Mutually unbiased bases in dimension six: covers and exact certificates

Source and verification

This article explains family 266 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Unbiased measurements and the dimension-six problem

An orthonormal basis of a finite-dimensional complex Hilbert space describes a projective measurement. If a unit vector is one of the basis vectors, that measurement has a certain outcome. Two bases are mutually unbiased when measuring a vector from either basis in the other produces the uniform distribution:

$$|⟨b,c⟩{|}^2=\frac{1}{d}(b\in \mathcal{B},c\in \mathcal{C}).$$

Let $N(d)$ be the largest size of a family of pairwise mutually unbiased orthonormal bases in ${\mathbb{C}}^d$, for $d\ge 2$. Family 266 contains two different upper-bound arguments. The main manuscript claims $N(6)=3$ through a complete computer-assisted exclusion of four bases, under explicit arithmetic and compiler conditions. The Fourier companion gives exact finite certificates leading to the weaker bound $N(6)\le 5$. Their verification obligations are different. (OpenAI, 2026) (OpenAI, 2026a)

The prerequisites are complex inner products, tensor products, orthogonal projections, elementary probability, and matrix norms. The certificate discussion also uses finite graphs and rational Gaussian elimination.

For a unit vector $b$, let $P_b=b{b}^{\ast }$ be its rank-one orthogonal projector. In the real vector space of Hermitian matrices, use the inner product $⟨X,Y⟩=\mathrm{tr}(XY)$. A direct calculation gives

$$\begin{array}{rl}& \mathrm{tr}((P_b-I/d)(P_c-I/d))\\ & =|⟨b,c⟩{|}^2-\frac{1}{d}.\end{array}$$

The traceless Hermitian matrices diagonal in a fixed basis form a subspace of dimension $d-1$. Unbiased bases give orthogonal such subspaces, by the displayed identity. They all lie in a space of dimension $d^2-1$, so $N(d)(d-1)\le d^2-1$, giving $N(d)\le d+1$. Thus the elementary dimension count allows seven bases in dimension six. Improving seven to five, and five to three, requires additional information.

🏷️ A direct construction of three bases

The lower bound uses ${\mathbb{C}}^6={\mathbb{C}}^2\otimes {\mathbb{C}}^3$. In dimension two take the columns of

$$\begin{array}{c}{A}_0=I_2,A_1=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right),\\ A_2=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ i& -i\end{array}\right).\end{array}$$

These are unitary, and all entries of $A_r^{\ast }{A}_s$ have squared modulus $1/2$ when $r\ne s$: the nontrivial comparison has entries $(1\pm i)/2$.

Put $\omega =e^{2\pi i/3}$ and define

$$\begin{array}{c}F=\frac{1}{\sqrt{3}}({\omega }^{jk}{)}_{j,k=0}^2,\\ D=\mathrm{diag}(1,\omega ,\omega ),\\ B_0=I_3,B_1=F,\\ B_2=DF.\end{array}$$

The identity $1+\omega +{\omega }^2=0$ proves unitarity of $F$. To compare the last two bases, compute

$$(F^{\ast }DF{)}_{jk}=\frac{1}{3}\sum _{x=0}^2{\omega }^{x^2+(k-j)x}.$$

The numerator is $1+2\omega$ when $j=k$ and $2+{\omega }^2$ otherwise; both have squared modulus three. Thus these three dimension-three bases are mutually unbiased. The columns of $U_r=A_r\otimes B_r$, for $r=0,1,2$, now give three bases in dimension six, since

$$U_r^{\ast }{U}_s=(A_r^{\ast }{A}_s)\otimes (B_r^{\ast }{B}_s),$$

and each squared entry modulus is $(1/2)(1/3)=1/6$. This proves attainment of three independently of either computational upper bound. (OpenAI, 2026b)

🏷️ Hadamard matrices and exhaustive phase covers

Apply a common unitary transformation to make the first proposed basis the coordinate basis. The second then has matrix $H/\sqrt{6}$, where $H$ is a complex Hadamard matrix:

$$|H_{ij}|=1,H^{\ast }H=H{H}^{\ast }=6I.$$

The factor $\sqrt{6}$ matters: the entries of the unscaled matrix have modulus one, while normalized basis vectors have entries of modulus $1/\sqrt{6}$. Row and column rephasings and permutations preserve the extension problem. They allow the first row and column of $H$ to be all ones, called dephasing. Conjugation and transposition also preserve extendibility; for transposition, interchange the first two bases using their unitary transition matrix and then conjugate.

Write $e(t)=e^{2\pi it}$ and $H=e(P)$ entrywise, with real phases measured in cycles. After dephasing, the first two rows and columns contain nine free phases. The main manuscript covers these by finitely many closed seed regions. Closed regions can overlap; this loses efficiency but preserves boundary configurations. (OpenAI, 2026b)

The remaining completion uses blocks of sizes $2+4$. The upper-right and lower-left blocks determine two-dimensional complementary spaces in ${\mathbb{C}}^4$. Compressing the unknown lower-right block between fixed orthonormal frames gives a $2\times 2$ matrix controlled by the Hadamard equations. A polar-decomposition estimate places it near $\sqrt{6}$ times a unitary matrix. A closed finite cover of the $2\times 2$ unitaries then covers all possible completions. The frames are fixed at an approximate seed center: the argument does not assume that center is an exact Hadamard matrix.

For the later refinement, define the orthogonality residual by

$$F_{ij}(P)=\frac{1}{2\pi i}\sum _{g=0}^{5}e(P_{ig}-P_{jg}).$$

Here $0\le j<i\le 5$. Then $F(P)=0$ exactly when $e(P)$ is Hadamard. Numerical directions are placed in a real matrix $T$. If an exact solution lies at displacement $\delta$ from the center, the certificate controls its component outside those directions:

$$\begin{array}{c}\eta =T^{\ast }\delta ,\nu =(I-T{T}^{\ast })\delta ,\\ \Vert \nu \Vert \le n_0+\iota \Vert \delta \Vert +\Gamma \Vert \delta {\Vert }^2,\\ \Vert T^{\ast }T-I\Vert \le \epsilon .\end{array}$$

Here $n_0,\iota ,\Gamma ,\epsilon$ are nonnegative certified bounds. The displacement is chosen after removing row and column rephasing directions by orthogonal projection. All displacement norms are Euclidean, with matrices viewed as real vectors.

The point of the normal bound is that searching only the $\eta$ coordinates does not discard $\nu$. Nor are the stored columns assumed exactly orthonormal. In fact, expansion gives

$$\begin{array}{rl}\Vert \nu {\Vert }^2& =\Vert \delta {\Vert }^2-2\Vert \eta {\Vert }^2+\Vert T\eta {\Vert }^2\\ & \ge \Vert \delta {\Vert }^2-(1+\epsilon )\Vert \eta {\Vert }^2.\end{array}$$

This converts bounds on tangent coordinates and the normal remainder into a bound on the whole displacement. It supports exhaustive refinement even where the exact solution set has singularities; no smooth parametrization of that set is asserted.

🏷️ Local exclusion as a two-graph problem

Fix a certified phase ball for the second basis. A vector unbiased to the coordinate basis has the form $e(\beta )/\sqrt{6}$, and its global phase can be fixed by ${\beta }_0=0$. Cover its other five phases by closed intervals. Necessary inequalities discard only regions that contain no vector unbiased to the second basis.

One elementary estimate illustrates the direction of these tests. With second-basis center $P$, vector-phase center $u$, and actual displacements $\delta ,x$, set

$$\begin{array}{c}{q}_{ji}=\frac{e(u_j-P_{ji})}{\sqrt{6}},S_i=\sum _j{q}_{ji},\\ S_i^{\prime }=\sum _j{q}_{ji}e(x_j-{\delta }_{ji}).\end{array}$$

Unbiasedness requires $|S_i^{\prime }|=1$. If $\Vert x{\Vert }_1\le L$ and $\Vert \delta {\Vert }_F\le r$, the chord inequality $|e(t)-1|\le 2\pi |t|$ and Cauchy–Schwarz give

$$\begin{array}{c}|S_i^{\prime }-S_i|\le \tau ,\\ \tau =2\pi (L/\sqrt{6}+r),\\ \max \{0,1-\tau \}\le |S_i|\le 1+\tau .\end{array}$$

Violation excludes a whole bin; satisfaction merely leaves it possible. Sharper tests retain the same second-basis displacement when comparing different candidate vectors. Giving each vector an unrelated second basis would lose this compatibility information.

Make the retained small vector balls vertices of two graphs. An edge in $\mathcal{O}$ means orthogonality has not been ruled out; an edge in $\mathcal{U}$ means unbiasedness has not been ruled out. The manuscript proves that twelve vectors in two additional bases would occupy distinct labels and produce two disjoint six-cliques in $\mathcal{O}$, with all 36 cross edges present in $\mathcal{U}$. Exhaustively excluding that pattern excludes the two additional bases throughout the entire Hadamard phase ball. (OpenAI, 2026b)

Search failures and arithmetic conditions

A failed local certificate, exhausted graph-search cap, or unresolved region is retained for further work. It is never counted as a nonexistence proof. The claimed upper bound also depends on the specified binary64 rounding model, compiler behavior, complete covering arguments, and accounting for every retained record.

The manuscript reports a fresh complete execution on October 4, 2026: 249,878 seed centers lead to 3,083 coarse records, then 600 intermediate records and 22,379 fine records. Initial exclusion leaves 163 records unresolved because of graph caps. Further refinement produces 1,127 records, all resolved by final exclusion. These are reported execution results, not computations reproduced for this post. The text explicitly declines to authenticate a separate September 7 run.

The published paper describes seven C++ source files and a computation archive, but its verification/ directory is absent from the paper directory in the pinned repository tree. Consequently those described artifacts were not available there for this review. We did not rerun the main exclusion or independently audit its full arithmetic appendix. The claimed conclusion remains attributed to the manuscript and its stated execution contract.

🏷️ Fourier moments and an exact alternative certificate

The companion studies normalized character sums of an unscaled Hadamard matrix. For an integer vector $a$ with ${\sum }_i{a}_i=0$, define

$$g_H(a)=\frac{1}{6}\sum _{k=0}^5\prod _{i=0}^5{H}_{ik}^{a_i}.$$

Negative powers are well defined because every entry has modulus one. The balancing condition cancels column phases; row phases multiply the sum by a scalar of modulus one. In particular, vanishing is preserved by rephasing. We have $|g_H(a)|\le 1$, $g_H(-a)=\overline{g_H(a)}$, and $g_H(e_i-e_j)=0$ for distinct standard coordinate vectors $e_i,e_j$.

The companion claims vanishing at every permutation of $\alpha =(1,1,1,-1,-1,-1)$, except for matrices equivalent to Tao’s cubic matrix. Here equivalence means row and column permutations and rephasings. A concrete representative of the exceptional matrix has an all-one first row and column, diagonal entries one, and five remaining indices arranged in a cycle. Off the diagonal its entries are $\omega$ for adjacent vertices and ${\omega }^{-1}$ for nonadjacent vertices. Both this matrix and its entrywise square are Hadamard. (OpenAI, 2026a)

The proof starts from injective moments. For an integer exponent array $A$ whose row and column sums vanish, average the real part of its corresponding product of entries over independently sampled distinct row and column labels, and also over $H$ and $H^{\mathsf{T}}$. Call the result $m_H(A)$. Every such moment lies in $[-1,1]$.

Suppose $A$ has $p\le 6$ rows and row $i$ contains one $+1$, one $-1$, and zeros elsewhere. Column orthogonality yields the contraction rule

$$(7-p)m_H(A)=-\sum _{\ell \ne i}{m}_H(A_{i\to \ell }),$$

where $A_{i\to \ell }$ adds row $i$ to row $\ell$ and deletes row $i$. To see the coefficient, first assign the other $p-1$ distinct row labels. Of six labels, $7-p$ remain free. The orthogonality sum over all six splits into those unused labels and the occupied labels; the latter merge the two corresponding exponent rows. Averaging proves the identity. With seven formal rows, all labels are occupied, so only the sum of six contractions survives. No nonexistent injection of seven labels into six is used.

Products of character sums expand into these moments by grouping independently sampled columns according to which labels coincide. For example, two samples agree with probability $1/6$, giving

$$⟨|g_{\tilde{H}}(a){|}^2⟩=\frac{1}{6}+\frac{5}{6}{m}_H([a,-a]).$$

The brackets average over row permutations and the two transpose orientations. The first term is the coincident-column contribution, where the exponents cancel.

The exact pair certificate uses the following explicit charges, with coordinates indexed from zero:

$$\begin{array}{rl}w& =(-3,-2,0,1,2,2),\\ a_1& =(-1,-1,0,0,1,1),\\ a_2& =(-2,0,0,0,1,1),\\ a_3& =(-1,-1,0,1,0,1),\\ a_4& =(-1,-1,0,1,1,0).\end{array}$$

Set $(c_1,c_2,c_3,c_4)=(2,1,1,1)$ and

$$\begin{array}{rl}{L}_H=& -g_H(w)\\ & +\sum _{j=1}^4{c}_j{g}_H(a_j)g_H(w-a_j).\end{array}$$

Every term has total charge $w$, so its absolute square is invariant under rephasing. The certificate expresses a nonnegative sum as zero:

$$3⟨|L_{\tilde{H}}{|}^2⟩+⟨|g_{\tilde{H}}(\alpha )g_{\tilde{H}}(q){|}^2⟩=0.$$

There are two choices of $q$: $2(e_0-e_1)$ within one sign triple of $\alpha$, and $2(e_0-e_3)$ across the triples. The identity is checked by expanding both sides in injective moments and applying the contraction equations. Every permutation in the finite average has positive weight. Hence if some permuted $g_H(\alpha )$ is nonzero, permutations within its two sign triples, together with their interchange, force $g_H(2(e_i-e_j))=0$ for every distinct $i,j$. This says exactly that the entrywise square of $H$ is Hadamard.

The companion then classifies this alternative. For any ratio $v$ of distinct rows, orthogonality of $H$ and its square gives $\sum v_i=\sum v_i^2=0$. Together with $|v_i|=1$, the associated degree-six polynomial has only its degree-six, degree-three, and constant terms. The entries therefore form two rotated triples of cube roots. Compatibility across rows and columns forces, after dephasing, the cubic matrix described above. This classification is an additional mathematical argument beyond the finite moment identity.

🏷️ From exact arithmetic to the five-basis bound

Assume seven bases exist. Their traceless diagonal subspaces exhaust the traceless matrix space, so the orthogonal-projection identity supplies further moment equations. The companion first excludes a cubic transition matrix by separate finite character counts. In the remaining case, every transition matrix satisfies the Fourier vanishing condition.

Sample the reference basis uniformly among all seven, independently permute its six coordinates uniformly, and uniformly label the six other bases to obtain transition matrices $H_0,\dots ,H_5$. Then sample a single column $z$ of $H_0$ and use it in every test function. All expectations below include this entire finite sampling:

$$\begin{array}{rl}{F}_0(a)& =g_{H_0}(a)z^{-a},\\ F_1(a)& =\left(\frac{1}{5}\sum _{s=1}^5{g}_{H_s}(a)\right)z^{-a}.\end{array}$$

Here $z^{-a}={\prod }_i{z}_i^{-a_i}$, and the five other transition matrices come from the other five nonreference bases. Averaging over this shared column turns Gram entries into three-factor moments: the extra factor is $g_{H_0}(v-a)$ when comparing charges $a,v$. For two $F_1$ factors, the 25 ordered choices of matrix labels split into five equal-label and twenty distinct-label cases, explaining the coefficients $1/5$ and $4/5$ in the certificate.

The source specifies six positive-definite integer matrices and finite combinations of these functions. Their positively weighted quadratic forms sum to a nonnegative number $Q$. Exact substitution gives

$$\begin{array}{c}Q=f+\sum _{\rho }{\beta }_{\rho }{X}_{\rho },|X_{\rho }|\le 1,\\ f=-\frac{613302797399911}{6480},\\ \sum _{\rho }|{\beta }_{\rho }|=\frac{179721388988719}{6480}.\end{array}$$

Consequently $Q\le f+\sum |{\beta }_{\rho }|=-2007321335237/30<0$, a contradiction. The matrices’ positive definiteness is checked through exact positive Schur-complement pivots. This excludes seven bases. The companion invokes Weiner’s completion theorem that $d$ mutually unbiased bases in dimension $d$ extend to $d+1$; hence six bases would produce seven, and the upper bound becomes five. This argument does not exclude four or five bases. (OpenAI, 2026a)

The use of modular arithmetic in the verifier does not assume the moments themselves are rational. If selected integer equations are $Wm=0$, a square pivot block $W_J$ invertible modulo a prime has nonzero integer determinant. Modular computations propose a rational substitution matrix $P$. The verifier then checks exactly that free coordinates are unchanged, pivot-input columns vanish, and $WP=0$. In pivot/free order this forces $P_{J,F}=-W_J^{-1}{W}_F$, so $Pm=m$ for every real solution of the selected equations. Extra unused equations and any modular underestimate of the full rank do not invalidate that implication.

The companion supplies a separate standard-library Python verifier and a checksum-checking wrapper in its pinned verification/ directory. We ran that wrapper with assertions enabled: it exited successfully and printed exactly the two rank tables and rational coefficients specified in the manuscript. This checks the two moment certificates implemented by that program. It does not check the separately specified cubic character counts, every derivation connecting the certificates to the original problem, or the main paper’s floating-point exclusion search.

🏷️ Key steps and proof outlines

A positive defect makes the completion cover exhaustive

The reduction to $2\times 2$ unitaries rests on an exact identity, rather than a numerical guess that completions are approximately unitary. Write a feasible Hadamard matrix in blocks

$$H^{\prime }=\left(\begin{array}{cc}{E}^{\prime }& B^{\prime }\\ C^{\prime }& D^{\prime }\end{array}\right),$$

with $E^{\prime }$ of size $2\times 2$. Let $K_B,K_C$ be orthonormal frames for $\ker B_0$ and $\ker C_0^{\ast }$, respectively, at the stored seed center. Put $P_C=I-K_C{K}_C^{\ast }$ and $Y=K_C^{\ast }{D}^{\prime }{K}_B$. Column orthogonality gives

$$\begin{array}{rl}6I-Y^{\ast }Y=& K_B^{\ast }{B}^{\prime \ast }{B}^{\prime }{K}_B\\ & +K_B^{\ast }{D}^{\prime \ast }{P}_C{D}^{\prime }{K}_B⪰0.\end{array}$$

Both terms are positive. Moreover, $B_0{K}_B=0$, so the first term is controlled by the seed displacement. For the second, the off-diagonal orthogonality equation yields

$$C_0^{\ast }{D}^{\prime }{K}_B=-E^{\prime \ast }(B^{\prime }-B_0)K_B-(C^{\prime }-C_0{)}^{\ast }{D}^{\prime }{K}_B.$$

The stored lower singular-value bound for $C_0$ then controls $P_C{D}^{\prime }{K}_B$.

Suppose the certified trace defect is at most $q<6$. Its eigenvalues ${\ell }_1,{\ell }_2$ are nonnegative with sum at most $q$, so $Y$ is nonsingular and its polar factor $V$ is unitary. Rationalizing the square roots gives

$$\Vert Y-\sqrt{6}V{\Vert }_F\le \frac{q}{\sqrt{6}+\sqrt{6-q}}.$$

Thus every exact completion lies near some member of a finite cover of $U(2)$. The frames never move with the unknown completion, and no exact solution is required at the numerical center. Outward rounding must certify the singular-value and defect inequalities; failure of either check leaves the seed unresolved (OpenAI, 2026b).

The cubic alternative is eliminated by a dimension contradiction

The Fourier certificate’s implication that $H^{\circ 2}$ is Hadamard still needs a classification argument. For a row ratio $v$, Newton’s identities and unit modulus give

$$\prod _{i=0}^5(z-v_i)=z^6-e_3{z}^3+e_6.$$

Consequently the cubed entries of $v$ either coincide or have two values, each occurring three times. Apply this to every row ratio and every column ratio.

After dephasing, a nonconstant row of $H^{\circ 3}$ has three entries equal to one and three equal to some $x\ne 1$. Two such rows cannot have different three-element supports: their intersection has size one or two, and the cubed ratio would either have more than two values or split into multiplicities two and four. Both contradict the previous paragraph. The same reasoning applies to columns. If any nonconstant row remains, permutations therefore produce

$$H^{\circ 3}=\left(\begin{array}{cc}{J}_3& J_3\\ J_3& x{J}_3\end{array}\right),x\ne 1,$$

where $J_3$ is the all-one matrix.

Now restrict the two classes of rows of $H$ to the last three columns. Opposite classes are orthogonal there: each ratio over these columns is one complete rotated cube-root triple, whose sum is zero. Each class must span at least two dimensions. Indeed, if its three restrictions spanned a line, full-row orthogonality and equality in Cauchy–Schwarz would make their restrictions to the other three columns proportional too. Three orthogonal full rows would then lie in a space of dimension at most two.

The two restricted spans would thus be orthogonal subspaces of ${\mathbb{C}}^3$, each of dimension at least two, an impossibility. Hence every dephased entry is a cube root. Counting the two occurrences of each cube root in each noninitial row and column reduces the remaining classification to a degree-two graph on five vertices, necessarily a five-cycle. This supplies the mathematical bridge from the exact pair-moment identity to Tao’s exceptional matrix; the finite verifier alone does not prove this bridge (OpenAI, 2026a).

🏷️ Connections

• Area laws and PEPS uses the same finite-dimensional quantum framework, with entanglement across spatial regions replacing compatibility of measurement bases.
• Circulant Hadamard matrices concerns an additional, much more restrictive matrix structure. Excluding matrices in a special class does not by itself cover all possible second bases in the present problem.

🏷️ Manuscripts and verification scope

• The maximum number of mutually unbiased bases in dimension six — pinned manuscript and companion source.
• Exact Fourier certificates for complex Hadamard matrices of order six — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The MUBSix comparator is described as proving an upper bound of five and the non-Tao Fourier vanishing statement for all coordinate permutations; it does not establish the main manuscript’s upper bound of three. The separate HadamardCubeFiber comparator concerns cancellation within either cube fiber of a row ratio when the ratio cubes take two distinct values. That supporting lemma alone does not establish the full Fourier theorem or a bound on the number of bases. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026a. Exact Fourier certificates for complex Hadamard matrices of order six.

🐻  OpenAI 2026b. The maximum number of mutually unbiased bases in dimension six.