A quantum surface, one boundary at a time

Let the lattice fluctuate, too.

In the ordinary Ising model, spins point up or down on a fixed lattice. Now imagine summing over the lattice itself: both the spins and the geometry fluctuate.

At criticality, the continuum theory is Ising matter coupled to two-dimensional gravity. Its minimal-string description is labelled (3, 4). These numbers specify the orders of two differential operators used to describe the model; the matter theory has central charge c = 1/2. We work in its zero-magnetic-field sector.

The theory organizes surfaces by their genus g, the number of handles, and their number of boundaries n. A disk has (g, n) = (0, 1). Add a handle while keeping one boundary and you get (1, 1): a torus with a disk removed.

Our target is the genus-one, one-boundary amplitude. It is a generating function for that topology, expressed in a spectral coordinate. It is not a probability density or a picture of a particular surface. Converting it into a function of physical boundary length requires a further transform.

A cubic curve with a memory.

The Ising background starts with the following energy polynomial, on the real branch x > 0:

E (p, x) = p^{3} - 3 x^{\frac{1}{3}} p

Before using it, let us separate three coordinates that play very different roles.

x: The background parameter. Changing it moves us through a family of classical curves. It is not a coordinate on the physical surface.
p: A parameter along a curve. It becomes the leading logarithmic derivative of an auxiliary wavefunction.
E: The spectral energy. For a fixed background, the equation above maps a point p to its energy.
ℏ: A formal expansion parameter. In the fields used below, each additional power of ℏ² advances one genus.

Choose the endpoint x = 1, where we will read off the amplitude. Then

E (p) = p^{3} - 3 p, Δ = \frac{d E}{d p} = 3 p^{2} - 3

The derivative vanishes at p = ±1. Near either point, the leading change in energy is quadratic rather than linear. Two local solutions for p meet there. These are the ramification points; their energy values are −2 and +2.

A small surprise: a third-order description of pure gravity, with no Ising matter, has exactly the same endpoint polynomial. Away from the endpoint its background scales as x^1/2, while Ising scales as x^1/3.

To describe the boundary observable we also need a disk function, y(p). Together, E(p) and y(p) form the spectral data for the disk differential ω_0,1 = y dE. Here are the two disks at x = 1, in the normalization used throughout this article:

Ising / (3, 4)

y_{I} = - \frac{1}{4} (p^{4} - 4 p^{2} + 2)

A quartic disk function.

Pure gravity / (3, 2)

y_{G} = \frac{(p^{2} - 2)}{3}

A quadratic disk function.

These small plots use the same axes: −2.2 ≤ p ≤ 2.2 horizontally and −1.8 ≤ y ≤ 1.2 vertically. The different disks distinguish the two theories even though their endpoint energy projections agree. We will see how the Ising disk follows from the moving background.

Replace a surface sum by an ODE.

The differential operator is an auxiliary device for calculating the amplitudes. On the Ising sector we use

Q = D^{3} + b D + \frac{ℏ}{2} b^{'}, D = ℏ \partial_{x}

Primes denote derivatives with respect to x. The last term comes from placing the field and derivative symmetrically: ½(bD + Db). It is fixed by this ordering, rather than being another adjustable field. In the full third-order problem there is also a constant coefficient a; we set it to zero in the Ising sector considered here.

b = - 3 x^{\frac{1}{3}} + ℏ^{2} b_{1} + ℏ^{4} b_{2} + \dots

The first term is known. The functions b₁, b₂, … are what we want to reconstruct. At leading order, replacing D by p in the spectral equation Qψ = Eψ gives our cubic energy polynomial.

From the operator and its adjoint one builds a normalized scalar resolvent, denoted R̂. For this calculation it is a formal series encoding the local spectral response of the operator. Its expansion begins

\hat{R} = \frac{1}{Δ} + ℏ^{2} {\hat{R}}_{2} + ℏ^{4} {\hat{R}}_{4} + \dots

The leading term is simply the reciprocal slope of the energy map. Higher terms follow recursively from the differential equation. They contain the unknown corrections to b.

STEP 1Specify the backgroundThe classical curve and its motion in x.

STEP 2Expand the resolventKeep the new field correction unknown.

STEP 3Find an allowed primitiveIts coefficients fix the correction.

A primitive means an antiderivative. At positive genus, seek Q̂_g whose derivative at fixed energy is the corresponding resolvent coefficient. Evaluating it at x = 1 gives the one-boundary differential:

𝒟_{E} {\hat{Q}}_{g} = {\hat{R}}_{2g}, ω_{g,1} = {\hat{Q}}_{g} d E

Here 𝒟_E means differentiation in x while keeping E fixed. The subscript 2g on R̂ labels the power of ℏ; the subscript g on Q̂ labels the genus.

Why “fixed energy” changes the derivative

If x changes, the point p must move to keep E(p, x) unchanged. Define V = ∂E/∂x at fixed p. The chain rule gives

Δ \frac{d p}{d x} + V = 0 \Rightarrow \frac{d p}{d x} = - \frac{V}{Δ}

Thus differentiating a function of both p and x requires an extra term, −(V/Δ)∂/∂p. This is the source of the stronger poles at ramification points.

Work through the disk integral

At genus zero, 𝒟_EQ̂₀ = 1/Δ. Since dx = −Δ dp/V along a fixed-energy trajectory,

{\hat{Q}}_{0} = - \int \frac{d p}{V}

For Ising, write E = p³ + b₀p and eliminate the background coefficient:

b_{0} = \frac{E}{p} - p^{2}, x = - \frac{{b_{0}}^{3}}{27}, V = - \frac{9 p}{{b_{0}}^{2}}

Then −1/V = E²/(9p³) − 2E/9 + p³/9. Integrate term by term, treating E as a constant:

{\hat{Q}}_{0} = - \frac{E^{2}}{18 p^{2}} - \frac{2 E p}{9} + \frac{p^{4}}{36} + C (E)

Only now substitute x = 1 and E = p³ − 3p. Choosing the common integration term C(E) = 0 gives the quartic Ising disk above. Adding a common function of E does not change the difference of the disk functions on the two local sheets used by spectral recursion.

How the ODE produces the resolvent

Set w = ℏψ′/ψ. Dividing Qψ = Eψ by ψ turns the linear equation into

w^{3} + 3 ℏ w w^{'} + ℏ^{2} w^{''} + b w + \frac{ℏ}{2} b^{'} = E

Expand w = p + ℏw₁ + ℏ²w₂ + ⋯. At each order, the new coefficient is multiplied by Δ. So it is found from known lower-order terms by division by Δ.

A paired solution χ of the adjoint equation has logarithmic derivative w̃ = ℏχ′/χ. A conserved bilinear pairing, the third-order analogue of a Wronskian, fixes the normalization and gives

Σ = w^{2} + {\tilde{w}}^{2} - w \tilde{w} + ℏ (w^{'} + {\tilde{w}}^{'}) + b, \hat{R} = \frac{1}{Σ}

The symmetric ordering implies w̃(ℏ) = −w(−ℏ). Substitution makes the normalized R̂ even in ℏ, although w itself still has odd terms. This scalar recursion supplies the source used in the next step.

Let the poles determine the correction.

We have supplied the classical Ising background, but we have not yet imposed its quantum string equation. Instead we ask the primitive to satisfy two analytic conditions:

Finite poles only where sheets meet. In the scaled coordinate below, the allowed locations are z = ±1. A pole is a point where the rational function diverges.
No pole of the amplitude differential at infinity. This restricts the numerator degree. It is a condition on Q̂₁ dE, not merely on Q̂₁.

Use the scale of the classical background to make the calculation dimensionless:

U = x^{\frac{1}{3}}, z = \frac{p}{\sqrt{U}}

The homogeneous Ising expansion fixes the first correction’s dependence on x. Only its numerical coefficient remains unknown:

b = - 3 U + ℏ^{2} d_{1} U^{-6} + O (ℏ^{4})

At genus one, the scalar recurrence bounds the resolvent’s poles by order seven. The fixed-energy derivative can raise a primitive’s pole order by two. With the Ising symmetry, the allowed primitive can therefore be written

\begin{matrix} {\hat{Q}}_{1} = U^{-5} q_{1} (z) \\ q_{1} (z) = \frac{c_{0} + c_{1} z^{2} + c_{2} z^{4} + c_{3} z^{6}}{{(z^{2} - 1)}^{5}} \end{matrix}

There are five unknown numbers: d₁ and four numerator coefficients. The numerator’s degree is at most six because multiplying the primitive by dE must leave a differential regular at infinity.

Insert this form into the primitive equation and clear denominators. Matching powers of z gives a small linear system. All coefficient conditions are satisfied by a unique answer:

d_{1} = \frac{1}{8}, q_{1} (z) = - \frac{3 z^{6} - 3 z^{4} + 17 z^{2} + 7}{216 {(z^{2} - 1)}^{5}}

The reconstructed field

b = - 3 x^{\frac{1}{3}} + \frac{ℏ^{2}}{8 x^{2}} + O (ℏ^{4})

The coefficient 1/8 is determined by the primitive conditions within the specified homogeneous Ising sector.

See the complete genus-one coefficient match

After removing a common factor U⁻⁸, the resolvent source with d₁ = 0 is

s_{1} (z) = \frac{25 z^{6} - 17 z^{4} + 99 z^{2} + 13}{324 {(z^{2} - 1)}^{7}}

The fixed-energy derivative and the new field contribution give the rational identity

- \frac{5}{3} q_{1} - \frac{z (z^{2} - 3)}{6 (z^{2} - 1)} {q_{1}}^{'} = s_{1} + \frac{d_{1} (z^{2} + 1)}{9 {(z^{2} - 1)}^{3}}

Here the prime on q₁ means d/dz. Multiply by 324(z² − 1)⁷. There are six potentially nonzero coefficients, at powers 0, 2, 4, 6, 8 and 10, and each must vanish. The odd powers vanish by symmetry. Solving the six linear conditions gives:

Coefficients of the primitive and field correction
d₁	c₀	c₁	c₂	c₃
1/8	−7/216	−17/216	1/72	−1/72

There are more conditions than unknowns, but they are consistent. For example, the coefficient of z¹⁰ is −324c₃ − 36d₁, which vanishes for these values. Substitution makes the entire polynomial zero.

Read off the surface amplitude

At x = 1, we have U = 1 and z = p. Multiplying the primitive by Δ converts dE to dp. Writing ω_1,1 = W_1,1(p) dp, the answer is

One handle · one boundary

W_{1,1} (p) = - \frac{3 p^{6} - 3 p^{4} + 17 p^{2} + 7}{72 {(p^{2} - 1)}^{4}}

This is the first positive-genus one-boundary amplitude, with the disk normalization fixed above.

Its only finite poles are at p = ±1. At large p, it behaves as −1/(24p²). Since dp/p² = −d(1/p), the differential is regular at infinity. The overall sign follows our spectral convention; this amplitude is not required to be a positive function.

Check against the Ising string equation.

Now bring in an independent description of the same model. Introduce the conventional potential 𝔲 = −b/3. The known Ising string equation, in our normalization, is

𝔲^{3} - \frac{3}{4} ℏ^{2} 𝔲 𝔲^{''} - \frac{3}{8} ℏ^{2} {(𝔲^{'})}^{2} + \frac{ℏ^{4}}{24} 𝔲^{''''} = x

Our reconstructed field predicts

𝔲 = - \frac{b}{3} = x^{\frac{1}{3}} - \frac{ℏ^{2}}{24 x^{2}} + O (ℏ^{4})

Substitute this into the equation. At order ℏ⁰ it reduces to (𝔲₀)³ = x. At order ℏ², the contribution from the correction to 𝔲 cancels the two derivative terms. The first quantum correction therefore passes a check that was not used to determine it.

The manuscript continues the same reconstruction through genus three in the homogeneous zero-magnetic-field sector. Those field corrections satisfy the string equation through ℏ⁶, and the one-boundary amplitudes agree with the known Ising results after accounting for the disk normalization. This is a reconstruction of established amplitudes, and a test of the method.

What changes when we add boundaries?

For one boundary, the normalized scalar pairing is enough. For several boundaries, its components are assembled into a rank-one matrix projector. Cyclic traces of products of this projector generate connected multi-boundary correlators.

The planar three-boundary result in the same normalization is particularly short:

W_{0,3} = - \frac{1}{6} \sum_{α = \pm 1}^{} \prod_{i = 1}^{3} \frac{1}{{(p_{i} - α)}^{2}}

Each p_i labels a boundary’s spectral coordinate. The two terms come from the two ramification points. Identifying the full matrix construction with topological recursion requires properties of the specified string system in addition to the scalar uniqueness argument.

What the example teaches

A boundary amplitude can constrain the quantum equation that produces it. In the Ising example, once the moving classical background, ordering and sector are specified, the allowed primitive determines the first correction. The disk keeps track of the background motion that an endpoint energy polynomial alone cannot reveal.

Sources & conventions

This article follows the Ising example in Resolvent reconstruction of minimal strings beyond KdV (arXiv:2609.28964), by Wasif Ahmed and Joydeep Naskar. It uses the monic spatial operator, symmetric ordering, endpoint x = 1 and quartic disk displayed above. The genus counts handles of the surfaces being summed over; the auxiliary classical curve is rational and has geometric genus zero.

M. Bergère, G. Borot and B. Eynard, Rational differential systems, loop equations, and application to the q-th reductions of KP. This reference is abbreviated BBE below. Section 6.3 provides the Ising string system and amplitudes. To match the conventions here, use t_B = 4x and ℏ_B = 4ℏ.
W. Ahmed, C. V. Johnson and K. Saraswat, Weil–Petersson volumes for extended JT supergravity from ordinary differential equations. Appendix C explains the descending primitive construction in the second-order setting.

The disk used here is one quarter of the BBE disk. Consequently stable amplitudes in these conventions are 4^2g−2+n times the corresponding BBE amplitudes. Keeping the disk and operator normalizations together is essential when comparing formulas.

A quantum surface,
one boundary at a time.

Let the lattice fluctuate, too.

A cubic curve with a memory.

Same snapshot. Different motion.

Ising / (3, 4)

Pure gravity / (3, 2)

Replace a surface sum by an ODE.

Why “fixed energy” changes the derivative

Let the poles determine the correction.

Read off the surface amplitude

Check against the Ising string equation.

What the example teaches

Sources & conventions

A quantum surface,one boundary at a time.

Let the lattice fluctuate, too.

A cubic curve with a memory.

Same snapshot. Different motion.

Ising / (3, 4)

Pure gravity / (3, 2)

Replace a surface sum by an ODE.

Why “fixed energy” changes the derivative

Let the poles determine the correction.

Read off the surface amplitude

Check against the Ising string equation.

What the example teaches

Sources & conventions

A quantum surface,
one boundary at a time.