Induced Gravity on the Selection–Stitch Lattice Framework

Induced Gravity on the Selection–Stitch Lattice:

Covariance from Vacuum Fluctuations, Strength from

Entanglement

Raghu Kulkarni

SSMTheory Group, IDrive Inc., Calabasas, CA 91302, USA

raghu@idrive.com

Abstract

In the Selection–Stitch Model (SSM), space is a lattice of entangled bonds. The vacuum’s

quantum ﬂuctuations, the ocean, induce the Regge action on this lattice if they are covariant.

We ask why the ocean is covariant and what ﬁxes Newton’s constant, and answer both with two

forms of one principle. First, the ocean is maximally mixed. At ﬁxed density, the maximally

mixed state of its positions is a Poisson process. If its couplings also keep no memory of an

earlier geometry, its vacuum energy depends on the metric only through the volume, and the

graviton stays massless. A sprinkled ocean responds to a shear at 1

001

002 of the covariant

value. The same operator on the lattice gives 1

23, and a memory

of an old geometry gives

exactly 1

−µ

. The ocean has no density halo around a defect. Second, maximal entanglement. If

each bond’s entanglement follows its share of area, entanglement equilibrium is the Hamiltonian

constraint. A solved defect slice satisﬁes it to within 5% once a ball spans two lattice spacings. If

the bond density maximizes the entanglement per unit area, that maximum is 1

. A two-qubit

bond model realizes this law. Since the bonds are frozen ocean, the two results describe one

substance.

Keywords: induced gravity; covariance; maximally mixed state; entanglement equilibrium; Regge

calculus; Newton’s constant

1 Introduction

Sakharov proposed that gravity is induced by the vacuum’s quantum ﬂuctuations [

]. In this

picture, the action of gravity is the energy cost of bending the space the ﬂuctuations live on. The

Selection–Stitch Model (SSM) gives this idea a concrete setting. Its space is a lattice of entangled

bonds. Its curvature sits on the lattice’s hinges as Regge deﬁcits [

]. Section 2 describes the model.

Ref. [

] showed three things. If the vacuum’s ﬂuctuations are covariant, they induce exactly

the Regge action on the lattice. The ﬂuctuations cannot be the lattice’s own ﬁelds, because those

give the graviton a mass. A defect’s veriﬁcation cost acts as a rest mass. As there, we call the

ﬂuctuations the ocean. Two questions were left open. One is why the ﬂuctuations are covariant.

The other is what ﬁxes the strength of gravity, Newton’s constant G.

This paper answers both questions with two forms of one principle. The principle is maximal

entropy in quantum form. The ocean’s state has the largest von Neumann entropy its density allows,

and the bonds carry the largest entanglement entropy per unit area.

The form of gravity comes from a maximally mixed ocean. The ocean is in the maximally

mixed state at ﬁxed density, and it keeps no memory of earlier geometries. Then its vacuum

energy depends only on the volume, which is covariance (Sections 3–6). The same ocean has

no density halo around a defect (Section 7).

The strength of gravity comes from maximal entanglement. The lattice’s bonds carry the vac-

uum’s entanglement. Their equilibrium with matter is the Hamiltonian constraint (Section 8).

Maximizing their entanglement per unit area ﬁxes 1/4G (Section 9).

In the SSM the bonds are made from the ocean, as ice is made from water. The two answers

therefore describe one substance (Section 10). Section 11 lists what remains open and grades each

claim (Table 5).

2 The SSM framework

This section summarizes the parts of the SSM that this paper uses. Each part rests on a published

or archived paper. This paper does not test those papers’ claims.

The vacuum assembles into a lattice. The basic object of the SSM is a bond, a Bell pair that

joins two nodes at a ﬁxed distance [

]. The network grows by two operations (Fig. 1a). A stitch

adds a node in the plane of an existing bond. A lift adds a node above a triangle. Stitches build ﬂat

sheets with six neighbors per node. Rare lifts build a foam of regular tetrahedra. That foam cannot

ﬁll space. Five tetrahedra around an edge leave a gap of 7

◦

. To relieve this strain, the foam turns

into stacked sheets. The result is the face-centered-cubic (FCC) lattice, with twelve neighbors per

node (Fig. 1b,c). Twelve is the most that can touch one sphere in three dimensions, and FCC is

the densest packing of equal spheres [

]. A random simulation of the two operations supports this

picture [5].

The lattice is a quantum code. Ref. [

] puts one qubit on each edge of the FCC lattice. It

deﬁnes two kinds of check (Fig. 2a,b). A

-check acts on the twelve bonds of a node. An

-check

acts on the twelve edges of an octahedral void. The two kinds commute, so they deﬁne a CSS

code [

]. On the smallest periodic box the code is [[192

130

3]], and its rate tends to 2

3. In the

SSM the vacuum is the code space of this code. A state that satisﬁes every check is a vacuum state.

In the SSM’s reading, a bond persists because the code keeps verifying it.

Matter is a trapped node. If crystallization is incomplete, a node of the foam can be left at

the center of a tetrahedral void [

]. It is bonded to the four corners of the void by four short bonds,

which we call spokes (Fig. 3c shows a two-dimensional cartoon). The node cannot leave the void.

To pass through a face, it would have to come closer to the three corners than the lattice allows.

Ref. [

] reads this as conﬁnement and identiﬁes the defect with a baryon. It also derives charges

and colors from the defect’s geometry. This paper uses only the geometry of the defect.

Mass is the cost of veriﬁcation. Ref. [

] proposes that a defect’s mass is the cost of checking it.

The cost is a count, the number of qubits the defect touches times the number of checks that can

detect it. Ref. [

] states the step from a count to an energy as a postulate for standing information.

Information that is held in place and never erased has energy. Each bit carries a ﬁxed amount

, so

a defect’s energy is its count times

. Landauer’s principle prices erasure and does not apply here.

Only ratios are used, so mass ratios are ratios of integers. The ﬁve costs that survive its selection

rules match the electron, muon, pion, proton, and neutron masses to within 0

12%. This paper does

stitch: new node

2 bonds, in plane

lift: new node, 3 bonds

out of plane,

√

2/3

(a) the two assembly operators

(b)

= 12 = 6 + 3 + 3

(FCC, ABC stacking)

6 in plane (stitch)

3 above (lift)

3 below (lift)

Bell pair

= 1

stitched sheet

= 6

tetrahedral foam

= 4

, frustrated

FCC crystal

= 12

, Kepler bound

stitch

lift,

lift

−

deficit

= 0.128

rad relieved

a node left in a tetrahedral void = matter

Figure 1: Assembly in the SSM [

]. (a) The two operations. A stitch adds a node in the plane, and

a lift adds a node above a triangle. (b) The ﬁnal coordination, with six neighbors in the node’s own

sheet and three in each sheet above and below. (c) The cascade from one bond to the FCC crystal.

A node left behind in a tetrahedral void is the SSM’s matter.

(a)

-check on a node (b)

-check on an octahedral void

Figure 2: The SSM code [

]. (a) A

-check on the twelve bonds of one node. (b) An

-check on

the twelve edges of one octahedral void. (c) The 13-site cluster of a node and its neighbors, whose

counts set the veriﬁcation costs of Ref. [9].

not use these values. It uses only one result of Ref. [

]. A veriﬁcation cost couples to geometry as a

rest mass, if veriﬁcation ticks measure proper time.

Gravity is Regge calculus on the lattice. In spacetime the SSM lattice is the

lattice,

whose slice at ﬁxed time is FCC [

]. Its geometry is ﬁxed by its edge lengths alone. Curvature sits

on the hinges as deﬁcit angles, and the Regge action is

πG

over the hinges [

]. On

this action reproduces the Einstein–Hilbert action exactly at second and third order. Its static ﬁeld

gives Newton’s law and the post-Newtonian parameters

= 1

000

002 [

]. Ref. [

] sets

the lattice scale. It matches one bit of entanglement per bond to the Bekenstein–Hawking entropy,

which puts the bond length at the Planck scale. We use the calibration of Ref. [

with

the

nearest-neighbor distance, so

= 1

665

ℓ

. Ref. [

] also treats black holes, as

regions in which the lattice has broken down.

(a) the ocean (b) frozen ocean: the lattice (c) a trapped node: the bonds bend

Figure 3: The SSM’s two phases, drawn in two dimensions. (a) The ocean, made of degrees of

freedom with no orientation. (b) The frozen phase, a lattice of bonds with the ocean around it. (c)

A trapped node (red) in a void, bonded by spokes. Nearby bonds change length, and this is the

curvature. The ocean is excluded only from the node’s own volume (Section 7).

The ocean induces the Regge action. The vacuum’s quantum ﬂuctuations are measured

directly today [

]. In the SSM they are older than the lattice. The bonds and defects were

stitched from them, and they persist around the ﬁnished lattice [

]. They are the ocean of Section 1

(Fig. 3). Ref. [

] shows that a covariant ocean induces the Einstein–Hilbert action. On the lattice’s

piecewise-ﬂat geometry this action is exactly the Regge action [

]. The check is done on the conical

hinges themselves, from the exact spectrum of a cone. Fields that live on the lattice itself fail.

Their vacuum energy depends on the shape of the lattice cells, and that gives the graviton a mass.

Covariance is the postulate this paper derives.

3 The covariance test

Covariance has a simple test on a lattice. Strain the lattice uniformly. The result is still ﬂat space,

written in other coordinates, and every Regge deﬁcit stays zero. A covariant vacuum energy may

therefore change with the volume, but not with the shape.

We place the ocean’s degrees of freedom at points

in a periodic four-dimensional box. The

points move with the lattice. We give the box a uniform metric

. The ocean’s ﬂuctuation operator

is a graph Laplacian L(g), and its one-loop vacuum term per degree of freedom is

f(g) =

ln det

′

L(g), (1)

where

det

′

omits the zero mode. The couplings are Gaussian in proper distance, with a smooth cut:

= e

−d

/2σ



1 − d



< r

), d

= ∆x

g ∆x

. (2)

Lengths are in half-cells of the D

lattice, and there is one point per D

lattice site on average.

Let

df/dϵ

be the volume response, with

= (1 +

)

. Let

f/dt

be the shear

response, with

, where

is traceless and

tr H

= 1. If

depends only on

det g

, then

= −κ

. We report

R =

−κ

, (3)

which is 1 for a covariant medium. If

R 

= 1, the vacuum energy depends on the shape of ﬂat space.

In the continuum this is a mass term for the graviton [

]. Gravitational waves bound the

graviton mass tightly [17], so R = 1 must hold almost exactly [4].

4 Placement from a maximally mixed ocean

Suppose we know only the mean density n of the ocean’s degrees of freedom. The placement with

the most entropy is then the Poisson process of intensity

, in which the points are independent

and uniform [

]. Any correlation, preferred spacing, or preferred direction would be extra

information. It would lower the entropy. In quantum terms, this placement is the maximally mixed

state of the ocean’s positions at ﬁxed density, the state of largest von Neumann entropy with that

constraint. Its position statistics are those of a Poisson process. The state need not be prepared.

Almost every pure state of a large system looks maximally mixed on a small part of it [20].

A Poisson process has the symmetry that covariance needs. Let

be a linear map with

det A

= 1. Moving every point by

turns a Poisson process into another Poisson process with

the same intensity [

]. The couplings depend only on proper distances. Moving the points by

and changing the metric to

−T

−1

therefore leaves the operator unchanged. Averaged over the

process,

f(A

gA) =

f(g) for every A with det A = 1. (4)

Any metric can be brought to a multiple of the identity by such a map. As a result,

depends

only through

det g

, and

= 1. This is exact for the average. A single sprinkling ﬂuctuates

around it, and the ﬂuctuation averages out over many points. The density is the only scale the

placement has, so it sets the ocean’s cutoﬀ. In spacetime, the same symmetry is the theorem that a

Poisson sprinkling picks out no frame [

]. Causal sets rest on this property [

]. Random lattices

were introduced in ﬂat space for a related reason, to restore rotation and translation symmetry on

average [23]. Ref. [24] built induced gravity from matter on a lattice of randomly placed points.

A regular lattice lacks the symmetry. A volume-preserving map moves its points onto a diﬀerent

lattice. Table 1 compares the two with the same operator. The sprinkled points pass the test. The

lattice points fail it, and they fail more as the kernel narrows and the operator sees the lattice’s

arrangement more closely.

Table 1: Shear response relative to the covariant value, Eq.

(3)

, for the same memoryless operator

on sprinkled points and on

lattice points. Box side 8 (2048 points). The sprinkled value is an

average over 10 sprinklings with random shear directions. The two lattice values are for a diagonal

and an oﬀ-diagonal shear.

points kernel width σ cut r

Poisson sprinkling 0.7 2.24 1.0009 ± 0.0015

lattice 0.7 2.24 1.233, 1.234

lattice 1.0 3.2 0.994, 0.995

5 Couplings without memory

The argument above assumed that the couplings depend only on the current metric. A medium with

no structure of its own has nothing else to depend on, except a second geometry, a reference metric

. A medium that formed in one geometry, and kept its couplings when the geometry changed,

carries such a g

. We call this memory.

Memory allows shape terms. With a reference metric, Eq.

(4)

holds only when

and

are

moved together. The vacuum term can then depend on the eigenvalues of

−1

. Terms such as

(

−1

) change under a shear at ﬁxed volume, and nothing forbids them. These are the potential

terms of massive gravity, which needs a reference metric for this reason [

]. A medium with

memory could still be covariant, but only if every shape term happened to cancel. Without memory,

covariance is forced by symmetry.

An exact example. Let every coupling see the blended metric

′

= (1

− µ

)

µg

, with

and memory

. If the memoryless medium is covariant, then

(

) =

(

det g

′

) for some function

A compression changes

det g

′

at ﬁrst order with weight 1

− µ

. A shear changes it at second order

with weight (1 − µ)

. The ratio is then

R(µ) = 1 − µ. (5)

Only

= 0 and

= 1 are covariant. At

= 1 the medium ignores the geometry completely. Then

and

both vanish, and the medium induces no gravity. Any value in between gives the graviton

a mass. The shape excess is

− µ

)

1,0

, where

1,0

0 is the memoryless volume

response.

Memory of neighbors. A milder memory keeps the coupling strengths current but remembers

which points are neighbors. In Eq.

(2)

, let the cut factor use the old distance

for a random

fraction

of the pairs. The other pairs use the current distance. The Gaussian factor always uses

the current distance. Table 2 and Fig. 4a show both kinds of memory on sprinkled points. The

blend follows Eq.

(5)

within sampling error. Remembering neighbors also breaks covariance, and

the break grows with the fraction remembered. When every pair remembers,

= 0

23. A ﬁxed

triangulation is another memory of neighbors. We built the Delaunay triangulation of a sprinkling,

held it ﬁxed, and computed the ﬁnite-element Laplacian from the current edge lengths. It gives

= 1

96 (box side 5, 625 points at twice that density, two sprinklings). Memory can push

either

way. Its sign and size depend on how the old structure enters the couplings.

Table 2: Shear response relative to the covariant value for two kinds of memory on a Poisson

sprinkling (box side 6, 648 points, 20 sprinklings). Blend:

= 1,

= 3

2. Remembered neighbors:

σ = 1, r

= 2.

memory µ R, blend fraction ϕ R, remembered neighbors

0 0.9995 ± 0.0009 0 0.999 ± 0.003

0.25 0.7497 ± 0.0007 0.25 0.942 ± 0.003

0.5 0.4998 ± 0.0005 0.5 0.845 ± 0.003

0.75 0.2499 ± 0.0002 0.75 0.675 ± 0.002

0.9 0.1000 ± 0.0001 1 0.232 ± 0.001

0.0 0.2 0.4 0.6 0.8 1.0

memory

0.0

0.2

0.4

0.6

0.8

1.0

shear response / covariant value

(a) memory breaks covariance

−

blend: metric

−

)

remembered neighbors, fraction

0.0 0.5 1.0 1.5 2.0 2.5

distance from the defect (half-cells)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

density / far density

defect

(b) no halo around a defect

maximally mixed ocean

halo, for comparison

Figure 4: (a) Shear response relative to the covariant value against memory, for the blend of Eq.

(5)

(circles, with the line 1

−µ

) and remembered neighbors (squares). (b) Density of a maximally mixed

ocean around an excluded ball of radius 0

6, relative to the far density, averaged over 400 sprinklings.

A 1/r proﬁle is shown for comparison.

6 Memory is veriﬁcation

In the SSM the lattice persists because the code veriﬁes it (Section 2). The ocean is the part of

the vacuum that no check veriﬁes. Its ﬂuctuations arise and decay without being recorded. We

identify these two kinds of persistence with the two kinds of memory. A veriﬁed structure keeps its

connections and its formation geometry, so ﬁelds that live on it carry a reference geometry. The

unveriﬁed ocean keeps no record, so its couplings can depend only on the geometry as it is now,

which is µ = 0.

This identiﬁcation explains the two main results of Ref. [

] as one result. Fields on the lattice

live on remembered structure. The symmetry of Section 4 does not protect them, and every coupling

rule tested in Ref. [

] gave a shape-dependent vacuum energy. The ocean carries no reference

geometry. Eq.

(4)

holds for it, and its induced action is covariant, so by Ref. [

] it is the Regge

action. The identiﬁcation is a modeling step. It ﬁts a model that deﬁnes persistence by veriﬁcation,

but this paper does not derive it from the code’s dynamics.

7 The ocean around a defect

A defect occupies a small volume

that the ocean cannot enter. Take the maximally mixed

placement again, with that volume excluded. The points are absent inside it and uniform outside

it. Their density outside is

(

V − v

). It is higher than without the defect by the fraction

v/V

and the same at every distance. There is no gradient. A Poisson process has no interactions, so

no point can sense that the defect is near. Fig. 4b shows the proﬁle. It is ﬂat within the sampling

noise, right up to the edge of the excluded ball.

A density halo would need an energy cost for gradients of the density,

|∇n|

. A maximally

mixed medium has no such cost. Even with one, and with a compressibility term

′′

δn

, the

density would relax as

−r/ξ

with

κ/f

′′

. That is short-ranged. A 1

tail would need

′′

= 0. Two consequences follow. First, the 1

ﬁeld of a mass is not a thinning of the ocean. It

is the curvature of the geometry, which the induced Regge action governs (Fig. 3c). Second, the

ocean’s density, and so its cutoﬀ, is the same near matter as far away. A Newton constant set by

that cutoﬀ therefore does not vary near masses, as tests of the equivalence principle require.

8 The strength of gravity from entanglement equilibrium

Covariance ﬁxes the form of the induced action. It does not ﬁx the coeﬃcient, 1

. For the

coeﬃcient we use the entanglement that the lattice carries. Each bond is a Bell pair, and the

vacuum’s entanglement across a surface is carried by the bonds the surface cuts. That Einstein’s

equation is a condition on the vacuum’s entanglement is Jacobson’s idea. He ﬁrst derived it as an

equation of state of local horizons [

], and later as the equilibrium of the entanglement in small

balls [

]. In holography the linearized Einstein equation follows in the same way from the ﬁrst law

of entanglement [

]. Others have sought geometry directly from the entanglement of abstract

degrees of freedom [

]. Here the degrees of freedom are the bonds. Two postulates deﬁne the

law.

Postulate A: entanglement follows area. The entanglement across any surface Σ in a slice is

(Σ) =

(Σ)

, where

is the area in the lattice’s geometry. In the ﬂat vacuum, each bond that

Σ cuts carries entropy s

and a share L

of its area. Then

G =

. (6)

With

2 this is the calibration of Section 2. When the lattice strains, a bond’s share of area

changes. Its entanglement must change with it, as

(0)

. Here

is the area of the bond’s

dual face.

Postulate B: the vacuum is in entanglement equilibrium. For every small ball at ﬁxed volume,

the change of entanglement from the geometry balances the change from the matter inside:

δA|

+ 2π δ⟨K⟩ = 0, K =

− r

dV. (7)

Here

is the modular Hamiltonian of a ball of radius

in ﬂat space [

]. On a time-symmetric

slice, to ﬁrst order in the curvature,

δA|

−

(

− r

)

R dV

. Equilibrium for every

ball then requires

= 16

πGρ

at every point. This is the Hamiltonian constraint of general

relativity. Two checks ﬁx the normalization. For a point mass, Schwarzschild initial data give

δA|

−

πGMR

, and 2

πδ⟨K⟩

πMR

. For a straight cosmic string of mass

per unit length,

the two sides balance when its deﬁcit angle is 8πGλ.

Three properties follow.

is ﬁxed by the area law, with no sum over particle species. A uniform

vacuum energy changes no entanglement, so it does not enter, and the cosmological constant is only

a constant of integration [

]. With boosts, equilibrium of every small causal diamond gives the

full Einstein equation [

]. We use only the time-symmetric slice. The source is the defect’s rest

mass, which by Ref. [4] is its veriﬁcation cost.

A test on the lattice. On a piecewise-ﬂat slice,

R dV

becomes 2

ℓ

on the edges. Equilib-

rium then becomes one equation per node:

e∋v

ℓ

= 16πG m

, (8)

the vertex form of the Hamiltonian constraint in 3+1 Regge calculus [

]. Here

is the rest mass

assigned to node

. We place the trapped node of Section 2 in the FCC lattice (Fig. 5a). Its rest

(a) trapped node: spokes (red), cage (blue)

0.6 1 2 3 4 6

distance of edge midpoint from void center (half-cells)

−

largest

| /

spoke

−

(b) curvature off the spokes (

= 0.1

)

Figure 5: (a) The trapped node

at the center of a tetrahedral void, joined by four spokes (red)

to the corners

–

. The void’s edges are blue. (b) Curvature oﬀ the spokes in the solved slice at

= 0

1 and box side 16. Points show the largest

|δ

among edges in each distance shell, in units of

the spoke deﬁcit. The dotted line shows r

−3

. The wall at r = 7.6 steepens the last points.

mass is spread over the four spokes, with mass

per unit length, and each spoke gives half its mass

to each end. We write

= 8

πGλ

. The slice is conformally ﬂat,

ℓ

(0)

. This is an assumption

about the initial data. It is exact for Schwarzschild initial data and at ﬁrst order. We impose Eq.

(8)

at every node inside a wall of radius

. A wall is needed, because a closed three-torus admits no

metric whose scalar curvature is nonnegative and somewhere positive [34, 35].

We solve the system by least squares on a periodic FCC box. At the trapped node only the

spokes meet. By the symmetry of the void, each spoke carries the deﬁcit

= 8

πGλ

exactly.

This is the deﬁcit of a cosmic string. Away from the defect, the conformal factor is

= 1 +

GM/

This is Schwarzschild initial data, with

equal to the source mass to 0

36% at box side 16. Fig. 5b

shows how the curvature falls oﬀ with distance.

We then test Eq.

(7)

directly on the solved slice, at box side 16, wall radius 7

6, and

= 0

01.

Each tetrahedron carries the constant metric ﬁxed by its six edge lengths. Through these metrics

we compute the area of a sphere of radius

and the volume inside it. We compare the area deﬁcit

at ﬁxed volume,

δA|

, with the matter side,

−

πG δ⟨K⟩

, from the spoke source. This comparison

checks consistency and does not test the postulates independently. The slice was built to satisfy

Eq.

(8)

, and at ﬁrst order equilibrium of every round ball is equivalent to it. The check tests whether

this equivalence survives on a lattice whose curvature sits on edges. Fig. 6 shows that it does.

The ratio of the two sides is 0

84, 0

96, 0

98, 0

99, and 0

99 at

= 1

5, 2

5, 3

5, 4

5, and 5

5. It

approaches one once the ball spans about two lattice spacings. Balls with no source inside are in

equilibrium to about 0

5%. Regions built from whole tetrahedra fail even without a source, because

their jagged boundaries pick up single edge strains. Equilibrium is a statement about round balls.

2 3 4 5

ball radius

(half-cells)

0.02

0.04

0.06

0.08

magnitude

(a) the two sides of the equilibrium condition

matter side,

⟨

⟩

geometry side,

−

2 3 4 5

ball radius

(half-cells)

0.80

0.85

0.90

0.95

1.00

ratio

two lattice

spacings

(b) geometry side / matter side

Figure 6: Entanglement equilibrium, Eq.

(7)

, on the solved trapped-node slice (box side 16, wall

radius 7

= 0

01, lengths in half-cells, lattice spacing

√

). (a) The geometry side and the matter

side. (b) Their ratio.

9 The bonds’ microscopic law from maximal entanglement

Postulate A asks each bond’s entanglement to follow its share of area. This section asks what can

supply that rule.

Counting bonds is not enough. One could take the entanglement across a surface to be

times the number of bonds it cuts. But when the edge lengths change, that number stays the

same while the area changes. A count measures geometry only if curvature comes from adding or

removing bonds at ﬁxed length. In the solved slices, curvature comes from changed edge lengths.

The code alone cannot supply the rule. In any stabilizer state, the entanglement across a cut

is a whole number of ebits. The checks that straddle the cut ﬁx this number [

]. No check involves

the edge lengths. The code, as deﬁned, therefore supplies at most the counting rule. The area rule

needs a bond whose entanglement changes with its length. For a pair of qubits this is the state

cos θ

|00⟩ + sin θ

|11⟩, with entropy

s(θ

) = −cos

ln cos

− sin

ln sin

, (9)

which rises to ln 2 as θ

→ π/4.

A minimal bond model. With

ﬁxed by Eq.

(6)

, the area rule requires

d ln s

/d ln a

= 1 at

the vacuum. Give each bond’s two qubits the Hamiltonian

= −J(a

) (X

+ Z

) − h (Z

+ Z

), J(a

) = J



(0)



, (10)

so that the coupling grows with the bond’s dual area. The ground state has the form of Eq.

(9)

Table 3 shows its entanglement and response for

= 1. Stretching a bond raises its entanglement,

as required. But the slope is one only at h/J

= 0.69.

Table 3: Bond model of Eq.

(10)

with

= 1. Vacuum entanglement and its response to the dual

area.

h/J

0.10 0.30 0.50 0.69 1.00 1.50 2.00

/ ln 2 0.97 0.80 0.60 0.45 0.30 0.17 0.11

d ln s

/d ln a

0.06 0.39 0.75 1.00 1.25 1.45 1.54

Maximal entanglement ﬁxes the slope. A second use of maximal entanglement selects this

point. It varies something diﬀerent from Eq.

(7)

. Jacobson’s condition varies the geometry of a

ball, holds the vacuum’s microscopic state ﬁxed, and takes 1

as given [

]. Here we hold the

geometry ﬁxed and vary the microscopic state. The ball and its boundary area

stay the same,

but the density of bonds crossing the boundary changes. The boundary is crossed by

A/a

bonds,

where

is the vacuum’s dual area per bond. The entanglement across it is then

= (

A/a

)

(

We postulate that the vacuum’s bond density maximizes this entanglement. Then

s/a

is stationary

at the vacuum:

s(a)



= 0 ⇐⇒

d ln s

d ln a



= 1 ⇐⇒

∂S

∂a



. (11)

This holds for any bond model in which

s/a

has a maximum. It makes 1

the largest entanglement

per unit area that the bonds can carry (Fig. 7a). Ref. [

] conjectured that the vacuum’s entanglement

per unit area equals 1

. Here the equality follows from the maximization of Eq.

(11)

, which we

postulate. In the bond model it ﬁxes the parameters. For

= 1 the maximum is at

h/J

= 0

691,

with

= 0

2 (Table 4). A coupling that follows the bond length instead,

= 1

2, has no

maximum.

This variation is allowed because the number of bonds is not conserved in the SSM. Bonds form

by selection and stitching [

], so a vacuum with a diﬀerent bond density is an allowed state of the

same geometry. The variation compares uniform vacua. Changing the density in one region requires

restitching. This postulate is an addition to Jacobson’s condition, which does not imply it.

Table 4: Bond model at the maximum of

s/a

, Eq.

(11)

, for three exponents

. The slope is one

in each case.

loc

is the local Newton constant at a 10% and a 49% stretch of the dual area,

under assumptions (i)–(iii) of the text. The last column is the stretch at which

loc

= 1

08. For

p = 1/2, s/a has no maximum.

p h/J

/ ln 2 G

loc

/G (10%) G

loc

/G (49%) stretch at 1.08

1 0.691 0.45 1.08 1.54 9.8%

1.5 0.450 0.65 1.20 2.62 4.4%

2 0.358 0.74 1.33 4.33 2.8%

Calibration. With

ﬁxed, Eq.

(6)

ties the bond length to

, as

∝ s

1/2

. The bond model’s

= 0

2 gives

= 1

121

ℓ

instead of 1

665

ℓ

. Every result stated in lattice units is

unchanged. This includes the deﬁcits, the equilibrium ratios, and the shear tests.

Strong ﬁelds. At a maximum,

s/a

curves downward. In any such model, a bond therefore follows

its area exactly at ﬁrst order and falls below it at second order. Reading this shortfall as a change

takes three assumptions. (i) The bonds hold the stretch, with no restitching. (ii) The stretch

0 1 2 3 4

dual area

0.4

0.5

0.6

0.7

0.8

0.9

1.0

entanglement per area,

(

)/(

)

(a) the vacuum sits at the maximum of

0 10 20 30 40 50

stretch of the dual area

(

)

1.0

1.2

1.4

1.6

1.8

2.0

loc

neutron-star bound

(b) local Newton constant if bonds hold the stretch

= 1

= 1.5

= 2

Figure 7: The bond model. (a) Entanglement per unit area against the dual area, for

= 1 at

h/J

= 0

691. The vacuum sits at the maximum, where

d ln s/d ln a

= 1. (b) The local Newton

constant if bonds hold a stretch without restitching, for three exponents

. The dashed line is the

bound from two-solar-mass neutron stars [38].

is uniform over a ball small enough for Eq.

(7)

to apply, so that

loc

G s

′

(

)

′

(

). (iii) The

stretch is measured from the vacuum edge lengths, which we estimate by

on a conformally ﬂat

slice. Under these assumptions, for

= 1,

loc

grows by 8% at a stretch of 10%. At a neutron star’s

surface (1

4 solar masses within 12 km) the stretch is about 49%, and

loc

grows by 54% (Fig. 7b).

Two-solar-mass neutron stars bound

inside them to at most 8% above its laboratory value at

95% conﬁdence, for a uniform change of

[

]. Applied to

loc

, this bound means that bonds do

not hold such stretches. The lattice must restitch once the stretch reaches a few percent to about

10%, depending on the model. How fully restitching relieves the stretch is not computed here. It is

where this law could depart from general relativity near compact objects.

10 One substance in two phases

The two halves of this paper describe one medium. In the SSM the bonds and defects are made

from the ocean, as ice is made from water [

]. Stitching ﬁxes the ocean’s correlations into Bell

pairs on the bonds. A trapped node is a frozen remnant of the same process (Fig. 3). The liquid

phase is covariant and has no orientation, and it gives gravity its form. The frozen phase carries the

entanglement whose equilibrium is the ﬁeld equation, and it gives gravity its strength. The ocean’s

entanglement across a surface and the bonds’ entanglement are therefore one quantity, counted once.

The SSM supports this reading, but no calculation here tests it.

This reading turns the induced Newton constant of Ref. [

] into a consistency condition. The

ﬁelds that froze into the bonds must together carry the bonds’ area law. Their induced coeﬃcient,

ind

, must equal 4

. Here

is a weighted count of the ﬁeld species. With one

ocean mode per

site, the cutoﬀ is Λ

= 4

216 (heat-kernel count) or 5

013 (Debye count) [

The condition reads

C =

48πs

(ΛL

)

(

5.88 or 4.16 (s

= ln 2),

2.66 or 1.88 (s

= 0.45 ln 2).

(12)

In the proper-time scheme, the SSM’s ﬁeld content gives

= 4 [

]. This is within a factor of 1

5 of

either requirement at

2, and within a factor of 2

1 at

= 0

2. But the weight of each

species depends on the regulator, not only on covariance. We also computed the weights with an

explicit covariant regulator, a sprinkled ocean with couplings in geodesic distance and path-ordered

spin transport. Scalars and gauge ﬁelds keep their proper-time signs, enlarged by factors of 2 to 4.

Each Weyl fermion gets a negative weight, from

−

6 to

−

12 in units of a minimal scalar, depending

on the fermion operator. With 48 Weyl fermions,

is then negative for this ﬁeld content. Whether

Eq.

(12)

holds therefore depends on how the ocean’s fermions couple to curvature at the cutoﬀ. In

this paper

is ﬁxed by the bonds, so the answer changes no result above. It decides only whether

the ocean’s own ﬁelds reproduce that value.

11 What remains open

The bonds’ law. Two inputs of the bond model are assumed, not derived from the code. One is

how the coupling depends on the dual area,

J ∝ a

with

= 1. The other is the maximization of

Eq. (11). How fully restitching relieves strong-ﬁeld stretches is not computed.

Memory and veriﬁcation. That veriﬁed structure carries a reference geometry, and the unveriﬁed

ocean does not, is a modeling step (Section 6). Deriving it needs an account of how veriﬁcation

records geometry.

The ocean’s own coeﬃcient. Whether the ocean’s ﬁelds reproduce the bonds’ area law, Eq.

(12)

depends on how fermions are regulated at the cutoﬀ. Covariance does not ﬁx this.

Lorentzian signature and higher orders. The tests are Euclidean, and equilibrium is applied

on a time-symmetric slice. In Lorentzian signature a Poisson sprinkling is a causal set, which

picks out no frame [

]. But couplings that depend only on proper intervals are nonlocal there.

Entanglement equilibrium has no action, so it gives no corrections beyond Einstein’s equation. The

cosmological constant remains a constant of integration whose value is not ﬁxed.

Table 5: Status of the claims.

claim status

SSM framework (Section 2) from Refs. [5, 7, 9, 10, 11, 12, 4]

maximally mixed placement at ﬁxed density is a Poisson process standard [18, 19]

Poisson placement with memoryless couplings gives

= 1, Eq.

(4) exact (ensemble average)

sprinkled ocean passes, D

lattice points fail (Table 1) computed

a reference geometry allows shape terms known (massive gravity) [16, 25]

blend memory gives R = 1 −µ; remembered neighbors also fail derived and computed

veriﬁed structure remembers, the unveriﬁed ocean does not identiﬁcation (SSM)

no density halo around a defect derived and computed

bonds’ entanglement follows area; the vacuum is in equilibrium postulates A and B

equilibrium = Hamiltonian constraint;

; Λ a constant of

integration

derived [27]

spoke deﬁcit 8πGλ exactly; Schwarzschild far ﬁeld computed

solved slice satisﬁes equilibrium ball by ball (Fig. 6) computed (consistency check)

stabilizer code gives whole ebits only known [36]

bond density maximizes entanglement, Eq. (11) postulate

bond model: slope one at the maximum; s

= 0.45 ln 2 for p = 1 computed

restitching above a model-dependent stretch estimate

bonds are frozen ocean; one quantity counted once interpretation (SSM)

the ocean’s own coeﬃcient reproduces the area law open

12 Conclusion

Induced gravity on the SSM lattice has a form and a strength. Both follow from two forms of

one principle. The ocean is maximally mixed at ﬁxed density, and it keeps no memory of earlier

geometries. Its vacuum energy therefore depends on the metric only through the volume. It sees the

lattice’s geometry but not its directions, the graviton stays massless, and its induced action is the

Regge action. A remembered geometry would allow the shape terms of massive gravity. This is why

ﬁelds on the veriﬁed lattice fail the test and the unveriﬁed ocean passes it. The ocean has no halo

around a defect, so the ﬁeld of a mass lives in the geometry. The bonds carry as much entanglement

per unit area as they can. Their equilibrium with matter is the Hamiltonian constraint, which the

solved defect slice satisﬁes ball by ball. Maximizing their entanglement per unit area makes 1

its largest value and ﬁxes the bonds’ microscopic law. Since the bonds are frozen ocean, the stiﬀness

of the liquid and the entanglement of the solid are one quantity. The remaining question is whether

the ocean’s own ﬁelds reproduce it.

Data availability

All results reproduce from the Python scripts archived at https:

//github.com/raghu91302/ssmtheory/blob/main/ssmtheory_form_strength_scripts.zip.

The archive’s README lists the script and the command for each table and ﬁgure. The scripts

need only NumPy, SciPy, and Matplotlib. All random numbers come from ﬁxed seeds.

Declarations

Funding: This research received no external funding.

Conﬂict of interest: The author declares no conﬂict of interest.

Use of artiﬁcial intelligence: The author used Claude (Anthropic) to help write and run the

numerical scripts and to draft and edit the manuscript. The author reviewed all results and text

and takes full responsibility for the content.

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