Minimal Right-Handed Singlet Candidate and the J^{01} Tension

Purpose

The previous screening note showed that

is still too small:

• it contains only weak doublets;
• it cannot literally supply the right-handed singlet completion;
• it still leaves J^{01} invisible in hypercharge once the left-handed T1 seed is fixed.

So the next finite question is:

what is the smallest algebraic repair that can even produce weak singlets?

This note answers that question and then checks its hypercharge consequences.

Minimal repair: add one more weak doublet factor

Under the current K = U(1) x SU(2) reading, the obstruction in the previous note came from one simple fact:

• T1 and T2 are both weak doublets;
• tensoring them only with color/internal factors on which the weak SU(2) acts trivially can never produce weak singlets.

So the smallest repair is to add one more SU(2) doublet factor, call it

At the purely algebraic level this is the first place where

can create weak singlets.

The current corpus already contains a natural parent-side source candidate for such a doublet factor: the visible quaternionic carrier S_{\mathrm{vis}} \cong \mathbf C^2_u inside the local slice H(u,v). This note does not claim that a second such factor is already physically derived. It only uses the representation algebra.

The candidate carrier

Take

with S_{\mathrm{aux}} neutral under J^{01} and Q7, and transforming as a weak doublet under the same SU(2).

Since

one gets

\(T1 \otimes S_{\mathrm{aux}} = (\mathbf 1,-1/2) \oplus (\mathbf 3,-1/2),\) \(T2 \otimes S_{\mathrm{aux}} = (\mathbf 1,+1/2) \oplus (\mathbf 3,+1/2).\)

So the weak-singlet part of \mathcal H_{\mathrm{cand}} is

This is the first carrier in the current static line that has the right color/weak shape for

• one down-type color singlet slot,
• one up-type color singlet slot,
• one charged lepton singlet slot,
• one neutral lepton singlet slot.

So one extra weak doublet factor is the minimal algebraic ingredient needed to create right-handed-style singlets at all.

Hypercharge on the singlet candidate

Keep the same hypercharge ansatz

with the same natural SU(3)-invariant traceless grading on \mathbf 3 \oplus \mathbf 1,

On the weak-singlet sector of \mathcal H_{\mathrm{cand}}, the four charge values are:

\(Y(\mathbf 3,\mathbf 1)_{-1/2} = -\frac a2 + \frac b3,\) \(Y(\mathbf 3,\mathbf 1)_{+1/2} = +\frac a2 + \frac b3,\) \(Y(\mathbf 1,\mathbf 1)_{-1/2} = -\frac a2 - b,\) \(Y(\mathbf 1,\mathbf 1)_{+1/2} = +\frac a2 - b.\)

Now match the standard one-generation right-handed target

\(d_R : (\mathbf 3,\mathbf 1)_{-1/3}, \qquad u_R : (\mathbf 3,\mathbf 1)_{+2/3},\) \(e_R : (\mathbf 1,\mathbf 1)_{-1}, \qquad \nu_R : (\mathbf 1,\mathbf 1)_0.\)

If we identify the -1/2 singlet branch with (d_R,e_R) and the +1/2 branch with (u_R,\nu_R), the equations are

Subtracting gives

and then averaging gives

The lepton singlets then come out automatically:

So this candidate does exactly realize the standard right-handed singlet hypercharges with

Up to the global orientation reversal swapping T1 <-> T2, this is the unique fit.

This is the first place in the static line where J^{01} becomes genuinely nontrivial and useful.

Comparison with the left-handed seed

This is where the next obstruction appears.

The earlier left-handed calculation on the bare seed

forced

But the minimal right-handed singlet candidate forces

up to the same global sign flip on a.

So:

• the Q7 normalization is stable;
• the J^{01} coefficient is not.

Therefore a single global coefficient pair (a,b) cannot simultaneously fit:

1. the naive bare left-handed seed T1 \otimes (\mathbf 3 \oplus \mathbf 1), and
2. the minimal singlet candidate (T1 \oplus T2) \otimes S_{\mathrm{aux}} \otimes (\mathbf 3 \oplus \mathbf 1).

That is the new tension.

What this means

The result is narrower than “the hypercharge ansatz fails.” In fact it shows something better:

1. the right-handed singlet sector is no longer mysterious at the representation-theory level;
2. one extra weak doublet factor is enough to make J^{01} enter with exactly the right charge splitting;
3. the real problem has moved to unifying the left-handed and right-handed embeddings under one global hypercharge operator.

So the live structural alternatives are now clear:

• the left-handed seed should also be embedded in a larger carrier before matching;
• or the hypercharge ansatz must be enlarged beyond a J^{01} + b Q7;
• or the present identification of the bare left-handed seed is only a partial projection of a larger static object.

What is now established

The following points are now closed within the current algebraic line:

1. one extra weak doublet factor is the minimal algebraic ingredient needed to produce weak singlets from the present T1/T2 framework;
2. on that minimal singlet candidate, the standard right-handed one-generation charges are reproduced exactly by \(Y = J^{01} + \frac12\,Q7\) up to the global orientation reversal J^{01} -> -J^{01};
3. the coefficient b = 1/2 is consistent with the earlier left-handed fit, but the coefficient of J^{01} is not;
4. so the first serious hypercharge problem is no longer “can J^{01} ever matter?” It can. The real problem is “how are the left-handed and right-handed sectors embedded so that one global Y works?”

What remains open

This note does not prove that S_{\mathrm{aux}} is already physically present in the framework. It only identifies the smallest kind of extra structure that would work algebraically.

So the next static/consistency question is now:

what is the smallest unified carrier containing both the left-handed doublet seed and the right-handed singlet candidate on which one global hypercharge operator is well defined?

That is the right next blocker.