Chollet’s Permanent Conjecture Through Order Six via Border Induction

1. Introduction

Context. Building on Oppenheim’s determinant inequality [1], Chollet proposed its permanental analogue in 1982 [2]. Gregorac and Hentzel established the conjecture in orders two and three [3]. Zhang developed an analytic maximizer approach to Chollet’s conjecture [4] and later surveyed its formulation and equivalent forms [5]. Hutchinson proved the real order-four case [6], and Rodtes subsequently proved the complex Hermitian order-four case [7]. Recent work treats parameterized annular classes and, in order five, matrices permutation-similar to ridge matrices [8]. A recent preprint proves the conjecture for symmetric Z-matrices with nonnegative diagonal and bipartite support, and for several graph-Laplacian families [9], while another establishes sharp rank-two and structured rank-three results for real order-five correlation matrices [10]. We prove the conjecture for arbitrary complex Hermitian positive semidefinite matrices in every order through six.

Several stronger permanent inequalities are known to fail. The Bapat–Sunder proposal [11], combined with Marcus’s permanent analogue of Hadamard’s determinant inequality [12], would imply Chollet’s conjecture. Drury found an order-seven counterexample to that proposal [13]. The permanent-on-top conjecture also fails, including for a complex order-five rank-two example [14]. These counterexamples do not decide Chollet’s conjecture.

Main result. For a $q\times q$ matrix $M=(m_{ij})$, write

$$per(M)=\sum _{\sigma \in {\mathfrak{S}}_q}{\displaystyle \prod _{i=1}^q}{m}_{i,\sigma (i)},per(⌀)=1.$$

For matrices of the same size, $A\circ B$ denotes their Hadamard product, ${(A\circ B)}_{ij}=a_{ij}{b}_{ij}$, and $\overline{A}$ denotes entrywise conjugation.

Theorem 1.1 

(Permanent inequality through order six). Let $1\le q\le 6$. For every pair of $q\times q$ complex Hermitian positive semidefinite matrices A and B,

$$per(A\circ B)\le per(A)per(B).$$

Equivalently, every $q\times q$ complex Hermitian positive semidefinite matrix C satisfies

$$per(C\circ \overline{C})\le per{(C)}^2.$$

Method. At a normalized scalar border, Proposition 3.2 uses the scalar Schur complement $H=B-x{x}^{*}$, without any invertibility or rank assumption, and encodes the two relevant permanents in ordinary and doubled occupation polynomials. An orthogonal decomposition by occupation profile expresses each coefficient as a sum of squared norms, while the doubled pencil lies on the parabolic slice $(a,b)=(t,t^2)$. Proposition 4.1 evaluates this slice at $t=\sqrt{2}$; the resulting coefficient comparisons close the self-conjugate induction through order six. Lemma 5.1 then passes from the self-conjugate formulation to arbitrary positive semidefinite pairs. The needed positivity facts are established in Section 2, the border calculus in Section 3, the induction in Section 4, and the transfer in Section 5.

2. Preliminaries

This section records the Gram arguments that supply all positivity statements used below. Inner products are linear in their first argument.

Lemma 2.1 

(Symmetric tensors and Gram closure). Let q be a positive integer. For vectors $v_1,\dots ,v_q$, define the unnormalized symmetric tensor

$$v_1\vee \dots \vee v_q=\sum _{\sigma \in {\mathfrak{S}}_q}{v}_{\sigma (1)}\otimes \dots \otimes v_{\sigma (q)}.$$

For arbitrary vectors $v_1,\dots ,v_q,w_1,\dots ,w_q$ in a finite-dimensional complex Hilbert space,

$$〈v_1\vee \dots \vee v_q,w_1\vee \dots \vee w_q〉=q!per\left({(\langle v_i,w_j\rangle )}_{i,j=1}^q\right).$$

(2.1)

Consequently, the permanent of every $q\times q$ complex Hermitian positive semidefinite matrix is real and nonnegative. If A and B are such matrices, then $\overline{A}$, $A\circ B$, and $A\circ \overline{A}$ are Hermitian positive semidefinite.

Proof. 

Sesquilinearity and the tensor-product inner product give

$$〈v_1\vee \dots \vee v_q,w_1\vee \dots \vee w_q〉=\sum _{\sigma ,\tau \in {\mathfrak{S}}_q}{\displaystyle \prod _{r=1}^q}\langle v_{\sigma (r)},w_{\tau (r)}\rangle .$$

For fixed $\sigma ,\tau$, reindex by $i=\sigma (r)$ and set $\pi =\tau \circ {\sigma }^{-1}$. The corresponding product is

$${\displaystyle \prod _{i=1}^q}\langle v_i,w_{\pi (i)}\rangle .$$

For each $\pi \in {\mathfrak{S}}_q$, every one of the $q!$ choices of $\sigma$ determines the unique choice $\tau =\pi \circ \sigma$. Thus every permanent monomial occurs exactly $q!$ times, proving (2.1).

Let M be Hermitian positive semidefinite. Unitary diagonalization gives

$$M=Qdiag({\lambda }_1,\dots ,{\lambda }_q)Q^{*},{\lambda }_i\ge 0.$$

Put

$$R=M^{1/2}=Qdiag(\sqrt{{\lambda }_1},\dots ,\sqrt{{\lambda }_q})Q^{*},$$

and let $u_i$ be row i of R. Since $R{R}^{*}=M$,

$$\langle u_i,u_j\rangle =\sum _k{R}_{ik}\overline{R_{jk}}=M_{ij}.$$

Applying (2.1) with $v_i=w_i=u_i$ yields

$$per(M)={\displaystyle \frac{1}{q!}}{\parallel u_1\vee \dots \vee u_q\parallel }^2\ge 0.$$

Zero eigenvalues have not been discarded, so this realization also covers singular and zero matrices.

Choose Gram vectors $a_i$ and $b_i$ for A and B. Then

$$\langle a_i\otimes b_i,a_j\otimes b_j\rangle =A_{ij}{B}_{ij},$$

so $A\circ B$ is a Gram matrix. Coordinatewise conjugation gives

$$\langle \overline{a_i},\overline{a_j}\rangle =\overline{\langle a_i,a_j\rangle },$$

so $\overline{A}$ is a Gram matrix. Finally, $a_i\otimes \overline{a_i}$ have Gram matrix $A\circ \overline{A}$. The permanent formula just proved applies to each of these matrices. □

Definition 2.2. 

For every positive integer q, let $({\mathrm{SC}}_q)$ denote the self-conjugate assertion

$$per(C\circ \overline{C})\le per{(C)}^2$$

for every $q\times q$ complex Hermitian positive semidefinite matrix C. Let $({\mathrm{PC}}_q)$ denote the pair assertion

$$per(A\circ B)\le per(A)per(B)$$

for every pair of $q\times q$ complex Hermitian positive semidefinite matrices A and B.

3. Border Occupation

The goal of this section is to encode a normalized positive semidefinite border by two permanent polynomials whose coefficients are nonnegative.

Setup 3.1. 

Fix $n\ge 2$, put $m=n-1$, and let $A\in {\mathbb{C}}^{n\times n}$ be Hermitian positive semidefinite with $a_{nn}=1$. Write

$$A=\left(\begin{array}{cc}B& x\\ x^{*}& 1\end{array}\right),x_i=a_{i,n}(1\le i\le m),$$

so that the entries of the last row are $\overline{x_j}$. Define

$$H=B-x{x}^{*},z_i={|x_i|}^2,$$

and

$$\begin{array}{cc}\hfill U& =H\circ \overline{H},\hfill \\ \hfill V_{+}& =H\circ \overline{x{x}^{*}},V_{-}=\overline{H}\circ x{x}^{*},\hfill \\ \hfill \Xi & =V_{+}+V_{-},W=z{z}^T.\hfill \end{array}$$

Thus

$${\Xi }_{ij}=h_{ij}\overline{x_i}{x}_j+\overline{h_{ij}}{x}_i\overline{x_j}=2Re(h_{ij}\overline{x_i}{x}_j).$$

For scalar indeterminates $t,a,b$, put

$$\begin{array}{cc}\hfill F(t)& =per(H+tx{x}^{*})={\displaystyle \sum _{k=0}^m}{\alpha }_k{t}^k,\hfill \\ \hfill \Gamma (a,b)& =per(U+a\Xi +bW)=\sum _{\begin{array}{c}r,s\ge 0\\ r+s\le m\end{array}}{\gamma }_{rs}{a}^r{b}^s.\hfill \end{array}$$

(3.1)

A coefficient outside the displayed ranges is understood to be zero.

Proposition 3.2 

(Border occupation identities). With the notation of Setup 3.1, the following assertions hold.

(i)

The matrices $H,U,V_{+},V_{-},\Xi ,W$ are positive semidefinite.

(ii)

All coefficients in (3.1) are nonnegative:

$${\alpha }_k\ge 0(0\le k\le m),{\gamma }_{rs}\ge 0(r,s\ge 0,r+s\le m).$$

(iii)

The permanent of the bordered matrix satisfies

$$per(A)=F(1)+F^{\prime }(1)={\displaystyle \sum _{k=0}^m}(k+1){\alpha }_k.$$

(iv)

We have

$$A\circ \overline{A}=\left(\begin{array}{cc}B\circ \overline{B}& z\\ z^T& 1\end{array}\right),K:=B\circ \overline{B}-W=U+\Xi ⪰0,$$

and

$$per(A\circ \overline{A})=\Gamma (1,1)+{\partial }_b\Gamma (1,1)=\sum _{\begin{array}{c}r,s\ge 0\\ r+s\le m\end{array}}(s+1){\gamma }_{rs}.$$

(v)

For real t, let $R(t)=H+tx{x}^{*}$. Then

$$U+t\Xi +t^{2}W=R(t)\circ \overline{R(t)},$$

and hence

$$\Gamma (t,t^2)=per(R(t)\circ \overline{R(t)}),F(t)=per(R(t)).$$

Moreover, $R(t)⪰0$ for $t\ge 0$.

Proof. 

For $y\in {\mathbb{C}}^m$, set $\eta ={(y,-x^{*}y)}^T$. Direct multiplication gives

$${\eta }^{*}A\eta =y^{*}(B-x{x}^{*})y=y^{*}Hy.$$

Thus $H⪰0$, without an inverse or a nonsingularity assumption. The matrices $x{x}^{*}$, $\overline{H}$, and $\overline{x{x}^{*}}$ are also positive semidefinite. Lemma 2.1 gives

$$U⪰0,V_{+}⪰0,V_{-}⪰0.$$

Hence $\Xi =V_{+}+V_{-}⪰0$. Since z is real, $W=z{z}^T⪰0$. This proves (i).

We now derive coefficient nonnegativity from complete occupation sectors. Let $u_i$ be row i of $H^{1/2}$, so $h_{ij}=\langle u_i,u_j\rangle$. Put

$$E={\mathbb{C}}^m\oplus \mathbb{C}e,$$

where e is a unit vector orthogonal to ${\mathbb{C}}^m$. For $S\subseteq \{1,\dots ,m\}$, define

$$y_i^S=\left\{\begin{array}{cc}e,\hfill & i\in S,\hfill \\ u_i,\hfill & i\notin S,\hfill \end{array}\right.$$

and, for $0\le k\le m$, put

$${\Sigma }_k=\sum _{\begin{array}{c}S\subseteq \{1,\dots ,m\}\\ |S|=k\end{array}}\left(\prod _{i\in S}{x}_i\right)y_1^S\vee \dots \vee y_m^S.$$

For $t\ge 0$, the vectors

$$v_i(t)=u_i+\sqrt{t}{x}_{i}e$$

have Gram matrix $H+tx{x}^{*}$. Multilinearity gives the exact expansion

$$v_1(t)\vee \dots \vee v_m(t)={\displaystyle \sum _{k=0}^m}{t}^{k/2}{\Sigma }_k.$$

Every summand of ${\Sigma }_k$ has exactly k tensor factors in $\mathbb{C}e$. Tensor words with different numbers of such factors are orthogonal, and permuting tensor positions preserves that number. Hence the ${\Sigma }_k$ are pairwise orthogonal. Lemma 2.1 therefore gives

$$F(t)={\displaystyle \frac{1}{m!}}{\displaystyle \sum _{k=0}^m}{t}^k{\parallel {\Sigma }_k\parallel }^2(t\ge 0).$$

Both sides are polynomials in t, so equality on the nonnegative half-line makes this a polynomial identity. Thus

$${\alpha }_k={\displaystyle \frac{1}{m!}}{\parallel {\Sigma }_k\parallel }^2\ge 0.$$

(3.2)

Here $\overline{E}$ denotes the conjugate Hilbert space: its elements are denoted $\overline{u}$, scalar multiplication is $\lambda \overline{u}=\overline{\overline{\lambda }u}$, and ${\langle \overline{u},\overline{v}\rangle }_{\overline{E}}=\overline{{\langle u,v\rangle }_E}$.

For the doubled pencil, consider the four pairwise orthogonal subspaces of $E\otimes \overline{E}$

$$\begin{array}{cccc}\hfill L_1& ={\mathbb{C}}^m\otimes \overline{{\mathbb{C}}^m},\hfill & \hfill L_2& ={\mathbb{C}}^m\otimes \mathbb{C}\overline{e},\hfill \\ \hfill L_3& =\mathbb{C}e\otimes \overline{{\mathbb{C}}^m},\hfill & \hfill L_4& =\mathbb{C}e\otimes \mathbb{C}\overline{e}.\hfill \end{array}$$

Define parameter-free labeled vectors

$$\begin{array}{cccc}\hfill q_{i,1}& =u_i\otimes \overline{u_i},\hfill & \hfill q_{i,2}& =\overline{x_i}(u_i\otimes \overline{e}),\hfill \\ \hfill q_{i,3}& =x_i(e\otimes \overline{u_i}),\hfill & \hfill q_{i,4}& =z_i(e\otimes \overline{e}).\hfill \end{array}$$

They belong respectively to $L_1,L_2,L_3,L_4$. Their same-label inner products are

$$\begin{array}{cc}\hfill \langle q_{i,1},q_{j,1}\rangle & =h_{ij}\overline{h_{ij}}=U_{ij},\hfill \\ \hfill \langle q_{i,2},q_{j,2}\rangle & =h_{ij}\overline{x_i}{x}_j,\hfill \\ \hfill \langle q_{i,3},q_{j,3}\rangle & =\overline{h_{ij}}{x}_i\overline{x_j},\hfill \\ \hfill \langle q_{i,4},q_{j,4}\rangle & =z_i{z}_j=W_{ij},\hfill \end{array}$$

and all cross-label inner products vanish.

For $a,b\ge 0$, put

$$g_i(a,b)=q_{i,1}+\sqrt{a}{q}_{i,2}+\sqrt{a}{q}_{i,3}+\sqrt{b}{q}_{i,4}.$$

The preceding inner products show entrywise that the Gram matrix of these vectors is $U+a\Xi +bW$.

For an occupation profile $\mathbf{d}=(d_1,d_2,d_3,d_4)\in {\mathbb{Z}}_{\ge 0}^4$ with $d_1+d_2+d_3+d_4=m$, define the complete profile sum

$$T_{\mathbf{d}}=\sum _{\begin{array}{c}(c_1,\dots ,c_m)\in {\{1,2,3,4\}}^m\\ \#\{i:c_i=j\}=d_j(1\le j\le 4)\end{array}}{q}_{1,c_1}\vee \dots \vee q_{m,c_m}.$$

The tensor-word spaces $L_{c_1}\otimes \dots \otimes L_{c_m}$ are mutually orthogonal: distinct words differ in a position whose two label spaces are orthogonal. Grouping these word spaces by their four label counts gives mutually orthogonal profile spaces. A permutation of tensor factors preserves every profile space, so the complete sums $T_{\mathbf{d}}$ for distinct profiles are orthogonal.

Multilinearity now gives

$$g_1(a,b)\vee \dots \vee g_m(a,b)=\sum _{\mathbf{d}}{a}^{(d_2+d_3)/2}{b}^{d_4/2}{T}_{\mathbf{d}}.$$

Lemma 2.1 and profile orthogonality yield

$$\Gamma (a,b)={\displaystyle \frac{1}{m!}}\sum _{\mathbf{d}}{a}^{d_2+d_3}{b}^{d_4}{\parallel T_{\mathbf{d}}\parallel }^2(a,b\ge 0).$$

Both sides are polynomials. Fixing $b\ge 0$ first shows equality of every coefficient in a; varying b then shows that those coefficient polynomials agree identically. Therefore

$${\gamma }_{rs}={\displaystyle \frac{1}{m!}}\sum _{\begin{array}{c}{d}_2,d_3\ge 0\\ d_2+d_3=r\end{array}}{∥T_{(m-r-s,d_2,d_3,s)}∥}^2\ge 0$$

(3.3)

whenever $r,s\ge 0$ and $r+s\le m$. The parameter monomials occur outside the complete sector vectors in both (3.2) and (3.3). Zero vectors simply give zero sector sums, so no limiting argument enters. This proves (ii).

Two border-calculus identities give the required derivative formulas. For an $m\times m$ matrix C, columns $u,v\in {\mathbb{C}}^m$, and $c\in \mathbb{C}$, partitioning permutations by the column occupied by the last row gives

$$\begin{array}{cc}\hfill per\left(\begin{array}{cc}C& u\\ v^T& c\end{array}\right)& =cper(C)+{\displaystyle \sum _{i,j=1}^m}{u}_i{v}_{j}per(C(i\mid j)),\hfill \\ \hfill {\displaystyle \frac{d}{dt}}per(C+tQ)& ={\displaystyle \sum _{i,j=1}^m}{Q}_{ij}per((C+tQ)(i\mid j)).\hfill \end{array}$$

(3.4)

Here $C(i\mid j)$ is obtained by deleting row i and column j. The second identity follows by differentiating each permutation monomial. The convention $per(⌀)=1$ covers $m=1$.

Since $B=H+x{x}^{*}$, apply (3.4) with $u=x$, $v=\overline{x}$, $c=1$, and $Q=x{x}^{*}$. We obtain

$$\begin{array}{cc}\hfill per(A)& =per(B)+{\displaystyle \sum _{i,j=1}^m}{x}_i\overline{x_j}per(B(i\mid j))\hfill \\ & =F(1)+F^{\prime }(1)={\displaystyle \sum _{k=0}^m}(k+1){\alpha }_k.\hfill \end{array}$$

This proves (iii).

For the doubled border,

$${(x{x}^{*})}_{ij}=x_i\overline{x_j},{\left(\overline{x{x}^{*}}\right)}_{ij}=\overline{x_i}{x}_j.$$

Consequently,

$$\begin{array}{cc}\hfill {(U+\Xi +W)}_{ij}& =h_{ij}\overline{h_{ij}}+h_{ij}\overline{x_i}{x}_j+\overline{h_{ij}}{x}_i\overline{x_j}+|x_i{|}^2{|x_j|}^2\hfill \\ & =(h_{ij}+x_i\overline{x_j})(\overline{h_{ij}}+\overline{x_i}{x}_j)={(B\circ \overline{B})}_{ij}.\hfill \end{array}$$

Thus $B\circ \overline{B}=U+\Xi +W$, and

$$K=B\circ \overline{B}-W=U+\Xi ⪰0.$$

The border formula in (3.4) now gives

$$per(A\circ \overline{A})=per(B\circ \overline{B})+{\displaystyle \sum _{i,j=1}^m}{z}_i{z}_{j}per((B\circ \overline{B})(i\mid j)).$$

Because

$$\Gamma (1,b)=per(U+\Xi +bW)=per(K+bW),$$

its value at $b=1$ is the first term, while its derivative at $b=1$ is the double sum. Therefore

$$per(A\circ \overline{A})=\Gamma (1,1)+{\partial }_b\Gamma (1,1)=\sum _{\begin{array}{c}r,s\ge 0\\ r+s\le m\end{array}}(s+1){\gamma }_{rs}.$$

This proves (iv).

For real t, distributivity of the Hadamard product gives

$$\begin{array}{cc}\hfill R(t)\circ \overline{R(t)}& =(H+tx{x}^{*})\circ \left(\overline{H}+t\overline{x{x}^{*}}\right)\hfill \\ & =U+t\left(H\circ \overline{x{x}^{*}}+\overline{H}\circ x{x}^{*}\right)+t^{2}W\hfill \\ & =U+t\Xi +t^{2}W.\hfill \end{array}$$

The definitions in (3.1) now give

$$\Gamma (t,t^2)=per(R(t)\circ \overline{R(t)}),F(t)=per(R(t)).$$

If $t\ge 0$, then $R(t)=H+tx{x}^{*}⪰0$, proving (v). □

4. The Square-Root-of-Two Induction

In this section, the border identities are converted into the induction $({\mathrm{SC}}_{n-1})\Rightarrow ({\mathrm{SC}}_n)$ for $2\le n\le 6$.

The coefficientwise step below requires only

$$t^k\le k+1(0\le k\le m),s+1\le t^{r+2s}(r,s\ge 0,r+s\le m).$$

For $t=\sqrt{2}$, the second family holds for all $r,s\ge 0$, while the first holds through $k=5$ and fails at $k=6$. Thus both families cover the required coefficient ranges exactly when $m\le 5$, or equivalently $n\le 6$.

Proposition 4.1 

(Rank-free induction step). Fix $2\le n\le 6$. If every $(n-1)\times (n-1)$ complex Hermitian positive semidefinite matrix M satisfies

$$per(M\circ \overline{M})\le per{(M)}^2,$$

then every $n\times n$ complex Hermitian positive semidefinite matrix A satisfies

$$per(A\circ \overline{A})\le per{(A)}^2.$$

Proof. 

Suppose first that $a_{ii}=0$ for some i. Positivity of the principal submatrix on indices $i,j$ gives

$$|a_{ij}{|}^2\le a_{ii}{a}_{jj}=0$$

for every j. Thus row and column i vanish in both A and $A\circ \overline{A}$. Every permutation monomial in their permanents is zero, so the desired inequality holds with equality.

Assume now that every diagonal entry is positive. Normalize the last diagonal entry with

$$D=Diag(1,\dots ,1,a_{nn}^{-1/2}),A^{\prime }=DAD.$$

The matrix $A^{\prime }$ is Hermitian positive semidefinite and $a_{nn}^{\prime }=1$. For each $\sigma \in {\mathfrak{S}}_n$,

$$\begin{array}{cc}\hfill {\displaystyle \prod _{i=1}^n}{(DAD)}_{i,\sigma (i)}& =\left({\displaystyle \prod _{i=1}^n}{D}_{ii}\right)\left({\displaystyle \prod _{i=1}^n}{D}_{\sigma (i),\sigma (i)}\right){\displaystyle \prod _{i=1}^n}{a}_{i,\sigma (i)}\hfill \\ & =a_{nn}^{-1}{\displaystyle \prod _{i=1}^n}{a}_{i,\sigma (i)}.\hfill \end{array}$$

Summing over $\sigma$ gives

$$per(A^{\prime })=a_{nn}^{-1}per(A).$$

Since D is real diagonal,

$$A^{\prime }\circ \overline{A^{\prime }}=D^2(A\circ \overline{A})D^2,per(A^{\prime }\circ \overline{A^{\prime }})=a_{nn}^{-2}per(A\circ \overline{A}).$$

The inequality for $A^{\prime }$ is therefore equivalent to the inequality for A. We may henceforth write A for the normalized matrix.

Apply Proposition 3.2 and put $m=n-1$. Its coefficients and pencil satisfy

$$\begin{array}{cc}\hfill per(A)& ={\displaystyle \sum _{k=0}^m}(k+1){\alpha }_k,\hfill \\ \hfill per(R(t))& ={\displaystyle \sum _{k=0}^m}{t}^k{\alpha }_k,\hfill \\ \hfill per(A\circ \overline{A})& =\sum _{\begin{array}{c}r,s\ge 0\\ r+s\le m\end{array}}(s+1){\gamma }_{rs},\hfill \\ \hfill per(R(t)\circ \overline{R(t)})& =\sum _{\begin{array}{c}r,s\ge 0\\ r+s\le m\end{array}}{t}^{r+2s}{\gamma }_{rs}.\hfill \end{array}$$

(4.1)

All coefficients in these sums are nonnegative.

Set $t=\sqrt{2}$. For $0\le k\le m\le 5$, the required inequalities $t^k\le k+1$ are

$$1\le 1,\sqrt{2}\le 2,2\le 3,2\sqrt{2}\le 4,4\le 5,4\sqrt{2}\le 6.$$

The last comparison is equivalently $32\le 36$ after squaring. For every integer $s\ge 0$, the binomial expansion $2^s={(1+1)}^s$ gives

$$s+1\le 2^s.$$

For $s\ge 1$, the two endpoint terms contribute 2, while the $s-1$ interior binomial coefficients are each at least 1; for $s=0$, equality holds. Since $t\ge 1$ and $r\ge 0$,

$$s+1\le 2^s=t^{2s}\le t^{r+2s}.$$

Coefficient nonnegativity in (4.1) therefore gives

$$0\le per(R(t))\le per(A)$$

and

$$per(A\circ \overline{A})\le per(R(t)\circ \overline{R(t)}).$$

Proposition 3.2 also gives $R(t)⪰0$. The induction hypothesis applies to this possibly singular $m\times m$ matrix, with no restriction on its diagonal. Combining these comparisons with the induction hypothesis at $t=\sqrt{2}$ gives

$$\begin{array}{cc}\hfill per(A\circ \overline{A})& \le \Gamma (\sqrt{2},2)\hfill \\ & =per(R(\sqrt{2})\circ \overline{R(\sqrt{2})})\hfill \\ & \le per{(R(\sqrt{2}))}^2\hfill \\ & \le per{(A)}^2.\hfill \end{array}$$

(4.2)

The scaling identities transfer (4.2) to the original matrix.

The argument does not require invertibility and therefore applies without modification to singular matrices and vanishing border coordinates. □

Remark 4.2 (Boundary of the fixed-weight method).

A uniform coefficientwise continuation of the same argument from order six to order seven would set $m=6$ and require $t^2\ge 2$ from the mixed coefficients, hence $t^6\ge 8$. The ordinary degree-six comparison would also require $t^6\le 7$. Together these conditions would require $8\le 7$, so the present one-parameter coefficientwise scheme does not directly extend to order seven. This method boundary gives no conclusion about order seven.

Theorem 4.3 

(Self-conjugate inequality through order six). For every integer q with $1\le q\le 6$, every $q\times q$ complex Hermitian positive semidefinite matrix C satisfies

$$per(C\circ \overline{C})\le per{(C)}^2.$$

Proof. 

The assertion $({\mathrm{SC}}_1)$ follows from the scalar case. A $1\times 1$ Hermitian positive semidefinite matrix is $M=[c]$ with $c\ge 0$, and

$$per(M)=c,per(M\circ \overline{M})=c^2.$$

Successive applications of Proposition 4.1 with $n=2,3,4,5,6$ give

$$({\mathrm{SC}}_1)⟹({\mathrm{SC}}_2)⟹({\mathrm{SC}}_3)⟹({\mathrm{SC}}_4)⟹({\mathrm{SC}}_5)⟹({\mathrm{SC}}_6).$$

□

5. Transfer from the Self-Conjugate Formulation to Pairs

This section derives the pair inequality from Cauchy–Schwarz on permutation monomials and records the converse.

Chollet observed this equivalence [2]; we include a proof in the notation used here.

Lemma 5.1 

(Equivalence of the two formulations). For every positive integer q, the assertions $({\mathrm{SC}}_q)$ and $({\mathrm{PC}}_q)$ are equivalent.

Proof. 

For a complex $q\times q$ matrix X and $\sigma \in {\mathfrak{S}}_q$, set

$$m_{\sigma }(X)={\displaystyle \prod _{i=1}^q}{X}_{i,\sigma (i)}.$$

For arbitrary complex $q\times q$ matrices X and Y, expansion gives

$$per(X\circ \overline{Y})=\sum _{\sigma \in {\mathfrak{S}}_q}{m}_{\sigma }(X)\overline{m_{\sigma }(Y)}.$$

Cauchy–Schwarz on these two permutation-monomial vectors yields

$$|per(X\circ \overline{Y}){|}^2\le per(X\circ \overline{X})per(Y\circ \overline{Y}),$$

(5.1)

because

$$per(X\circ \overline{X})=\sum _{\sigma \in {\mathfrak{S}}_q}{|m_{\sigma }(X)|}^2.$$

Set $X=A$ and $Y=\overline{B}$. Then $X\circ \overline{Y}=A\circ B$ and $Y\circ \overline{Y}=B\circ \overline{B}$. Lemma 2.1 shows that $per(A\circ B)$ is real and nonnegative, so (5.1) becomes

$$per{(A\circ B)}^2\le per(A\circ \overline{A})per(B\circ \overline{B}).$$

(5.2)

Assume $({\mathrm{SC}}_q)$. Applying it separately to A and B gives

$$per{(A\circ B)}^2\le per(A\circ \overline{A})per(B\circ \overline{B})\le per{(A)}^{2}per{(B)}^2.$$

All quantities whose square roots are taken are nonnegative by Lemma 2.1. Taking nonnegative square roots proves $({\mathrm{PC}}_q)$.

Conversely, assume $({\mathrm{PC}}_q)$ and let $C⪰0$. Lemma 2.1 gives $\overline{C}⪰0$ and $per(C)\in {\mathbb{R}}_{\ge 0}$. Hence termwise conjugation gives

$$per(\overline{C})=\overline{per(C)}=per(C).$$

Applying $({\mathrm{PC}}_q)$ to $(C,\overline{C})$ yields

$$per(C\circ \overline{C})\le per(C)per(\overline{C})=per{(C)}^2.$$

Thus $({\mathrm{PC}}_q)$ implies $({\mathrm{SC}}_q)$. □

6. Proof of the Main Theorem

The self-conjugate induction and the permutation-monomial transfer now complete the proof stated in the introduction.

Proof of Theorem 1.1. 

Fix q with $1\le q\le 6$. Theorem 4.3 proves $({\mathrm{SC}}_q)$, and Lemma 5.1 then proves $({\mathrm{PC}}_q)$. This is the asserted inequality, and Lemma 5.1 also gives the stated equivalence. □

7. Conclusions

We have proved Chollet’s permanent inequality for complex Hermitian positive semidefinite matrices in every order through six. The argument combines a rank-free border induction for the self-conjugate inequality with a permutation-monomial Cauchy–Schwarz transfer to arbitrary pairs. The failure of the next fixed-weight coefficient comparison limits this method but gives no conclusion about order seven.