Unified Carrier Test: The Neutral-S_aux No-Go

Purpose

The previous note identified the first algebraic carrier that can host right-handed-style weak singlets:

with S_{\mathrm{aux}} \cong \mathbf 2.

But that still left the real question open:

can one embed both the left-handed doublet seed and the right-handed singlet sector into one small unified carrier so that a single global \(Y = a J^{01} + b Q7\) works on all of them at once?

The first obvious unified test carrier is

This note computes that case exactly.

Assumptions of the test

This test uses the smallest natural extension of the previous notes:

1. T1 and T2 carry the usual J^{01} eigenvalues -1/2 and +1/2;
2. Q7 is the same SU(3)-invariant traceless grading on \mathbf 3 \oplus \mathbf 1, \(Q7 = \mathrm{diag}\!\left(\frac13,\frac13,\frac13,-1\right);\)
3. S_{\mathrm{aux}} is a weak SU(2) doublet;
4. S_{\mathrm{aux}} is neutral under J^{01} and Q7.

The fourth point is the crucial test assumption. If the no-go below appears, that is where the next modification must enter.

Decomposition of the unified carrier

Because

each T sector contributes:

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

So \mathcal H_{\mathrm{unif}} contains:

• left-handed-shaped weak doublets from the \mathbf 1 part of \mathbf 1 \oplus S_{\mathrm{aux}},
• right-handed-shaped weak singlets from the S_{\mathrm{aux}} singlet channel,
• and extra weak triplets as spectators.

At the level of representation shape alone, this is the first small carrier that contains both types of slots simultaneously.

Left-handed equations

If the left-handed quark/lepton doublets are taken from one fixed J^{01} branch, their charges are either

or

Matching the usual left-handed targets

forces, in either branch choice,

So the left-handed sector again kills the J^{01} contribution if it is embedded as a pure T1 or pure T2 doublet branch.

Right-handed equations

Now take the weak singlets from the S_{\mathrm{aux}} singlet channel.

Their charges are:

\(Y_R^{(-)}(\mathbf 3,\mathbf 1) = -\frac a2 + \frac b3, \qquad Y_R^{(-)}(\mathbf 1,\mathbf 1) = -\frac a2 - b,\) \(Y_R^{(+)}(\mathbf 3,\mathbf 1) = +\frac a2 + \frac b3, \qquad Y_R^{(+)}(\mathbf 1,\mathbf 1) = +\frac a2 - b.\)

To reproduce the standard one-generation right-handed pattern, one branch must carry

and the opposite branch must carry

Those equations force

with the sign determined by which branch is identified as the up-type/neutral branch.

So the right-handed sector requires a nonzero J^{01} coefficient, exactly as in the previous note.

The no-go

Combine the two results:

• left-handed pure-branch doublets force \(a = 0, \qquad b = \frac12;\)
• right-handed singlets force \(a = \pm 1, \qquad b = \frac12.\)

Therefore:

on the unified carrier \((T1 \oplus T2) \otimes (\mathbf 1 \oplus S_{\mathrm{aux}}) \otimes (\mathbf 3 \oplus \mathbf 1),\) with S_{\mathrm{aux}} neutral under J^{01} and Q7, there is no single global coefficient pair (a,b) for which \(Y = a J^{01} + b Q7\) reproduces both the standard left-handed doublet charges and the standard right-handed singlet charges.

This is a clean algebraic no-go.

What fails, exactly

The important point is what the no-go does not say.

It does not say:

• the unified carrier is too small to contain the right representation shapes;
• J^{01} can never help;
• Q7 was the wrong normalization.

In fact:

• the carrier is large enough to contain both doublets and singlets;
• J^{01} is exactly what makes the right-handed singlet fit work;
• b = 1/2 is stable across all successful local fits.

What fails is only this:

the left-handed sector, when embedded as a pure J^{01} branch with neutral auxiliary factor, does not tolerate the same nonzero J^{01} coefficient that the right-handed sector needs.

So the obstruction is now sharply localized.

Consequences

There are three immediate ways the framework could escape this no-go:

1. Enlarge the embedding of the left-handed sector. The bare left-handed doublets may need to live in a larger carrier rather than in the pure \mathbf 1 part of \mathbf 1 \oplus S_{\mathrm{aux}}.

2. Give S_{\mathrm{aux}} extra grading data. If the auxiliary factor carries its own nontrivial charge under a new commuting grading, then Y may need a third term beyond a J^{01} + b Q7.

3. Treat the visible left-handed seed as a projection of a larger operator-level structure. Then the charge fit on the bare seed would no longer be the full global constraint.

The current note does not choose among these. It only shows that the neutral-S_{\mathrm{aux}} unified carrier is not enough.

What is now established

The following points are now closed:

1. the smallest unified carrier containing both left-handed-shaped doublets and right-handed-shaped singlets is \((T1 \oplus T2) \otimes (\mathbf 1 \oplus S_{\mathrm{aux}}) \otimes (\mathbf 3 \oplus \mathbf 1);\)
2. on that carrier, the standard left-handed fit still forces \(a = 0, \qquad b = \frac12;\)
3. on the same carrier, the standard right-handed singlet fit forces \(a = \pm 1, \qquad b = \frac12;\)
4. therefore the neutral-S_{\mathrm{aux}} unified carrier gives a finite no-go for one global \(Y = a J^{01} + b Q7.\)

What remains open

The next blocker is now explicit:

what is the smallest modification of the unified carrier or of the hypercharge operator that preserves the successful b = 1/2 structure while removing the a = 0 versus a = \pm 1 conflict?

That is a much narrower problem than the original hypercharge question.