Hilbert 12, SICs and AI

Hilbert's 12th Problem, SIC-POVMs, and AI

Imran Zaidi · 22 October 2026

How a language model works

  • Text is split into tokens; each token becomes a long vector of numbers
  • Attention: each token looks back at earlier tokens and pulls in what's relevant
  • Dozens of layers of this; at the end, predict the next token
  • Trained on enormous amounts of text, then trained further to reason and use tools
ThecapitalofFranceis→ Paris

How Claude added numbers (March 2025)

Anthropic's attribution graph for 36 + 59: one path approximates the sum, another computes the last digit

Anthropic, "Tracing the thoughts of a large language model" (March 2025)

July 2026: Claude disproves the Jacobian conjecture

Keller (1939): a polynomial map \(F:\mathbb C^n\to\mathbb C^n\) whose Jacobian determinant is a nonzero constant must be invertible.
  • Open for 87 years
  • Claude Fable 5 found a counterexample in dimension 3 (posted by Levent Alpöge, Anthropic)
  • Jacobian determinant \(-2\), yet three points have the same image:
    \((0,0,-\tfrac14),\ (1,-\tfrac32,\tfrac{13}2),\ (-1,\tfrac32,\tfrac{13}2)\ \mapsto\ (-\tfrac14,0,0)\)
March 2025: adding two-digit numbers by rough guesses. July 2026: a counterexample to an 87-year-old conjecture.

Why: neural + symbolic

Neural: the language model
  • Learned from data: recognizes patterns, suggests ideas and approaches
  • Approximate: can be confidently wrong
Symbolic: exact tools
  • Running code, computer algebra, proof checkers (Lean)
  • Never wrong about what they check
Model proposes→ Tools compute and check→ Model revises↺
  • The loop runs thousands of times, often with many agents in parallel
  • Models are also trained on problems whose answers can be checked

A quantum state is a unit vector

\(\psi=(a,b)\in\mathbb C^2,\qquad |a|^2+|b|^2=1\)
  • A qubit: spin up/down, or a photon's polarization
  • In general: a unit vector in \(\mathbb C^d\), for a system with \(d\) distinguishable outcomes
  • Multiplying \(\psi\) by a phase \(e^{i\theta}\) gives the same physical state

The Born rule

\(\text{Prob}(\text{outcome }\varphi)=|\langle\varphi|\psi\rangle|^2\)
  • Qubit \(\psi=(a,b)\): outcome 0 with probability \(|a|^2\), outcome 1 with probability \(|b|^2\)
  • That is why states have length 1: the probabilities add up to 1
  • It is the only place probability enters quantum theory

Order matters: noncommutativity

Position and momentum (Heisenberg, 1925)

\(xp-px=i\hbar\)

Uncertainty follows (1927)

\(\Delta x\,\Delta p\ \ge\ \hbar/2\)

  • Measuring one disturbs the other; no state has both sharp
  • Discrete version in \(\mathbb C^d\): shift \(X\) and clock \(Z\), with \(ZX=\omega\,XZ\), \(\omega=e^{2\pi i/d}\)
  • Together they generate the Weyl–Heisenberg group: all the "shift and phase" moves

The measurement problem

Rule 1 · between measurements

The state evolves smoothly and deterministically (Schrödinger's equation)

Rule 2 · at a measurement

One random outcome; the state "collapses" to match it

  • Which rule applies when? What counts as a measurement?
  • Schrödinger's cat (1935): is the cat alive and dead until someone looks?
  • Wigner's friend (1961): the friend sees an outcome; for Wigner outside the lab, has it happened?

Two standard answers

Copenhagen (Bohr, Heisenberg)
  • Split the world: quantum system, classical apparatus
  • Collapse happens at the "cut"
  • Problem: where is the cut? What makes an apparatus classical?
Many-worlds (Everett, 1957)
  • No collapse: Rule 1 always
  • Every outcome happens, in branching worlds
  • Problem: if everything happens, what does "probability \(|a|^2\)" mean?

QBism's answer

A quantum state is not a property of the system. It is an agent's beliefs about what they will experience.
  • Probabilities are personal (Bayesian, in de Finetti's sense)
  • Collapse is just updating beliefs after a new experience: nothing physical jumps
  • Wigner and his friend each have their own state assignment: no contradiction
  • The Born rule is a consistency rule for beliefs, like the rules of probability

Caves, Fuchs, Schack (2002 onward); Mermin

Is a quantum state just a list of probabilities?

  • If a state is an agent's beliefs, it should be writable as plain probabilities
  • Goal: replace the state vector by its outcome probabilities for one fixed reference measurement
  • The reference measurement is still built from vectors, but it is fixed once and for all, like a ruler. Only the state becomes probabilities
  • Then the Born rule becomes a rule for combining probabilities (Fuchs–Schack)

How many outcomes do you need?

\(d^2\) outcomes
  • A general state (density matrix) has \(d^2-1\) real parameters
  • A measurement with \(n\) outcomes gives \(n-1\) independent probabilities
  • So you need \(n\ge d^2\). The most symmetric choice: a SIC-POVM

SIC-POVM

\(d^2\) unit vectors in \(\mathbb C^d\)
\(|\langle\psi_j|\psi_k\rangle|^2=\dfrac{1}{d+1}\)
  • POVM: a general measurement, positive operators \(E_j\) with \(\sum_j E_j=I\). Here \(E_j=\tfrac1d|\psi_j\rangle\langle\psi_j|\)
  • All pairs equally far apart
  • \(d^2\) equiangular lines: the most possible in \(\mathbb C^d\)
  • \(d=2\): four points on the Bloch sphere forming a regular tetrahedron
  • "SICs" for short

The Urgleichung

\(p(i)\) = probability of SIC outcome \(i\);   \(r(j\,|\,i)\) = probability of outcome \(j\) of your experiment, if the system were in SIC state \(i\)

Classical: total probability
\[q(j)=\sum_i p(i)\,r(j\,|\,i)\]
Quantum: the Urgleichung (Fuchs–Schack)
\[q(j)=\sum_i\Big[(d+1)\,p(i)-\tfrac1d\Big]\,r(j\,|\,i)\]
The Born rule becomes one consistency condition on probabilities.
This only works if a SIC exists in dimension \(d\).

Zauner's conjecture (1999)

A SIC exists in every dimension \(d\)

  • Built from one vector by the \(d^2\) "shift and phase" operations (Weyl–Heisenberg group)
  • Numerical solutions: every \(d\le193\), some up to \(d=39604\)
  • Exact solutions: every \(d\le53\), some much larger
  • For twenty-five years, found one dimension at a time by computer

SICs are made of number theory

\(K=\mathbb Q\big(\sqrt{(d+1)(d-3)}\big)\)
  • Exact SICs are built from algebraic numbers that generate class fields of \(K\) (Appleby et al.)
  • Their overlaps are essentially Stark units: conjectured by Gene Kopp (checked in \(d=5,11,17,23\)), extended to every \(d\) by Appleby–Flammia–Kopp
  • This is Hilbert's 12th problem for real quadratic fields

Appleby–Flammia–Kopp (2025)

Stark units
special values of Faddeev's
quantum dilogarithm
→ a ghost SIC
same algebra, real numbers
→ a Galois symmetry
a different embedding
→ a SIC
numbers on the unit circle
  • An explicit candidate SIC in every dimension
  • Built from number theory alone
  • Checked against known SICs; new ones in \(d=100\)
  • Needs two conjectures: their Twisted Convolution Conjecture and Stark's conjecture
  • Theorem: the two together imply Zauner's conjecture

Where I started: A SIC factory

1 · Ghost
AFK's formula, numerically
→2 · Refine
Newton's method, thousands of digits
→3 · Recognise
exact algebraic numbers via the Galois structure
→4 · Certify
check mod primes, then every embedding
  • Every overlap written exactly, not just a few invariants: no guessing or optimisation step at the end
  • Height control (the published bottleneck): a good basis of the number field and automatic rescaling keep the integers small
  • A cheap exactness check: rebuild everything modulo primes and verify the SIC equations there
Every dimension from 4 to 90, every SIC in each
About 1 to 3 hours per dimension near \(d=90\)

A SIC in dimension 54

A dimension-54 SIC rendered by SIC Atlas

Two obstacles to Zauner's conjecture

1 · Twisted Convolution Conjecture

\(\tilde\Pi^2=\tilde\Pi\)

The ghost must be a projector

2 · A Galois symmetry

real line → unit circle

AFK: follows from the Stark conjectures

AFK, Theorem 1.4: Stark Conjecture + Twisted Convolution Conjecture ⇒ Zauner's conjecture

The Twisted Convolution Conjecture

  • The ghost is a matrix built from the cocycle values: \(\tilde\Pi=\sum_{\mathbf p}c_{\mathbf p}D_{\mathbf p}\), one coefficient for each shift-and-phase move \(D_{\mathbf p}\)
  • Conjecture (AFK): it is a projector, \(\tilde\Pi^2=\tilde\Pi\)
  • Moves multiply up to a phase, \(D_{\mathbf q}D_{\mathbf p-\mathbf q}=\chi(\mathbf q,\mathbf p-\mathbf q)\,D_{\mathbf p}\), so this is one identity for each label \(\mathbf p\):
\[\sum_{\mathbf q}\chi(\mathbf q,\mathbf p-\mathbf q)\,c_{\mathbf q}\,c_{\mathbf p-\mathbf q}\;=\;c_{\mathbf p}\]
A convolution of the special values with themselves, twisted by the phase \(\chi\)

The Radchenko–Wheeler breakthrough

18 September 2026: the Stark units of every real quadratic field are algebraic
  • Their values: Faddeev's modular quantum dilogarithm at the real quadratic fixed point \(\tau=\gamma\tau\), one value per label \[\Phi_\gamma(z;\tau)=\frac{\big(e(z);\,e(\tau)\big)_\infty}{\big(e\big(\tfrac{z}{c\tau+d}\big);\,e(\gamma\tau)\big)_\infty}\]
  • These values satisfy an overdetermined system of polynomial equations, including a pentagon identity
  • The equations build fusion categories (Izumi's near-group categories); Ocneanu rigidity makes the values algebraic
  • The pentagon gives AFK's Twisted Convolution Conjecture: the first obstacle, gone

What was left: one step

1 · Twisted Convolution Conjectureproved from a pentagon identity
The numbers are algebraicproved via fusion categories
2 · A Galois symmetry onto the unit circleremaining
"…the only remaining obstacle to proving Zauner's conjecture in full is to show that there exists \(\sigma\) for which the values all land on the unit circle."
Radchenko–Wheeler, arXiv:2609.21892 (September 2026)

Three problems with an AI researcher

ProblemWhat I built
No real working memory
each session forgets; old claims come back
Ledgers: the proof, every withdrawn claim, a north-star document read before every task
No sense of its own uncertainty
guesses sound like theorems
Every claim labelled: proved, checked, numerical, heuristic, speculative; predictions written down before computing
Errors compound undetected
a small mistake becomes a foundation
Councils of agents that review and vote; hostile reviewers; independent re-computation; Lean

Filament: the proof as a ledger

  • Every lemma is a node: a statement, what it depends on, and how it is justified
  • Its status is computed, never typed in: proved, conditional, checked in Lean
  • The proof changes only through reviewed patches; the system shows what each change affects
  • Every explanation is pinned to the exact version of the statement it explains, so stale text is caught
Lean checks truth. Filament keeps the structure: what is claimed, why, and how sure we are.

Where Filament sits

Natural language

The paper, the reader website, these slides. Written by people and agents.

↕ every sentence is pinned to an exact version of a node; if the node changes, the sentence is flagged

Filament ledger

Each node: statement · what it uses · how it is justified · status, computed from the dependencies

integrality.theorem_u  every value is a p-adic unit
uses: line_product, frobenius_defect, ineq_outer, rf …
status: proved · conditional on RW eq. (6)
↕ records what Lean has checked, and under which hypotheses

Lean

Machine-checked proofs. Checks truth, but says nothing about meaning or importance.

How the proof changes
A patch is proposed
e.g. PC-0052: restate the integrality proof directly
Independent review
My approval
Applied as a new version (F11); every affected explanation is flagged

The second attempt: what was left

  • 18 September: Radchenko–Wheeler prove the convolution identity and algebraicity
  • Left: a Galois symmetry that puts the values on the unit circle
  • My route: a reciprocity law, predicting the values modulo primes \(p\)
  • Model: the CM case, where Frobenius at \(p\) acts by \(p\)-th powers (Kronecker: \(j(p\tau)\equiv j(\tau)^p\)); Chebotarev then pins down the Galois action
  • Everything came down to one property:
Integrality: no value has the prime \(p\) in its denominator

The wall: a global identity, a local question

The question

Is each single value free of \(p\) in its denominator?

The strongest tool: RW's pentagon

Products of three values, summed over every label at once

  • Use it exactly: every combination re-sums the same global information, never one value
  • Use only how many factors of \(p\) each term has: bad terms can cancel, and fake examples pass every test

Late September: many agents, many approaches, the same wall. "It needs a new idea."

Six agents, one target

Me · 29 September, 12:36 am
Spawn six subagents and work together on figuring out the crucial new idea that will let us solve the core difficulty. We need a way through the pentagon dilemma.
Claude · 29 September, 2:47 am

This could be important. L4 claims a route that doesn't use the pentagon at all.

The idea: don't use the global identity at all. Use only local tools: products of values along a line, and RW's Fourier identity on one line at a time.

The proof in three steps

STEP 1

No \(p\) in the denominator

So remainders mod \(p\) make sense

STEP 2

Mod \(p\): a neighbour = the \(p\)-th power

This is what Frobenius does

STEP 3

Chebotarev + a sign picks the symmetry

Two primes above \(p\); a real place decides

Step 1 · The Bruhat–Tits tree

  • One vertex for each lattice in the plane, up to scaling
  • An edge when one lattice is \(p\) times as dense as the other. Every vertex has \(p+1\) neighbours (picture: \(p=2\))
  • Axis (crater): the most symmetric lattices. Tube: everything within distance \(R\) of it

Step 1 · Where the values live

  • RW's values belong to matrices; each matrix has a lattice, so each value sits at a vertex
  • Tube lemma: all the matrices we need live inside the tube, \(R=v_p(\varepsilon_0-1)\)
  • Inside the tube: one neighbour points inward (toward the axis), \(p\) point outward
  • On the outer sphere (distance exactly \(R\)): no outward neighbours
Goal (integrality): no value anywhere in the tube has \(p\) in its denominator

Step 1 · Lines link neighbours

  • A matrix's values are indexed by its labels (the torsion points): a grid
  • A line: \(p\) evenly spaced labels
  • RW's line product formula:
\(\displaystyle\prod_{\text{line}}E\;=\;\zeta\cdot E(\text{neighbour})\)

\(\zeta\) a root of unity. So the factors of \(p\) along a line add up to the neighbour's.

0 1 2 3 4 5 6
A line of 7 labels, \(p=7\)

Step 1 · The floor

Frobenius defect \(\;\varphi=\dfrac{E(\text{inward neighbour})}{E^{\,p}}\)
  • \(A\) = worst pole: the most factors of \(p\) in any denominator
  • At the worst value: \(\varphi\) has at least \((p-1)A\) factors of \(p\)
  • Defect identity: a lattice's count is the average of its outward neighbours'
  • So step to the largest; repeat to the outer sphere
4 7 5 8 5 9 4 6 7 × 6 = 4 + 7 + 5 + 8 + 5 + 9 + 4 = 42 outward neighbours (one layer farther out)
Number in each circle: factors of \(p\) in that lattice's Frobenius defect, in units of \(A\). Illustrative, \(p=7\).

Step 1 · The ceiling

  • RW's Fourier identity relates the values to their discrete Fourier transform. On the outer sphere it has \(p^R\) in its denominator
  • That forces the values along each line into a low-degree pattern (a cyclic code)
  • Result: a pole \(a\) on the outer sphere makes the inward neighbour's pole at least \(3a\)
  • Since no pole exceeds \(A\): poles on the outer sphere are at most \(A/3\)
On the outer sphere: \(\varphi\) has at most \(\big(1+\tfrac p3\big)A\) factors of \(p\)

Step 1 · A floor above a ceiling

forced at the worst pole

\(\ge (p-1)A\)

allowed at the edge

\(\le (1+\tfrac p3)A\)

  • Recall: \(A\) = the worst pole, the most factors of \(p\) in any denominator
  • The Frobenius defect never shrinks moving outward (an exact average)
  • At the edge, RW's Fourier identity caps it
\(p-1\gt 1+\tfrac p3\) for \(p\ge5\ \Rightarrow\ A=0\)

Frobenius, then a choice

\(E(\text{neighbour})\equiv E^{\,p}\pmod p\)
  • The real-quadratic analogue of Kronecker's congruence
  • Two primes above \(p\): locally you can't tell which Frobenius you see
  • A sign at the second real place decides; Chebotarev assembles all primes

Onto the unit circle

The reciprocity law says what every Galois symmetry does to every value

  • At the other real embedding, complex conjugation sends each value to its inverse, so it has absolute value 1
  • That is Radchenko–Wheeler's remaining step, and with it, SICs in every dimension \(d\ge5\)

How this answers Hilbert's 12th problem

1 · The values are in the right place

The reciprocity law puts every value in \(K^{\mathrm{ab}}\), the maximal abelian extension of \(K\), and says exactly how each Galois symmetry moves it

2 · There are enough of them

Shintani's product formula pairs the values with characters: \(\sum_B\bar\chi(B)\log W(B)=-2n\,L'(0,\bar\chi)\ne0\). So no nontrivial Galois symmetry fixes them all

\(K^{\mathrm{ab}}=K\big(i,\ \text{the values}\big)\): explicit generators, as \(j\) and torsion points give for imaginary quadratic fields

What the reciprocity law gives

ResultAlso uses
Hilbert's 12th problem for real quadratic \(K\): \(K^{\mathrm{ab}}=K(i,\ \text{the values})\)Shintani's product formula
Zauner SICs in every dimension \(d\ge5\)RW's quadratic relation (Thm. 7)
Rank-one abelian Stark conjecture at a real place, \(|S|\ge3\), with Tate's square-root conditionin one configuration, the Faddeev–Kashaev–Volkov identity
The values are units at every finite prime—

All four use Radchenko–Wheeler's Fourier identity (their eq. (6)).