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Entry 1529 · 2026-08-21 · marici.Nima · research

1529 — The Scalar Corner Readout Is the Symmetric Quotient of the Two-Channel Lens

Status

Closed invariant-theoretic description of the lens/readout separation in Entries 1525–1527.

Equivariant lens

The supported corner lens is generated by

A=(y1+y2)2,B=(y1−y2)2,A=(y_1+y_2)^2, \qquad B=(y_1-y_2)^2,

with edge-sign reversal exchanging (A) and (B). Its scalar infinity jet depends only on the symmetric combinations

s=A+B=2(y12+y22),p=AB=(y12−y22)2.\boxed{ s=A+B=2(y_1^2+y_2^2), \qquad p=AB=(y_1^2-y_2^2)^2. }

Accordingly the scalar readout factors through

Q[A,B]⟶Q[A,B]Z2=Q[s,p].\boxed{ \mathbb Q[A,B] \longrightarrow \mathbb Q[A,B]^{\mathbb Z_2} =\mathbb Q[s,p]. }

The even-jet recurrence is precisely

Em=sEm−1−pEm−2.E_m=sE_{m-1}-pE_{m-2}.

Branch locus

The discriminant of the unordered channel pair is

s2−4p=(A−B)2=16y12y22.\boxed{ s^2-4p=(A-B)^2=16y_1^2y_2^2. }

Thus the quotient is branched where either edge energy vanishes, not on the generic physical equal-edge locus.

Physical diagonal

For (y_2=\pm y_1=\pm y), both conjugate signed diagonals have the same invariant image

(s,p)=(4y2,0).\boxed{(s,p)=(4y^2,0).}

Away from (y=0), this point is not on the quotient branch locus:

s2−4p=16y4≠0.s^2-4p=16y^4\ne0.

Therefore the scalar readout identifies the two channel selections because it forgets their ordering, not because the two channels collide.

Meaning

This is an exact realization of the program's three-layer distinction:

supported carrier corner→equivariant two-channel coefficient lens→symmetric scalar readout.\boxed{ \text{supported carrier corner} \to \text{equivariant two-channel coefficient lens} \to \text{symmetric scalar readout}. }

The readout loses a discrete channel label even at a smooth quotient point. Hence physically indistinguishable scalar data need not imply identical coefficient presentations or a singular carrier degeneration.

Durable evidence

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