The Growth Direction and the Chiral Choice in the E
6
of the
Selection–Stitch Model
Raghu Kulkarni
∗
SSMTheory Group, IDrive Inc., Calabasas, CA 91302, USA
October 2026
Abstract
In the Selection–Stitch Model, the vacuum is a face-centered-cubic lattice, and matter sits in
its tetrahedral voids. A recent paper showed that labeling each matter weight by the orientation
of its void, up or down, builds the 72 roots of E
6
. The result contains trinification, su(3)
C
⊕
su(3)
L
⊕ su(3)
R
. The orientation flip F fixes the color factor and swaps the other two. That
paper fixed the electric charge Q. It left open which factor holds the weak su(2), and so what
the hypercharge is. This paper answers the algebraic part of that question and states what a
physical answer still needs. (1) Of the twelve roots that commute with color, exactly four raise
Q by one unit, two in each of the swapped factors. If the weak su(2) is generated by a root,
a hypercharge Y = Q − T
3
that commutes with color and with the weak su(2) can be built in
only two ways, up to relabeling. (2) The choice in the factor that acts on the charge-
2
3
quarks
gives the hypercharges of one Standard Model generation, plus the usual extra states of an E
6
27, with no free parameter. (3) The flip swaps the two choices and maps the weights of the
27 onto those of the conjugate 27. The choice is a single bit, and nothing symmetric under
the flip can set it. A hop of a defect applies the flip. On the quarks and the charged lepton,
it reproduces the charge differences of a Dirac mass term. (4) Reversing the direction in which
the lattice grew acts on the roots exactly as the flip. If the growth direction leaves a lasting
record that couples to the weak sector, its sign could set the bit. Whether such a record exists
is open, because the finished static crystal does not keep one. The paper states the condition
a selection mechanism must meet. It needs a lasting quantity that changes sign under the flip
and couples to the weak sector. Also open are the dynamics of the unused factor, the source
of the Higgs quantum numbers, and the link between void orientation and Lorentz chirality. A
table gives the status of each claim, and a public script checks every computed one.
Keywords: E
6
; trinification; chirality; hypercharge; parity violation; left–right symmetry; orien-
tation flip; Selection–Stitch Model
1 Introduction
The weak interaction acts only on left-handed fermions. Why it does so has been an open question
since parity violation was proposed and observed [1, 2]. Left–right symmetric models restore parity
at high energy and break it through the vacuum values of scalar fields [3, 4, 5]. Trinification,
SU (3)
C
×SU(3)
L
×SU(3)
R
inside E
6
[6, 7, 8, 9], has the same feature. A discrete symmetry swaps
L and R, and something must break it.
∗
ORCID: 0009-0008-8364-7155. Email: raghu@idrive.com
1
In the Selection–Stitch Model (SSM), the vacuum is a face-centered-cubic (FCC) lattice. It
grows by in-plane stitches and rare out-of-plane lifts [10], it carries a stabilizer code [11], and it
holds matter as nodes trapped in tetrahedral voids [10, 12]. A recent paper [13] showed that the
two orientations of these voids cannot be told apart in the static crystal, but become a real label
on the path of a moving defect. Giving each matter weight of the vacuum’s f
4
two copies, one per
orientation, gives the root system of E
6
. The orientation flip F reverses the label and folds e
6
back
to f
4
. It fixes one su(3), taken as color, and swaps the other two. Three conditions pick out a
charge Q, and on the 27 it gives the Standard Model charges [13].
That paper ended with one open question, the chiral projection. Its comparison with one Stan-
dard Model generation listed the weak doublets as “absent (kinematic home only)” and hypercharge
as “kinematic: Y = Q−T
3L
awaits the chiral projection” [13]. This paper takes up that question in
two parts. What exactly must be chosen to place the weak su(2) and fix hypercharge? And what
in the model makes the choice?
Main results. The logic runs in one direction. The earlier paper fixes Q [13]. This paper
classifies where the weak su(2) can sit, and the chosen place then fixes Y . (i) Of the twelve roots of
e
6
that commute with color, exactly four raise Q by one unit, two in each of the swapped factors.
If the weak su(2) is generated by a root, a hypercharge Y = Q −T
3
that commutes with color and
with the weak su(2) can be built in only two ways, up to a Weyl relabeling (Theorem 1). (ii) The
choice in the factor that acts on the charge-
2
3
quarks gives the hypercharges of one Standard Model
generation, plus the usual E
6
extra states, with no free parameter. The anomalies of the 27 cancel
(Theorem 2 and Proposition 1). (iii) The flip swaps the two choices and maps the weights of the
27 onto those of the 27. The choice is one bit, and the static crystal, which is symmetric under the
flip, cannot set it (Theorem 3 and Section 5). A hop of a defect applies the flip. On the quarks and
the charged lepton it reproduces the charge differences of a Dirac mass term (Proposition 2). (iv)
Reversing the direction in which the lattice grew acts on the roots exactly as the flip (Theorem 4).
This is a statement of covariance. It says which placement goes with which growth direction. It
becomes a selection mechanism only if the growth direction leaves a lasting physical record that
couples to the weak sector, and that is open (Section 6).
What is standard and what is new. Trinification, the 27 of E
6
, its hypercharges, its
anomaly freedom, and its left–right partners are standard parts of unified model building [6, 14, 15].
They are not claimed as new. What is new is how they arise here. The swapped factors come from
the up and down orientations of tetrahedral voids. The charge comes from the earlier construction,
and the choice between the factors is tied to the flip of those orientations and to the growth direction
of the lattice.
Scope. The results are statements about root systems, one representation, and a lattice ge-
ometry. No gauge dynamics is built. The paper does not explain why the charged bosons of the
unused factor are heavy or what supplies the Higgs quantum numbers. It does not show that void
orientation is Lorentz chirality, and it does not build the left-handed charged-current coupling of
the Standard Model. Physical names are labels for algebraic objects, as in Ref. [13]. Each claim
is marked derived (proved), computed (evaluated by the script), argued, or open. A public script
checks every computed claim (see Data Availability).
2 The Construction Used
This section lists, with explicit vectors, everything taken from Ref. [13]. The 72 roots live in
R
4
bond
⊕ R
2
class
. The 24 roots ±e
i
± e
j
of D
4
are the vacuum bonds. Each of the 24 short matter
weights s of f
4
is lifted twice, to (s, ϵ v
[s]
) with ϵ = ±1. Here v
[s]
is one of three unit vectors at
2
120
◦
, one for each of the three nonzero classes of D
∗
4
/D
4
, and the sign ϵ is the void orientation.
The orientation flip and the extended inversion are
F = diag(1, 1, 1, 1, −1, −1), X = diag(−1, −1, −1, 1, 1, −1), (1)
and both map the 72 roots to themselves. With the ordering of Ref. [13], the simple roots are
α
1
= (0, 1, −1, 0; 0, 0), α
2
= (0, 0, 1, −1; 0, 0),
α
3
= (
1
2
, −
1
2
, −
1
2
,
1
2
; −
1
2
,
√
3
2
), α
4
= (
1
2
, −
1
2
, −
1
2
,
1
2
;
1
2
, −
√
3
2
),
α
5
= (−
1
2
,
1
2
,
1
2
,
1
2
; −
1
2
, −
√
3
2
), α
6
= (−
1
2
,
1
2
,
1
2
,
1
2
;
1
2
,
√
3
2
),
(2)
with marks (2, 3, 2, 2, 1, 1). Removing the mark-3 node of the extended diagram leaves three or-
thogonal A
2
systems. Color is the one fixed by the flip, with roots ±(e
2
− e
3
), ±(e
1
+ e
2
) and
±(e
1
+ e
3
). The other two are generated by {α
4
, α
6
} and {α
3
, α
5
}, and F swaps them. The charge
chosen in Ref. [13] is
Q =
2
3
(−e
1
+ e
2
+ e
3
). (3)
It commutes with color, is odd under X, and is even under F . On the 27 it takes the values +
2
3
(3
states), −
1
3
(6), +
1
3
(6), −
2
3
(3), +1 (2), 0 (5) and −1 (2). We use the 27 whose weights are listed
in Table A3 of Ref. [13]. In it, the color triplet carries charge +
2
3
.
Naming the two factors. Reference [13] named the swapped pair su(3)
L
and su(3)
R
by
convention. Here the 27 fixes the names. su(3)
L
is the factor that acts on the color triplet of
charge +
2
3
, and su(3)
R
is the other one. With the simple roots of Eq. (2), su(3)
L
is generated by
α
4
and α
6
, and su(3)
R
by α
3
and α
5
. Under (su(3)
C
, su(3)
L
, su(3)
R
) the 27 splits into three blocks
of nine states. The quark block is colored and charged under L. The antiquark block is colored
and charged under R. The color-singlet block is charged under both. [computed]
Fields and conjugates. We use the usual convention [14]. The 27 holds left-handed Weyl
fields, so the conjugate fields u
c
, d
c
and e
c
appear in it. Under the left placement of Section 4 they
have Y = −
2
3
, +
1
3
and +1. Their charge conjugates are the right-handed fields u
R
, d
R
and e
R
,
which transform in the 27. The two representations are not equivalent. Write W
27
for the set of
weights of the 27. The weights of the 27 are then −W
27
. In this paper, Lorentz chirality enters
only through this field convention. That the void orientation is Lorentz chirality is not shown
(Section 8).
3 Exactly Two Ways to Place the Weak Algebra
A hypercharge of Standard Model form needs three things: a weak su(2) that commutes with color,
its diagonal generator T
3
, and a diagonal element Y = Q − T
3
that commutes with both color and
the weak su(2). The last condition makes the charge split as Q = T
3
+ Y , with Y the same on all
members of a weak multiplet. Take an su(2) generated by a root β, with T
3
(µ) = µ · β/2. The
condition then reads Y · β = 0, that is, Q ·β = β · β/2 = 1. The raising generator must raise Q by
exactly one unit, so the charged weak bosons carry charge ±1.
Theorem 1 (Two placements). Let the weak su(2) be generated by a root β, with T
3
(µ) = µ ·β/2.
Twelve roots of e
6
commute with color, the six of su(3)
L
and the six of su(3)
R
. Exactly four of
them satisfy Q ·β = 1. Two are in su(3)
L
, namely α
6
and −α
4
, and two are in su(3)
R
, namely α
5
and −α
3
. For each of the four, Y = Q −β/2 commutes with color and with the su(2) of β. For the
four roots with Q ·β = 0, Q −β/2 does not commute with that su(2). The two choices in su(3)
L
are
related by the Weyl reflection in α
4
+ α
6
, which fixes Q. The same holds in su(3)
R
with α
3
+ α
5
.
3
α
4
Q·β = −1
α
6
Q·β = +1
α
4
+ α
6
Q·β = 0
−α
4
Q·β = +1
−α
6
Q·β = −1
−(α
4
+ α
6
)
Q·β = 0
su(3)
L
α
3
Q·β = −1
α
5
Q·β = +1
α
3
+ α
5
Q·β = 0
−α
3
Q·β = +1
−α
5
Q·β = −1
−(α
3
+ α
5
)
Q·β = 0
su(3)
R
F
F α
4
= α
3
F α
6
= α
5
Figure 1: The twelve roots that commute with color, drawn as the two A
2
root systems of su(3)
L
and su(3)
R
. Each root is labeled with its value of Q · β. Only the four filled roots raise Q by one
unit, so only they can generate a weak su(2) with charged bosons of charge ±1 (Theorem 1). The
two in each factor are related by a Weyl reflection. The flip F maps α
4
to α
3
and α
6
to α
5
, so it
swaps the two pictures (Theorem 3).
Proof. A root that commutes with color is orthogonal to the color roots, so its bond part is or-
thogonal to e
2
− e
3
and e
1
+ e
2
. Checking the 72 roots, these are exactly the twelve roots of
su(3)
L
⊕ su(3)
R
. The values of Q on the simple roots are (0,
2
3
, −1, −1, 1, 1) [13]. In su(3)
L
this
gives Q ·α
4
= −1, Q ·α
6
= 1 and Q ·(α
4
+ α
6
) = 0. The roots with Q ·β = 1 are then α
6
and −α
4
,
and su(3)
R
works the same way. If Q·β = 0, then (Q −β/2)·β = −1 = 0. The reflection in α
4
+α
6
maps α
6
to −α
4
, and it fixes Q because Q · (α
4
+ α
6
) = 0. The script checks each statement.
The two Weyl-related choices in one factor give the same set of hypercharge values. The
reflection swaps the two charge-(−
1
3
) triplets of the quark block. In the color-singlet block it moves
six states. It swaps the two neutral singlets with the neutral members of the two doublets with
Y = −
1
2
, and it swaps e
c
with the charged member of the doublet with Y = +
1
2
. Ma found the
same kind of ambiguity in the right factor [16]; this is its counterpart in the left factor. It does not
affect anything below. Up to this relabeling, there are exactly two placements. The weak su(2) sits
in su(3)
L
, giving Y
L
= Q − T
3L
, or in su(3)
R
, giving Y
R
= Q − T
3R
. [derived]
The classification is conditional. It assumes the charge Q and the color factor of Ref. [13], and
a weak su(2) generated by a single root. It is not a classification of every possible weak embedding.
Figure 1 shows the twelve roots and their values of Q · β.
4 The Left Placement Gives One Generation
Theorem 2 (The spectrum of Y
L
). Take β = α
6
, T
3L
(µ) = µ · α
6
/2 and Y
L
= Q − α
6
/2. On the
27, with no free parameter:
1. The quark block holds, for each color, one weak doublet with charges (
2
3
, −
1
3
) and Y
L
=
1
6
, and
one weak singlet with Q = Y
L
= −
1
3
.
4
Table 1: The 27 under the left placement. T
3L
and Y
L
are computed from Eq. (3) and β = α
6
.
The field names are the standard ones for a 27 of E
6
[15].
Block States Q T
3L
Y
L
Fields
quark 3 + 3
2
3
, −
1
3
+
1
2
, −
1
2
1
6
quark doublet (u, d)
3 −
1
3
0 −
1
3
singlet quark D
antiquark 3 −
2
3
0 −
2
3
u
c
6 +
1
3
0 +
1
3
d
c
, D
c
color singlet 2 + 2 0, −1 +
1
2
, −
1
2
−
1
2
lepton doublet, second doublet
1 + 1 +1, 0 +
1
2
, −
1
2
+
1
2
doublet with Y = +
1
2
1 +1 0 +1 e
c
2 0 0 0 two neutral singlets
2. The antiquark block holds only weak singlets, with Y
L
= Q. The value is −
2
3
for three states
and +
1
3
for six.
3. The color-singlet block holds two weak doublets with Y
L
= −
1
2
, one with Y
L
= +
1
2
, and three
singlets with Y
L
= +1, 0 and 0.
In all, the 27 holds six weak doublets and fifteen singlets, and Q = T
3L
+ Y
L
on every state.
Proof. Direct evaluation of Q, T
3L
and Y
L
on the 27 weights. The script prints each value.
Table 1 and Figure 2(a) compare the result with the fields of one generation [14, 15]. The Stan-
dard Model states appear with their usual hypercharges. The quark doublet has
1
6
, the antiquarks
have −
2
3
and
1
3
, the lepton doublet has −
1
2
, and the charged antilepton has 1. The other states are
the usual extra states of a 27. They are a weak-singlet quark D of charge −
1
3
and its antiquark, a
pair of doublets with Y = ∓
1
2
that can carry Higgs quantum numbers, and two neutral singlets [15].
[computed]
Proposition 1 (Anomalies). For one left-handed 27 with Y = Y
L
, the four anomaly sums that
involve hypercharge vanish:
P
Y = 0,
P
Y
3
= 0,
P
colored
Y = 0 (the SU(3)
2
Y sum), and
P
T
2
3L
Y = 0 (the SU(2)
2
Y sum).
Proof. From Table 1,
P
Y = 6·
1
6
−3·
1
3
−3·
2
3
+6·
1
3
−4·
1
2
+2·
1
2
+1 = 0. The other sums follow in the
same way, and the script evaluates all four in exact arithmetic. The result is expected, because E
6
has no cubic anomaly [17] and the Standard Model group sits inside it. The computation confirms
that the placement of Theorem 2 is such an embedding.
The other anomaly conditions hold too. The colored states of the 27 form three triplets and
three antitriplets, so the SU (3)
3
anomaly cancels. The 27 holds six weak doublets, an even number,
so the global SU(2) anomaly is absent. The mixed gravity–hypercharge anomaly is proportional
to
P
Y , the first sum of Proposition 1. [computed]
No choice of hypercharge is made. Once the weak su(2) sits in su(3)
L
, Q fixes Y
L
, and Q was
fixed in Ref. [13] before the Standard Model spectrum was used. This confirms the claim of Ref. [13]
that Y = Q − T
3L
follows once the weak group is fixed. [derived]
5
Table 2: The flip on two weights. Each µ is a left-handed field of the 27. Its image F µ equals −ν,
where ν is a weight of the 27 with opposite charges. So F µ is a weight of the 27, and it carries the
charge, color and hypercharge of the right-handed partner of µ. Charges are listed as (Q, T
3L
, Y
L
).
All charges, including those of F µ, are evaluated with the fixed generators Q, T
3L
and Y
L
of the left
placement. Since F is an orthogonal involution, Y
L
(F µ) = (F Y
L
) · µ = Y
R
(µ), so the last column
also gives Y
R
of the original weight.
Field Weight µ Charges of µ ν = −F µ in the 27 Charges of ν F µ is the
weight of
u
L
(−1, 0, 0, 0; 0,
√
3
3
) (
2
3
,
1
2
,
1
6
) u
c
: (1, 0, 0, 0; 0,
√
3
3
) (−
2
3
, 0, −
2
3
) u
R
, with
Q =
2
3
, Y
L
=
2
3
e
L
(
1
2
, −
1
2
, −
1
2
,
1
2
; −
1
2
, −
√
3
6
) (−1, −
1
2
, −
1
2
) e
c
: (−
1
2
,
1
2
,
1
2
, −
1
2
; −
1
2
, −
√
3
6
) (1, 0, 1) e
R
, with
Q = −1,
Y
L
= −1
5 The Flip Swaps the Two Placements
Theorem 3 (One bit). The orientation flip maps the left placement onto the right one. The root
F α
6
lies in su(3)
R
and has Q · F α
6
= 1, and F Y
L
= Y
R
. On weights, F (W
27
) = −W
27
, the weight
set of the 27. Under the right placement, with T
3R
(µ) = µ · F α
6
/2, the quark block holds no weak
doublet, and the antiquark block holds six doublet states. No hypercharge of Standard Model form
is unchanged by F.
Proof. F swaps su(3)
L
and su(3)
R
, fixes color, and fixes Q, because Q has no component in the
class plane. Applied to Y
L
= Q −α
6
/2, it gives Q −F α
6
/2, the hypercharge of the right placement.
The image under F of each weight of the 27 is minus a weight of the 27, and it is not itself a
weight of the 27. So F maps W
27
onto −W
27
, the weights of the 27. By Theorem 1 there are
four hypercharges of Standard Model form, two for each placement. F maps the two of the left
placement onto the two of the right placement, and these differ, so F fixes none of the four. The
script checks each statement.
Two examples. F keeps Q, while conjugation reverses every charge. The two facts fit together
because F µ = −ν for a weight ν of the 27 whose charges are opposite to those of µ. Table 2
shows this for one quark and one lepton. Take the up quark u
L
of one color. Its weight µ =
(−1, 0, 0, 0; 0,
√
3
3
) is the first weight in Table A3 of Ref. [13]. Its image is F µ = (−1, 0, 0, 0; 0, −
√
3
3
).
This equals −ν, where ν = (1, 0, 0, 0; 0,
√
3
3
) is the weight of u
c
in the 27, with Q = −
2
3
and the
conjugate color. So F µ is the weight of the charge conjugate of u
c
, which is u
R
. It has Q = +
2
3
,
the color of u
L
, and Y
L
= +
2
3
. The electron works the same way. The weight of e
L
goes to minus
the weight of e
c
, which is the weight of e
R
, with Q = −1 and Y
L
= −1. In both cases F keeps
charge and color and takes a left-handed field of the 27 to the right-handed field of the same charge,
which sits in the 27. Here “left-handed” and “right-handed” are the labels of the field convention
of Section 2. [computed]
These charges, and those of Proposition 2 below, are evaluated with one fixed set of generators,
Q, T
3L
and Y
L
. Moving the weight and keeping the generators is the same as keeping the weight
and moving the generators. Since F is an orthogonal involution, Y
L
(F µ) = Y
R
(µ) for every weight
µ. The statement F Y
L
= Y
R
of Theorem 3 and the charges in Table 2 are two views of the same
fact. [computed]
6
The right placement is the mirror image of the left one. With the weak su(2) in su(3)
R
, the
charge-
2
3
quarks of the 27 are weak singlets, and the weak doublets sit in the antiquark block.
The weak interaction would then act on u
c
and d
c
, whose charge conjugates are the right-handed
fields u
R
and d
R
. The chiral projection is a single yes-or-no choice, between the left and the right
placement. [derived]
Nothing symmetric under F can make this choice. In the static crystal, the two void orientations
are exact inversion partners (Theorem 1 of Ref. [13]), and the algebra built there is f
4
, which the
flip leaves unchanged. Spatial inversion acts on the full algebra as F X, not as F (Section 8.3 of
Ref. [13]). Since X preserves su(3)
L
, F X also swaps the two factors. It maps (Q, Y
L
) to (−Q, −Y
′
R
),
where Y
′
R
is a hypercharge of the right placement. Inversion cannot make the choice either. No
static structure that is symmetric under inversion can prefer Y
L
to Y
R
. Whatever sets the bit must
change sign under F. [derived]
The flip on a moving defect. The flip also acts along the path of a defect. A trapped node
moves by hopping between neighboring voids, and each hop reverses its orientation label [18]. So
each hop applies F [13]. Reference [13] used this to require that a charge conserved along the path
be even under F , and Q is. Theorem 3 shows that Y
L
and T
3L
are not. The next result says what
a hop does to them.
Proposition 2 (A hop). On every state of the 27, F keeps Q and the color weight. It shifts Y
L
and T
3L
by opposite amounts, ∆Y
L
= −∆T
3L
. The shift is ±
1
2
on 16 states, 0 on 9, and ±1 on
2. The doublet quark of charge
2
3
goes to Y
L
=
2
3
, and the doublet quark of charge −
1
3
goes to
Y
L
= −
1
3
. One lepton of charge −1 goes from Y
L
= −
1
2
to Y
L
= −1. Of the two doublets with
Y
L
= −
1
2
, exactly one has both members shifted by
1
2
. In the other, the shifts are 0 and 1.
Proof. F fixes Q and the color roots, so it keeps Q and the color weight. Then ∆T
3L
= −∆Y
L
follows from Q = T
3L
+ Y
L
. The script evaluates the counts and values.
The hop picks out the lepton doublet as the doublet whose two members both shift by
1
2
. The
values reached,
2
3
, −
1
3
and −1, are the hypercharges of u
R
, d
R
and e
R
. In the convention above,
these fields sit in the 27, which is where F sends the weights. So a hop takes the weight of each
of these left-handed states to a weight with the color, charge and hypercharge of its right-handed
partner. The change in (T
3
, Y ) is (−
1
2
, +
1
2
) for u and (+
1
2
, −
1
2
) for d and e. In the Standard Model,
Dirac masses come from the gauge-invariant Yukawa terms
L
Y
= −y
u
Q
L
e
Hu
R
− y
d
Q
L
H d
R
− y
e
L
L
H e
R
+ h.c., (4)
where H is the Higgs doublet with Y = +
1
2
and
e
H = iσ
2
H
∗
has Y = −
1
2
. When the neutral
member H
0
of H, with (T
3
, Y ) = (−
1
2
, +
1
2
), takes its vacuum value, each term links f
L
to f
R
.
The fermion’s (T
3
, Y ) then changes by (−
1
2
, +
1
2
) in the u term and by (+
1
2
, −
1
2
) in the d and e
terms, and the vacuum absorbs the difference. These are the shifts found in Proposition 2. The 27
contains doublets with Y = ±
1
2
that could carry them (Table 1). This paper builds no such term,
and it does not say which degree of freedom of the lattice supplies them. The hop result is a match
to the charge differences of a Dirac mass term. It is not a derived mass mechanism. [computed; the
comparison is argued] Figure 2(b) shows the hop for the quarks and the lepton doublet.
Theorem 3 and Proposition 2 agree with the condition used in Ref. [13]. A moving defect
conserves Q, as a massive fermion does below the electroweak scale. In general it does not conserve
Y or T
3
on their own. The bit and the orientation of a defect are also different things. The bit
fixes, once, which factor holds the weak su(2). The orientation of a moving defect changes at every
hop, much as a Dirac mass term mixes the two chiralities of a massive fermion. [argued] Which
degree of freedom of the lattice supplies the missing (T
3
, Y ) is open. [open]
7
−
1
2
0
1
2
T
3L
−1
−
2
3
−
1
3
0
1
3
2
3
1
Y
L
3
3
3
6
3
2
1
2
1
1
2
(a) the 27 under Y
L
gray lines: constant Q
quark block
antiquark block
color-singlet block
−
1
2
0
1
2
T
3L
e
L
e
R
u
L
u
R
d
L
d
R
ν
L
S
(b) a hop applies F
open: µ; filled: F µ
arrows run along constant Q
Figure 2: (a) The 27 in the plane of T
3L
and Y
L
, with the number of states at each point. Gray
lines are lines of constant Q = T
3L
+ Y
L
. Weak doublets are the pairs at T
3L
= ±
1
2
with equal
Y
L
(Theorem 2, Table 1). (b) A hop applies F . Open circles are weights µ of left-handed fields,
and filled circles are their images F µ, evaluated with the same generators. Each arrow keeps Q
and changes (T
3
, Y ) by one half in opposite directions. The quarks and the electron reach the
hypercharges of u
R
, d
R
and e
R
. The neutral member of the lepton doublet reaches a neutral singlet
S with Y
L
= 0 (Proposition 2).
6 The Growth Direction and the Bit
The orientation of a tetrahedral void is defined relative to the direction in which the lattice grew.
The vacuum is built by in-plane stitches within a close-packed sheet and by rare lifts that start the
next sheet on one side only, so growth has a direction, normal to the sheets [10]. An up-void has
its apex along that direction, and a down-void has its apex against it [13] (Figure 3).
Theorem 4 (Covariance under the growth direction). Reversing the growth direction relabels every
tetrahedral void, up to down and down to up. In the construction of Section 2, relabeling every
orientation, ϵ → −ϵ, maps the 72 roots exactly as F does. So reversing the growth direction maps
Y
L
to Y
R
.
Proof. Take growth along [111]. A void with three vertices on the lower sheet and one on the upper
sheet has its apex along the growth direction. A void with one vertex below and three above has its
apex against it. Reversing the direction swaps “below” and “above”, so every void changes label.
The script checks this on all 216 voids of a finite patch, 108 of each kind. The lifted roots are
(s, ϵv
[s]
). Replacing ϵ by −ϵ in all of them gives (s, −ϵv
[s]
), which is the image under F , and the
bond roots are unchanged in both cases. Theorem 3 then gives the last statement.
The first statement of the theorem holds by definition, since the void labels are defined relative
to the growth direction. The content is in the second statement. The growth direction enters the
8
sheet A
sheet B
up down
growth
(a) growth up
sheet A
sheet B
down up
growth
(b) growth reversed
Figure 3: Side view, in schematic form, of two neighboring close-packed sheets and the two kinds
of tetrahedral void between them. (a) With growth upward, the void with three vertices on sheet
A and its apex on sheet B is “up”, and the other is “down”. (b) Reversing the growth direction
swaps every label, while the lattice itself is unchanged (Theorem 4).
algebra exactly as the flip does, so it changes sign under F . [derived]
What Theorem 4 does and does not show. Theorem 4 is a covariance statement. It
says that the placement and the growth direction change together. It does not by itself show that
the growth direction selects a placement in nature. For that, the growth direction must leave a
lasting physical record, and the static lattice does not keep one. By Theorem 1 of Ref. [13], the
finished crystal read upward is equivalent, by inversion, to the same crystal read downward. Two
growth histories with opposite directions can end in the same lattice. If nothing else remembers
the direction, reversing it only relabels the same physical state. [derived]
A selection mechanism needs a lasting quantity χ with F : χ 7→ −χ that enters a weak-
interaction operator. The construction offers two candidates, and neither is shown to work. The
first is the growth front. In the model, growth continues at a moving surface [10], and that
surface faces along the growth direction. If the front persists, its normal is a present-day record
of the direction. The second is the time order of events on the path of a defect, which carries the
generator odd under P and even under P T (Section 4 of Ref. [13]). Neither has been shown to give
a χ that couples to the weak sector. [open]
We find no other global quantity in the static construction that changes sign under F . The
bonds form a lattice that is symmetric under inversion. The stabilizer checks of the code sit on
vertices and octahedral voids, a set that is also symmetric under inversion [11]. Trapped matter
occupies voids of both orientations [10]. A single defect in a given void also changes sign under F ,
but its label is defined relative to the growth direction and changes as it moves (Proposition 2).
This claim rests on the list just given. [argued]
Figure 4 sums up the steps. What is derived is the link between the growth direction and the
placement. The opposite direction goes with the mirror placement. The growth rules of Ref. [10]
produce a single-domain crystal, with one stacking order and one growth direction throughout. If
a lasting record of the growth direction sets the bit, the bit has one value everywhere in such a
domain, and the weak interaction has the same handedness at every point. [argued]
7 Relation to Left–Right Symmetric Models
In left–right symmetric models, the discrete symmetry that swaps L and R is exact in the La-
grangian. Scalar fields break it when they take vacuum values [4, 5]. In trinification, the swap of
SU (3)
L
and SU(3)
R
combined with conjugation plays the same role [8, 9]. Theorem 3 shows that
the flip F is a symmetry of this kind, since it swaps the two factors and maps the weights of the
27 onto those of the 27.
The proposed source of the breaking is different here. In left–right models a field settles into one
9
su(3)
C
fixed by F
su(3)
L
weak su(2) ∋ α
6
su(3)
R
mirror: F α
6
flip F (Theorem 3)
Y
L
= Q − T
3L
one generation (Table 1)
Y
R
= Q − T
3R
mirror assignment
growth direction (Theorem 4):
one sign goes with Y
L
, the other with Y
R
Figure 4: The chiral choice. Of the roots that commute with color, only those of the swapped pair
can carry a charged weak su(2) (Theorem 1). The flip swaps the two placements (Theorem 3). The
growth direction, which the flip reverses, goes with one placement for each sign (Theorem 4). A
lasting record of it would select one placement; whether one exists is open.
of two equal vacua. Here the breaking would come from how the vacuum was made, the direction in
which it grew, if that direction leaves a lasting record that couples to the weak sector (Section 6).
This is like the thermodynamic arrow of time, which breaks time-reversal symmetry through a
starting condition, while the laws stay symmetric. If so, the sign of parity violation would be tied
to a large-scale fact shared by the whole domain.
The construction does not supply what left–right models must also supply. The hypercharge
Y
L
has components along the diagonal generators of su(3)
R
, as in every trinification model [8, 9].
That factor contributes to hypercharge, even though its charged bosons must be heavy. How they
become heavy, at what scale, and how the two extra neutral directions of the rank-6 algebra are
broken are questions of dynamics. This paper does not address them. [open]
8 What Is Derived and What Is Not
Table 3 lists each claim with its status. The entries marked argued and open mark the limits of the
result.
A note on chirality. On the path of a defect, the orientation doublet carries a generator that
is odd under P and even under P T (Section 4 of Ref. [13]). A chirality has the same symmetry.
Reference [13] does not identify the label with Lorentz chirality, and neither does this paper.
Chirality and helicity also differ for massive particles; the property needed here is Lorentz chirality.
To explain parity violation, the model must produce the charged-current coupling
L
CC
=
g
√
2
ψ
L
γ
µ
T
+
ψ
L
W
+
µ
+ h.c., (5)
with no light right-handed charged current. Placing states in an su(2) multiplet is needed for
this. It does not give their Lorentz chirality, their gauge coupling, or the absence of right-handed
currents. These are open. The results of Sections 3–6 do not depend on them. They concern which
hypercharge goes with which placement and which growth direction, whatever the label means
physically. How the construction relates to the hypotheses of the Nielsen–Ninomiya theorem [19] is
10
Table 3: Status of each claim. “Derived” means proved from the stated construction. “Computed”
means evaluated by the script.
Claim Status Where
exactly four roots that commute with color raise Q
by one; two placements up to relabeling
derived Theorem 1
Y
L
gives one generation and the E
6
extra states,
with no free parameter
computed Theorem 2, Table 1
the anomalies of the 27 cancel computed Proposition 1, Section 4
F swaps the placements and maps W
27
onto
−W
27
, the weights of the 27
derived Theorem 3
worked examples u
L
→ u
R
and e
L
→ e
R
;
Y
L
(F µ) = Y
R
(µ)
computed Table 2, Section 5
nothing symmetric under F , and not inversion
F X, can set the bit
derived Section 5
a hop keeps Q and color; the quarks and charged
lepton move to right-handed hypercharges
computed Proposition 2
the hop matches the charge differences of a Dirac
mass term
argued Section 5
a hop is a gauge-invariant mass term open Section 5
reversing the growth direction acts as F
(covariance)
derived Theorem 4
a lasting record of the growth direction selects the
placement
open Section 6
no other global quantity of the static construction
is odd under F
argued Section 6
one handedness throughout a single-domain
crystal, if the record exists
argued Section 6
what supplies the missing (T
3
, Y ) in a hop open Section 5
void orientation is Lorentz chirality; left-handed
charged currents with no light right-handed ones
open Section 8
the charged bosons of su(3)
R
are heavy open Section 7
couplings, masses, and the breaking of the two
extra U(1) directions
open Section 7
not analyzed, and no way around that theorem is claimed. Future work should test the construction
against it.
9 Predictions and Falsifiers
Three statements can fail. First, the hypercharges of Table 1 are fixed with no free parameter. A
Standard Model state with a different hypercharge, or quark weak doublets built from the antiquark
block, would rule out the placement. The Standard Model values are known, so this is a consistency
test, but passing it is not automatic. The right placement gives the mirror assignment, and a
hypercharge built on a neutral root fails to commute with its weak algebra (Theorem 1). Second,
a hop must take the weight of each left-handed quark and charged-lepton state to a weight with
11
the color, charge and hypercharge of its right-handed partner (Proposition 2). Any later account
of the Higgs-like vacuum in which a hop changes Q would contradict the construction. Third, if
a lasting record of the growth direction sets the bit, the handedness of the weak interaction is the
same throughout a single-domain crystal. A region of the observable universe where charged weak
currents couple with the opposite handedness would rule out either the single-domain growth of
Ref. [10] or the link made here.
10 Conclusions
The chiral projection left open in Ref. [13] has an algebraic part and a physical part. The algebraic
part is now answered, under the stated assumptions. Only two placements of a root-generated weak
su(2) fit the chosen charge. The left one gives the hypercharges of one Standard Model generation
with no free parameter, and the flip swaps it with its mirror. The physical part comes down to one
bit that nothing symmetric under the flip can set. Reversing the growth direction acts exactly as
the flip, so the growth direction goes with one placement for each sign. It selects a placement only
if it leaves a lasting record that couples to the weak sector, and finding that record is the main
open step. On the quarks and the charged lepton, a hop of a defect matches the charge differences
of a Dirac mass term. Also open are the dynamics of the unused factor, the source of the Higgs
quantum numbers, and the link between void orientation and Lorentz chirality.
Data Availability
All computations are specified in the text. A Python 3 script (NumPy only) that rebuilds the E
6
root system of Ref. [13] and checks every computed claim of this paper, with 33 labeled checks,
and a second script that draws Figures 1–3 from the same computation, are available at https:
//github.com/raghu91302/ssmtheory/raw/main/chiral_choice_scripts.zip.
Funding
This research received no external funding.
Conflicts of Interest
The author is employed by SSMTheory Group, IDrive Inc., Calabasas, USA. The author declares
that the research was conducted in the absence of any commercial or financial relationships that
could be construed as a potential conflict of interest.
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13
