Geometry · Wall crossing · Open FJRW theory

When walls
collide.

Take two paths.
Get two answers.
Find the missing wall.

A wall changes the answer when you cross it. Cross two walls in a different order and the changes need not agree. A new wall can repair the disagreement. But consistency is only the first test.

Follow the first path ↓

01 · One beginning, one destination

The order matters.

Follow the upper route, then the lower. Watch two coefficients change at the crossings. With just the two incoming walls, the routes disagree.

The crossing

Two walls
Σ₁ · first wallΣ₂ · second wall

The point travels through a local affine chart. It changes the potential only at a crossing. The two routes avoid the joint.

The live ledger

Before the first crossing

Coefficients of the two mixed terms

t₁t₂ xy⁵12
t₁t₂ x⁵y0

At the forward endpoints

Upper route(2, 10)
Lower route(8, 4)

The paths disagree.

The question

The start and finish are the same. Should the answer depend on which side of the joint you take? The missing information sits exactly where the two operations interact.

Open FJRW theory studies invariants of marked disks for a polynomial such as x⁴ + y⁴. Those invariants depend on boundary conditions. Changing the conditions acts on a potential, a polynomial whose coefficients record the invariants with their prescribed signs.

Here we follow one exact example with two distinct labels. Their variables satisfy t₁² = t₂² = 0, while t₁t₂ survives. A single label sees each wall separately. The product t₁t₂ detects their interaction.

The exact example, operators, and path conventions

Work in Q[x,y,t₁,t₂]/(t₁²,t₂²). The labels have twists (1,0) and (0,1), both with descendent degree one. Put Xa,b = xᵃyᵇ((b+1)x∂x − (a+1)y∂y), Σ₁ = exp(−t₁X₁₀/2), Σ₂ = exp(−t₂X₀₁/2), and Uc = exp(ct₁t₂X₁₁). Each exponential terminates after its linear term in this ring.

Wmin = x⁴ + y⁴ − 4t₁xy⁴ − 2t₂y⁵ + 12t₁t₂xy⁵. At the forward endpoints, write V(a,b) = x⁴ + y⁴ − 2t₁x⁵ − 4t₂x⁴y + t₁t₂(a xy⁵ + b x⁵y). The two numbers are potential coefficients, not an unqualified count of disks.

The supports have normals (3,4), (3,2), and (1,1); the mixed support is a ray in s₂ ≥ 0. The horizontal routes have s₂ = ±1/10, with common endpoints (−1/4,0) and (1/4,0). Their end connectors miss every support. These are shortened versions of the paper’s local test routes, with the same crossings and signs. The drawn axes use a common scale.

The ribbon lists crossings chronologically. An operator word acts rightmost first: the corrected upper route is Σ₂UcΣ₁, and the lower is Σ₁Σ₂. Reverse travel starts at that route’s own endpoint and applies the inverse crossings, recovering Wmin.

Current potential:


 

02 · Let consistency choose

Find the missing wall.

Add a ray at the joint. Change its coefficient until the two paths give the same operation—not just the same value on one polynomial.

The mixed-wall coefficient

0

This control also changes the ray and both endpoint ledgers above.

Replay the paths with this wall ↑

The return test

Exact arithmetic
Difference on x—
Difference on y—

Upper answer—
Lower answer—

Consistency forces a correction. Its coefficient is determined by the failure of the original operations to commute. In a larger truncation, further interactions can force further walls.

Why checking x and y is enough here

The transformations fix t₁ and t₂ and are algebra automorphisms. Agreement on x and y therefore gives agreement on the entire coefficient ring. The comparison uses exact rational coefficients; no numerical tolerance is involved.

[X₁₀, X₀₁] = −3X₁₁, and Σ₂Σ₁ = U3/4Σ₁Σ₂. Thus Σ₂U−3/4Σ₁ = Σ₁Σ₂. On Wmin, the upper endpoint is V(2−8c,10+8c), and the lower is V(8,4).

Paper 2 organizes transformations into factors ordered by the slopes of their boundary rays in each finite truncation. The boundary ray for Xk₁,k₂ is spanned by (k₁+1,k₂+1); it is not automatically a direction in an affine base. The quartic paper supplies the cofactor placement used here. This explorer verifies the stated two-label model, not an arbitrary all-order scattering diagram.

03 · Beyond the local joint

The surface remembers.

A local return test can close while a journey around the whole surface changes a lattice frame. Follow one specified core loop of the quartic annulus.

The annular cover

Topological schematic

The loop separates Ã, B₀ from B₁, C̃. The cut-open view is topological, not an affine coordinate chart.

The returned frame

0 circuits

Both arrows use the same drawing scale. The numbers are the exact integer coordinates. We track an exponent frame in F-coordinates.

Local versus global

The correction wall settles a local disagreement. It does not make the surface’s global holonomy disappear. For this core loop, K has trace 18 and eigenvalues 9 ± 4√5.

What comes from the annular construction

The affine base disk has singular points A, B, C. The parity character is 1 around A and C, and 0 around B. Cutting from A to C and gluing two copies crosswise gives an annulus. Its four singular points are Ã, B₀, B₁, C̃; the two lifts of B are focus–focus, while the branched germs at à and C̃ are different. The source figure’s relative placement is retained in the annulus view.

The selected core-loop class is γCγA−1, based in sheet S₀. On the exponent lattice in the source’s F-coordinates it acts by K = [[−1,2],[−10,19]]. Its determinant is one. Other loops in the punctured annulus can have different holonomies, even if they become freely homotopic after forgetting the singular points.

The animation, where enabled, marks progress around the loop. The returned matrix is applied at the end of a circuit; no interpolated affine connection is being computed. A tangent-frame representation would use the inverse transpose. The two boundaries and the marked core curve are drawn schematically, not as a Euclidean metric model of the affine structure.

04 · Consistency was only the first test

The test the periods miss.

The locally consistent wall has coefficient −3/4. The Gross–Kelly–Tessler (GKT) transition needs 1/4. The two answers differ, but their relevant period classes agree.

A fixed comparison, independent of your coefficient dial above.

Annular transport

The consistent wall

Mixed coefficient −3/4

The GKT transition

The required endpoint

Mixed coefficient 1/4

A correction can be imposed.

Apply the algebraic normal form to carry V(8,4) to V(0,12). The finite operation is explicit. Its intrinsic geometric origin is the open question.

Where the explorer reaches the research frontier

Can the surface supply its own correction?

The missing construction is an operation intrinsic to the annulus, at its boundary, that produces the required normal form. Replacing a wall coefficient by the GKT value does not by itself solve the annular consistency problem.

The discrepancy, period class, and imposed normal form

Annular transport gives Wtr = V(8,4), whereas Wmax = V(0,12). The two-label GKT transition is Σ₂U1/4Σ₁; the annular path word is Σ₂U−3/4Σ₁. They differ by U₁, so U₁Wtr = Wmax.

The relevant t₁t₂ period class of V(a,b) is (6 − (a+b)/2)[xy], which vanishes for both endpoints. Here [xy] abbreviates the class of xy dx∧dy for the twisted differential ℏd + d(x⁴+y⁴)∧, after extracting the t₁t₂ coefficient of the exponential deformation. In the paper, the discrepancy has the relative primitive β = t₁t₂d(x²y²). This exact finite comparison explains why the period test loses information; it does not say that every observable or every period problem is identical.

UcV(a,b) = V(a−8c,b+8c), and the maximal normal form is Ua/8V(a,b) = V(0,a+b). The final control applies this algebraic operation externally. It does not implement the missing boundary construction.

A companion experimentWhy begin with the quartic?

Select a square or use the exponent controls. Pale squares have boundary-independent primary invariants; coloured squares admit primary wall crossing. The outlined square is your selection.

Insertion type
Central charge ĉ = 2 − 2/r − 2/s1

The main explorer uses two descendent labels. Descendents allow wall crossing even below the primary threshold. The theorem concerns the stated Fermat theories and symmetric canonical boundary conditions.

From picture to paper

Three steps toward a geometric home.

Exact finite computations support each experiment. The papers distinguish the threshold, formal factorization, annular construction, and the remaining geometric problem.

I · Where it beginsThe central-charge threshold for wall-crossing in two-variable open FJRW theory II · How it factorsSlope-ordered factorization of wall-crossing in two-variable open FJRW theory III · Where it livesAnnular scattering in quartic open FJRW theory

The “walls” in At the Wall concern bundle slopes. Here they describe transformations of open FJRW invariants. The shared vocabulary does not identify the two theories.