Using topology to reveal all mutually unbiased bases in dimension six

The paper in detail

Take the first basis to be the identity. A second basis \(H/\sqrt6\) is then unbiased to it exactly when \(H\) is a complex Hadamard matrix: its entries have modulus one and \(H^\dagger H=6I_6\). A triple adds a matrix \(K\) such that \(K\) and \(H^\dagger K/\sqrt6\) are Hadamard as well. Permuting and rephasing rows and columns gives equivalent sets. Earlier work excluded quartets containing the Fourier family and showed that Tao's matrix admits no third basis. Numerical searches found only two kinds of triple, but nothing ruled out others.

The first step is a reconstruction. After dephasing, the top-left \(3\times3\) corner \(E(\theta)\) of \(H\) has four free phases. Orthogonality fixes \(BB^\dagger=6I_3-EE^\dagger\) for the side block \(B\), and leaves two candidate blocks, distinguished by the sign of the imaginary part of an invariant \(\Psi(B)\). Their squared separation is a trigonometric polynomial \(R(\theta)\). Theorem 1 states that for every seed with \(R(\theta)>0\) the matched completion \(H_+(\theta)\) exists, is unique up to equivalence, and varies continuously, and that every order-six Hadamard class except Tao's and \(F_6\) arises this way.

Theorem 2 states that all classes except Tao's form a connected space \(\mathcal X\). Every component of \(\{R>0\}\) contains a positive local maximum of \(R\). There are exactly eighteen maxima, all related by symmetry, and explicit paths with \(R>0\) join them. A one-parameter family of Hadamard matrices attaches \(F_6\). Connectedness then lets 2,775 four-dimensional phase balls cover \(\mathcal X\), once every boundary point of each ball has been shown to lie inside another ball.

Inside each ball, interval arithmetic encloses every vector unbiased to both \(I_6\) and \(H/\sqrt6\). Either no third basis exists, or the triple is invariant under an order-three unitary \(T\). Solving the triples with this symmetry exactly leaves the two known types, and neither extends to a fourth basis. Further consequences: no vector is unbiased to all three bases of any triple, and some smaller configurations exist, including one of type \(\{5,3,3,3\}_6\) that had been believed impossible. The next open cases are dimensions 10 and 12, and dimension 12 is known to support five bases.

Transferable methods

1. Connectedness turns a finite cover into a global proof

   A computer can check finitely many regions, and the proof must show that these regions miss no Hadamard matrix. The authors prove that the space of non-Tao classes is connected, so a finite union of certified regions that is both open and closed must be the whole space.

   Let \(W\) be the union of the certified closed regions, together with their complex conjugates, transposes and conjugate transposes. A finite union of closed sets is closed. If every class on the boundary of a region lies in the interior of another region, \(W\) is also open. A connected space has no clopen subsets other than \(\varnothing\) and itself, so a nonempty \(W\) equals \(\mathcal X\), and every statement certified region by region holds on all of \(\mathcal X\).

   Connectedness of the strict seeds \(\Omega=\{\theta:R(\theta)>0\}\) reduces to a finite check. The closure of a component of \(\Omega\) in the torus \(\mathbb T^4\) is compact, so \(R\) attains a maximum there. The maximum is positive and \(R=0\) on the boundary, so it lies inside the component and is a local maximum of \(R\). Joining all local maxima by paths along which \(R>0\) therefore joins all components. Since the reconstruction \(\theta\mapsto[H_+(\theta)]\) is continuous, its image is connected too.

   Editorial check: we evaluated the closed formula for \(R\) (Eq. 16 of the paper). At \(\theta^*=(\alpha,2\alpha,-\alpha,-2\alpha)\) with \(\alpha=\arccos\tfrac14\) it gives \(389.2390\ldots=3^{13}/2^{12}\). Along \(\theta_c(\varphi)=(\varphi,\varphi+\pi,\varphi+\pi,\varphi)\) it equals 256 exactly, and 200,000 random seeds never exceeded \(3^{13}/2^{12}\). Every corner of \(F_6\) gives \(R=0\). The family \(H(t)\) of Eq. (9) is Hadamard at each \(t\) we tried, and \(H(0)=F_2\otimes F_3\).

   How to reuse it

   When a space of solutions has a continuous parametrisation, proving it connected lets a computer-assisted proof run on overlapping local charts instead of a global grid. Here the cover shrank from OpenAI's 249,878 nine-dimensional centres to 2,775 balls in four phases. Every point the parametrisation misses, here Tao's matrix and \(F_6\), needs its own argument.

2. An invariant that separates the two completions

   A fixed corner determines the side block only up to a two-way ambiguity. The invariant \(\Psi\), which is unchanged by rephasing and permuting the block, labels the two solutions. Its imaginary part is fixed up to sign, so \(R=4(\operatorname{Im}\Psi)^2>0\) says the two branches stay apart and each can be followed continuously.

   For a unit-modulus \(3\times3\) block \(B\) set \[\Psi(B)=\frac{\operatorname{perm}(B)\,\det(B)^2}{\prod_{j,k}B_{jk}}.\] Multiplying a row by \(e^{i\phi}\) multiplies the permanent by \(e^{i\phi}\), \(\det^2\) by \(e^{2i\phi}\), and the product of the entries by \(e^{3i\phi}\), so \(\Psi\) is unchanged; the same holds for columns, and permutations change none of the three factors (editorial verification). The corner fixes \(g=|\det B|^2\), \(\eta=|\operatorname{perm}B|^2\) and \(\chi=-2\operatorname{Re}\Psi(B)\). Hence \(|\Psi|^2=g^2\eta\) and \[R=4g^2\eta-\chi^2=4(\operatorname{Im}\Psi)^2,\] and only the sign of \(\operatorname{Im}\Psi\) is left free.

   Editorial check: on randomly permuted and dephased corners of \(H(0.3)\) with \(R>0\), the closed-form \(g\) and \(\eta\) agreed with \(|\det B|^2\) and \(|\operatorname{perm}B|^2\) of the actual side block, and \(R\) agreed with \(4(\operatorname{Im}\Psi)^2\), to all printed digits (for example \(g=11.732\), \(\eta=3.9107\), \(R=48.0817\)).

   The symmetry that closes the triple classification works in a similar way. Let \(T=\operatorname{diag}(1,1,\omega,\omega,\omega^2,\omega^2)\) with \(\omega=e^{2\pi i/3}\). Any \(v\) unbiased to \(I_6\) has \(|v_i|^2=\tfrac16\), so \[\langle v,Tv\rangle=\tfrac16\,(2+2\omega+2\omega^2)=0,\] and likewise for \(T^2\). Thus \(\{v,Tv,T^2v\}\) is automatically orthonormal, each further basis is fixed by two vectors, and a symmetric triple by twelve phases, few enough to solve the unbiasedness equations exactly.

   How to reuse it

   To make a discrete reconstruction continuous, look for an invariant of the ambiguous piece that the known data fix up to a sign. The squared separation of the two roots is then a polynomial whose positivity marks where the branches can be followed. Diagonal symmetries with repeated eigenvalues give orthogonality for free on vectors whose entries all have the same modulus.

Source: Sections II.1–II.3 and V.1 (reconstruction, connectedness and cover); Section III.1 (symmetric triples). Based on the linked paper version. Derivations marked as editorial, and suggested uses in other problems, are editorial guidance rather than results of the paper.

Context & limitations

Why it matters. Six is the first dimension that is not a prime power in which all sets of mutually unbiased bases are known. The proof turns a global question into finitely many local ones through a topological fact: the order-six complex Hadamard matrices other than Tao's form a connected space. Protocols built on three complementary measurements in dimension six can now be optimised over an explicit two-parameter set, and protocols that need four are ruled out. The classification also proves a conjecture of Matolcsi, Matszangosz, Varga and Weiner, and its identities give an analytical proof of the Fourier-vanishing conjecture.

Limitations. The proof is computer-assisted. Interval arithmetic certifies 2,775 phase balls, a chart around \(F_6\), and 117,391 boundary caps. The reconstruction, connectedness and quartet exclusion are formalised in Lean and checked by two independent certifiers (Arb and Julia), but the triple classification rests on a single certifier in Arb and exact arithmetic. Nothing carries over to dimensions 10 or 12. OpenAI had already reported a computer-assisted proof of \(N(6)=3\). This proof does not use its certificates, though it reuses its graph-search test with the underlying inequalities reproved in Lean. The authors state that ChatGPT 6 Astra and Claude Opus 5.5 assisted significantly in developing the proof, which the authors directed and refined.

Description based on Sections I–V (main text and Methods) and the AI-assistance statement of the linked arXiv version, read 2026-11-05. The supplementary proofs (Sections .6–.10) and the certificates were not checked.