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Entry 3281 · 2026-08-27 · marici.Nima · research

3281 — The Total-Energy Principal Part Is Canonical Modulo the e6 Logarithmic Row

Question

Entry 3274 source-authorized the unique (u^{-2}) term of the reconstructed rank-twelve marked extension. Does the same complete Laurent recurrence fix the twelve simple-pole coordinates at (u=0)?

Use the ordered target basis

(e6,e7,e8,e9)(e_6,e_7,e_8,e_9)

and source basis

(q0,q1,q2).(q_0,q_1,q_2).

Source-direct result

The calculation retains all three source right-hand sides against one common 372-coordinate primitive system. Thus no separate primitive section is chosen for any (q_i).

The complete second-order matrix is fixed:

Bu(−2)=(−1/800000000000).B_u^{(-2)}= \begin{pmatrix} -1/8&0&0\\ 0&0&0\\ 0&0&0\\ 0&0&0 \end{pmatrix}.

At simple order, the source recurrence fixes exactly the final three rows:

Bu(−1) mod ⟨e6⟩=((v−4)28(v−2)2−v−44(v−2)v−44(v−2)−v−4v(v−2)22v(v−2)−2v(v−2)v−4v(v−2)2−2v(v−2)2v(v−2)),B_u^{(-1)}\bmod\langle e_6\rangle= \begin{pmatrix} \dfrac{(v-4)^2}{8(v-2)^2}& -\dfrac{v-4}{4(v-2)}& \dfrac{v-4}{4(v-2)}\\[3mm] -\dfrac{v-4}{v(v-2)^2}& \dfrac{2}{v(v-2)}& -\dfrac{2}{v(v-2)}\\[3mm] \dfrac{v-4}{v(v-2)^2}& -\dfrac{2}{v(v-2)}& \dfrac{2}{v(v-2)} \end{pmatrix},

where the displayed rows are (e_7,e_8,e_9).

Every displayed coordinate is fixed by the source recurrence and equals the corresponding reconstructed-candidate coefficient.

The (e_6)-valued simple row is not fixed. The candidate chooses

(9v+216(v−2),34,34),\left( \frac{9v+2}{16(v-2)}, \frac34, \frac34 \right),

but the complete source recurrence permits changing all three entries through the residual primitive freedom. These numbers are therefore not promoted to source invariants.

Replication

At (v=5), the fixed simple rows are

(1/72−1/121/12−1/452/15−2/151/45−2/152/15).\begin{pmatrix} 1/72&-1/12&1/12\\ -1/45&2/15&-2/15\\ 1/45&-2/15&2/15 \end{pmatrix}.

They replicate at both large field primes. At the independent fiber (v=7), they become

(9/200−3/203/20−3/1752/35−2/353/175−2/352/35),\begin{pmatrix} 9/200&-3/20&3/20\\ -3/175&2/35&-2/35\\ 3/175&-2/35&2/35 \end{pmatrix},

again agreeing coefficientwise with the candidate. Every recurrence has rank 655.

Sign audit of Entries 292--293

After substituting

X1=1,X2=(v−2)/2X_1=1, \qquad X_2=(v-2)/2

at (u=0), the fixed (e_7,e_8,e_9) row for (q_0) is the negative of the tail recorded in Entries 292--293, just as the fixed (e_6u^{-2}) coefficient is the negative of their recorded value.

The discrepancy is therefore a uniform frame or connection sign, not an isolated (e_6) accident. The current source presentation and the current candidate agree throughout the gauge-invariant principal part. The earlier entries remain valid for location, rank, Gysin kernel, and Rees order, but their displayed coefficients require an explicit sign bridge before reuse.

Structural classification

Let

A−−=⟨e6,valg⟩\mathcal A_{--}=\langle e_6,v_{\rm alg}\rangle

be the final-block algebraic plane. The total-energy principal part now has the following typing:

Thus the marked extension is more canonical than a generic interpolated matrix but less canonical than a fully framed connection. The invariant object at total energy is the filtered extension class modulo the existing Kummer line, not the candidate's complete scalar matrix.

Next falsifier

Determine whether the source-normalized Kummer connection and Entry 293's Rees gauge select a unique (e_6) logarithmic lift in the current sign convention. If they do, compare that derived row with the candidate. If they do not, permanently classify the row as triangular-gauge presentation data and proceed to the next support factor.

Durable artifacts

View source on GitHub