur positivity through $r$ rows therefore makes
every minor of order $k\le r$ nonnegative.
Conversely, for a partition $\lambda$ with $k$ rows, choose
$a_i=i-1$ and $b_j=\lambda_{k+1-j}+j-1$.
These strictly increasing indices give the minor
$s_\lambda[\rho]$, proving necessity.
\end{proof}

\subsection{Matrix-ball coefficient spaces}
\begin{proof}[Proof of Theorem~\ref{thm:ball}]
The generating function of the specialization $\Bqt$ gives
$H_n=h_n[\Bqt]$.
Lemma~\ref{lem:pf-schur} therefore identifies the first condition
with
$c_\lambda(q,t)\ge0$ for every $\ell(\lambda)\le r$.
The central-operator proof of Theorem~\ref{thm:kernel}, applied
in every degree, identifies the second condition with the same
inequalities. Its hook sign test identifies these inequalities
with the fourth condition, and proves them also for $t\le0$.
It remains to establish the equivalent matrix-ball condition.

Near the origin, the Schur Cauchy identity gives the homogeneous
expansion
\begin{equation}\label{eq:ball-schur}
 \mathscr H_{q,t}(U,V)
 =\nlsum_{n\ge0}\nlsum_{\lambda\vdash n}
 c_\lambda(q,t)s_\lambda(UV^*).
\end{equation}
For a linear map $U:\C^p\to\C^d$, its tensor power induces a linear
map on Schur modules,
$S_\lambda(U):S_\lambda(\C^p)\to S_\lambda(\C^d)$.
The tensor inner products induce inner products for which this
construction commutes with composition and adjoints.
The character of $S_\lambda$ is $s_\lambda$, so
\[
 s_\lambda(UV^*)
 =\tr\!\left(S_\lambda(U)S_\lambda(V)^*\right).
\]
In orthonormal bases, write the polynomial entries of $S_\lambda(U)$
as $f_{\lambda,ab}(U)$. Then
\[
 s_\lambda(UV^*)
 =\nlsum_{a,b}f_{\lambda,ab}(U)
 \overline{f_{\lambda,ab}(V)}.
\]
The right side is a positive semidefinite kernel. It is nonzero exactly when
$\ell(\lambda)\le r$, because both Schur modules must be nonzero.

To prove necessity, one must separate the coefficient spaces of
different partitions. The degree-$n$ polynomials on
$\C^{d\times p}$ have the Cauchy decomposition
\[
 \operatorname{Sym}^n((\C^d)^*\otimes\C^p)
 \cong
 \bigoplus_{\substack{\lambda\vdash n\\\ell(\lambda)\le r}}
 S_\lambda((\C^d)^*)\otimes S_\lambda(\C^p).
\]
One way to obtain it is to realize the symmetric power as the
invariants under the diagonal $S_n$ action on
$((\C^d)^*)^{\otimes n}\otimes(\C^p)^{\otimes n}$.
Schur--Weyl decomposition in the two factors leaves
$(V_\lambda\otimes V_\mu)^{S_n}$, whose dimension is one for
$\lambda=\mu$ and zero otherwise. The matrix entries
$f_{\lambda,ab}$ span the indicated $\lambda$ summand: their span
is a nonzero equivariant image of that irreducible module.
In particular, these polynomial spaces form a direct sum.

Every linear functional on degree-$n$ polynomials can be written as
$L=\nlsum_{|\gamma|=n}b_\gamma\partial^\gamma|_0$.
If an analytic kernel is positive semidefinite near zero, then
\[
 L_U\overline{L_V}\,\mathscr H(U,V)\big|_{U=V=0}\ge0,
\]
since derivatives at zero are limits of finite-difference
functionals, and applying the same one in both arguments gives
nonnegative quadratic forms. Fix a nonzero $\lambda$ summand and choose $L$ to vanish
on every other degree-$n$ summand while remaining nonzero on this
one. Derivatives of order $n$ at zero annihilate homogeneous terms
of other degrees. Applying $L$ to \eqref{eq:ball-schur} therefore gives
\[
 0\le c_\lambda(q,t)
 \nlsum_{a,b}|Lf_{\lambda,ab}|^2.
\]
The sum is strictly positive, so
$c_\lambda(q,t)\ge0$ is necessary for every partition with
$\ell(\lambda)\le r$.

For sufficiency, it remains to justify the infinite positive sum.
Suppose first that all the coefficients under consideration are
nonnegative. On $\|U\|_{\mathrm{op}}\le a<1$, the eigenvalues of
$UU^*$ lie in $[0,a^2]$, and at most $r$ of them are nonzero.
The tableau expansion of $s_\lambda$ has nonnegative coefficients,
so
\[
 s_\lambda(UU^*)\le a^{2|\lambda|}s_\lambda(1^r).
\]
The diagonal majorant is finite:
\[
 \nlsum_\lambda c_\lambda(q,t)a^{2|\lambda|}s_\lambda(1^r)
 =\nlprod_{j\ge0}
 \left(\frac{1-tq^ja^2}{1-q^ja^2}\right)^r<\infty.
\]
To justify this equality without assuming convergence of
\eqref{eq:ball-schur}, apply its formal identity to the scalar
variable $z$ in $zI_r$. The product is analytic for $|z|<1$, and
its Taylor series converges at $z=a^2$.
For two arguments, the Hilbert--Schmidt Cauchy--Schwarz inequality
bounds each $|s_\lambda(UV^*)|$ by
$\sqrt{s_\lambda(UU^*)s_\lambda(VV^*)}$.
The same diagonal majorant therefore gives normal convergence of
the positive kernel sum on smaller balls.

The expansion also converges normally at the other real
parameters. The hook formula gives
$|c_\lambda(q,t)|\le c_\lambda(q,-|t|)$, and the right side has
nonnegative coefficients. Applying the preceding bound at
$-|t|$ proves absolute convergence of \eqref{eq:ball-schur}.
Its sum agrees near zero with the determinant product and hence
agrees throughout the ball. In the claimed parameter range, finite
partial sums are positive semidefinite, and so are their limits,
which proves sufficiency.
\end{proof}
For $\alpha>0$, the specialization $t=q^\alpha$ has the locally
uniform limit
\begin{equation}\label{eq:ball-jack}
 \mathscr H_{q,q^\alpha}(U,V)
 \longrightarrow\det(I-UV^*)^{-\alpha}
 \qquad(q\uparrow1).
\end{equation}
with the determinant power defined by the logarithm that vanishes
at $UV^*=0$: since $0<(1-q^{\alpha k})/(1-q^k)\le\max(1,\alpha)$
with limit $\alpha$, the logarithmic series is dominated on a
ball of radius $a<1$ by $d\max(1,\alpha)\nlsum_{k\ge1}a^{2k}/k$,
and dominated convergence proves \eqref{eq:ball-jack}.

The substitution $t=q^\alpha$ translates $t\in\mathcal R_r(q)$
into $\alpha\in\mathcal W_r$, and the same range is sharp for the
limiting determinant kernel: applied to \eqref{eq:jack-schur},
the cellwise sign test gives it directly, since positive integers
kill partitions with too many rows, $\alpha\ge r-1$ makes every
relevant factor nonnegative, and a hook gives a negative
coefficient between consecutive integers. The determinant kernel
and its parameter range belong to the classical theory of matrix
balls and bounded symmetric domains \citep{faraut1990function};
see also \citet{sra2026hua} for the contraction-matrix
formulation.

\section{Scalar witness proofs and Gaussian estimates}\label{app:scalar-proofs}

The scalar theorems of Section~\ref{sec:positivity} are proved there
from a few lemmas, each stated next to the mechanism it serves. Their
proofs are calculations rather than ideas, and we collect them here
so that the main text can follow the construction of negative
matrices without interruption. The one estimate with content of its
own is the Gaussian deletion bound of Lemma~\ref{lem:gaussian-deletion}:
its constant is independent of the size, and this uniformity is what
lets Theorem~\ref{thm:rank-budget} perturb $t$ in every size.

\subsection{The rank of a classical scalar witness}

The classical complete-monotonicity obstruction supplies the
rank information needed by the across-deformation theorem.

\begin{lemma}\label{lem:classical-scalar-witness}
If $\alpha>0$ is not an integer and
$s=\lfloor\alpha\rfloor+2$, there is a complex Hermitian
positive semidefinite matrix $A$, of some finite size and rank
at most $s$, such that $\per_\alpha(A)<0$.
\end{lemma}
\begin{proof}
Choose rank-one positive semidefinite matrices
$B_1,\ldots,B_d$ spanning the real vector space of Hermitian
$s\times s$ matrices, and put
\[
 p(x)=\det\Bigl(\nlsum_{j=1}^d x_jB_j\Bigr).
\]
Since $\alpha$ is not an integer and $\alpha<s-1$, the complex
Scott--Sokal criterion, in the form stated in
\citet[Theorem~2.6(2)]{branden2012solutions}, says that
$p^{-\alpha}$ is not completely monotone on $(0,\infty)^d$.
Consequently, for some $y\in(0,\infty)^d$ and some
$\nu\in\mathbb N^d$, the coefficient of $x^\nu$ in
$p(y-x)^{-\alpha}$ is negative.

Set $C=\nlsum_j y_jB_j$. Since the $B_j$ are positive
semidefinite and span the Hermitian matrices, $C$ is positive
definite. Write
$C^{-1/2}B_jC^{-1/2}=u_ju_j^*$ with $u_j\in\C^s$, and put
$U=[u_1\ \cdots\ u_d]$ and $X=\operatorname{diag}(x)$.
Sylvester's identity and the classical master theorem give
\[
 p(y-x)^{-\alpha}
 =\det(C)^{-\alpha}\det(I-XU^*U)^{-\alpha},
 \qquad
 [x^\nu]p(y-x)^{-\alpha}
 =\frac{\det(C)^{-\alpha}}{\nu!}
   \per_\alpha\bigl((U^*U)[\nu]\bigr).
\]
Thus $A=(U^*U)[\nu]$ is a Gram matrix of repeated vectors in
$\C^s$ and satisfies $\per_\alpha(A)<0$. This construction is that of
\citet[Theorem~2.3]{branden2012solutions}, with its Gram dimension
tracked.
\end{proof}
The last assertion of Theorem~\ref{thm:deformation-wallach}
is uniform on compact sets of noninteger
exponents: for a compact $J\subset(m,m+1)$, each matrix remains
negative on a neighborhood of $(q,\alpha)=(1,\alpha_0)$, and a
finite cover of $J$ supplies finitely many matrices of rank at
most $m+2$ and a common bound $q_J<1$. It gives nothing at a
prescribed $q$, and no uniformity as $\alpha$ approaches an
integer.

\subsection{The root-deletion average}

We complete the calculation in the proof of
Lemma~\ref{lem:scalar-rank-two-ratio}, using its polynomial $p$,
Gaussian norm, and quantity $Q(p)$.

\begin{proof}[Root-average calculation]
To compute $\E Q(p)$ we change variables on the zero set.
For a projective zero $z$ of $p$, let
$F(p,z)=\|p/\ell_z\|_G^2/\|p\|_G^2$, where $\ell_z$ is any
unit linear form vanishing at $z$; its phase does not affect $F$.
The quantity $F$ is invariant under simultaneous unitary changes
of the polynomial and the zero, and $0\le F\le1$, because after a
unitary change of variables $\ell_z=z_2$, and multiplying each
Gaussian-orthogonal monomial by $z_2$ multiplies its squared norm
by a positive integer. Consequently the finite measure
\[
 B\longmapsto\E\nlsum_{z:\,p(z)=0}F(p,z)\,\mathbf1_B(z)
\]
on $\mathbf{CP}^1$ is unitarily invariant, hence a constant
multiple of the probability measure
\[
 d\nu(z)=\frac{d^2z}{\pi(1+|z|^2)^2}
\]
in the affine chart $(z_1,z_2)=(1,z)$, and the constant is
$\E Q(p)$.

We compute the density of this measure at $z=0$. Write
$P_c(z)=p(1,z)$. The incidence map
\[
 (z,c_1,\ldots,c_n)\longmapsto
 \left(-\nlsum_{j=1}^n c_j\binom nj^{1/2}z^j,c_1,\ldots,c_n\right)
\]
parametrizes coefficient vectors together with a distinguished
zero; off the null set of multiple zeros it has one preimage for
each zero, and its real Jacobian is $|P_c'(z)|^2$.
The area formula, applied to the Gaussian coefficient density and
the bounded weight $F$, therefore gives the density of the
measure. At $z=0$ the constraint reads $c_0=0$ and
$P_c'(0)=\sqrt n\,c_1$, so the density there is
\[
 \frac n\pi\E\bigl[|c_1|^2F(p,0)\mid c_0=0\bigr],
\]
where conditioning on $c_0=0$ means that $c_0$ is set to zero
while the other coefficients retain their independent Gaussian
laws.
The density is continuous near zero by dominated convergence,
since on compact sets of $z$ the squared Jacobian is bounded by a
constant times $\nlsum_{j=1}^n|c_j|^2$ and $F\le1$.
Comparing with $d\nu/d^2z=1/\pi$ at zero yields
\begin{equation}\label{eq:scalar-root-average}
 \E Q(p)=n\E\bigl[|c_1|^2F(p,0)\mid c_0=0\bigr].
\end{equation}

Gaussian monomial orthogonality gives, when $c_0=0$,
\[
 \|p\|_G^2=n!\nlsum_{j=1}^n|c_j|^2,
 \qquad
 \|p/z_2\|_G^2=n!\nlsum_{j=1}^n\frac{|c_j|^2}{j}.
\]
The variables $X_j=|c_j|^2$ are independent standard exponentials,
so with $S=\nlsum_{j=1}^nX_j$, which has mean $n$, the vector
$(V_j)=(X_j/S)$ is independent of $S$ and uniform on the simplex.
Thus
\[
 \E V_1^2=\frac{2}{n(n+1)},\qquad
 \E V_1V_j=\frac{1}{n(n+1)}\quad(j\ne1).
\]
Substituting into \eqref{eq:scalar-root-average} gives
\[
 \E Q(p)
 =n\E\left[\frac{X_1\nlsum_{j=1}^nX_j/j}{S}\right]
 =\frac{n(\mathsf H_n+1)}{n+1},
\]
and by \eqref{eq:scalar-standard-character} some realization
satisfies \eqref{eq:scalar-rank-two-ratio}.
\end{proof}

\subsection{Haar averages and exponential rates}
\begin{proof}[Proof of Lemma~\ref{lem:haar-immanant-average}]
Let $\Pi_\lambda$ be the central orthogonal projection onto the
$\lambda$-isotypic component of $(\C^n)^{\otimes n}$, and put
$e=e_1\otimes\cdots\otimes e_n$.
The central idempotent formula gives
\[
 \Imm_\lambda(P_n)
 =\frac{n!}{f^\lambda}
     \langle e,\Pi_\lambda P_n^{\otimes n}e\rangle,
 \qquad
 \langle e,\Pi_\lambda e\rangle
 =\frac{(f^\lambda)^2}{n!}.
\]
For the second, only the identity permutation has a nonzero
matrix coefficient at $e$.

Under Schur--Weyl decomposition, $Q_r^{\otimes n}$ acts on the
$\lambda$-isotypic component as $S_\lambda(Q_r)\otimes I$, where
$S_\lambda(\C^n)$ is the irreducible polynomial representation of
$\mathrm U(n)$ with character $s_\lambda$.
Averaging its unitary conjugates and applying Schur's lemma gives
\[
 \E P_n^{\otimes n}
 =\nlsum_{\lambda\vdash n}
   \frac{\tr S_\lambda(Q_r)}{\dim S_\lambda(\C^n)}\Pi_\lambda
 =\nlsum_{\lambda\vdash n}
   \frac{s_\lambda(1^r)}{s_\lambda(1^n)}\Pi_\lambda.
\]
Hence the first equality in \eqref{eq:haar-immanant-average}
holds. The hook-length and hook-content formulas give
\[
 \frac{f^\lambda}{s_\lambda(1^n)}
 =\frac{n!}{\nlprod_{(i,j)\in\lambda}(n+j-i)},
\]
which proves the second equality.
Finally, $s_\lambda(1^r)=0$ when $\ell(\lambda)>r$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:scalar-Haar-strip}]
For a partition $\lambda$ with at most $r$ rows, append zero parts
to write $\lambda=(\lambda_1,\ldots,\lambda_r)$.
The hook-content product \eqref{eq:schur} can be written as
\[
 c_\lambda(q,t)
 =q^{\nlsum_{i=1}^r(i-1)\lambda_i}
   \frac{\displaystyle\nlprod_{i=1}^r
               \nlprod_{j=1}^{\lambda_i}(1-tq^{j-i})}
        {\displaystyle\nlprod_{(i,j)\in\lambda}(1-q^{h_{ij}})}.
\]
Under the hypotheses on $t$, the only possible negative numerator
factor is the factor for the cell $(r,1)$.
Thus $c_\lambda(q,t)>0$ when $\lambda_r=0$ and
$c_\lambda(q,t)<0$ when $\lambda_r>0$.

The same product gives bounds uniform in the partition: for each
fixed $i\le r$, the partial products
$\nlprod_{j=1}^{l}(1-tq^{j-i})$ have no zero factor and converge as
$l\to\infty$ to a nonzero limit, so their absolute values lie
between positive constants, including at $l=0$.
Since $h_{ij}\ge\lambda_i-j+1$, the denominator satisfies
\[
 (q;q)_\infty^r
 \le\nlprod_{i=1}^r(q;q)_{\lambda_i}
 \le\nlprod_{(i,j)\in\lambda}(1-q^{h_{ij}})
 \le1.
\]
It follows that constants $a,b>0$, depending only on $q,t,r$,
satisfy
\begin{equation}\label{eq:haar-coefficient-bounds}
 a\,q^{\nlsum_{i=1}^r(i-1)\lambda_i}
 \le |c_\lambda(q,t)|
 \le b\,q^{\nlsum_{i=1}^r(i-1)\lambda_i}.
\end{equation}

To compare the terms in the Haar average, define
\[
 \mathcal D_r
 =\left\{p\in[0,1]^r:p_1\ge\cdots\ge p_r,
                          \ \nlsum_{i=1}^r p_i=1\right\},
 \qquad
 \varphi(x)=(1+x)\log(1+x)-x,
\]
and
\begin{equation*}
 \Phi(p)=(\log q)\nlsum_{i=1}^r(i-1)p_i
                         -\nlsum_{i=1}^r\varphi(p_i).
\end{equation*}
For $p_i=\lambda_i/n$ and $n\ge2r$, comparison with a Riemann
sum gives, uniformly in $\lambda$,
\[
 \log\nlprod_{(i,j)\in\lambda}(n+j-i)
 =n\log n+n\nlsum_{i=1}^r\varphi(p_i)+O_r(1).
\]
Indeed, replacing $j-i$ by $j$ inside each logarithm changes the
sum by $O_r(1)$, and the function $\log(1+x)$ has bounded
derivative on $[0,1]$.
Stirling's formula and the dimension formula
\[
 1\le s_\lambda(1^r)
 =\nlprod_{1\le i<j\le r}
       \frac{\lambda_i-\lambda_j+j-i}{j-i}
 \le(n+r)^{\binom r2}
\]
then show that constants $A,B>0$, depending only on $q,t,r$,
satisfy, for every $\lambda\vdash n$ with $\ell(\lambda)\le r$
and every $n\ge2r$,
\begin{equation}\label{eq:haar-term-bounds}
 A\sqrt n\,e^{n(\Phi(p)-1)}
 \le |c_\lambda(q,t)|\,\E\Imm_\lambda(P_n)
 \le B n^{\binom r2+1/2}e^{n(\Phi(p)-1)}.
\end{equation}

The function $\Phi$ is strictly concave, with Hessian
$-\diag((1+p_i)^{-1})$.
Set
\[
 K=\frac{r+1}{\nlsum_{j=0}^{r-1}q^j},\qquad
 p_i^*=Kq^{i-1}-1\quad(1\le i\le r).
\]
Condition~\eqref{eq:haar-strip-threshold} gives
$p_1^*>\cdots>p_r^*>0$, and $\nlsum_i p_i^*=1$.
At $p^*$ all coordinates of $\nabla\Phi$ equal $-\log K$, so
$p^*$ is the unique maximizer of $\Phi$ on $\mathcal D_r$.
Compactness and uniqueness imply the strict gap
\begin{equation}\label{eq:haar-rate-gap}
 \Phi_\partial
 :=\max_{\substack{p\in\mathcal D_r\\p_r=0}}\Phi(p)
 <\Phi(p^*)=:\Phi_*.
\end{equation}

For every sufficiently large $n$, choose
$\lambda_i^{(n)}=\lfloor np_i^*\rfloor$ for $2\le i\le r$
and $\lambda_1^{(n)}=n-\nlsum_{i=2}^r\lambda_i^{(n)}$.
These parts form a partition with $\lambda_r^{(n)}>0$ and
$\lambda^{(n)}/n=p^*+O_r(n^{-1})$.
The gradient of $\Phi$ is bounded on $\mathcal D_r$, so
\eqref{eq:haar-term-bounds} gives a constant $A'>0$ with
\[
 |c_{\lambda^{(n)}}(q,t)|\,
      \E\Imm_{\lambda^{(n)}}(P_n)
 \ge A'\sqrt n\,e^{n(\Phi_*-1)}.
\]
Every positive term has $\lambda_r=0$, and there are at most
$(n+1)^{r-1}$ such partitions, so their sum is at most
\[
 B' n^{\binom r2+r-1/2}e^{n(\Phi_\partial-1)}
\]
for a constant $B'>0$. Keeping only the negative term of
\eqref{eq:schur} indexed by $\lambda^{(n)}$ therefore gives
\[
 \E\Pqt(P_n)
 \le B'n^{\binom r2+r-1/2}e^{n(\Phi_\partial-1)}
       -A'\sqrt n\,e^{n(\Phi_*-1)}<0
\]
for every sufficiently large $n$, by the strict gap
\eqref{eq:haar-rate-gap}, and a negative expectation forces a
negative value at some projection in the orbit.

If $t$ ranges over a compact subinterval of
$(q^{r-1},q^{r-2})$, the row products used in
\eqref{eq:haar-coefficient-bounds} converge uniformly and stay
uniformly away from zero.
All constants in \eqref{eq:haar-term-bounds} can therefore be
chosen uniformly on that subinterval.
The rate function and the gap \eqref{eq:haar-rate-gap} depend only
on $q,r$, which proves the asserted uniform choice of $n_0$.
\end{proof}

\subsection{Gaussian deletion and an explicit rank-three instance}
\begin{proof}[Proof of Lemma~\ref{lem:gaussian-deletion}]
If some $u_i=0$, both sides vanish. Otherwise let $z\in\C^d$
collect the coordinates of $g_1,\ldots,g_k$, and put
$p=\nlprod_iL_i$, $\rho=\|z\|^2$, and
$\Delta=\nlsum_\nu\partial_{z_\nu}\partial_{\bar z_\nu}$. Direct
differentiation and $c_a^2\le(\max_ac_a)c_a$ give
\[
 \Delta\log L_i\ge\frac{\beta a_{ii}}{L_i}.
\]
Let $\E_p$ be the Gaussian measure weighted by $p/\E p$.
Bihomogeneity and integration by parts give
$\E_p(\Delta p/p)=n$, while $\rho\sim\Gamma(n+d,1)$ under $\E_p$
and $\|\partial\log p\|^2\ge n^2/\rho$ by Euler's identity, so that
\[
 \beta\frac{\nlsum_ia_{ii}F_C(A[[n]\setminus\{i\}])}{F_C(A)}
 \le\E_p\Delta\log p
 \le n-\frac{n^2}{n+d-1}
 =\frac{n(d-1)}{n+d-1}.
\]
Integration by parts is applied only to the polynomial $p$; the
nonnegative logarithmic terms, after multiplication by $p$, are
bounded in sum by $\Delta p$, hence integrable. If
$T_a=\nlsum_{|S|=a}\bigl(\nlprod_{i\in S}a_{ii}\bigr)F_C(A[S^c])$,
then \eqref{eq:delete} gives $(a+1)T_{a+1}\le KT_a$, hence
$T_a\le K^aF_C(A)/a!$, and expanding the product gives
\eqref{eq:shift}. Finally, $u_i^*Yu_i\le(\tr Y)a_{ii}$ proves
the last assertion.
\end{proof}

In rank three, the constants in the proof of
Theorem~\ref{thm:rank-budget} give an explicit pair $(q,t)$.
\begin{rmk}[A rank-three instance]\label{rem:rank-three-instance}
At $q=1/10$ and $t=21/2000$ every factor of $D_\lambda(t)$ with
$\ell(\lambda)\le3$ has absolute value at most one on $[0,1/50]$,
so $|D_\lambda'|\le B_3=370/3$, and $M_3\le729/64$; with
$\delta=1/2000$ the constant is
$1-(370/3)(729/64)/2000=3809/12800$, and
\[
 \per_{1/10,21/2000}(A)
 \ge\frac{3809}{12800}\per_{1/10,1/100}(A)\ge0
 \qquad(\rank A\le3)
\]
in every size. As $21/2000>q^2$, the region
$\mathcal R_3(1/10)=[0,1/100]\cup\{1/10\}$ is strictly smaller
than the scalar region in rank three, whereas the two coincide for
$q>(1+\sqrt{13})/6$ (Corollary~\ref{cor:scalar-Haar-classification}).
\end{rmk}

\section{The exact scalar regions in sizes three and four}
\label{app:size-three}

Holding the size fixed, instead of the rank, gives a third scalar
regime, and in small sizes it can be settled completely. The answers
show how far the fixed-size regions are from the all-size ones of
Section~\ref{sec:positivity}: the thresholds in sizes three and four
both lie strictly above $t=q$, and their witnesses give necessary
conditions in every larger size.

In size three the scalar region can be found exactly, because
once the diagonal is normalized the permutation sum depends on
the matrix through two real numbers, which positive
semidefiniteness confines to an explicit region. The argument
works for arbitrary nonnegative cycle weights $w=(w_k)_{k\ge1}$,
for which we write
\[
 \per_w(A)=\nlsum_{\sigma\in S_n}
 \nlprod_{C\in\Cyc(\sigma)}w_{|C|}
 \nlprod_{i=1}^na_{i,\sigma(i)}.
\]
For constant weights and real symmetric matrices, the minimum
below was found by \citet[Theorem~6.1]{osogami2013remarks}; the
argument extends to independent cycle weights and complex
matrices.

\begin{theorem}[Sharp positivity in size three]
\label{thm:scalar-three}
Let $w_1,w_2,w_3\ge0$. For every Hermitian positive
semidefinite matrix $A=(a_{ij})_{i,j=1}^3$,
\begin{equation}\label{eq:scalar-three-lower}
 \per_w(A)
 \ge\left(w_1^3+
       \min\left\{0,\frac{3w_1w_2-w_3}{4}\right\}\right)
       a_{11}a_{22}a_{33}.
\end{equation}
The constant is sharp among matrices with prescribed positive
diagonal entries. If a diagonal entry vanishes, both sides of
\eqref{eq:scalar-three-lower} are zero. Consequently,
\begin{equation}\label{eq:scalar-three-criterion}
 \per_w(A)\ge0\text{ for every Hermitian }A\succeq0
 \quad\Longleftrightarrow\quad
 4w_1^3+3w_1w_2-w_3\ge0.
\end{equation}
For Macdonald weights with $0\le q,t<1$, this condition is
equivalent to
\begin{equation}\label{eq:scalar-three-threshold}
 t\le\tau_3(q):=
 \frac{2+3q+3q^2}{3+3q+2q^2}.
\end{equation}
\end{theorem}
\begin{proof}

If a diagonal entry of $A$ vanishes, the corresponding row and
column vanish by positive semidefiniteness, every permutation
monomial is zero, and \eqref{eq:scalar-three-lower} holds with
both sides zero. Otherwise a diagonal congruence, which multiplies
the weighted permanent by $a_{11}a_{22}a_{33}$, reduces the assertion
to a correlation matrix $R=(r_{ij})$. Put
\[
 s=|r_{12}|^2+|r_{23}|^2+|r_{31}|^2,
 \qquad z=\operatorname{Re}(r_{12}r_{23}r_{31}).
\]
The permutation expansion and determinant inequality give
\[
 \per_w(R)=w_1^3+w_1w_2s+2w_3z,
 \qquad \det R=1-s+2z\ge0.
\]
If $z\ge0$, the weighted permanent is at least $w_1^3$.
Suppose $z<0$ and write $u=-z$. The arithmetic--geometric
mean inequality and the determinant inequality imply
\begin{equation}\label{eq:scalar-three-feasible}
 0<s<1,
 \qquad
 u\le\min\left\{
       \left(\frac{s}{3}\right)^{3/2},\frac{1-s}{2}\right\}.
\end{equation}
The two bounds agree at $s=3/4$.

On $0\le s\le3/4$ the resulting lower bound
\[
 w_1^3+w_1w_2s-2w_3(s/3)^{3/2}
\]
is concave in $s$, so its minimum is attained at an endpoint,
where it equals $w_1^3$ and $w_1^3+3w_1w_2/4-w_3/4$.
For $3/4\le s\le1$, the other bound in
\eqref{eq:scalar-three-feasible} gives
\[
 \per_w(R)
 \ge w_1^3-w_3+(w_1w_2+w_3)s
 \ge w_1^3+\frac34w_1w_2-\frac14w_3.
\]
Together these cases prove \eqref{eq:scalar-three-lower}.
The value $w_1^3$ is attained by $I_3$, and the other endpoint
value by
\[
 R_*=
 \begin{pmatrix}
 1&-1/2&-1/2\\
 -1/2&1&-1/2\\
 -1/2&-1/2&1
 \end{pmatrix}\succeq0,
\]
with eigenvalues $3/2,3/2,0$; diagonal congruence transfers both
equality cases to any positive diagonal, which proves sharpness,
and the lower bound with these examples proves
\eqref{eq:scalar-three-criterion}.

For Macdonald weights, $w_1>0$ and
\[
 \frac{4w_1^3+3w_1w_2-w_3}{w_1}
 =\frac{4(1-t)^2}{(1-q)^2}
  +\frac{3(1-t^2)}{1-q^2}
  -\frac{1+t+t^2}{1+q+q^2}.
\]
The right side factors as
\begin{equation*}
 \frac{3(1-qt)
  \bigl(2+3q+3q^2-(3+3q+2q^2)t\bigr)}
 {(1-q)^2(1+q)(1+q+q^2)},
\end{equation*}
and all factors other than the last linear one are positive on
$0\le q,t<1$, which proves \eqref{eq:scalar-three-threshold}.
\end{proof}
\begin{example}[Scalar failure above the exact size-three threshold]\label{ex:negative}
The rank-two matrix $R_*$ in Theorem~\ref{thm:scalar-three}
satisfies $\Pqt(R_*)<0$ exactly when $t>\tau_3(q)$.
For example,
\[
 \per_{0,3/4}(R_*)=-\frac3{64},\qquad
 \per_{1/10,3/4}(R_*)=-\frac{250}{8019}.
\]
\end{example}

The threshold exceeds $q$, since
\[
 \tau_3(q)-q
 =\frac{2(1-q^3)}{3+3q+2q^2}>0
 \qquad(0<q<1),
\]
so the size-three scalar region is an interval that contains
$\mathcal R_3(q)=[0,q^2]\cup\{q\}$ and extends beyond $t=q$; at
$q=0$ the threshold is $t=2/3$. In the Jack limit the criterion
becomes $\beta(4\beta-1)(\beta+1)\ge0$, so for the
$\beta$-permanent with $\beta>0$ the size-three threshold is
$\beta\ge1/4$.

Padding $R_*$ with an identity block carries the size-three
obstruction to every larger size.
\begin{corr}[A necessary condition in every larger size]
\label{cor:scalar-three-necessary}
Fix $n\ge3$ and $0\le q,t<1$. If $\Pqt(A)\ge0$ for every
$n\times n$ Hermitian positive semidefinite matrix, then
$t\le\tau_3(q)$. More precisely, if $t>\tau_3(q)$, the real
positive semidefinite matrix $R_*\oplus I_{n-3}$ satisfies
\[
 \Pqt(R_*\oplus I_{n-3})
 =w_1^{n-3}\left(w_1^3+\frac34w_1w_2-\frac14w_3\right)<0.
\]
\end{corr}
\begin{proof}
Every nonzero permutation monomial respects the two diagonal
blocks. The singleton blocks contribute $w_1^{n-3}>0$, and
Theorem~\ref{thm:scalar-three} gives the sign of the remaining
factor.
\end{proof}

In size four the same reduction to correlation matrices leaves
four cycle sums, whose values have a convex hull with only seven
vertices, because seven test matrices decide
every class-function inequality on $4\times4$ positive
semidefinite matrices \citep[Theorem~2(a)]{barrett1999cone}. The
sign of $\per_w$ on the cone is therefore decided by seven
matrices, three of which give the exact region.

\begin{theorem}[Sharp positivity in size four]
\label{thm:scalar-four}
Put $S(q)=1+q+q^2$, $B(q)=2+q+q^2+2q^3$, and
\begin{equation*}
 \tau_4(q)=\frac{2S(q)}{B(q)+\sqrt{B(q)^2-4qS(q)^2}}.
\end{equation*}
For $0\le q,t<1$, the following are equivalent:
\begin{enumerate}[label=(\roman*),leftmargin=2.2em]
\item $\Pqt(A)\ge0$ for every Hermitian positive semidefinite
$A\in M_4(\C)$;
\item $Q_q(t):=S(q)-B(q)t+qS(q)t^2\ge0$;
\item $t\le\tau_4(q)$.
\end{enumerate}
In particular, $\tau_4(0)=1/2$. If $t>\tau_4(q)$, a negative
witness is the rank-two correlation matrix
\begin{equation}\label{eq:H-four}
 H=
 \begin{pmatrix}
 1&s&s&s\\
 s&1&\mathrm is&-\mathrm is\\
 s&-\mathrm is&1&\mathrm is\\
 s&\mathrm is&-\mathrm is&1
 \end{pmatrix},
 \qquad s=\frac1{\sqrt3}.
\end{equation}
\end{theorem}

We identify the seven test matrices directly, using the cited
theorem only for their number.

\begin{lemma}[Seven cycle-sum vectors]\label{lem:seven}
For a correlation matrix $A$ of order four, put
\[
 x=\nlsum_{i<j}|a_{ij}|^2,\qquad
 y=|a_{12}a_{34}|^2+|a_{13}a_{24}|^2+|a_{14}a_{23}|^2,
\]
and let $z$ and $v$ be the sums of the permutation monomials of
cycle types $(3,1)$ and $(4)$. The convex hull of $(x,y,z,v)$
over all complex correlation matrices of order four has exactly
the following seven vertices, where $J_m$ is the all-ones matrix,
$R_m=\frac{m}{m-1}I_m-\frac1{m-1}J_m$, and $H$ is the matrix
\eqref{eq:H-four}:
\[
\begin{array}{c|rrrr}
 A&x&y&z&v\\ \hline
 I_4&0&0&0&0\\
 J_4&6&3&8&6\\
 J_3\oplus1&3&0&2&0\\
 J_2\oplus J_2&2&1&0&0\\
 R_3\oplus1&3/4&0&-1/4&0\\
 R_4&2/3&1/27&-8/27&2/27\\
 H&2&1/3&0&-2/3
\end{array}
\]
\end{lemma}
\begin{proof}
Let $P$ be the convex hull, a compact set. A real class-function
evaluation on a correlation matrix is an affine function of
$(x,y,z,v)$; invertible diagonal congruence multiplies every
permutation monomial by the same positive factor, and a positive
semidefinite matrix with a zero diagonal entry has a zero row.
The seven-test theorem \citep[Theorem~2(a)]{barrett1999cone}
with convex separation therefore makes $P$ the convex hull of at
most seven points, and it remains to exhibit the seven listed
points as distinct vertices. In the face arguments below, each
inequality is applied first to generating correlation matrices
and then to their convex combinations, and equality forces
equality in every component with positive coefficient.

Every correlation matrix has $0\le x\le6$. At $x=0$ the matrix
is $I_4$; at $x=6$, $\tr(A^2)=4+2x=16=(\tr A)^2$ forces rank one,
and a rank-one correlation matrix is diagonally unitarily
equivalent to $J_4$. These two values give unique cycle-sum
vectors, hence vertices.

The face $y=0$ consists of matrices whose graph of nonzero
off-diagonal entries has no two disjoint edges, so the graph lies
in a star or in a triangle with an isolated vertex. A star has
$z=v=0$; a triangle has $v=0$ and $z=2\operatorname{Re}p$ with
$p=a_{12}a_{23}a_{31}$ after relabelling. The bound $z\le2$ has
equality only at the cycle vector of $J_3\oplus1$. If
$\operatorname{Re}p=-a<0$, the $3\times3$ determinant and the
AM--GM inequality give
$3a^{2/3}\le3|p|^{2/3}\le x\le1-2a$, so $a\le1/8$ and
$z\ge-1/4$, with equality forcing $x=3/4$ and the cycle vector of
$R_3\oplus1$. The two extrema on this face are two more vertices.

The affine functional $\det A=1-x+y+z-v$ is nonnegative, and its
zero face is generated by matrices of rank at most three, on
which $4+2x=\tr(A^2)\ge(\tr A)^2/3=16/3$. Equality forces the
spectrum $(4/3,4/3,4/3,0)$, so $A=\frac43(I-uu^*)$ for a unit
vector $u$ with $|u_i|=1/2$, and diagonal unitary conjugation
gives $R_4$, the fifth vertex.

Finally, $e_3(A)=\nlsum_{|T|=3}\det A[T]=4-2x+z\ge0$ vanishes
exactly at rank at most two, where $4+2x\ge16/2$ with equality
exactly for the spectrum $(2,2,0,0)$. Then $A^2=2A$, $z=0$, and
$v=y-1$, and the diagonal equations of $A^2=2A$ give
$|a_{12}|^2=|a_{34}|^2=a$, $|a_{13}|^2=|a_{24}|^2=b$,
$|a_{14}|^2=|a_{23}|^2=c$ with $a+b+c=1$, so
$y=a^2+b^2+c^2\in[1/3,1]$. The endpoints of this segment are
attained by $H$, which satisfies $H^2=2H$ and so is positive
semidefinite of rank two with the listed cycle sums, and by
$J_2\oplus J_2$. These are the last two vertices.
\end{proof}

Evaluating $\per_w$ at the seven vertices turns Lemma~\ref{lem:seven}
into inequalities on the weights; four vertices give positive values,
and the other three give the conditions below.
\begin{corr}[Positive cycle weights in size four]
\label{cor:scalar-four-weights}
For $w_1,w_2,w_3,w_4>0$, $\per_w$ is nonnegative on every
Hermitian positive semidefinite matrix of order four if and only
if
\begin{align}
 4w_1^3+3w_1w_2-w_3&\ge0,\label{eq:four-triangle}\\
 27w_1^4+18w_1^2w_2+w_2^2-8w_1w_3+2w_4&\ge0,\label{eq:four-simplex}\\
 3w_1^4+6w_1^2w_2+w_2^2-2w_4&\ge0.\label{eq:four-tetra}
\end{align}
\end{corr}
\begin{proof}
For a correlation matrix,
$\per_w(A)=w_1^4+w_1^2w_2x+w_2^2y+w_1w_3z+w_4v$, which is
positive on the first four matrices of Lemma~\ref{lem:seven};
its values on $R_3\oplus1$, $R_4$, and $H$, multiplied by
positive factors, are the left sides of
\eqref{eq:four-triangle}--\eqref{eq:four-tetra}.
\end{proof}

\begin{proof}[\textbf{Proof of Theorem~\ref{thm:scalar-four}}]
A brief calculation gives
\begin{equation}\label{eq:H-factor}
 \Pqt(H)
 =w_1^4+2w_1^2w_2+\frac{w_2^2-2w_4}{3}
 =\frac{8(1-t)(1-qt)Q_q(t)}
 {3(1-q)^4(1+q)^2(1+q^2)},
\end{equation}
with every factor other than $Q_q(t)$ positive on the square,
so $Q_q(t)\ge0$ is necessary. Since $Q_q(0)=S(q)>0$,
$Q_q(1)=-(1-q)^2(1+q)<0$, and
$B(q)^2-4qS(q)^2=(1-q)^2(4q^4+8q^3+9q^2+8q+4)>0$, the quadratic
$Q_q$ has exactly one root in $(0,1)$, namely $\tau_4(q)$, and
is nonnegative on $[0,1)$ exactly to its left; the formula also
holds at $q=0$.

For sufficiency, apply Corollary~\ref{cor:scalar-four-weights}.
Inequality \eqref{eq:four-triangle} is $t\le\tau_3(q)$ by
Theorem~\ref{thm:scalar-three}, and with $D=3+3q+2q^2$,
\[
 D^2Q_q(\tau_3(q))
 =-(1-q)^2(1+q)(3q^4+8q^3+12q^2+8q+3)<0,
\]
so $\tau_4(q)<\tau_3(q)$ and \eqref{eq:four-triangle} follows
from $t\le\tau_4(q)$. Inequality \eqref{eq:four-simplex} holds
everywhere: $\Pqt(R_4)$ equals
$8(1-t)(1-qt)\bigl(a(q)-b(q)t+c(q)t^2\bigr)$ over
$27(1-q)^4(1+q)^2(1+q^2)(1+q+q^2)$, where
\begin{align*}
 a(q)&=5+12q+20q^2+20q^3+16q^4+8q^5,\\
 b(q)&=12+28q+41q^2+41q^3+28q^4+12q^5,\\
 c(q)&=8+16q+20q^2+20q^3+12q^4+5q^5,
\end{align*}
and the quadratic in $t$ is positive for every real $t$ because
$c(q)>0$ and
\[
 b(q)^2-4a(q)c(q)
 =-(1-q)^2\bigl(16+64q+152q^2+240q^3+271q^4
              +240q^5+152q^6+64q^7+16q^8\bigr)<0.
\]
Inequality \eqref{eq:four-tetra} is $Q_q(t)\ge0$ by
\eqref{eq:H-factor}.
\end{proof}

Since $Q_q(q)=(1-q)(1-q^4)>0$, the thresholds satisfy
$q<\tau_4(q)<\tau_3(q)$ for $0\le q<1$: the size-four region is
strictly larger than the all-size rank-two region $t\le q$ and
strictly smaller than the size-three region; at $q=0$ they are
$t\le1/2$ and $t\le2/3$. Padding the witness by singleton blocks,
$\Pqt(H\oplus I_{n-4})=w_1^{n-4}\Pqt(H)$, so $t\le\tau_4(q)$ is
necessary for scalar positivity in every fixed size $n\ge4$,
which sharpens Corollary~\ref{cor:scalar-three-necessary}.

\section{Partition coefficients for the block inequalities}\label{app:frenkel}

The proof of Theorem~\ref{thm:block-inequalities} expands $\Pqt$ in
products of permanents of principal blocks and needs the signs of the
coefficients. At $t=q^M$ these coefficients are sums over assignments
of distinct colors $q^r$ to the blocks, which makes them divided
differences of $x\mapsto x^\alpha$. Their signs are then those of the
falling factorial $\alpha(\alpha-1)\cdots(\alpha-k+1)$, and this is where the size
hypothesis $t\in\mathcal R_n(q)$ comes from.
Let $\Pi_n$ denote the set partitions of $[n]$, and put
$z^{\underline k}=z(z-1)\cdots(z-k+1)$, with
$z^{\underline0}=1$.

For positive integers $s_1,\ldots,s_k$ and a symmetric-function
specialization $\rho$, define
\begin{equation}\label{eq:augmented-monomial}
 \widetilde m_s[\rho]
 :=\nlsum_{\theta\in\Pi_k}
    \nlprod_{C\in\theta}
       (-1)^{|C|-1}(|C|-1)!\,
       p_{\nlsum_{i\in C}s_i}[\rho].
\end{equation}
On a finite alphabet $z=(z_1,\ldots,z_M)$, this is
\begin{equation}\label{eq:augmented-distinct}
 \widetilde m_s(z)
 =\sum_{\substack{j_1,\ldots,j_k\in[M]\\
                   j_1,\ldots,j_k\text{ distinct}}}
       z_{j_1}^{s_1}\cdots z_{j_k}^{s_k}.
\end{equation}
Indeed, expanding the power sums in \eqref{eq:augmented-monomial}
and fixing the indices $j_1,\ldots,j_k$ chosen for the $k$
factors, the coefficient of $z_{j_1}^{s_1}\cdots z_{j_k}^{s_k}$
factors over the groups of equal indices, and for a group of size
$r$ the factor is $\nlsum_{\sigma\in S_r}(-1)^{r-c(\sigma)}$, which
equals one for $r=1$ and zero for $r>1$. In the usual monomial
notation,
$\widetilde m_s=(\nlprod_j m_j(s)!)m_{\operatorname{sort}(s)}$,
where $m_j(s)$ is the number of entries of $s$ equal to $j$.

For pairwise distinct real nodes $x_0,\ldots,x_k$, write
\[
 [x_0,\ldots,x_k]f
 :=\nlsum_{j=0}^k\frac{f(x_j)}
                         {\nlprod_{\ell\ne j}(x_j-x_\ell)}
\]
for the divided difference of order $k$. Its sign determines the
sign of every coefficient in the expansion.

\begin{lemma}[Signs of the partition coefficients]
\label{lem:augmented-sign}
Let $0<q<1$, $\alpha>0$, $k\ge1$, and
$s=(s_1,\ldots,s_k)\in\N^k$ with $s_i\ge1$.
For $\tau\in S_k$, set $a_i=s_{\tau(i)}$,
$S_j=\nlsum_{i=j+1}^k a_i$ for $0\le j\le k$, and $x_j=q^{S_j}$. Then
\begin{equation}\label{eq:augmented-divided-difference}
 \widetilde m_s[B_{q,q^\alpha}]
 =\nlsum_{\tau\in S_k}
    q^{\nlsum_{i=1}^k(i-1)a_i}
    [x_0,\ldots,x_k](x^\alpha),
\end{equation}
and
\begin{equation}\label{eq:augmented-sign}
 \operatorname{sgn}\widetilde m_s[B_{q,q^\alpha}]
 =\operatorname{sgn}\alpha^{\underline{k}}.
\end{equation}
If $t\in\mathcal R_n(q)$ and $\nlsum_i s_i=n$, then
$\widetilde m_s[B_{q,t}]\ge0$.
\end{lemma}

\begin{proof}
First take $\alpha=M$ to be an integer with $M\ge k$.
The specialization $B_{q,q^M}$ is the finite alphabet
$(1,q,\ldots,q^{M-1})$. In \eqref{eq:augmented-distinct},
sort the selected indices as
$0\le r_1<\cdots<r_k\le M-1$ and let $\tau\in S_k$ be the
permutation for which $r_i$ carries the exponent
$a_i=s_{\tau(i)}$. Put
\[
 g_0=r_1,\qquad
 g_j=r_{j+1}-r_j-1\quad(1\le j<k),\qquad
 g_k=M-1-r_k.
\]
These gaps are nonnegative and sum to $M-k$, and the term is
\[
 q^{\nlsum_i a_i r_i}
 =q^{\nlsum_i(i-1)a_i}\nlprod_{j=0}^k x_j^{g_j}.
\]
Summing over the gaps therefore gives
$q^{\nlsum_i(i-1)a_i}h_{M-k}(x_0,\ldots,x_k)$.
Partial fractions in the generating function
$\nlprod_{j=0}^k(1-x_jz)^{-1}$ give
\[
 h_{M-k}(x_0,\ldots,x_k)
 =\nlsum_{j=0}^k
    \frac{x_j^M}{\nlprod_{\ell\ne j}(x_j-x_\ell)}
 =[x_0,\ldots,x_k](x^M).
\]
Summing over $\tau$ proves \eqref{eq:augmented-divided-difference}
for these integer values of $\alpha$.

For fixed $q$, both sides are polynomials in $t=q^\alpha$ of
degree at most $\nlsum_i s_i$, the right side because
$x_j^\alpha=t^{S_j}$, and they agree at the infinitely many
points $t=q^M$, hence identically.

The nodes satisfy $0<x_0<\cdots<x_k=1$, and for $f\in C^k$ on
their interval the mean value theorem for divided differences
gives
\[
 [x_0,\ldots,x_k]f=\frac{f^{(k)}(\xi)}{k!}
 \qquad\text{for some }\xi\in(x_0,x_k).
\]
For $f(x)=x^\alpha$, $f^{(k)}(x)=\alpha^{\underline{k}}x^{\alpha-k}$,
so every summand in \eqref{eq:augmented-divided-difference} has
the sign of $\alpha^{\underline{k}}$; if the falling factorial
vanishes, $x^\alpha$ is a polynomial of degree less than $k$ and
every divided difference is zero. Either way
\eqref{eq:augmented-sign} follows.

Finally, for $0<t<1$, membership of $t=q^\alpha$ in
$\mathcal R_n(q)$ means $\alpha\in\{1,2,\ldots\}$ or
$\alpha\ge n-1$, and then $\alpha^{\underline{k}}\ge0$ because
$k\le n$; at $t=0$ the conclusion follows by continuity.
\end{proof}

The positive coefficients let Lieb's inequality act on each
principal block separately.

\begin{prop}[Expansion into products of permanents]
\label{prop:positive-partition-expansion}
Let $n\ge1$, $0<q<1$, $t\in\C$, and $A\in M_n(\C)$.
For $\pi\in\Pi_n$, put
$a_\pi(q,t)=\widetilde m_{(|B|:B\in\pi)}[B_{q,t}]$.
Then
\begin{equation}\label{eq:positive-partition-expansion}
 \Pqt(A)
 =\nlsum_{\pi\in\Pi_n}a_\pi(q,t)
       \nlprod_{B\in\pi}\per(A[B]).
\end{equation}
If $t\in\mathcal R_n(q)$, every coefficient $a_\pi(q,t)$
is nonnegative.
\end{prop}

\begin{proof}
Take $t=q^M$ with an integer $M\ge n$, so that
$w_{|C|}=\nlsum_{r=0}^{M-1}q^{r|C|}$ and each cycle $C$ of $\sigma$
chooses a color $r\in\{0,\ldots,M-1\}$ with weight $q^{r|C|}$. The indices whose cycles
share a color form a block $B$ of a set partition $\pi$. For
prescribed blocks and distinct colors, summing the permutations
within $B$ gives $\per(A[B])$, and summing over the distinct
colors gives $\widetilde m_{(|B|:B\in\pi)}(1,q,\ldots,q^{M-1})$.
Thus \eqref{eq:positive-partition-expansion} holds at $t=q^M$,
and since both sides are polynomials in $t$ of degree at most
$n$, it holds for every $t$.
Lemma~\ref{lem:augmented-sign} proves the final assertion.
\end{proof}

\section{Boundary formulas used in the positivity proofs}\label{app:boundary}

Several proofs use $\Pqt$ at the edges of the parameter square,
where its formulas simplify: the case $q=0$ of
the rank-two theorem uses the Hall--Littlewood formula, and the
matrix-ball comparison the Jack limit. We also prove the two-type
positivity mentioned after Theorem~\ref{thm:scalar-rank-two}.

\Needspace{6\baselineskip}
\begin{prop}[Limits and boundary formulas]\label{thm:limits}
For $A\in M_n(\C)$, the following hold.
\begin{enumerate}
\item If $q=e^{-\varepsilon}$ and
$t=e^{-\alpha\varepsilon}$ with $\alpha>0$, then
$\Pqt\to\per_\alpha$ as $\varepsilon\downarrow0$, with Jack
parameter $1/\alpha$.
\item On $q=t$, one has $\Pqt(A)=\per(A)$.
\item At $q=0$,
\begin{align}\label{eq:HL}
 \per_{0,t}(A)
 &=\nlsum_{S\subseteq[n]}(-t)^{|S|}
 \det A[S]\per A[S^c]\nonumber\\
 &=(1-t)\nlsum_{j=0}^{n-1}(-t)^j
 \Imm_{(n-j,1^j)}(A),
 \qquad n\ge1.
\end{align}
The subset identity also holds for $n=0$, with value $1$.
\item At $t=0$, the Schur coefficients are
\[
 c_\lambda(q,0)=
 \frac{q^{\nlsum_i(i-1)\lambda_i}}
 {\nlprod_{(i,j)\in\lambda}(1-q^{h_{ij}})}.
\]
\item For an integer $m\ge1$,
\begin{equation}\label{eq:colors}
 \per_{q,q^m}(A)=
 \nlsum_{S_0\sqcup\cdots\sqcup S_{m-1}=[n]}
 \nlprod_{r=0}^{m-1}q^{r|S_r|}\per A[S_r].
\end{equation}
The blocks are ordered and may be empty.
\end{enumerate}

\end{prop}
\begin{proof}
For fixed $k$,
\[
 \frac{1-e^{-\alpha k\varepsilon}}
 {1-e^{-k\varepsilon}}\longrightarrow\alpha.
\]
The permutation sum has finitely many terms, so its limit is
$\per_\alpha(A)$. The Macdonald power-sum norm tends to
$z_\mu\alpha^{-\ell(\mu)}$; the Jack scalar product with parameter
$\theta$ has norm $z_\mu\theta^{\ell(\mu)}$.
Thus $\theta=1/\alpha$. On $q=t$ every $w_k=1$.

At $q=0$, the determinant product reduces to
$\det(I-tXA)/\det(I-XA)$. The principal-minor expansion of the
numerator and the ordinary-permanent instance of the master theorem
give the subset identity in \eqref{eq:HL}. For its immanant form,
take $q\downarrow0$ in \eqref{eq:schur}. A partition that is not a
hook contains the cell $(2,2)$, whose numerator is $q(1-t)$; every
other numerator stays bounded and every denominator tends to one.
Its coefficient therefore vanishes. In the hook $(n-j,1^j)$, the
cell $(1,1)$ contributes $1-t$, the other first-row cells contribute
one, and the remaining $j$ first-column cells contribute $-t$.
The product of these factors is the stated hook coefficient. Setting $t=0$ in
\eqref{eq:schur} gives the fourth assertion directly.

Finally,
$w_k(q,q^m)=\nlsum_{r=0}^{m-1}q^{rk}$.
Choosing a term in this sum colors a $k$-cycle with a color $r$ and
assigns it weight $q^{rk}$. Let $S_r$ contain the labels in cycles of
color $r$. For a fixed ordered partition $(S_0,\ldots,S_{m-1})$,
summing the permutations within each block gives $\per A[S_r]$,
while the color weight is $q^{r|S_r|}$. Every colored permutation
occurs once, which proves \eqref{eq:colors}.
\end{proof}

As $q\uparrow1$, each cell factor
$(q^{i-1}-q^\alpha q^{j-1})/(1-q^{h_{ij}})$ of the hook product tends
to $(\alpha+j-i)/h_{ij}$, and since $n$ has finitely many partitions,
the immanant expansion becomes
\begin{equation}\label{eq:jack-schur}
 \per_\alpha(A)=\nlsum_{\lambda\vdash n}
 \left(\nlprod_{(i,j)\in\lambda}
 \frac{\alpha+j-i}{h_{ij}}\right)\Imm_\lambda(A).
\end{equation}

With two distinct Gram vectors, each repeated, every permutation
monomial is nonnegative.
\begin{corr}[Two-type positivity]\label{cor:two-type-positive}
Let $K=\left(\begin{smallmatrix}a&b\\\overline b&d\end{smallmatrix}\right)
\succeq0$ and $0\le q,t<1$. If $K[p,m]$ repeats its first index
$p$ times and its second index $m$ times, then
$\Pqt(K[p,m])\ge0$ for every $p,m\in\N$.
\end{corr}
\begin{proof}
A permutation of the repeated labels crosses from the first type
to the second as many times as it crosses back. If that number is
$j$, its matrix monomial is $a^{p-j}d^{m-j}|b|^{2j}\ge0$.
Every cycle weight is positive, so the defining sum is nonnegative.
\end{proof}

\end{document}