Let sim_1 and sim_2 be equivalence relations on [n]=1,…,n, let P_1 and P_2 be the groups of permutations preserving their respective equivalence classes, and let A be an n× n Hermitian positive semidefinite matrix. Define f_sim_1,sim_2(A)=∑_sigma_1∈ P_1∑_sigma_2∈ P_2∏_i=1^n A_sigma_1(i),sigma_2(i). Matrix-analysis conjecture. Under these hypotheses, operatornameRe(f_sim_1,sim_2(A))≥| P_1 cap P_2|∏_i=1^n A_ii. The paper states that this conjecture implies the tree real-part conjecture; no proof or resolution is supplied.

Matrix-analysis inequality implying the tree real-part conjecture

About 1 year old · traced to 2025

Let $\sim_1$ and $\sim_2$ be equivalence relations on $[n]=\{1,\ldots,n\}$ , let $P_1$ and $P_2$ be the groups of permutations preserving their respective equivalence classes, and let $A$ be an $n\times n$ Hermitian positive semidefinite matrix. Define

f_{\sim_1,\sim_2}(A)=\sum_{\sigma_1\in P_1}\sum_{\sigma_2\in P_2}\prod_{i=1}^n A_{\sigma_1(i),\sigma_2(i)}.

Matrix-analysis conjecture. Under these hypotheses,

\operatorname{Re}(f_{\sim_1,\sim_2}(A))\geq\lvert P_1\cap P_2\rvert\prod_{i=1}^n A_{ii}.

The paper states that this conjecture implies the tree real-part conjecture; no proof or resolution is supplied.

References

Primary source

Joseph Malkoun, “Finite graphs and configurations of points”, arXiv:2508.13472 (2026).

matrix-analysis positive-semidefinite-matrices

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

An unverified posted construction claims the matrix conjecture is false in dimension forty-two, while the paper itself supplies no proof or resolution.

Joseph Malkoun’s 2025 paper formulates this matrix inequality and states that it would imply the tree real-part conjecture, but supplies neither a proof nor a resolution.

Posted attempt

A complete counterexample is claimed using a strictly positive definite $3\times3$ block, repeated $14$ times to form $n=42$ : the resulting value has negative real part, whereas $\lvert P_1\cap P_2\rvert\prod_i A_{ii}=1$ . The construction is not independently verified, so it establishes only a claimed disproof, not a settled result.

Current status (as of August 2026): The matrix conjecture has a complete but unverified counterexample claim; no verified resolution is recorded, and the tree real-part conjecture remains open.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

For equivalence relations ~₁,~₂ on [n], let P₁,P₂ be their class-preserving permutation groups and define f(A)=Σ_{σ∈P₁,τ∈P₂} ∏{j=1}^n A{σ(j),τ(j)}. Consider the Hermitian matrix B = [[1, 1/2, 7i/10], [1/2, 1, 1/2], [-7i/10, 1/2, 1]]. Its leading principal minors are 1, 3/4, and 1/100, so B is strictly positive definite. On one three-element block let ~₁ have classes {1,2},{3} and ~₂ have classes {1},{2,3}. Then P₁={e,(12)}, P₂={e,(23)}, P₁∩P₂={e}, and direct evaluation of the four terms gives f(B)=1+1/4+1/4+7i/40=(60+7i)/40.

Now let n=42 and take A=B⊕⋯⊕B with fourteen blocks. On each block use the same two equivalence relations, independently. Their class-preserving groups are products of the block groups, their intersection is the identity, every diagonal entry of A is 1, and the defining double permutation sum factors: f(A)=((60+7i)/40)^14. Exact integer arithmetic gives Re (60+7i)^14 = −475148137808619635065249, 40^14 = 26843545600000000000000. Consequently Re f(A)=−475148137808619635065249/26843545600000000000000 < 0 < 1 = |P₁∩P₂| ∏{j=1}^{42} A{jj}. Thus the conjecture fails even for a strictly positive definite Hermitian correlation matrix. The source separately discusses a related disconnected-power mechanism for graph amplitudes, but its matrix conjecture has no connectedness restriction; this counterexample concerns that matrix conjecture and does not disprove the separate conjecture about trees.