Carlos Toledo
sandbox draft · not reviewed · page state: open / unsigned
Sandbox · survival tracker · opened 2026-07-30

Three attacks on one thesis — what survives

Equilibrium is a geometry — and refusal is how you measure it

the experiment is the FORM

Three independent attacks on the same thesis, recorded SIDE BY SIDE so we can watch which positions survive contact with evidence and whether a hybrid rises. The experiment is the FORM: every position carries a decidable test and a status that may change ONLY by evidence, never by preference. A strategy document that cannot be falsified is a mood board.

In plain words — the same thing said twice

The left column is what is actually done. The right column is what it means, for someone who does not work in validated numerics. Neither column is the marketing one — if the analogy claims more than the technical step, the analogy is wrong and gets rewritten.

What a certificate normally does
A validated-numerics certificate proves: a true solution exists within radius r of the computed one, and is unique there. Both halves are local, and both assume the answer is a POINT.
in plain words
It is a GPS that can only say “you are at this exact street address.” Useful almost always — and useless the moment you are standing in the middle of a park, because there is no address to give.
Where it stops working
When the coupling is monotone but NOT strictly, the solution set is a positive-dimensional face. The Jacobian is singular along it, contraction fails, and the certificate refuses.
in plain words
The park. There is no single right answer, so a machine built to give one just fails. The thesis is that the failure is not empty: which way it fails tells you which way the park extends.
The instance we picked
Wardrop S1: two populations on one network, edge cost c = j¹ + j² — the TOTAL flow, identical for both, with no population index.
in plain words
A toll that counts cars, not owners. Two taxi firms share the roads; each road charges by how many cars are on it. Swap which firm's cars use which road, keep each road's total the same, and every cost is identical. So “how many of firm A's cars are on road X” has no single answer.
What we computed
The equilibrium face's tangent space is the null space of both conservation laws restricted to the shared edges. Exactly, over the rationals: k = 6.
in plain words
Six independent swaps. Each is a loop — firm A takes a bit more of one road, gives it back round the loop — and no loop can be built from the others. The six pictures above are those loops.
The one that explains itself
Direction 6 is supported on edges (7,8) and (7,10), the two exit arcs, which carry no conservation row.
in plain words
Nobody said which firm leaves by which exit — only how many cars leave. So the split between the two exits is free, and it is free for exactly the reason exits have no balance to satisfy.
The check that mattered
Four independent solves, differenced, projected onto the predicted span: residual 2.4e-14 against a movement of 18.8. Non-shared edges move by 4.3e-14.
in plain words
We worked out the shape of the park from the rules alone, without ever asking the GPS. Then we asked it four times from four starting points. Every answer landed inside the shape — including that it never wanders onto roads only one firm can use.
The second problem: the wall
As the parameter A grows, Z₁ → 1 and the contraction argument stops closing. Measured: closes to A = 5, refuses at A = 6.
in plain words
A bridge with a weight limit stamped on it by the inspector's method, not by the bridge. Past some load the technique can no longer conclude — which is not the same as the bridge failing.
The objection
Z₁ = 1 depends on the preconditioner, the radius policy and the Newton point. It is a property of the METHOD, not of the equations — raised independently by two of three reviews.
in plain words
Whose limit is it? If a different inspector with a different technique stamps a different number, the number was about the inspector.
Locating it without crossing it
(1−Z₁) falls LINEARLY in A, so a fit over the last four successful certificates predicts where it reaches zero — landing inside the measured bracket every time.
in plain words
You do not have to drive onto the bridge to find where it collapses. Watch the safety margin shrink as you add load, draw the line, read off where it would hit zero.
Is the limit real, or an artifact?
Refine the grid: A*(N) over N = 10…40 converges, first order, A*(N) ≈ A − 6.4/N.
in plain words
Measure with a finer ruler. If the answer keeps changing, you were measuring the ruler. If it settles, you were measuring the bridge. It settles — so the wall is not a drawing artifact.
What is still open
Convergence in N rules out a discretization artifact. It does NOT rule out a preconditioner artifact — every grid in the family shares one preconditioning strategy. That test is not run.
in plain words
The rulers settled — but every ruler came from the same factory. We have ruled out that the answer depends on how FINE the ruler is. We have not ruled out that it depends on how the factory makes rulers.
And the honest limit
Two extrapolation models give A ≈ 5.846 and 5.763. They disagree by 0.08.
in plain words
We know it settles. We do not know exactly where. The amber band on the chart is that ignorance, drawn to scale rather than hidden behind a decimal point.

Proof-status ledger — every claim, by what backs it

Four badges, no blurring between them, and each names a MECHANISM rather than a status this project awards itself. MUTATION-TESTED — a headless battery in this repo checks it and is mutation-tested: revert the check and it goes red. STANDARD — off-the-shelf or cited, not re-derived here. PROSPECTIVE — pre-registered: the shape of the claim is fixed in advance, the number is not yet computed. OPEN — the honest frontier, known and not done.

Deliberately absent: the words proved and certified as a status. Both are rungs on this project's evidence ladder — rung 5 needs a named human signature, rung 4 needs a ledger certificate record — and no such record covers any row below. What backs the green rows is a mutation-tested battery, which is what MUTATION-TESTED says and all it says. Most of this ledger is OPEN, and that is the honest shape of a programme one day old.

direction 1
12934567810
3 edges
direction 2
12934567810
4 edges
direction 3
12934567810
3 edges
direction 4
12934567810
3 edges
direction 5
12934567810
4 edges
direction 6
12934567810
2 edges

Teal = population 1 gains, oxblood = population 1 gives up; population 2 moves by exactly the opposite. Amber nodes are the exits, which carry no conservation row. Every direction is a cycle: flow goes round and comes back, so both conservation laws and every edge total are untouched. Direction 6 is the two populations swapping between exit 8 and exit 10 — it exists for the same reason the exits carry no row.

ClaimStatusWhat backs it
S1's cost is the edge total j1+j2, identical for both populations, with no population indexMUTATION-TESTEDread from the SHIPPING kernel at run time, not transcribed — test-s1-face.py G1–G1c
OCCUPANCY: the phenomenon is known at TITLE level — unique totals with non-unique class splitsSTANDARDBoyce & Xie 2013, "Assigning user class link flows uniquely", TR Part A 53:22–35, DOI 10.1016/j.tra.2013.06.002. An entire programme (proportionality, entropy maximisation, PAS) exists to REPAIR this non-uniqueness
The dimension count is FOLKLORE — ker of a node–arc incidence matrix IS the circulation space, so k is the cycle rank m − n + cSTANDARDthree independent derivations reach it in under a page: linear algebra (2 lines), graph theory (textbook), and the cycle-swap argument that Bar-Gera's paired-alternative-segment machinery is built from. Not a rediscovery, not unclaimed
All six directions are attained at the tested equilibrium — each movable in BOTH signs without a flow going negativeMUTATION-TESTEDtest-attainment.py 5/5. Narrowest room [−7.18, +1.59], widest [−42.32, +57.68]; no shared edge at zero, so the point is relatively interior. Raised as a defect in our claim and resolved by TESTING rather than softening — "dimension exactly 6" is earned AT THAT POINT and must always carry the point
The S1 equilibrium face has tangent dimension k = 6MUTATION-TESTEDexact rational null space. Two mutants fire: treating the exits as ordinary nodes gives k=5, losing exactly the exit-swap direction. A node-count shortcut (11 − 5) agrees on this instance — but see the row below: it is not a formula
The node-count shortcut shared − conservation nodes is NOT general — it held on the paper instance by luckMUTATION-TESTEDtest-face-general.py — a random layered DAG with 19 shared edges has rank 10, so k = 9 while the shortcut predicts 8. Found by a test written to break our own conjecture. The publishable quantity is the RANK-based dimension; the shortcut must never be stated as a formula
The face computation now runs on an arbitrary two-population networkMUTATION-TESTEDface_general.py, exercised on 12 grids and random DAGs; every basis vector satisfies both conservation laws with residual exactly 0 over ℚ. This is the precondition for pointing it at a public benchmark rather than at our own 15 edges
Each of the six directions satisfies both Kirchhoff laws, and they are independentMUTATION-TESTEDresidual exactly 0 over ℚ — not a tolerance. G3, G4
This is not an enclosure. Exact arithmetic on an affine system is algebraic correctness, not validated numericsSTANDARDthe literature gate's own terminology contract, applied to us
k = 6 holds at a relatively interior point of the face; at a boundary point the local cone is smallerSTANDARDstandard polyhedral geometry; any claim must name which point
A monotone VI has a convex solution set, so two distinct equilibria give a whole segmentSTANDARDclassical — and the cheapest available upgrade for the nonlinear cases
Bordered Krawczyk closes at two points of a face, giving that segment with certificatesPROSPECTIVEshape fixed in advance: F̃(x) = (F(x), v̂ᵀx − t) at t₁ ≠ t₂. Not computed
Verified rank + the constant-rank theorem gives the dimension for a NONLINEAR instanceOPENthe honest frontier. S1 is affine and needed none of this

The three attacks

approach A

Four-phase pipeline

Formalise the Krawczyk breakdown as a theorem, then build a 4-step automated diagnostic engine.

First move: Prove M = I − Y[J] has an eigenvalue ≈ 1 with eigenvector in ker(J(x0)).

approach B

Five directions + the manifold zoo

Reframe the vocabulary, then get EMPIRICAL evidence before any theorem.

First move: Generate thousands of synthetic systems with KNOWN manifolds and measure angle(predicted tangent, true tangent).

approach C

Lyapunov–Schmidt frame

One object unifies all three claims: a verified enclosure of ker(Df). Everything else is classical theory or a corollary.

First move: Check whether S1's cost is AFFINE. If so the equilibrium set is a polyhedral face, exact in rational arithmetic — no intervals at all.

Where they agree — the signal

Three readers, no coordination. Convergence here is worth more than any single argument.

A, B, C

C2 is the core and must be attacked first

Unanimous, and it matches the thesis's own self-assessment. The strongest signal in the set — three independent readers converging on the same claim without coordination.

A, B, C

The computational primitive is a verified null-space / singular-value enclosure

A reaches it via interval SVD separation bounds, C via Rump-style verified SVD, B assumes it. Nobody proposes a different primitive. This is what to build first whatever else survives.

A, B, C

The directional split (tangent vs transverse) is the mechanism

A calls it V_∥ / V_⊥; B calls it anisotropic contraction; C calls it Lyapunov–Schmidt. Same object, three vocabularies.

B, C

The thesis as currently titled is too broad to defend

B says so explicitly ('avoid making Equilibrium is a geometry the central mathematical claim'); C implies it by narrowing everything to the kernel enclosure. Two of three want the headline changed.

Where they disagree — the decisions

Each divergence is resolved by argument, not preference, and the resolution is dated. Two of these overturn parts of the original thesis.

D1 · Is REFUSAL a legitimate mathematical object to build on?
AYES — builds directly on Z₁ ≥ 1 and bisects on ρ(I − Y(A)[J(A)]) − 1.
BNO — 'refusal is algorithm-dependent'. Reframe as ANISOTROPIC CONTRACTION, which 'fits interval analysis much better'.
CNO, and sharper — 'Z₁ = 1 is a property of the METHOD, not of the equations. It depends on the preconditioner, the radius policy, the Newton point. As stated, the wall isn't a mathematical object.'
Verdict: TWO OF THREE REJECT THE THESIS'S CENTRAL WORD, and C's version is a referee-grade objection that would be raised by the first reader. This is the single most consequential disagreement in the set.
Consequence: The word 'refusal' survives as the NARRATIVE hook and dies as the mathematical object. What gets certified is anisotropic contraction against a FIXED, DECLARED policy. C's split of C3 into Version A (about equations) and Version B (about the certificate) is the disciplined form.
RESOLVED AGAINST THE ORIGINAL FRAMING · 2026-07-30
D2 · What is the FIRST move?
AProve the eigenvalue/eigenvector theorem about M = I − Y[J].
BRun the synthetic manifold zoo (line, circle, torus, intersecting planes, saddle, cusp, fold) and measure tangent angles. 'Forget C1. Forget C3.'
CCheck whether S1's cost c = j¹+j² is affine. If it is, solve the complementarity system exactly over ℚ and enumerate the face — settling C1 for S1 AND yielding free ground truth for C2.
Verdict: C's is cheapest and most decisive; B's is the strongest FALSIFIER; A's presupposes what the other two want tested first.
Consequence: Do C then B. C is ~a day and may settle C1-for-S1 outright; B is the experiment that can kill C2 before any theorem is attempted. A's theorem is what you write if both survive.
RESOLVED — order is C, then B, then A · 2026-07-30
D3 · Is C1 hard, or partly free?
AHard — interval SVD with strict separation bounds σ_{n−k} > ε > 0 and the rest enclosing zero.
BDoes not engage; explicitly defers C1.
CThe LOWER bound is FREE: a monotone VI has a CONVEX solution set, so two certified distinct equilibria imply a certified segment — dimension ≥ 1 at no extra computational cost. Only the upper bound needs the SVD.
Verdict: C is strictly stronger and nobody contradicts it. It also reuses the mfg-cap disjoint-balls shape already built here.
Consequence: C1 splits. Lower bound by convexity (cheap, and it UPGRADES Wardrop S1 from DEMONSTRATED to PROVED). Upper bound by verified rank + the constant-rank theorem.
RESOLVED IN FAVOUR OF C · 2026-07-30
D4 · How much of C2 is actually NEW?
AThe whole directional decomposition is the contribution.
BThe contribution is empirical first, theorem second.
CThe tangency half is a COROLLARY, not a computation: once the rank enclosure holds, the tangent space EQUALS ker(Df) by the constant-rank theorem. Classical. The only new object is the two-sided directional bound — contraction < 1 on v̂^⊥, identity-like ≥ 1 − O(ε) on span(v̂).
Verdict: C's narrowing is the most valuable single observation in the three documents. It shrinks the claimable novelty to one lemma — which is exactly what makes it defensible.
Consequence: Do not claim 'we certify the tangent'. Claim the two-sided directional bound, and cite the constant-rank theorem for the rest.
RESOLVED IN FAVOUR OF C · 2026-07-30
D5 · Does C3 survive at all?
AYes — bisect g(A) = ρ(I − Y(A)[J(A)]) − 1.
BReframe as singular perturbation: study Z(ε) as ε → 0, making it a bifurcation theorem rather than an observation.
COnly in two separated versions. Version A certifies the TRUE fold via an extended system (occupancy HIGH — van den Berg–Lessard school). Version B fixes a certificate policy and brackets Z₁(A) = 1 (weaker, but literally what C3 asks). THE PRIZE is a connecting theorem: Z₁ ≥ f(σ_min, Lipschitz), Renegar-style, bounding the distance from the method's wall to the equations' singularity.
Verdict: A's version is C's Version B without noticing the method-dependence problem, so D1 damages it as stated. C's distance theorem is the most novel item ANY of the three proposes, and C flags it as 'the least occupied ground in the whole program'.
Consequence: C3 is demoted to last and split. The distance theorem is promoted to the real prize — IF it survives an occupancy check, which it has NOT yet had.
RESOLVED — split, demoted, with one item promoted · 2026-07-30

Open positions — each with the test that kills it

Status may change only by evidence. Nothing here is claimed.

P1 · from C (the lemma), A (the mechanism) · OPEN — the publishable core
Two-sided directional bound: contraction certified < 1 on v̂^⊥ and identity-like ≥ 1 − O(ε) on span(v̂), both in interval arithmetic.
Test / falsifier: Implement on a synthetic system with a KNOWN 1-D solution manifold; both bounds must close simultaneously. Falsifier: a system where the transverse bound closes but the tangential one does not behave as identity-like — which would mean the refusal is not measuring the tangent.
P2 · from C · OPEN — and it is the cheapest route to C2
The bordering direction that restores contraction IS operationally the tangent.
Test / falsifier: Border F̃(x) = (F(x), v̂ᵀx − t) at two values t₁ ≠ t₂. If Krawczyk closes for both, the certified points are distinct and convexity gives the segment. Falsifier: bordering with a direction NOT in the kernel also restores contraction — which would break the identification.
P3 · from B · OPEN — this is the FALSIFIER for the whole thesis
Predicted tangent matches true tangent to near machine precision across a zoo of known manifolds.
Test / falsifier: line · circle · torus · intersecting planes · saddle · cusp · fold. Measure angle(predicted, true). B's stated bar: consistently ~1e-3 degrees or better. Falsifier: angles that do not concentrate near zero, or that degrade systematically with curvature — in which case C2 is dead and it was found in days, not months.
P4 · from C · OPEN — highest prize, NO occupancy check yet
Distance theorem — Z₁ ≥ f(σ_min, local Lipschitz constants), bounding the method's wall against the equations' singularity.
Test / falsifier: Occupancy check FIRST: Renegar's condition-number theory and the perturbation literature are the obvious incumbents. Only then attempt. Falsifier: the inequality already exists in the conditioning literature under another name — which is exactly how the tropical and sheaf proposals died.
P5 · from B · OPEN — the deliverable shape, not a theorem
A 'Geometry Certificate' is a new certificate OBJECT: dimension, codimension, tangent basis, contraction per direction, curvature, reach.
Test / falsifier: Can the object be emitted for ONE instance with every field either certified or explicitly absent? Falsifier: fields that can only be estimated, not enclosed — those must be dropped rather than shipped soft.
P6 · from C · OPEN — one day of work, and it gates everything else
S1's cost is affine, so its equilibrium face is exactly computable over ℚ.
Test / falsifier: Read the S1 cost. c = j¹ + j² is linear in the flows; if the whole VI is affine, solve the complementarity system as an LP over rationals and enumerate the face. Outcome either way is useful: exact ground truth for P3, or confirmation that intervals are genuinely required.

The hybrid — RISING — and it is not any one of the three

Read together, the three do not select a winner. They compose into something none of them proposed alone.

frame
C's Lyapunov–Schmidt split. It concentrates the entire certification burden on ONE object — a verified enclosure of ker(Df) — and makes C1, C2 and C3 three readings of it rather than three problems.
vocabulary
B's ANISOTROPIC CONTRACTION, which removes the algorithm-dependence objection that two of three raised against the word 'refusal'.
first move
C's affinity check, then B's manifold zoo.
core claim
C's two-sided directional bound — the only part not already a corollary of the constant-rank theorem.
falsifier
B's angle measurement across the manifold zoo. This is what makes the programme honest: it can die in a week.
output
B's Geometry Certificate.
the prize
C's distance theorem, pending an occupancy check it has not had.
narrative
C's closing sentence is the one that survives: a point-certificate applied to a geometric equilibrium DOESN'T FAIL, IT PROJECTS — and the projection's kernel is the geometry. 'Refusal' keeps its place as the way in, not as the object.

What this costs the thesis

Recorded because a comparison that only finds encouragement is worthless.

The headline 'refusal is how you measure it' does not survive as MATHEMATICS. It survives as the way in. Two of three readers rejected it independently and C's objection is referee-grade.

C2's tangency half is classical (constant-rank theorem). The claimable novelty shrinks to one lemma — which is what makes it defensible rather than dismissible.

C1's lower bound is free and UPGRADES Wardrop S1 from DEMONSTRATED to PROVED. That is the cheapest real win available in the whole programme.

C3 is demoted and split; only the distance theorem is worth the name, and it has had no occupancy check.