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  pdftitle={Complex and real counterexamples to an inequality arising from Drury's conjecture},
  pdfauthor={Yair Lavi}
}

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\title{Complex and real counterexamples to an inequality arising from Drury's conjecture}
\author{Yair Lavi}
\date{19 October 2026}

\begin{document}
\maketitle
\begin{abstract}
We give counterexamples to the following inequality, stated by Zhang
as a consequence of Drury's 2016 permanent conjecture: if
$A=(a_{jk})_{j,k=1}^n$ is a Hermitian positive semidefinite matrix of order
$n\geq2$, then
\[
 \sum_{j=1}^{n-1}|a_{jn}|^2\per A[\,\hat{\jmath},\hat{n}\,]\leq\per A
\]
where $A[\,\hat{\jmath},\hat{n}\,]$ is the principal submatrix obtained by
deleting indices $j$ and $n$.
Our complex example has order twelve and rank two; our real example has
order twenty-one and rank three.
\end{abstract}

\section{Introduction and main results}\label{sec:intro}

For a square matrix $A=(a_{jk})_{j,k=1}^n$, its permanent is
\[
 \per A=\sum_{\sigma\in S_n}\prod_{j=1}^n a_{j,\sigma(j)},
\]
where $S_n$ is the set of permutations of $\{1,\ldots,n\}$; the permanent
of the empty matrix is defined to be one.
For distinct indices $j,r$, write $A[\,\hat{\jmath},\hat{r}\,]$ for the
principal submatrix obtained by deleting indices $j,r$, and write
$A[\,\hat{r}\,]$ for deletion of index $r$ alone. Set
\begin{equation}\label{eq:q}
 q_r(A)=\sum_{j\ne r}|a_{jr}|^2\per A[\,\hat{\jmath},\hat{r}\,],
 \qquad q(A)=q_n(A).
\end{equation}

In his 2016 survey, Zhang \cite{Zhang} recorded a conjecture attributed
to Drury (private communication, 2016): for every Hermitian positive
semidefinite matrix $A$ of order $n\geq2$,
\begin{equation}\label{eq:stronger}
 \bigl(a_{nn}\per A[\,\hat{n}\,]\bigr)^2+q(A)^2\leq(\per A)^2.
\end{equation}
Zhang stated that this would imply the following inequality, equation
(21) of his survey:
\begin{equation}\label{eq:target}
 q(A)\leq\per A.
\end{equation}
This implication uses the nonnegativity of $q(A)$ and $\per A$ for
positive semidefinite $A$, which follows from Lemma~\ref{lem:norm} below.
We put the distinguished
index last; simultaneous permutation gives the source's first-index
formulation, or the equivalent assertion $q_r(A)\leq\per A$ for every $r$.

Hutchinson \cite{Hutchinson} disproved \eqref{eq:stronger} with a complex
matrix of order four. Chalmers \cite{Chalmers} subsequently gave a real
counterexample of order eight. Both displayed matrices nevertheless
satisfy \eqref{eq:target} at every distinguished
index: their largest ratios $q_r(A)/\per A$ are, respectively,
approximately $0.37547$ and $0.47963$.\footnote{Exact evaluation of the
defining permanent sums gives minimum gaps $\min_r(\per A-q_r(A))$ of exactly
$3.124761868597335$ and $626908608$, respectively, for the matrices
as printed.}
These examples therefore did not settle \eqref{eq:target}.

Our main result disproves \eqref{eq:target} over both fields.

\begin{theorem}\label{thm:main}
There is a complex positive semidefinite matrix of order twelve and
rank two for which $q(A)>\per A$. There is also a real positive
semidefinite matrix of order twenty-one and rank three for which
$q(A)>\per A$.
\end{theorem}

The constructions are explicit. Their gaps $\per A-q(A)$ are
$900(5183-4608\sqrt2)$ in the complex case and
$-1\,152\,963\,189\,145\,600$ in the real case.
These orders are upper bounds on the least orders of counterexamples;
we do not assert that they are minimal.

The strict violations also give the following consequence.

\begin{corollary}\label{cor:extensions}
For every $n\geq12$, there is a complex positive definite matrix of order
$n$, with every diagonal entry one, for which $q(A)>\per A$.
For every $n\geq21$, such a matrix can be chosen real.
\end{corollary}

For vectors $v_1,\ldots,v_n\in\mathbb C^d$, we use the Gram convention
$a_{jk}=\sum_r v_{j,r}\overline{v_{k,r}}$. Thus $A=VV^*$ is positive
semidefinite, where $V$ has rows $v_j$.
The proof expresses the permanent of $A$ as a sum of squared polynomial
coefficients with factorial weights. In the complex construction,
two regular pentagons leave only
three monomials in the full product. In the real construction, two
hexagons and eight equatorial directions give sparse products, and
rotation cancels the cross terms in the sums of deleted products.

Section~\ref{sec:polynomials} establishes the polynomial identity.
Sections~\ref{sec:complex} and \ref{sec:real} give the two constructions
and their exact evaluations. Section~\ref{sec:completion} completes the
proofs of Theorem~\ref{thm:main} and Corollary~\ref{cor:extensions}.
Section~\ref{sec:author-conjecture} proves that inequality \eqref{eq:target}
at every distinguished index implies the standard-representation instance
of the author's quadratic conjecture \cite{Lavi}, and shows that the two
counterexamples do not disprove the conjecture in the standard block.

\section{A polynomial formula for Gram permanents}\label{sec:polynomials}

For multiindices $\alpha\in\mathbb Z_{\geq0}^d$ and a polynomial
$F(z)=\sum_\alpha f_\alpha z^\alpha$, define
\[
 \|F\|^2=\sum_\alpha \alpha!\,|f_\alpha|^2,
 \qquad z^\alpha=\prod_{r=1}^d z_r^{\alpha_r},\quad
 \alpha!=\prod_{r=1}^d\alpha_r!.
\]
The associated inner product is
$\langle F,G\rangle=\sum_\alpha\alpha!f_\alpha\overline{g_\alpha}$;
in particular, distinct monomials are orthogonal. Put
$\ell_v(z)=\sum_r v_rz_r$.
This is the coefficient form of a standard Gaussian moment norm: if
$Z=(Z_1,\ldots,Z_d)$ has independent coordinates with density
$\pi^{-1}e^{-|z|^2}$ on $\mathbb C$, then the mixed moments
$\mathbb E[Z^\alpha\overline{Z^\beta}]$ vanish unless
$\alpha=\beta$, when they equal $\alpha!$. Thus
$\|F\|^2=\mathbb E|F(Z)|^2$. We use the coefficient form for the
exact evaluations below.

\begin{lemma}\label{lem:norm}
For vectors $v_1,\ldots,v_m,w_1,\ldots,w_m\in\mathbb C^d$, let
$B=(b_{ij})_{i,j=1}^m$ with
$b_{ij}=\sum_{r=1}^d v_{i,r}\overline{w_{j,r}}$. Then
\[
 \left\langle\prod_{i=1}^m\ell_{v_i},
                 \prod_{j=1}^m\ell_{w_j}\right\rangle=\per B.
\]
In particular, for the Gram matrix $A$ of $v_1,\ldots,v_n$,
\[
 \per A=\left\|\prod_{j=1}^n\ell_{v_j}\right\|^2.
\]
This polynomial norm is invariant under substitutions $z\mapsto Uz$
with $U^*U=I$ (unitary substitutions). If $v_n=e_1=(1,0,\ldots,0)$ and
$P=\prod_{j<n}\ell_{v_j}$, then
\begin{equation}\label{eq:norms}
 \per A=\|z_1P\|^2,\qquad
 q(A)=\sum_{j<n}|v_{j,1}|^2\|P/\ell_{v_j}\|^2.
\end{equation}
Here division means omission of the specified factor.
\end{lemma}

\begin{proof}
The mixed identity is Wick's formula for the complex Gaussian linear
forms, written in coefficient form. We verify it directly.
Let $F=\prod_{j=1}^m\ell_{v_j}$ and
$G=\prod_{j=1}^m\ell_{w_j}$. An assignment
$\phi:\{1,\ldots,m\}\to\{1,\ldots,d\}$ chooses one coordinate from
each factor; write
$\operatorname{mult}(\phi)=(|\phi^{-1}(1)|,\ldots,|\phi^{-1}(d)|)$
for its coordinate multiplicities. Expanding the coefficients gives
\[
 \langle F,G\rangle
 =\sum_\alpha\alpha!
   \sum_{\substack{\phi,\psi\\
      \operatorname{mult}(\phi)=\operatorname{mult}(\psi)=\alpha}}
   \prod_{j=1}^m v_{j,\phi(j)}\overline{w_{j,\psi(j)}}.
\]
For two assignments with common multiplicities $\alpha$, the relation
$\psi\circ\sigma=\phi$ requires $\sigma\in S_m$ to map each fibre
$\phi^{-1}(r)$ bijectively onto $\psi^{-1}(r)$. These bijections can be
chosen independently in $\alpha_r!$ ways for each $r$, giving exactly
$\alpha!=\prod_r\alpha_r!$ such permutations.
Replacing $\alpha!$ by this count and reindexing the conjugated factors
therefore gives
\[
\begin{aligned}
\langle F,G\rangle
 &=\sum_{\sigma\in S_m}\sum_\phi
   \prod_{j=1}^m v_{j,\phi(j)}
       \overline{w_{\sigma(j),\phi(j)}}\\
 &=\sum_{\sigma\in S_m}\prod_{j=1}^m b_{j,\sigma(j)}
 =\per B.
\end{aligned}
\]
Taking $w_j=v_j$ gives the Gram norm identity.
For unitary $U$, the random vectors $UZ$ and $Z$ have the same law, so
the Gaussian representation above gives $\|F(Uz)\|=\|F(z)\|$.
Applying the Gram norm identity to the full product and to each double
deletion proves \eqref{eq:norms}; the weights are
$|a_{jn}|^2=|v_{j,1}|^2$.
\end{proof}

\section{The complex counterexample}\label{sec:complex}

In this section write $(x,y)=(z_1,z_2)$ in the polynomial norm.

\begin{proposition}\label{prop:complex}
Let $\zeta=e^{2\pi\ii/5}$. Take the ten vectors $(1,-\rho)$ with
\[
 \rho\in\{\zeta^j/\sqrt2,-2\zeta^j:0\leq j<5\},
\]
one further vector $(0,1)$, and distinguished vector $(1,0)$.
Their Gram matrix $A$ is positive semidefinite of order twelve and rank two,
and
\[
 \per A-q(A)=900(5183-4608\sqrt2)<0.
\]
\end{proposition}

\begin{proof}
The assertion can also be checked by exact symbolic computation from
the listed vectors. We give a direct derivation that makes the sparse
product and the sign of the gap transparent.
The two coordinate vectors establish the rank. The two regular pentagons
make the product especially sparse. Let $P$ be the product of the linear
forms associated with the eleven non-distinguished vectors. The identity
\[
 \prod_{j=0}^4(x-c\zeta^j y)=x^5-c^5y^5,
\]
applied with $c=1/\sqrt2$ and $c=-2$, gives
\[
 P=y(x^5-\frac{\sqrt2}{8}y^5)(x^5+32y^5).
\]
The three monomials in $xP$ are orthogonal, so
\begin{align*}
 \per A
 &=11!+(32-\sqrt2/8)^2(6!)^2+32\cdot11!\\
 &=33\cdot11!+(1024+1/32-8\sqrt2)(6!)^2\\
 &=1\,848\,112\,200-4\,147\,200\sqrt2.
\end{align*}
The extra vector $(0,1)$ contributes zero to $q(A)$.
Deleting a root $\rho$ from its fivefold factor leaves
\[
 \frac{x^5-\rho^5y^5}{x-\rho y}
   =\sum_{i=0}^4\rho^{4-i}x^iy^{4-i}.
\]
After multiplication by the other fivefold factor and by $y$, the two
resulting ranges of $x$-degrees, $0,\ldots,4$ and $5,\ldots,9$, are
disjoint. The ten resulting monomials are distinct and hence orthogonal,
so each deletion norm depends on $\rho$ only through $|\rho|^2$.
Set $s=1/2$ and
$t=4$, the squared root radii on the first and second pentagons.
The squared constants in their fivefold factors are $s^5$ and $t^5$,
respectively. Formula \eqref{eq:norms} now gives
\begin{align*}
 q(A)&=5\sum_{i=0}^4\Big[(s^{4-i}+t^{4-i})(5+i)!(5-i)!\\
 &\hspace{5em}+(s^{4-i}t^5+t^{4-i}s^5)i!(10-i)!\Big]\\
 &=5(6\,375\,420+362\,314\,080)
 =1\,843\,447\,500.
\end{align*}
The two numbers in parentheses are the sums of the first and second
terms inside the brackets, respectively. Hence
\begin{align*}
 \per A-q(A)
 &=(1\,848\,112\,200-1\,843\,447\,500)-4\,147\,200\sqrt2\\
 &=4\,664\,700-4\,147\,200\sqrt2\\
 &=900(5183-4608\sqrt2).
\end{align*}
Since $(9/8)^2=81/64<2$, we have $\sqrt2>9/8$, and therefore
$4608\sqrt2>4608\cdot9/8=5184>5183$. The gap is negative.
\end{proof}

\section{The real counterexample}\label{sec:real}

In this section write $(x,y,z)=(z_1,z_2,z_3)$ in the polynomial norm.

\begin{proposition}\label{prop:real}
For $0\leq j<6$, put
\begin{align*}
 a_j&=(1,\sqrt{2/3}\cos(j\pi/3),\sqrt{2/3}\sin(j\pi/3)),\\
 b_j&=(1,2\cos(j\pi/3+\pi/6),2\sin(j\pi/3+\pi/6)).
\end{align*}
For $0\leq j<8$, put
\[
 c_j=(0,\sqrt2\cos(j\pi/8),\sqrt2\sin(j\pi/8)).
\]
The Gram matrix $A$ of these twenty real vectors and the distinguished
vector $w=(1,0,0)$ is positive semidefinite of rank three and satisfies
\[
 \per A-q(A)=-1\,152\,963\,189\,145\,600<0.
\]
\end{proposition}

\begin{proof}
The assertion can also be checked by exact symbolic computation from
the displayed vectors. We give a direct derivation: rotations make the
full product sparse and cancel cross terms in the sums of deleted
products, leaving finite coefficient sums.
Positive semidefiniteness follows from the Gram construction, and
$w,c_0,c_4$ establish rank three. Use the unitary coordinates
\[
 u=(y+\ii z)/\sqrt2,\qquad v=(y-\ii z)/\sqrt2.
\]
Set $s=1/3$, $t=2$, and for $r>0$ define
\begin{align}
 C_r&=x^6-6r x^4uv+9r^2x^2u^2v^2-r^3(u^3+v^3)^2,\label{eq:C}\\
 D_r&=x^6-6r x^4uv+9r^2x^2u^2v^2+r^3(u^3-v^3)^2.\label{eq:D}
\end{align}
For $L_\theta=e^{-\ii\theta}u+e^{\ii\theta}v$, let
$\theta_j=j\pi/3$, $\phi_j=\theta_j+\pi/6$ and $\psi_j=j\pi/8$.
Since $L_\theta=\sqrt2(\cos\theta\,y+\sin\theta\,z)$, the linear forms
of the listed vectors are
\[
 \ell_{a_j}=x+\sqrt{s}L_{\theta_j},\qquad
 \ell_{b_j}=x+\sqrt{t}L_{\phi_j},\qquad
 \ell_{c_j}=L_{\psi_j}.
\]
Opposite factors pair to $x^2-rL_\theta^2$.
For the first hexagon, the phases $e^{-2\ii\theta_k}$, $0\leq k<3$,
run through the cube roots of unity. With $\omega=e^{2\pi\ii/3}$, the
identity
\[
 \prod_{k=0}^2(h-\omega^{-k}p-\omega^kq)
 =h^3-p^3-q^3-3hpq,
\]
applied with $h=x^2-2suv$, $p=su^2$, and $q=sv^2$, gives
$\prod_j\ell_{a_j}=C_s$.
The unitary substitution
\[
 \tau:(u,v)\mapsto(e^{-\ii\pi/6}u,e^{\ii\pi/6}v)
\]
sends $(u^3+v^3)^2$ to $-(u^3-v^3)^2$, and hence sends $C_r$ to $D_r$.
Thus $\prod_j\ell_{b_j}=D_t$.
Also
\[
 \prod_{j=0}^7\ell_{c_j}
 =\prod_{j=0}^7(e^{-\ii j\pi/8}u+e^{\ii j\pi/8}v)
 =\ii(u^8-v^8).
\]
Indeed, $e^{2\ii j\pi/8}$ runs through the eighth roots of unity and
the product of the scalar phases is $e^{-7\pi\ii/2}=\ii$.
The scalar $\ii$ can be dropped from every norm, so the product of the
twenty non-distinguished factors is represented by $P=(u^8-v^8)C_sD_t$.

For $F=\sum f_{ijk}x^iu^jv^k$, define
\begin{equation}\label{eq:H}
 H(F)=\sum_{i,j,k}|f_{ijk}|^2i!\big[(j+8)!k!+j!(k+8)!\big]
     =\|u^8F\|^2+\|v^8F\|^2.
\end{equation}
Every monomial of $xC_sD_t$ has $j-k$ divisible by six.
A cross term between $u^8F$ and $v^8F$ would require two such
differences to differ by sixteen, which is impossible. Thus
\begin{equation}\label{eq:realper}
 \per A=H(xC_sD_t).
\end{equation}

To treat the deletions, put $L=u+v$, $M=uv$ and
\begin{equation}\label{eq:R}
 R_r=(x-\sqrt r L)
       \big[x^4+r(L^2-6M)x^2+r^2(L^2-3M)^2\big].
\end{equation}
Multiplication gives $C_r=(x+\sqrt r L)R_r$.
Here $x+\sqrt sL=\ell_{a_0}$, so $R_sD_t$ is exactly the hexagon
product with $a_0$ deleted.
By rotation through $\theta$ we mean the substitution
\[
 (u,v)\mapsto(e^{-\ii\theta}u,e^{\ii\theta}v).
\]
Both $C_r$ and $D_r$ are invariant under rotations through $\pi/3$.
The six products obtained by deleting one factor from $C_sD_t$ on
the first hexagon are obtained from
$R_sD_t$ by the substitutions
$(u,v)\mapsto(e^{-\ii j\pi/3}u,e^{\ii j\pi/3}v)$.
These substitutions preserve $H$. Rotation through $j\pi/3$ multiplies
a monomial $x^a u^b v^c$ by $e^{-\ii(b-c)j\pi/3}$.
A cross term from multiplication by $u^8-v^8$ pairs monomials whose
values of $b-c$ differ by sixteen, so it acquires a factor
$e^{\pm16\ii j\pi/3}$.
Its sum over $j=0,\ldots,5$ vanishes, since
\[
 \sum_{j=0}^5e^{16\ii j\pi/3}=0.
\]
Hence the six squared deleted norms sum to $6H(R_sD_t)$.
For the second hexagon, apply $\tau$.
It preserves $H$, since it changes each coefficient only by a unit phase.
It sends the $b_j$ form to $x+\sqrt t L_{\theta_{j+1}}$ and the
$a_j$ form to $x+\sqrt s L_{\phi_j}$.
Thus, after $\tau$, the hexagon product with $b_j$ deleted is a rotation
of $R_tD_s$ through $(j+1)\pi/3$. The equatorial factor becomes
\[
 \tau(u^8-v^8)=e^{-4\pi\ii/3}
       (u^8-e^{2\pi\ii/3}v^8).
\]
Its cross terms carry a fixed unit phase and one of the factors
$e^{\pm16\ii(j+1)\pi/3}$, which again sum to zero over the six values
of $j$. The second deletion sum is therefore $6H(R_tD_s)$.
The twelve hexagon vectors have first coordinate one, while the
equatorial vectors have first coordinate zero. Therefore
\begin{equation}\label{eq:realq}
 q(A)=6H(R_sD_t)+6H(R_tD_s).
\end{equation}

The explicit polynomials \eqref{eq:C}--\eqref{eq:R} now give
\begin{align*}
 H(xC_sD_t)&=647\,582\,321\,598\,464\,000,\\
 H(R_sD_t)&=94\,164\,692\,010\,188\,800,\\
 H(R_tD_s)&=13\,957\,855\,454\,412\,800.
\end{align*}
\ref{sec:coefficients} supplies the grouped coefficient sums
for these finite evaluations. Equations \eqref{eq:realper} and
\eqref{eq:realq} yield
\[
 q(A)=648\,735\,284\,787\,609\,600,
\]
and hence the asserted strict negative gap.
\end{proof}

\section{Completion of the proofs}\label{sec:completion}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Proposition~\ref{prop:complex} gives the complex matrix of order twelve
and rank two. Proposition~\ref{prop:real} gives the real matrix of order
twenty-one and rank three. Both are positive semidefinite Gram matrices,
and both have strictly negative gaps $\per A-q(A)$.
\end{proof}

\begin{proof}[Proof of Corollary~\ref{cor:extensions}]
Start with the appropriate matrix in Theorem~\ref{thm:main} and adjoin
an identity block of the required size, possibly empty. Retain the
original distinguished coordinate, and simultaneously permute that
coordinate to the last position.
This preserves the gap: terms involving the added indices have zero weight,
and identity blocks have permanent one. Since the gap is a continuous
polynomial in the real and imaginary parts of the matrix entries, adding
a sufficiently small positive multiple of the identity preserves its strict
negativity and makes the matrix positive definite.

For a positive diagonal matrix $D$, both $\per(DAD)$ and $q(DAD)$ equal
the corresponding quantity for $A$ multiplied by $\prod_jD_{jj}^2$.
Indeed, in a term of $q$, the squared entry supplies the factors for the
two deleted indices, and the permanent supplies all the remaining factors.
Choosing $D_{jj}=a_{jj}^{-1/2}$ makes the diagonal one and preserves
the sign. Each operation preserves real entries when the initial matrix
is real.
\end{proof}

\section{Relation to the author's quadratic conjecture}\label{sec:author-conjecture}

In this section we show that inequality \eqref{eq:target},
imposed at every distinguished index, implies the standard-representation
instance of the author's quadratic conjecture \cite{Lavi}.
We therefore test the two counterexamples of Theorem~\ref{thm:main}
against that instance; the bounds below show that they do not disprove
the author's quadratic conjecture for the standard representation.

Let $A=(a_{ij})_{i,j=1}^n$ be Hermitian positive semidefinite of order
$n\geq1$, and, for $\sigma\in S_n$, write
\[
 a_\sigma(A)=\prod_{k=1}^n a_{k,\sigma(k)}.
\]
For a partition $\lambda\vdash n$, let $\chi_\lambda$ be the corresponding
irreducible character of $S_n$, with degree $d_\lambda=\chi_\lambda(e)$.
The author's quadratic permanental dominance conjecture \cite{Lavi} is
\begin{equation}\label{eq:author-conjecture}
 \sum_{\sigma,\tau\in S_n}\chi_\lambda(\sigma\tau)
       a_\sigma(A)a_\tau(A)\leq d_\lambda(\per A)^2
 \qquad(\lambda\vdash n).
\end{equation}
The author proved the conjecture for $n\leq5$ and also proved that it
implies both the $S_n$ immanant case of Lieb's permanental dominance
conjecture and Chollet's permanent conjecture \cite{Lavi}.

For $n=1$, \eqref{eq:author-conjecture} is an equality. For $n\geq2$,
we restrict attention to the standard representation.
Let $\rho_{\mathrm{nat}}(\sigma)e_j=e_{\sigma(j)}$ be the natural
permutation representation on $\mathbb C^n$, and let
$\rho_{\mathrm{std}}$ be its restriction to the subspace
$\mathbf1^\perp$ of vectors whose coordinates sum to zero.
This is the irreducible representation of shape $(n-1,1)$ and dimension
$n-1$. Put
$F_{\mathrm{std}}(A)=\sum_\sigma a_\sigma(A)
\rho_{\mathrm{std}}(\sigma)$, and let $\mathrm{fix}(\pi)$ denote the
number of fixed points of $\pi\in S_n$.
Since $\chi_{(n-1,1)}(\pi)=\mathrm{fix}(\pi)-1$, the standard-instance
form of \eqref{eq:author-conjecture} is
\begin{equation}\label{eq:standard-conjecture}
 \operatorname{tr}F_{\mathrm{std}}(A)^2
 =\sum_{\sigma,\tau\in S_n}
   \bigl(\mathrm{fix}(\sigma\tau)-1\bigr)
   a_\sigma(A)a_\tau(A)
 \leq(n-1)(\per A)^2.
\end{equation}

We call the following equivalent cofactor inequality the
\emph{natural representation form}. Let $A(i|j)$ be the matrix obtained
from $A$ by deleting row $i$ and column $j$, and set
\[
 \Gamma=(\gamma_{ij})_{i,j=1}^n,\qquad
 \gamma_{ij}=a_{ij}\per A(i|j).
\]
Expansion of the permanent along each row and each column gives
\[
 \Gamma\mathbf1=\Gamma^{\mathsf T}\mathbf1=(\per A)\mathbf1.
\]
Write $N(A)=\sum_{\sigma\in S_n}a_\sigma(A)
\rho_{\mathrm{nat}}(\sigma)$. Its $(i,j)$ entry is
\[
 \sum_{\sigma:\,\sigma(j)=i}a_\sigma(A)
   =a_{ji}\per A(j|i)=\gamma_{ji},
\]
so $N(A)=\Gamma^{\mathsf T}$. The natural representation
splits into the trivial representation and the standard one. Its
trivial block is $\per A$, and $\Gamma$ is Hermitian because $A$ is
Hermitian. Consequently
\[
 \sum_{i,j=1}^n|\gamma_{ij}|^2
 =\operatorname{tr}\Gamma^2
 =(\per A)^2+\operatorname{tr}F_{\mathrm{std}}(A)^2.
\]
Thus \eqref{eq:standard-conjecture} is equivalent to the natural form
\begin{equation}\label{eq:natural-conjecture}
 \boxed{\displaystyle
 \operatorname{tr}\Gamma^2
 =\sum_{i,j=1}^n |a_{ij}|^2\,|\per A(i|j)|^2
 \leq n(\per A)^2.}
\end{equation}

If $A$ is the Gram matrix of $v_1,\ldots,v_n$ and
$f_i=\prod_{k\ne i}\ell_{v_k}$, the mixed version of
Lemma~\ref{lem:norm} gives
\[
 \gamma_{ij}
 =\langle v_i,v_j\rangle\langle f_i,f_j\rangle
 =\langle v_i\otimes f_i,v_j\otimes f_j\rangle.
\]
Thus $\Gamma$ is the Gram matrix of these tensors with the product
inner product, and is positive semidefinite.
The matrix $F_{\mathrm{std}}(A)$ is the restriction of
$N(A)=\Gamma^{\mathsf T}=\overline{\Gamma}\succeq0$ to
$\mathbf1^\perp$. Its eigenvalues are therefore nonnegative and sum to
$\operatorname{tr}\Gamma-\per A$. Their squared sum is at most the square
of their sum, giving
\begin{equation}\label{eq:trace-bound}
 \operatorname{tr}F_{\mathrm{std}}(A)^2
 \leq\bigl(\operatorname{tr}\Gamma-\per A\bigr)^2.
\end{equation}

\begin{proposition}\label{prop:drury-natural}
Let $A$ be Hermitian positive semidefinite of order $n\geq2$.
If $q_i(A)\leq\per A$ for every distinguished index $i$, then
\eqref{eq:natural-conjecture} holds.
\end{proposition}

\begin{proof}
Fix an index and permute it to the last position. Write
\[
 A=\begin{pmatrix}B&c\\c^*&a\end{pmatrix},\qquad
 \Pi_{jk}=\per B(k|j)\quad(1\leq j,k<n).
\]
Here $B(k|j)$ deletes row $k$ and column $j$ from $B$.
If $B$ is the Gram matrix of $v_1,\ldots,v_{n-1}$, set
$f_j=\prod_{\ell\ne j}\ell_{v_\ell}$.
The assignment count in the proof of Lemma~\ref{lem:norm}, applied to
the two products with factors $k$ and $j$ omitted, gives
$\Pi_{jk}=\langle f_k,f_j\rangle$. Hence $\Pi$ is positive semidefinite.
Put $d=a\per B$ and $\eta=c^*\Pi c$. Expanding along the last row and
then the last column, and using the definitions, gives
\[
 \per A=d+\eta,\qquad q_n(A)=\sum_{j<n}|c_j|^2\Pi_{jj},\qquad
 \gamma_{nj}=\overline{c_j}(\Pi c)_j\ (j<n),\quad \gamma_{nn}=d.
\]
By Cauchy--Schwarz for the positive semidefinite matrix $\Pi$,
$|(\Pi c)_j|^2\leq\Pi_{jj}\eta$. Hence
\[
 \sum_{j=1}^n|\gamma_{nj}|^2
 \leq d^2+\eta q_n(A)
 \leq d^2+\eta\per A
 =(\per A)^2-d\eta\leq(\per A)^2.
\]
Here $d,\eta\geq0$. The same bound holds at every index, and summing
its rows proves \eqref{eq:natural-conjecture}.
\end{proof}

The two matrices of Theorem~\ref{thm:main} violate \eqref{eq:target}
at the distinguished index, but neither disproves
\eqref{eq:standard-conjecture}. To evaluate their traces, we use
Lemma~\ref{lem:norm} on each principal deletion:
\[
 \operatorname{tr}\Gamma
 =\sum_{i=1}^n\|v_i\|^2
   \left\|\prod_{k\ne i}\ell_{v_k}\right\|^2.
\]
Exact coefficient evaluation of the principal-deletion products gives
the following trace values. For the complex matrix of order twelve,
\[
\begin{aligned}
 \per A&=1\,848\,112\,200-4\,147\,200\sqrt2, &q(A)&=1\,843\,447\,500,\\
 \operatorname{tr}\Gamma&=7\,447\,658\,400-1\,382\,400\sqrt2.
\end{aligned}
\]
In particular,
$\frac{81}{20}\per A-\operatorname{tr}\Gamma
=37\,196\,010-15\,413\,760\sqrt2>0$, since $\sqrt2<3/2$.
By \eqref{eq:trace-bound},
\[
 \frac{\operatorname{tr}F_{\mathrm{std}}(A)^2}{11(\per A)^2}
 \leq\frac{(\operatorname{tr}\Gamma-\per A)^2}{11(\per A)^2}
 <\frac{3721}{4400}<1.
\]
For the real matrix of order twenty-one, the corresponding evaluation gives
\[
\begin{aligned}
 \per A&=647\,582\,321\,598\,464\,000, &q(A)&=648\,735\,284\,787\,609\,600,\\
 \operatorname{tr}\Gamma&=\frac{10\,242\,155\,794\,291\,097\,600}{3}.
\end{aligned}
\]
Again by \eqref{eq:trace-bound},
\[
 \frac{\operatorname{tr}F_{\mathrm{std}}(A)^2}{20(\per A)^2}
 \leq\frac{(\operatorname{tr}\Gamma-\per A)^2}{20(\per A)^2}
 =\frac{523\,670\,800\,175\,527\,514\,454\,169}
       {573\,887\,169\,546\,878\,472\,124\,500}<1.
\]
Hence neither counterexample to \eqref{eq:target} disproves the author's
conjecture in the standard block.

\section*{Acknowledgments}

The proofs in this paper were carried out by GPT-5.6-sol, GPT-6 Astra,
Claude Fable 5, and Claude Fable 5.1, under the guidance of the author. The author has reviewed the
resulting proof arguments. Responsibility for the final text rests with the
author.

\appendix
\renewcommand{\thesection}{Appendix \Alph{section}}

\section{Exact coefficient sums for the real construction}\label{sec:coefficients}

Here $s=1/3$, $t=2$ and $g=2\,867\,200=2^{14}\cdot5^2\cdot7$ is a
convenient common factor. Each row collects monomials
of $x$-degree $i$ in the polynomial at the top of its column.
For the first column, the factorial weight uses $i+1$ because of the
additional factor $x$ in $H(xC_sD_t)$. The two deletion columns use
the weight in \eqref{eq:H} directly. All entries are divided by $g$.
The factors in \eqref{eq:R} contain $\sqrt r$ only in the even
$x$-degree part; $D_r$ has only even $x$-degrees. Hence their products
have rational squared coefficients. Some entries divided by $g$ retain
a denominator of three.

\begin{center}
\begin{tabular}{r r r r}
\toprule
$i$ & $C_sD_t$ (append $x$) & $R_sD_t$ & $R_tD_s$\\
\midrule
12 & 175134960 & 0 & 0\\
11 & 0 & 1122660 & 1122660\\
10 & 1980372240 & 340200 & 2041200\\
9 & 0 & 16433928 & 12941208\\
8 & 6835774680 & 6941088 & 24712128\\
7 & 0 & 72538935 & 69766515\\
6 & 27872814620 & 40765522 & 161003052\\
5 & 0 & 579632760 & $247708930/3$\\
4 & 31090176000 & 562398336 & $762812608/3$\\
3 & 0 & 1364065056 & $97793008/3$\\
2 & 8900720640 & 1868892480 & 150162880\\
1 & 0 & 4640822784 & 128911744\\
0 & 149003796480 & 23688084480 & 3948014080\\
\midrule
Sum & 225858789620 & 32842038229 & 4868113649\\
\bottomrule
\end{tabular}
\end{center}

\begin{samepage}
For example, the leading monomial $x^{12}$ of $C_sD_t$ contributes
$2\cdot13!\cdot8!/g=175134960$ to the first column.
The table follows by expanding the displayed degree-six and degree-five
polynomials and applying \eqref{eq:H}; it includes every $x$-degree.
It gives the compact comparison
\[
 \frac{\per A}{4g}=56\,464\,697\,405
 \quad<\quad
 \frac{q(A)}{4g}=56\,565\,227\,817.
\]
\end{samepage}

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\end{document}