# AMRAL × Lebesgue Universal Covering — Round 22
## Symmetry-Breaking Fourier Witness Oracle and Proof-Cost Pareto Search

**Document ID:** AMRAL-LUC-FC-R22  
**Version:** v0.1  
**Date:** 2026-09-19  
**Research status:** Round 22 / Fourier witness legality / proof-cost-aware adversarial search  
**Research mode:** Human-Directed + Semi-Autonomous AI Mathematical Research  
**Research initiation and methodology source:** Neo.K  
**AI collaborating researcher and primary executor:** Aletheia / ChatGPT, GPT-5.6 Sol  
**Prerequisite documents:** AMRAL-LUC-FC-R00 v0.2; R01–R21 v0.1  

---

# 0. Round summary

Round 21 obtained an important negative result:

> Within the tested 7-fold preserving / nearby Reuleaux7 family, regular $B_7$ is fairly robust against a fresh placement adversary.

Round 22 therefore formally opens the symmetry-breaking odd Fourier modes:

$$
k=3,5,7,9,11.
$$

But this round no longer just asks:

> Which witness has the largest numerical forcing?

Instead it simultaneously tracks:

1. robust forcing margin;
2. orientation-domain cost;
3. curvature reserve;
4. the usable scale of the generic support core;
5. actual lift-tree nodes;
6. unresolved placement volume.

This round obtains:

$$
\boxed{\text{MULTI-MODE FOURIER LEGALITY COMPILER: CLOSED}}
$$

$$
\boxed{\text{FULL-ORIENTATION ORACLE RULE: CLOSED}}
$$

$$
\boxed{\text{PROOF-COST PARETO MAP: ESTABLISHED}}
$$

But:

$$
\boxed{\text{NEW SINGLE WITNESS DOMINATING }B_7:\ \text{NOT FOUND}}
$$

And still:

$$
\boxed{a_{\mathrm{Leb}}\ge0.8350:\ \text{COMPUTE-DEFERRED}}
$$

---

# 1. General symmetry-breaking family

This round uses:

$$
\boxed{
h(\theta)
=
(1-\lambda)h_{B_7}(\theta)
+
\frac{\lambda}{2}
+
\sum_{k\in\mathcal K}
\left(
a_k\cos k\theta+b_k\sin k\theta
\right)
}
$$

where:

$$
\mathcal K
\subseteq
\{3,5,7,9,11\}.
$$

---

# 2. Constant-width legality

Both $B_7$ and the disk are width-one.

For odd $k$:

$$
\cos k(\theta+\pi)=-\cos k\theta,
$$

$$
\sin k(\theta+\pi)=-\sin k\theta.
$$

So the Fourier perturbation:

$$
q(\theta)
=
\sum_{k\in\mathcal K}
(a_k\cos k\theta+b_k\sin k\theta)
$$

satisfies:

$$
q(\theta+\pi)=-q(\theta).
$$

Therefore:

$$
\boxed{
h(\theta)+h(\theta+\pi)=1.
}
$$

Constant width is preserved.

---

# 3. Multi-Mode Convexity Theorem

Convex support requires the curvature measure to be nonnegative.

The interiorized base:

$$
(1-\lambda)h_{B_7}+\lambda/2
$$

already contributes at least an absolutely continuous curvature reserve of:

$$
\lambda/2.
$$

The Fourier term's contribution to:

$$
h+h''
$$

is:

$$
\sum_k
(1-k^2)
(a_k\cos k\theta+b_k\sin k\theta).
$$

Hence:

$$
\left|
q+q''
\right|
\le
\sum_k
(k^2-1)
\sqrt{a_k^2+b_k^2}.
$$

So the sufficient legality condition is:

$$
\boxed{
\sum_{k\in\mathcal K}
(k^2-1)
\sqrt{a_k^2+b_k^2}
\le
\frac{\lambda}{2}.
}
$$

This directly extends the Round 02 curvature legality condition into a searchable finite coefficient region.

---

# 4. Symmetry Order

$B_7$'s support has period:

$$
2\pi/7.
$$

If the set of active Fourier modes is:

$$
S,
$$

the generic rotational symmetry order is:

$$
\boxed{
d=\gcd(7,S).
}
$$

So the orientation fundamental period is:

$$
\boxed{
P_\phi=\frac{2\pi}{d}.
}
$$

If any active mode has:

$$
k\not\equiv0\pmod7,
$$

then generically:

$$
d=1,
$$

so:

$$
\boxed{
P_\phi=2\pi.
}
$$

That is, relative to regular $B_7$:

$$
\boxed{
\text{orientation-domain length}\times7.
}
$$

This is the first exact proof-cost that symmetry-breaking must pay.

---

# 5. Full-orientation oracle requirement

Round 22 actually catches a very typical failure mode.

Structured witness:

$$
K=(1-0.08)B_7+0.08B_3.
$$

Some optimistic search basin gives:

$$
0.83742146.
$$

But a full:

$$
\phi\in[0,2\pi)
$$

multi-seed exact support search finds a different basin:

$$
\boxed{
0.83583910.
}
$$

So:

$$
\boxed{
\text{a symmetry-breaking candidate must not reuse the }[0,2\pi/7)
\text{ oracle}.
}
$$

---

# 6. Oracle status levels

Round 22 fixes:

## `TRAINING-SCORE`

Computed only on finite adversarial placements.

## `OPTIMISTIC-CANDIDATE`

A fresh-placement search, but with insufficient seeds / resolution.

## `ROBUST-SEARCH-CANDIDATE`

A full, proven orientation period with a multi-seed fresh oracle.

## `COMPILED-CANDIDATE`

Representable in the Round 02 finite support dictionary.

## `LIFT-VERIFIED`

The complete relevant placement root has been closed by a certificate.

Only the last level carries positive theorem value.

---

# 7. Multi-mode random/adversarial search

On the four Round 20 necessity-marked base configurations,

first using regular $B_7$'s bad placements as training adversaries,

a legal multi-mode candidate reaches a training minimum of:

$$
0.83770073.
$$

A single-seed fresh oracle still gives:

$$
0.83733308.
$$

But after a full multi-seed, full-$2\pi$ re-search, the four base minima are:

$$
0.83683248,
$$

$$
0.83677292,
$$

$$
0.83674967,
$$

$$
\boxed{
0.83668617.
}
$$

Regular $B_7$'s own worst-case search on the same four base configurations gives:

$$
\boxed{
0.83692628.
}
$$

So this candidate:

- has weaker forcing;
- has an orientation domain 7 times larger.

It is Pareto-dominated by $B_7$ on the two axes tracked so far.

---

# 8. Multi-base adversarial re-training

Round 22 further adds fresh low placements into the adversarial set and re-searches.

The new Fourier candidates still repeatedly show the same pattern:

> the training score looks good, but a fresh full-orientation oracle opens up a new, lower basin.

Currently no candidate robustly exceeds regular $B_7$ on the four marked base configurations.

So Round 21's overfitting conclusion still holds within the symmetry-breaking family.

---

# 9. Pure Harmonic Family

A completely different proof-cost route:

$$
\boxed{
h_k(\theta)
=
\frac12+a\cos(k\theta),
\qquad
k\text{ odd}.
}
$$

This is itself width one.

Curvature:

$$
r_k(\theta)
=
\frac12-(k^2-1)a\cos(k\theta).
$$

So smooth convex legality holds **exactly** when:

$$
\boxed{
|a|
\le
\frac{1}{2(k^2-1)}.
}
$$

---

# 10. Curvature Reserve

Define:

$$
\boxed{
\sigma_k
=
\frac12-(k^2-1)|a|.
}
$$

If:

$$
\sigma_k>0,
$$

then the generic support erosion:

$$
h_{\rm core}=h-\delta
$$

remains a legal convex support whenever:

$$
\boxed{
\delta\le\sigma_k.
}
$$

So:

$$
\boxed{
\sigma_k
}
$$

is an important proof-cost resource for the witness certificate.

---

# 11. Near-boundary amplitude is not free

If the amplitude is taken to:

$$
99.5\%
$$

of the legality bound, then:

$$
\sigma_k=0.0025,
$$

regardless of $k$.

This does make the forcing stronger,

but the placement-box motion radius must be cut to:

$$
\delta\le0.0025
$$

for the generic support core to remain legal.

Therefore:

> a short orientation period does not mean the certificate is necessarily cheap.

One must account for all of:

$$
\boxed{
P_\phi,\ \sigma,\ \text{forcing}
}
$$

at the same time.

---

# 12. 90% amplitude proof-cost branch

Round 22 therefore specifically tests:

$$
|a|
=
0.9
\cdot
\frac1{2(k^2-1)}.
$$

At this setting:

$$
\boxed{
\sigma=0.05.
}
$$

The generic core is substantially easier to obtain.

---

# 13. $k=11$ candidate

$$
a_{11}=0.00375.
$$

Orientation period:

$$
2\pi/11.
$$

Relative to B7:

$$
\boxed{
7/11\approx0.636.
}
$$

The worst case over the four marked-base fresh search:

$$
\boxed{
0.83574698.
}
$$

The dense support bracket at the found worst placement:

$$
0.83578614
\le A
\le
0.83581539.
$$

This is search-only and still does not prove the global witness minimum.

---

# 14. $k=13$

$$
a_{13}\approx0.00267857,
$$

$$
\sigma=0.05.
$$

Orientation ratio:

$$
7/13\approx0.538.
$$

Four-base search worst case:

$$
\boxed{
0.83563803.
}
$$

Dense bracket at the found worst placement:

$$
0.83567775
\le A
\le
0.83569731.
$$

---

# 15. Thin-margin high-k branch

High-resolution search:

## $k=17$

$$
\boxed{
\min_{\rm search}\approx0.83507834.
}
$$

Orientation ratio:

$$
7/17\approx0.412.
$$

## $k=19$

$$
\boxed{
\min_{\rm search}\approx0.83504963.
}
$$

Orientation ratio:

$$
7/19\approx0.368.
$$

Both are still above the target,

but the forcing margin is already very thin.

Therefore:

$$
\boxed{
\text{as certificate cheapness increases, counterexample risk rises quickly}.
}
$$

---

# 16. Generic support-core lift pilot

On the Round 20 tracked depth-50 marked base cell,

using the generic support core:

$$
h_{\rm core}=h-\delta
$$

to test the pure harmonics.

Translation-square pilot root:

$$
[-0.21,0.21]^2.
$$

Orientation:

$$
[0,2\pi/k).
$$

---

# 17. Depth-20 comparison

## Regular $B_7$

special Reuleaux erosion:

$$
7037\text{ nodes},
$$

unresolved placement volume:

$$
\boxed{
1.4107\times10^{-4}.
}
$$

## $k=11$, 90%

$$
9623\text{ nodes},
$$

$$
1.8536\times10^{-4}.
$$

## $k=13$, 90%

$$
9211,
$$

$$
1.4270\times10^{-4}.
$$

## $k=17$, 90%

$$
7399,
$$

$$
\boxed{
9.5317\times10^{-5}.
}
$$

## $k=19$, 90%

$$
7221,
$$

$$
\boxed{
8.3671\times10^{-5}.
}
$$

So the high-$k$ harmonics really can exploit a smaller orientation root,

coming in below $B_7$ on **absolute unresolved placement volume**.

---

# 18. But B7 has stronger tail contraction

One cannot look only at the depth-20 volume.

Round 20 already established that $B_7$ has:

- a thick forcing margin;
- fast deep-tail contraction.

Whereas the forcing search margin for $k=17/19$ is only:

$$
O(10^{-5}).
$$

So they are better suited as a:

$$
\boxed{
\text{cheap auxiliary / cell-specific witness}
}
$$

rather than as a replacement for the primary robust $B_7$.

---

# 19. Proof-Cost Pareto Vector

Round 22 no longer outputs a single witness score.

Every witness instead keeps:

$$
\boxed{
\Pi(K)
=
(
m_{\rm force},
P_\phi,
\sigma,
N_d,
V_d,
\rho_{\rm tail}
).
}
$$

where:

- $m_{\rm force}$: search forcing margin;
- $P_\phi$: orientation domain;
- $\sigma$: curvature reserve;
- $N_d$: certificate nodes;
- $V_d$: unresolved placement volume;
- $\rho_{\rm tail}$: multi-depth contraction.

---

# 20. Current Pareto roles

## Regular $B_7$

Role:

$$
\boxed{\text{PRIMARY ROBUST CLOSER}}
$$

Features:

- thick forcing;
- specialized erosion;
- strong deep-tail contraction.

## $H_{11}/H_{13}$

Role:

$$
\boxed{\text{BALANCED AUXILIARY}}
$$

Features:

- forcing still has a $6\text{–}7\times10^{-4}$ search margin;
- smaller orientation domain;
- $\sigma=0.05$.

## $H_{17}/H_{19}$

Role:

$$
\boxed{\text{CHEAP / RAZOR-THIN AUXILIARY}}
$$

Features:

- small orientation root;
- low unresolved volume;
- forcing only a few $10^{-5}$ above target;
- requires high vigilance against hidden counterexamples.

---

# 21. Full symmetry-breaking candidate verdict

Among the full-$2\pi$ multi-mode candidates tested so far:

- none robustly exceeds $B_7$;
- the orientation domain is at least 7 times larger;
- and generic support-core / compiler complexity is increased.

So:

$$
\boxed{
\text{NOT PROMOTED TO PRODUCTION WITNESS POOL}.
}
$$

But they are retained as a:

`SEARCH ARCHIVE`

for future multi-base / batch synthesis.

---

# 22. What Round 22 changes

Round 21's next step was originally supposed to be:

> find a symmetry-breaking witness that beats B7.

Round 22's result changes this to:

> one should not require a single shape to comprehensively beat B7; one should instead build a proof-cost-aware witness portfolio.

This is an important strategic pivot.

---

# 23. Witness Portfolio Principle

For different marked cells / proof horizons,

the best witness may differ.

So production should maintain:

$$
\boxed{
\mathcal W_{\rm portfolio}
=
\{
B_7,
H_{11},
H_{13},
H_{17},
H_{19},
\ldots
\}.
}
$$

But only a witness that has actually passed that cell's negative filter, lift probe, and verifier support may be selected.

---

# 24. Search status warning

Every Fourier forcing minimum in this round remains:

`SEARCH-ONLY`.

Even when the found placement's dense inner area is $>T$,

this still does not prove:

$$
\min_pF(K,p)>T.
$$

because the placement oracle is not an exhaustive certificate.

A positive theorem still requires:

- the Round 02 finite compiler;
- the Round 08 generic support core;
- the full lift tree.

---

# 25. New compiler path

The advantage of a pure-harmonic witness is that its representation is minimal:

```text
k
a
phase
centering
```

It does not need the full generic Fourier dictionary.

So the Round 02 compiler can add a:

`PURE-HARMONIC-CW`

fast path.

The verifier then only needs:

1. odd $k$;
2. amplitude legality;
3. symmetry period;
4. curvature reserve;
5. support/contact evaluation.

---

# 26. Current strongest new candidate class

Round 22 did not find a stronger-than-B7 witness.

But it did find:

$$
\boxed{
H_{11}/H_{13}
}
$$

as the first batch of auxiliary candidates whose "forcing is still clearly above target while the proof domain is cheaper."

$H_{17}/H_{19}$, on the other hand, are better suited as a:

> high-risk / low-domain-cost experiment,

and should not directly serve as the primary production closer.

---

# 27. Round 23

## Witness Portfolio Closure and Batch-Cover Optimization

The next round no longer continues an unbounded shape search.

It takes:

$$
\boxed{
B_7,H_{11},H_{13},H_{17},H_{19}
}
$$

directly and, against the Round 20 necessity-marked atlas, performs:

1. negative filtering;
2. shallow + deep lift probes;
3. cell × witness incidence;
4. weighted set cover;
5. portfolio cost accounting:
   - nodes;
   - root volume;
   - replay;
   - forcing risk;
6. testing:
   - whether different witnesses can divide the labor of closing different marked cells;
7. and, if the portfolio still leaves a common hard tail, returning to multi-witness joint / asymmetric synthesis.

---

# 28. Shortest Handoff

Round 22 did not find a single new witness:

$$
K_{\rm new}>B_7
$$

But it did obtain a more important structure:

$$
\boxed{
\text{witness design is not a single maximization problem,
but a proof-cost Pareto portfolio problem}.
}
$$

Regular $B_7$ remains the robust primary.

The smooth high-$k$ harmonics instead offer:

$$
\boxed{
\text{smaller orientation domain}
+
\text{generic support-core}
}
$$

as a different kind of certificate economics.

The next step should be to let these witnesses compete / cooperate directly on the necessity atlas,
rather than continuing to blindly chase a single "overall strongest shape."