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How do five places organize matter?

Turn five internal places on or off. Keep choices with zero, two or four occupied places. Their charges add, producing sixteen components grouped into one family in this construction.

Step 1 of 8

Start with three colour slots and two weak slots

The construction supplies five internal slots, marked as a group of three and a group of two. The marking tells us which transformations must preserve each group.

1. The specified five dimensional complex carrier with its marked 3+2 split.. Keep the three colour directions and two weak directions as separate blocks.. A carrier C⁵ = C³ ⊕ C² with a fixed marking.. 2. The five distinct slots from the first step.. Enumerate each subset and retain subsets with even size.. The three counts are 1, 10 and 5.. 3. One empty state, ten degree two states and five degree four states.. Take the direct sum of the three even exterior powers.. S⁺ has sixteen components.. 4. The marked 3+2 carrier, its block preserving unitary transformations and the charge convention X = 6Y.. Solve 3x_c + 2x_w = 0 and choose the primitive integer pair in the stated orientation; divide by six.. Each colour slot carries Y = −1/3 and each weak slot Y = +1/2.. 5. An even degree two state with one colour slot and one weak slot occupied, and the charge operator from the previous branch.. Add the two occupied slot hypercharges.. This component has hypercharge Y = 1/6.. 6. All sixteen even selections and the normalized slot charge operator.. For every mask, count occupied colour and weak slots, add their charges, and group equivalent occupation types.. The groups are νᶜ, uᶜ, Q, eᶜ, L and dᶜ, with sixteen components in total.. 7. The six sector multiplicities and integer charges X = (0, −4, 1, 6, −3, 2).. Sum nX², then divide by 36 to obtain the trace of Y² on one family.. Tr X² = 120; Tr Y² = 10/3 on the sixteen component family.. 8. The one-family charge trace, three Weyl families, one complex Higgs doublet and the stated carrier normalization.. Apply one loop weights 2/3 for Weyl fermions and 1/3 for the complex scalar, then multiply by 3/5.. b₁ = 4 + 1/10 = 41/10; M = 10b₁ = 40 + 1 = 41..
Arrows carry the named inputs into later calculations. Added inputs stay marked. Reduced motion is on. Use Next step or choose a stage below.
Carry forward
The specified five dimensional complex carrier with its marked 3+2 split.
Added at this step
The marked five dimensional carrier and the stated unitary structure are supplied inputs to this chapter.
Operation
Keep the three colour directions and two weak directions as separate blocks.
Result
A carrier C⁵ = C³ ⊕ C² with a fixed marking.
C=C3⊕C2,dim⁡C=3+2=5\begin{gathered}C=\mathbb C^3\oplus\mathbb C^2,\\[.4em]\dim C=3+2=5\end{gathered}

In words: Keep the three colour directions and two weak directions as separate blocks.

Why this step works and what it assumes

The five directions are internal carrier slots. They are not five spacetime directions. The marked carrier and its physical identification belong to the stated TFPT construction; this journey begins with them.

Read the original calculation
See where this fits in the whole theory
Try the individual controls

Try this. Turn on one colour place and one weak place. Compare their added charge, then explore the other allowed choices.

Choose slots. Add their numbers.

The model supplies five slots. Each selected colour slot adds −1/3; each selected weak slot adds +1/2. Try one of each.

Three colour slots

Two weak slots

Choose a slot to begin.

Before0
Selected slot charge0

0 selected · an even pattern in the set below.

0 selected · 1 pattern

2 selected · 10 patterns

4 selected · 5 patterns

One empty pattern + ten pairs + five four-slot patterns = sixteen even choices. These sums are hypercharges in the specified slot dictionary. The full calculation connects them to the particle representations.

Choose the occupied places

The five places are internal labels. Three have colour labels and two have weak labels. Each toggle changes which places are occupied.

Keep the even choices

Keep zero, two or four occupied places. There are one, ten and five such choices, giving sixteen in total.

Add their charges

The display adds the charge contributions of the occupied places. Compare the sum before and after changing your selection.

Explore the full interactive calculation

Build a component of one family

Select occupied colour and weak slots. Their charges add. Choose a complete sector to see where it sits in the sixteen component representation.

Three colour slots · Y = −1/3 each

Two weak slots · Y = +1/2 each

Y = 1 × (−1/3) + 1 × (1/2) = 1/6

Q · Quark doublet · 6 components

Electric charge Qₑ = T₃ + Y = 1/2 + (1/6) = 2/3

Exterior degree 2 · in S⁺ = Λ⁰ ⊕ Λ² ⊕ Λ⁴

Choose a sector, or select the slots above.

Colour slot Weak slot
Λ⁰Empty
1
Λ²Pairs
10
Λ⁴Four slots
5
1 + 10 + 5 = 16 components in one family
The even exterior representation of the given five slot carrier. These are internal components.

The outlined selection locates this component in the whole family. The superscript c denotes the conjugate field convention.

See how the whole family contributes 120

For X = 6Y, each sector contributes its number of components times X². Changing the selected component above highlights its sector here. The full family sum stays fixed.

Tr X² over all sixteen components
SectorComponentsX = 6YContribution
νᶜ101 × 0² = 0
uᶜ3-43 × (-4)² = 48
Q616 × 1² = 6
eᶜ161 × 6² = 36
L2-32 × (-3)² = 18
dᶜ323 × 2² = 12
Total16120
What is assumed in this demonstration?

This reader starts with the five dimensional carrier, its 3+2 marking, the normalized hypercharge operator and the even exterior representation. It demonstrates the charge readout. The source and physical selection of that marking are separate questions.

Inspect the charge claims and their inputs
Apply the operations that build the representation

Try the operation that builds the spinor.

c₁c₂c₃w₁w₂

c₁ ∧ w₁

Start with c₁ and w₁ occupied: an even degree two basis state.

Each insertion or removal changes parity. A pair of these operations returns to the original parity. The even 16 dimensional space is preserved by the quadratic spin action, rather than by each individual gamma operator.

Single insertion or removal flips even and odd degree. Quadratic spin generators preserve parity. The even sixteen dimensional subspace is the specified D₅ half spinor.

Where do the three family directions enter?
1i−1−i
Read loops around the marks.

On the punctured sphere, going around all four punctures is homologically trivial. Thus:

ℓ₁ + ℓᵢ + ℓ₋₁ + ℓ₋ᵢ = 0

One relation among four loops leaves three independent cycles. This fixes rank H₁ = 3.

Read functions with simple poles.

Allow a simple pole at each marked point and require regularity elsewhere, including infinity. Every such function has the form:

f(z)=b0+∑a∈Dbaz−af(z)=b_0+\sum_{a\in D}\frac{b_a}{z-a}

One constant and four independent pole coefficients give five modes.

An explicit basis of H⁰(ℙ¹, O(D)):

  • 11
  • 1z−1\frac{1}{z-1}
  • 1z−i\frac{1}{z-i}
  • 1z+1\frac{1}{z+1}
  • 1z+i\frac{1}{z+i}
Which pairing turns the three directions into A₃?

The family dictionary supplies the residue and Cartan pairing. In four formal coordinate directions, the differences e₁ − e₂, e₂ − e₃ and e₃ − e₄ span the sum zero subspace. Their Gram matrix is:

(2−10−12−10−12),det⁡=4\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix},\qquad\det=4

This is A₃: its twelve roots are eᵢ − eⱼ for i ≠ j. The eᵢ are formal basis vectors. This specified pairing is distinct from geometric chord distances between the complex marks. Three cycles alone would not determine it.

Read the family dictionary in its source context

These are two exact invariants of the same divisor D = {1, i, −1, −i}. Identifying this five dimensional function space with the physical, marked 3 + 2 carrier is the conditional carrier reading. Riemann–Roch does not assign colour to particular poles or derive the marked split.

Read the exact scope of the carrier reading in v189

How this fits into the whole theory

These five internal places supply the carrier used by the E₈ and alpha calculations. The three colour labels and the three family directions have different jobs.

What this experiment establishes

The charge reader assumes the marked carrier, normalized hypercharge and even exterior representation. Anomaly cancellation checks that spectrum; a full interacting chiral realization also requires its physical fields, source and continuum construction.