PROGRESS0 / 00 attemptsloading…

Paper course · the remaining gaps

Five Derivations

Each one is a calculation you can finish with a pencil. Together they close everything between the labs and reading the tutorials at full strength — except the last, which is honest about being a mountain.

What this is

A worksheet, not a result. The mathematics on this page is standard topological quantum computation, roughly 1984–2009. None of it is new. “It was simply new to my hands. I had read it for years and never computed with it, and those turn out to be different states of knowing.” These sheets exist to move a reader from the first state to the second, and they make no claim of their own.

The method is predict–then–check. Every input box, drill and gate here is built for one move: write down what you think the number is before you reveal it. Numbers are sourced to a paper where a paper says them, and marked unchecked where none does — a sheet whose whole method is check it yourself cannot ask you to take its own arithmetic on trust.

If you are coming from an operator formalism, you may want to ask of every quantity here what the declared observable is, and what mapping puts it into the domain before the algebra acts. That is a fair question, and these sheets sit upstream of it — the arithmetic of d² = d + 1 is the same arithmetic whichever way it is answered. They take no position on it. The full note is on the index.

Prerequisites: the three labs. Derivations 1–4 are a few days of paper between them; derivation 5 is a project of its own size, and the page says so.

Assumed background — not repeated here

φ and its identities · ℤ[φ] arithmetic · powers via Fibonacci coefficients · 𝒟 = √(2+φ) · quantum dimension as growth rate · the 2cos(π/n) ladder · fusion rules · matrix multiplication · eigenvalues · Verlinde mechanically · integrality as rigidity · eiθ as the unit circle in the complex plane.

Derivation 1Galois conjugationhours

The single highest-value item, and it operates entirely inside the ℤ[φ] system set up in the Golden Arithmetic lab.

The map. φ was built from √5. There is a second, equally valid square root: −√5. Swapping them is a field automorphism — call it σ. It leaves every ordinary rational number alone and moves only the √5 part.

φ  = (1 + √5)/2
σ(φ) = (1 − √5)/2 = 1 − φ

check:  1 − 1.618034 = −0.618034 = −1/φ   ✓

σ sends φ to the other root. That is the whole content.

On a general element

σ(a + bφ) = a + b(1 − φ) = (a + b) − bφ
On paper — derivation 1
0 / 0
Gate — derivation 1

Why does Galois conjugation leave the fusion ring untouched?

σ(φ) equals:

A quantum dimension of −1/φ tells you the theory is:

Why this is the load-bearing derivation

Fusion coefficients Nabc are integers. σ fixes every integer. Therefore σ leaves the entire fusion ring untouched — one line, no machinery.

But σ sends dτ = φ to −1/φ. Same ring, different quantum dimension. That is exactly Fibonacci → Yang–Lee, and it is why Rowell–Stong–Wang can write "the Galois conjugate of the Fibonacci theory is the Yang-Lee theory" as a plain statement of fact.

"One modulus, two arguments" has a mechanism, and this is it.

Derivation 2The modular group, and why S existsa day

⚠ Notation collision — read before anything else

τ means two entirely different things in this literature. The Fibonacci anyon is τ. The modular parameter of a torus is also τ. They are unrelated and appear in the same papers. Watch the context; there is no fixing it.

A torus has two independent loops — call them a (around the hole) and b (through it). Nothing distinguishes them intrinsically. So there must be an operation that swaps them, and that operation is S. A second operation shears one along the other: T. Together they generate SL(2,ℤ).

On the modular parameter:

S :  τ → −1/τ        swap the two cycles
T :  τ → τ + 1       shear
On paper — derivation 2
0 / 0
Gate — derivation 2

What does the S-matrix physically represent?

In S² = I, why do the off-diagonal entries vanish?

S² = (1/𝒟²) ⎡ 1·1 + φ·φ    1·φ + φ·(−1) ⎤
            ⎣ φ·1 + (−1)·φ  φ·φ + (−1)(−1) ⎦

   = (1/𝒟²) ⎡ 1 + φ²     0     ⎤
            ⎣    0    φ² + 1   ⎦

and  1 + φ² = 𝒟²   so

S² = ⎡ 1  0 ⎤   =  the identity
     ⎣ 0  1 ⎦

The off-diagonal entries cancel before you divide by anything: φ − φ = 0. The diagonal ones are 𝒟²/𝒟². Nothing was approximated.

What you have just shown

Swapping the torus's two cycles twice returns you to where you started — and the anyon S-matrix knows this. S is not an arbitrary array of numbers; it is a geometric operation on a doughnut, written down. That is why tutorial 4 opens with a torus.

Derivation 3Deriving the central chargea day

Here you produce one. c = 14/5 falls out of arithmetic you can do by hand.

The Gauss sum:

Σa da² θa = 𝒟 · e2πi c/8

Fibonacci has two terms. The twists are θ1 = 1 and θτ = e2πi(2/5) = ei·144°.

The exact step that makes it hand-doable

cos 144° = −cos 36° = −φ/2       ← because 2cos36° = φ
sin 144° =  sin 36°  = 0.587785
On paper — derivation 3
0 / 0
Gate — derivation 3

In the Gauss sum, which quantity supplies the magnitude and which the phase?

The Gauss sum's modulus came out 1.902113. That number is:

A negative central charge is the signature of:

What just happened

You took quantum dimensions (magnitudes) and twists (phases), added two complex numbers, and out came the central charge of a conformal field theory living on the boundary of the material. Two arrows and some golden-ratio arithmetic.

That is the Gauss sum being "the weld." Not a slogan — six lines of arithmetic.

Derivation 4Lee–Yang zeros, on two sitesa day

The circle theorem is usually presented as a deep result. On two spins you can derive it in ten lines and see exactly why a transition is zeros pinching the real axis.

⚠ Read the case of the letters

Z (capital) is the partition function. z (lowercase) is the fugacity, e2βh. They are different objects and they appear in the same equation. Statistical mechanics does this constantly — it is the same trap as τ meaning both the anyon and the torus modulus. Writing them visibly differently on paper is a legitimate defence.

Two Ising spins in a field: E = −J s₁s₂ − h(s₁+s₂), each s = ±1. Four states.

↑↑ :  E = −J − 2h
↑↓ :  E = +J
↓↑ :  E = +J
↓↓ :  E = −J + 2h

Z = eβ(J+2h) + 2e−βJ + eβ(J−2h)

write  K = eβJ,  z = e2βh

Z = K z + 2/K + K/z
On paper — derivation 4
0 / 0
Gate — derivation 4

Why does the 1D Ising model have no phase transition?

Physically, what happens when partition-function zeros pinch the real axis?

Complex zeros are argument; a divergent susceptibility is magnitude. Criticality is where:

You have just proved the Lee–Yang circle theorem for N = 2

|z| = 1 exactly. The zeros of the partition function lie on the unit circle, for every temperature. That is the theorem, and on two sites it is four lines of algebra.

And now the part that teaches the physics

Track where on the circle the zeros sit as temperature changes:

K = eβJtemperaturezangle from +real axis

Physical magnetic fields are real, which means z sits on the positive real axis — angle 0. A phase transition happens only if a zero reaches it.

The punchline, and it is a real result

As T → 0 the zeros march only as far as ±90°. They never reach the positive real axis. Therefore the 1D Ising model has no phase transition at any finite temperature — which is correct, and famous, and you just derived it by watching two complex numbers fail to arrive.

In 2D they do pinch. That is the whole difference.

Derivation 5Berry phaseweeks — and that is honest

This one gates tutorial 2 and nothing else. With graduate QM the on-ramp is real, but a worksheet does not close it, and this page will not pretend otherwise.

The canonical calculable example, and the right place to start: a spin-½ in a magnetic field, with the field direction traversing a closed loop on the sphere. The Berry phase comes out as

γ = −½ Ω        Ω = solid angle enclosed by the path

Do that one first. It is in every graduate text, it is fully computable, and it establishes the shape: a phase that depends only on the geometry of the path, not on how fast you traverse it. That is holonomy, and once it is in your hands the ASW calculation is the same move with a different Hilbert space.

The sequence, honestly
0 / 0
Gate — derivation 5

The relationship between Aharonov–Bohm and Berry phase is:

In ASW, what produces the extra phase beyond the Aharonov–Bohm contribution?

A Berry phase depends on:

Where this leaves you

Derivations 1–4 are a few days of paper and they open tutorials 1, 3 and 4 at full strength. Derivation 5 is a separate project of its own size, and it buys exactly one tutorial. Do 1–4 first and the ratio stays in your favour.

Built 29 August 2026, ordered by leverage rather than by difficulty. Every number here was computed, not recalled; the check boxes verify against live arithmetic rather than stored answers, so the page cannot quietly agree with a wrong number.
Sources

Checked against the paper itself: Rowell, Stong & Wang (arXiv:0712.1377) for the Galois statement quoted in derivation 1; Fisher (1978) for the Yang–Lee edge singularity.

Stated on reputation, not checked here: Verlinde (1988) and Moore–Seiberg (1989) for the fusion formula; Yang & Lee (1952) for the circle theorem; Berry (1984) for the geometric phase. All are standard and none was opened while writing this page — treat the attributions, not the mathematics, as the soft part.

Every numerical answer the check boxes grade against is computed in your browser at the moment you press the button, from the definitions on the page.