QX–NEPHRIEL / Seventh-String Closure
A quantitative order-seven, rank-three Chollet inequality
Prepared for Maciej Nowicki, 3 October 2026.
Research status. This is an AI-generated mathematical derivation with an executed exact-arithmetic certificate. It has not received independent mathematical peer review or proof-assistant formalization. The argument proves the stated restricted theorem, not the unrestricted Chollet conjecture. Priority of the restricted result has not been established. All hypotheses, reductions, and finite checks used for the main theorem are included below.
1. The target and the claim
For an n by n complex matrix A, its permanent is
[ \operatorname{per} A=\sum_{\sigma\in S_n}\prod_{i=1}^n a_{i,\sigma(i)}. ]
Chollet's 1982 conjecture asks whether
[ \operatorname{per}(A\circ B)\le\operatorname{per}A,\operatorname{per}B ]
for all complex Hermitian positive semidefinite matrices A and B of the same order. The product symbol means the entrywise, or Hadamard, product, not matrix multiplication. Conjugation below is entrywise.
An author-maintained progress report consulted on 3 October 2026 explicitly lists complex rank at most three as covered in every order except seven. A recent preprint supplies general inequalities for estimating the same permanent ratio using spherical integrals and eigenvalues. The result below combines these ingredients with a new-to-this-derivation scalar-envelope closure at order seven. These observations support targeting the case; they do not constitute a comprehensive priority audit. See SOURCES.md.
Theorem 1
For every pair of complex Hermitian positive semidefinite 7 by 7 matrices A and B, each of rank at most three,
[ \boxed{\operatorname{per}(A\circ B)\le\frac{24}{25}\operatorname{per}A\operatorname{per}B.} ]
In particular, the ordinary Chollet inequality holds on this entire class. The coefficient 24/25 is a certified convenient coefficient; it is not claimed optimal.
The stronger self-conjugate statement is
[ \boxed{\operatorname{per}(C\circ\overline C)\le\frac{24}{25}\operatorname{per}(C)^2} \tag{1} ]
for every order-seven positive semidefinite C of rank at most three.
2. Reduction to correlation matrices and self-conjugate pairs
Positive semidefiniteness implies |a_ij|^2 <= a_ii a_jj. Consequently a zero diagonal entry forces its entire row and column to vanish. The permanent and the permanent of any Hadamard product containing that matrix both vanish. Thus such cases are immediate.
For positive diagonal entries, write A=D C D where D=diag(sqrt(a_ii)) and C has diagonal one. Then C is a correlation matrix, has the same rank, and with h=product_i a_ii,
[ \operatorname{per}A=h\operatorname{per}C,\qquad \operatorname{per}(A\circ\overline A)=h^2\operatorname{per}(C\circ\overline C). ]
Therefore the self-conjugate ratio is unchanged by diagonal normalization.
For arbitrary A,B, Cauchy–Schwarz applied to their permutation monomials gives
[ |\operatorname{per}(A\circ B)|^2 \le \operatorname{per}(A\circ\overline A) \operatorname{per}(B\circ\overline B). \tag{2} ]
Permanents of positive semidefinite matrices are nonnegative, as also follows from the Gaussian identity proved in Section 4. The Hadamard product is positive semidefinite because, for Gram representations A_ij=<v_i,v_j> and B_ij=<w_i,w_j>, it is the Gram matrix of v_i tensor w_i. Applying (1) to both factors in (2) proves Theorem 1. Notice that both factors are assumed rank at most three; the Hadamard product need not have rank at most three.
It suffices from here onward to consider an order-seven correlation matrix C of rank at most three. Set
[ P=\operatorname{per}C,\qquad Q=\operatorname{per}(C\circ\overline C). ]
3. A combinatorial upper bound for Q
Define b_k=(k!)^(1/k) for integers k>=1, and let b be linear between successive integer arguments.
Lemma 2: the interpolated row bound
If N is an n by n matrix with entries in [0,1] and row sums s_i>=1, then
[ \operatorname{per}N\le\prod_i b(s_i). \tag{3} ]
Proof. First suppose N is binary, with d_i ones in row i. Bregman's theorem gives per(N)<=product_i(d_i!)^(1/d_i), with a zero row treated separately. Here is the standard entropy argument for that binary result. If N has no perfect matching, the conclusion is immediate. Otherwise choose a perfect matching uniformly, and expose its chosen columns in a uniformly random order of rows, independent of the matching. The chain rule bounds the matching's entropy by the sum of the logarithms of the numbers of still-available neighbors. For a fixed full matching and row i, the rows matched to its d_i neighboring columns have uniformly random relative order. The number remaining when i is exposed is therefore uniform on {1,...,d_i}. The expected logarithm is log(d_i!)/d_i. Taking expectations and exponentiating proves the binary bound. This is the established entropy proof of Bregman's theorem, not a new theorem claimed here.
Now fix each row sum of a matrix in [0,1]^(n by n). The row polytope is compact; the permanent is affine in each row separately. Starting with a maximizer, replace its rows one by one by vertices of their row polytopes without decreasing the permanent. A row vertex has at most one fractional entry: two fractional entries would admit a nontrivial opposite perturbation preserving the row sum. A row of sum k+t, 0<=t<1, is therefore a convex combination of binary rows of sums k and k+1. Multilinearly expanding these combinations and applying the binary bound gives product_i[(1-t_i)b_k_i+t_i b_(k_i+1)], which is (3). QED.
The exact root brackets in Section 7 certify that b is increasing and concave on [1,7]. For N=C circle conjugate(C), its row sums are
[ s_i=\sum_j|c_{ij}|^2=(C^2)_{ii}\ge1. ]
Each s_i<=7 because all entries of C have modulus at most one. Consequently (3), the arithmetic–geometric mean inequality, and concavity imply
[ Q\le\left(\frac{1}{7}\sum_i b(s_i)\right)^7 \le b\left(\frac{\operatorname{tr}C^2}{7}\right)^7. \tag{4} ]
Pad the nonzero eigenvalues to three, writing them as t>=lambda_2>=lambda_3>=0. Since tr(C)=7,
[ 7/3\le t\le7,\qquad \lambda_2+\lambda_3=7-t. ]
For fixed t, the sum lambda_2^2+lambda_3^2 is maximized by making these two numbers as unequal as their bounds permit. Thus
[ \frac{\operatorname{tr}C^2}{7}\le M(t), \qquad M(t)=\begin{cases} \dfrac{2t^2+(7-2t)^2}{7},&7/3\le t\le7/2,\[6pt] \dfrac{t^2+(7-t)^2}{7},&7/2\le t\le7. \end{cases} \tag{5} ]
For clarity, in the first range the extremal remaining pair is (t,7-2t), and in the second it is (7-t,0). Both are admissible in their stated ranges. Because b is increasing, (4) becomes
[ \boxed{Q\le b(M(t))^7.} \tag{6} ]
This is the first certificate: a global upper bound requiring only one eigenvalue.
4. Two Gaussian-sphere lower bounds for P
The following spherical estimates are known ingredients; the spectral-tilt version appears in the recent work recorded in SOURCES.md. Their derivation is supplied to expose every normalization.
Write C_ij=<v_i,v_j> with unit vectors v_i in C^3, padding zero coordinates when the rank is smaller. Let g be a standard complex Gaussian vector in C^3, so E g_j conjugate(g_k)=delta_jk. Expansion into complex Gaussian monomials yields
[ P=\mathbb E\prod_{i=1}^7|\langle v_i,g\rangle|^2. \tag{7} ]
One can see (7) directly by pairing each of the seven unconjugated linear factors with a conjugated factor; the possible pairings are precisely the permutations in the permanent.
Write g=R U, with U uniform on the complex unit sphere and independent of R. Then R^2 has the Gamma(3,1) distribution and
[ D:=\mathbb E R^{14}=(3)_7=\frac{9!}{2}=181440. ]
Set T_i=|<v_i,U>|^2 and X=product_i T_i. Equation (7) becomes
[ P=D\mathbb E X. \tag{8} ]
For every unit vector v_i, T_i has density 2(1-x), 0<x<1, hence E log(T_i)=-3/2. Jensen's inequality gives
[ \boxed{P\ge D e^{-21/2}.} \tag{9} ]
This is valid without independence of the seven T_i.
For the second bound, let w be a unit top eigenvector of S=sum_i v_i v_i*, whose nonzero eigenvalues equal those of C. Define
[ T_w=|\langle w,U\rangle|^2,\qquad d\nu=3T_w,d\mu. ]
This is a probability measure since E_mu T_w=1/3. Write gamma_i=|<w,v_i>|^2. Then sum_i gamma_i=t. Conditional rotational symmetry in the two-dimensional orthogonal complement gives
[ \mathbb E(T_w\mid T_i=x)=\gamma_i x+(1-\gamma_i)\frac{1-x}{2}. ]
The elementary integrals
[ \mathbb E(T_i\log T_i)=-5/18,\qquad \mathbb E((1-T_i)\log T_i)=-11/9 ]
therefore imply
[ \mathbb E_\nu\log T_i=\gamma_i-11/6, \qquad \mathbb E_\nu\log T_w=-5/6. ]
The logarithms are integrable; the density is bounded and the relevant zero sets have spherical measure zero. Changing measure in (8) and applying Jensen once more yields
[ P=\frac D3\mathbb E_\nu(X/T_w) \ge\frac D3\exp\left(\sum_i\mathbb E_\nu\log T_i-\mathbb E_\nu\log T_w\right) =\frac D3e^{t-12}. \tag{10} ]
Consequently
[ \boxed{P\ge D\max{e^{-21/2},\tfrac13e^{t-12}}.} \tag{11} ]
In particular, P is strictly positive for correlation matrices in the stated class. Equation (11) is the second certificate, a lower bound using the same eigenvalue as (6).
5. The scalar problem
Combining (6) and (11),
[ \frac Q{P^2}\le R(t) :=\frac{b(M(t))^7}{D^2\max{e^{-21},\tfrac19e^{2t-24}}}, \qquad7/3\le t\le7. \tag{12} ]
The crossing point of the two denominator branches is
[ t_0=\frac32+\log3. ]
Thus
[ R(t)=\begin{cases} e^{21}b(M(t))^7/D^2,&t\le t_0,\ 9e^{24-2t}b(M(t))^7/D^2,&t\ge t_0. \end{cases} \tag{13} ]
The only interior knot of b(M(t)) between t_0 and 7/2 is M(t)=3, occurring at
[ t_1=\frac{7+\sqrt7}{3}. ]
We now prove the exact global reduction
[ \boxed{\sup_{7/3\le t\le7}R(t) \le\max{R(t_0),R(t_1),R(7/2)}.} \tag{14} ]
This is a continuous-domain argument, not a numerical sample.
6. Why three points suffice
The finite root certificate below establishes the following elementary properties of b:
- b is increasing and concave on [1,7];
- its slopes are less than 5/12 throughout [1,7];
- its slopes are less than 2/5 on [3,7];
- b(x)>=(x+2)/3 on [3,7].
6.1 Before t_0
M(t) is increasing on [7/3,7]. On [7/3,t_0], the denominator branch in (13) is constant, so R increases. Its maximum there is R(t_0).
6.2 Between t_0 and 7/2
Put u=t-7/3. In this interval,
[ M(t)=7/3+6u^2/7,\qquad 0\le u\le7/6. ]
Within any linear segment of b, write b(M)=aM+c, with 0<a<5/12. Since log R differs from 7 log b(M)-2t only by a constant, its second derivative has the sign of
[ b(M)-2a(M-7/3). ]
That expression satisfies
[ b(M)-2a(M-7/3) \ge1-2\frac5{12}\frac76=\frac1{36}>0. \tag{15} ]
To check the derivative explicitly: if B(u)=d+(6a/7)u^2, then B''B-(B')^2=(12a/7)[d-(6a/7)u^2], which is the stated sign condition. Therefore log R is convex separately on [t_0,t_1] and [t_1,7/2]. A convex function is bounded above by the larger of its endpoint values on a closed interval. This leaves only t_0,t_1,7/2 as possible maxima in the middle range. No convexity across the knot itself is assumed.
6.3 After 7/2
Put v=t-7/2, so 0<=v<=7/2 and
[ M(t)=7/2+2v^2/7. ]
Away from the finitely many knots, the derivative is
[ \frac d{dt}\log R=\frac{4a v}{b(M(t))}-2. ]
Using a<2/5 and b(M)>=(M+2)/3,
[ \frac d{dt}\log R \le \frac{24v}{5(11/2+2v^2/7)}-2 \le\frac{28}{15}-2=-\frac2{15}<0. \tag{16} ]
The second inequality follows because v/(11/2+2v^2/7) increases on [0,7/2]: its derivative has numerator 11/2-2v^2/7>=2>0. Its largest value occurs at v=7/2. R is continuous at all knots; hence it decreases on the entire tail. This completes the proof of (14).
7. The exact finite certificate
No floating-point estimate is needed in the proof. Define lower and upper rational brackets by dividing the following vectors by 10000:
[ (\ell_1,\ldots,\ell_7)= \frac1{10000}(10000,14142,18171,22133,26051,29937,33800), ]
[ (h_1,\ldots,h_7)= \frac1{10000}(10000,14143,18172,22134,26052,29938,33801). ]
Exact integer arithmetic verifies ell_k^k<=k!<=h_k^k. It also verifies:
[ \ell_{k+1}>h_k\quad(1\le k\le6), ] [ 2\ell_k>h_{k-1}+h_{k+1}\quad(2\le k\le6), ] [ h_{k+1}-\ell_k<5/12\quad(1\le k\le6), ] [ h_{k+1}-\ell_k<2/5\quad(3\le k\le6), ] [ \ell_k\ge(k+2)/3\quad(3\le k\le7). ]
These prove all properties of b used in Sections 3 and 6; the last property extends by interpolation.
Let E=27183/10000 and q=2599/1000. A rational Taylor tail yields
[ e\le\sum_{j=0}^9\frac1{j!}+\frac{1/10!}{1-1/11}<E<3. ]
Also
[ \sum_{j=0}^8\frac{(1099/1000)^j}{j!}>3 ]
implies log3<1099/1000, so t_0<q. The inequality e<3 gives log3>1, so t_0>5/2>7/3. Finally (13/5)^2<7 gives t_1>16/5; these observations verify the asserted ordering of the intervals.
Write
[ m_q=M(q)=\frac{2q^2+(7-2q)^2}{7},\qquad \beta=(3-m_q)h_2+(m_q-2)h_3. ]
Here 2<m_q<3. Increasingness of M and b gives
[ R(t_0)\le\frac{E^{21}\beta^7}{D^2}<\frac{24}{25}. \tag{17} ]
For the knot t_1, since t_1>16/5,
[ e^{24-2t_1}<e^{88/5}=\frac{e^{18}}{e^{2/5}} <\frac{25}{37}E^{18}, ]
where e^(2/5)>1+2/5+(2/5)^2/2=37/25. Thus
[ R(t_1)<\frac{9h_3^7 E^{18}}{D^2}\frac{25}{37}<\frac{24}{25}. \tag{18} ]
At the spectral midpoint,
[ R(7/2)\le \frac{9[(h_3+h_4)/2]^7E^{17}}{D^2}<\frac{24}{25}. \tag{19} ]
Every quantity in (17)–(19) is rational. Cross-multiplication proves the inequalities using integers. For readable reporting only, the three upper bounds are approximately 0.954617923041444, 0.793755502346409, and 0.891681828686875. These decimals are not premises.
The program verify_q7.py executes these checks and writes exact_certificate.json containing each full numerator, denominator, and exact positive margin. It reported 49 successful exact checks. Four of those checks concern the optional analytic-tail extension; the rest cover this theorem's finite obligations.
8. Completion
Equations (14) and (17)–(19) imply R(t)<24/25 for every admissible top eigenvalue t. Equations (6) and (11) therefore give Q<(24/25)P^2 for every correlation matrix in the theorem. Diagonal normalization extends the non-strict form to all positive semidefinite matrices of the stated order and rank, including zero-diagonal cases. Cauchy–Schwarz as in (2) then proves Theorem 1 for arbitrary pairs A,B. QED.
If both matrices have positive diagonal, their permanents are positive and the proof in fact gives strict inequality with coefficient 24/25. Equality with that coefficient can occur only through the trivial zero-permanent cases. The coefficient is deliberately not presented as extremal.
9. What has and has not been established
The argument above is a complete proof of a precisely stated rank-restricted theorem, with explicit finite verification. It is not a redefinition of the conjecture or an algorithm assuming a global feasibility oracle. The essential new step in this derivation is the order-seven scalar-envelope closure; the permanent estimates and the entropy argument have precedents.
The proof does not cover arbitrary ranks at order seven. It does not settle unrestricted Chollet, the permanent-on-top conjecture, or other stronger permanent proposals. It does not establish a new optimal constant. A search found an explicit report listing the targeted case as missing, but could have missed other sources. Publication-level novelty remains unconfirmed.
The appendices supply an independent derivation for all orders n>=10 and exact interval certificates for n=8,9. Coupled with the cited preprint theorem for all matrices of orders at most six, this would give the complete rank-at-most-three case in every order. That last small-order input is attributed, not independently re-proved by this package.
10. Relation to the Qingxiao-inspired design
The operational structure that survives is the dual-certificate gate: the numerator is controlled from above and the denominator from below, using the same spectral coordinate. A switch between two lower bounds replaces a heuristic change of search mode. The final certificate replaces an optimization guess with a globally valid statement.
The actual domains used are combinatorial matching entropy, complex Gaussian moments, spherical change of measure, eigenvalue geometry, one-variable convexity, and exact rational arithmetic. Persistent homology, braid groups, and symplectic integration are not used in this proof and are not claimed as hidden ingredients. The inspiration helped organize a search; it is not itself mathematical evidence.