BLOCH'S LAW FOR FINITE-RANGE HEISENBERG
FERROMAGNETS IN THREE DIMENSIONS
a complete proof in plain words
for curious readers
================================================================
CONTENTS
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0. PREFACE: WHAT WE ARE PROVING AND WHY
1. MAGNETS, SPINS, AND THE HEISENBERG HAMILTONIAN
2. PRESSURE, MAGNETIZATION, AND THE STATEMENT OF THE LAW
3. WHERE THE T^{3/2} COMES FROM: SPIN WAVES
4. THE STRATEGY IN PLAIN ENGLISH
5. SLOTS, SWAPS, AND THE LOOP REPRESENTATION
6. PINS AND THE RETURN PROBABILITY
7. EXPOSING THE FREE CYCLES AND THE FRESH WALK
8. GEOMETRY: THREE BONDS MAKE A CUBIC LATTICE
9. GEOMETRY: THE NASH INEQUALITY
10. GEOMETRY: FLOWS AND SPARSE SOURCES
11. THE SPARSE-PIN BOOTSTRAP
12. LOCALIZATION, DIFFUSION, AND THE HITTING ENERGY
13. LONG RETURNS AND THE SINGLETON BOUND
14. THE IDEAL WALK AND ITS HEAT KERNEL
15. THE COUPLING THAT RECOVERS THE COEFFICIENT
16. FROM PINNED RETURNS TO SPONTANEOUS MAGNETIZATION
17. PUTTING IT ALL TOGETHER
APPENDIX A. NOTATION CHEAT SHEET
APPENDIX B. BACKGROUND FACTS WE RELY ON
REFERENCES
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0. PREFACE: WHAT WE ARE PROVING AND WHY
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A magnet is a piece of matter whose atoms carry tiny arrows called
spins. Below a certain temperature the arrows line up and the magnet
is strong. Heat makes them wobble, so the magnet gets weaker. Bloch
observed in 1930 that for a three-dimensional ferromagnet the
magnetization drops, at very low temperature, by an amount
proportional to T^{3/2}, where T is the absolute temperature.
That statement had been believed, checked by experiment, and computed
to leading order by physicists for decades, but a fully rigorous
derivation for the *quantum* Heisenberg ferromagnet was a genuinely
open problem. This paper proves it, for every value of the quantum
spin, and for every finite-range interaction whose support generates
the whole lattice. The coefficient in front of T^{3/2} is not
mysterious: it is exactly the T^{3/2} piece that one gets by summing
the energies of noninteracting spin waves. The work is to show that
the interacting system has the same leading term.
Read this paper the way you would read a detective story. The
criminal is (S - m(β))β^{3/2}; the question is what its limit is.
The proof builds the identity of the criminal out of small pieces,
and every piece is proved and every error term is bounded. You can
skip the technical proofs on a first read (they are clearly marked)
and still understand the argument. But you should not skip them twice:
the whole point of the paper is that every step is honest.
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1. MAGNETS, SPINS, AND THE HEISENBERG HAMILTONIAN
================================================================
1.1 Lattice and spin.
We work on the cubic lattice Z^3 = {(a,b,c) : a,b,c integers}. Points
of Z^3 are called sites. Each site carries a quantum spin of size S,
where S is one of 1/2, 1, 3/2, 2, .... At each site the spin can be
measured along any axis. Along the z-axis it has 2S+1 possible values
-S, -S+1, ..., S. The state space at one site is C^{2S+1}, with
basis vectors e_m (m = -S, ..., S). The spin operators there act as
S^z e_m = m e_m,
S^+ e_m = sqrt(S(S+1) - m(m+1)) e_{m+1} (m < S),
S^+ e_S = 0,
S^- = (S^+)^*,
S^x = (S^+ + S^-)/2,
S^y = (S^+ - S^-)/(2i).
S^+ raises the z-spin by one and S^- lowers it. S^x and S^y are the
measured spins along the x and y axes. The bold symbol
S_x = (S^x_x, S^y_x, S^z_x) is the spin-vector at site x.
If you have never seen this before, one thing matters. The operators
satisfy the standard commutation relations of angular momentum. The
Hamiltonian below is the simplest possible rotation-invariant way to
couple nearest spins, and it is the actual physical model of a
Heisenberg magnet.
1.2 The interaction.
Fix a function J : Z^3 -> [0, infinity) that is symmetric, has
finite support, and generates the lattice. That means
(I1) J(0) = 0, J(z) = J(-z),
(I2) {z : J(z) > 0} generates Z^3 as an additive semigroup.
The second condition rules out a model in which the lattice splits
into decoupled sublattices. It also guarantees that the one-magnon
dispersion below is quadratic and strictly positive for all
non-zero k (see section 3).
Intuitively J(z) measures the coupling strength between a site and
the site z away. When J is nonzero only for z at distance 1, we
recover the classical nearest-neighbour Heisenberg model. But our
result also allows longer-range couplings, and even anisotropic
couplings, provided the support still generates the whole lattice.
1.3 The Hamiltonian.
Let Λ = Λ_N = [-N, N]^3 ∩ Z^3 be a cube of side (2N+1). On this
cube we set
H_{S,N} = - (1/2) sum_{x,y in Λ} J(y-x) S_x . S_y,
M_{S,N} = sum_{x in Λ} S^z_x.
The dot product S_x . S_y means S^x_x S^x_y + S^y_x S^y_y + S^z_x S^z_y.
The factor of 1/2 is because the sum counts each unordered pair {x,y}
twice. M_{S,N} is the total z-magnetization of the block. The
Hamiltonian is real symmetric, and it commutes with M_{S,N}, because
the interaction conserves total S^z.
For the whole paper we set the Planck constant, the Boltzmann
constant, and the lattice spacing equal to one. Then temperature is
T = 1/β, and higher β means colder.
1.4 Free boundary.
The sum in H_{S,N} runs only over pairs x, y that are both inside Λ.
That is the free-boundary convention. A bond from an interior site
to an exterior site is simply not there. Almost every serious
statement below is about this free model, and the boundary is the
whole reason the word "pin" matters later.
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2. PRESSURE, MAGNETIZATION, AND THE STATEMENT OF THE LAW
================================================================
2.1 Pressure.
For a real number h, called the external magnetic field, define the
partition function
Z_{Λ_N}(β, h) = Tr exp[ -β ( H_{S,N} - h M_{S,N} ) ].
This is a single real number: the trace of a matrix exponential.
Physically it counts all the microscopic configurations weighted by
their Boltzmann factor. The pressure is the log of Z divided by
β times the volume:
p_N(β, h) = (1 / (β |Λ_N|)) ln Z_{Λ_N}(β, h).
It is a theorem (we recall the proof in section 16.1) that p_N has a
limit as N -> infinity for every β > 0 and every real h. Call that
limit p(β, h) = p_β(h). The function h -> p_β(h) is convex and
S-Lipschitz, hence has a right-derivative at h = 0. Define
m_{S,J}(β) = (d/dh)^+ p_β(h) |_{h = 0}.
This m is the spontaneous magnetization at inverse temperature β.
Intuitively, m is the leftover z-magnetization per site when you
first take the block infinitely large, and then let a positive
magnetic field shrink to zero from above. Because p_β is even in h
(rotation by pi about the x-axis flips the sign of the field but
leaves the energy invariant) one automatically has 0 <= m <= S.
2.2 The diffusion matrix.
The central quantity in the coefficient of T^{3/2} is the 3x3 matrix
D_{S,J} = (S/2) sum_{z in Z^3} J(z) z z^T,
where z z^T is the 3x3 outer product of the vector z with itself.
The sum is finite by assumption. This matrix is symmetric positive
definite because the support of J generates Z^3: any nonzero vector
v with v^T D v = 0 would give J(z) v.z v.z = 0 for all z, and so
v would be orthogonal to the whole generating set.
The matrix D is exactly the quadratic coefficient of the one-magnon
dispersion:
ε_{S,J}(k) = S sum_z J(z) ( 1 - cos(k . z) )
= k^T D_{S,J} k + O(|k|^4). (2.1)
You can check this by Taylor expanding cos(k.z) = 1 - (k.z)^2/2 + ...:
the k^0 and k^1 terms cancel because of J(0)=0 and symmetry.
2.3 The theorem.
THEOREM 2.1 (Bloch's law). For every fixed S in {1/2, 1, 3/2, ...}
and every J satisfying (I1)-(I2),
lim_{β -> infinity} β^{3/2} ( S - m_{S,J}(β) )
= ζ(3/2) / ( 8 π^{3/2} sqrt( det D_{S,J} ) ). (2.2)
Here ζ(s) is the Riemann zeta function and det denotes the 3x3
determinant.
The order of limits matters and is part of the theorem: for each N,
first take N -> infinity at fixed β and h, then h -> 0 at fixed β,
then β -> infinity.
2.4 Checking the coefficient.
For the nearest-neighbour model J(±e_j) = 1, J(other) = 0, one has
D_{S,J} = S I (the identity matrix) because sum_j e_j e_j^T = I.
Then sqrt(det D) = S^{3/2}, and the right side of (2.2) is
ζ(3/2)/(8 π^{3/2} S^{3/2}). This is the formula in every physics
textbook. Our theorem reduces to it in the standard case and covers
many more interactions besides.
================================================================
3. WHERE THE T^{3/2} COMES FROM: SPIN WAVES
================================================================
The rest of the paper is a rigorous proof. Before diving in, we owe
the reader a plain-English explanation of the coefficient. It comes
from spin waves, also called magnons.
3.1 Magnons.
At absolute zero, all spins point fully up: the ground state is the
tensor product e_S ⊗ e_S ⊗ ... of the top spin at each site, and its
total magnetization is S times the number of sites.
The lowest excited states are made by applying S^- at one site, so
one spin drops from S to S-1. But S^- at a fixed site is not an
eigenstate of H. The eigenstates are superpositions
Σ_x e^{i k · x} S^-_x, one for each momentum k in the first
Brillouin zone [-π, π]^3. These are the one-magnon states. Direct
calculation gives their energies relative to the ground state:
ε_{S,J}(k) = S Σ_z J(z) ( 1 - cos(k·z) ). (3.1)
As k → 0, ε → 0; the dispersion is quadratic, ε(k) ≈ k^T D_{S,J} k,
by the same Taylor expansion as in (2.1). Magnons are gapless,
long-wavelength modes. This gap closing is what causes the T^{3/2}
law; if the dispersion had a finite gap, the correction would be
exponentially small in β instead of power-law.
3.2 Ideal magnon gas.
Now consider many magnons. Magnons are bosons. At temperature β^{-1}
the average number of magnons at momentum k in the ideal (free) case
is
<n_k> = 1 / (e^{β ε(k)} - 1), (3.2)
the standard Boltzmann distribution with zero chemical potential.
The chemical potential is zero because, at low temperature, the
dilute gas never hits the constraint of having at most 2S V total
magnons; magnons are free to appear and disappear without cost.
Each magnon reduces the z-magnetization of the ground state by 1
(because it has lowered one spin from S to S-1). So
M(β) = S V - Σ_k <n_k>,
m(β) = M(β)/V = S - (1/V) Σ_k <n_k>. (3.3)
The deficit S - m(β) is exactly (1/V) Σ_k <n_k> per site.
3.3 The expansion that produces ζ(3/2).
At large β, the discrete sum (1/V) Σ_k becomes the integral
∫_{[-π,π]^3} d^3k / (2π)^3 times a function. Use the identity
1 / (e^x - 1) = Σ_{n ≥ 1} e^{-n x} (3.4)
for every x > 0. Then
(1/V) Σ_k <n_k> = Σ_n ∫ d^3k / (2π)^3 e^{-n β ε(k)}.
For large nβ, the integral is dominated by small k, where
ε(k) ≈ k^T D k. Extending the integral from the Brillouin zone to
all of R^3 changes only higher-order corrections. The Gaussian
integral in three dimensions is
∫ d^3k e^{-n β k^T D k} = π^{3/2} (n β)^{-3/2} / sqrt(det D).
Dividing by (2π)^3 = 8π^3 and summing over n:
(1/V) Σ_k <n_k> ≈ (1/(8π^{3/2} sqrt(det D))) β^{-3/2} Σ_n n^{-3/2}
= ζ(3/2) β^{-3/2} / (8π^{3/2} sqrt(det D)). (3.5)
Substituting into m(β) = S - (1/V) Σ_k <n_k> and multiplying by
β^{3/2}, we get exactly (2.2).
3.4 A useful rearrangement: sum over "periods".
The expansion (3.4) is doing real work here. The n-th term is the
heat kernel after elapsed time nβ. In the rigorous proof, the
primitive object is a Markov walk whose natural time scale is the
"period" β of the time circle. The return probability after n
periods is roughly κ (nβ)^{-3/2}, and the total down probability is
its dyadic (in fact, n-adic) sum. So the ζ(3/2) emerges naturally
as the sum κ Σ_{n ≥ 1} n^{-3/2} of "return at every multiple of β"
events.
This sum-over-periods structure is what sections 12-13 of this paper
organize. The Bloch coefficient is literally a sum over n of
(n β)^{-3/2}, which converges to κ ζ(3/2). It is not an artifact of
wanting a neat answer; it is the natural expression in the random
interchange framework.
3.5 Why interactions do not change the coefficient.
In the interacting model, magnons scatter off each other. Dyson
computed these corrections in 1956 and found that their total
contribution to the magnetization is of order T^4, not T^{3/2}.
So the leading T^{3/2} coefficient is untouched, and equals the
free-gas value (3.5). That was the classical calculation behind
Bloch's law.
What is not obvious (and what made this an open problem rather
than a homework exercise) is how to make the argument rigorous
for the interacting quantum spin system, where the "magnon gas" is
not dilute in the sense of a free Bose gas: magnons can be as
numerous as SV, and they interact strongly at high temperatures
near the phase transition. The proof below handles all of this
by working with exact return probabilities and proving bounds that
are uniform in the interaction and in the temperature.
The strategy below turns each step of the physics argument into a
theorem about the exact quantum system.
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4. THE STRATEGY IN PLAIN ENGLISH
================================================================
The proof is long, but the strategy is short. There are six steps.
4.1 Turn magnetization into a return probability.
The random interchange representation (Tóth 1993, Nachtergaele 1994)
represents each spin-S as ℓ = 2S quantum spin-1/2 slots. Over a
"time period" β, adjacent slots randomly exchange; at the end of
the period, a uniform permutation at each site (the "seam")
closes the period. All swaps together form a random permutation
of the (site, slot, time) points.
A cycle of this permutation can be colored "up" or "down"
independently. The partition function becomes a sum over these
colored loops. Crucially, the down-probability at a slot-time
point i has a clean formula:
d(i) = r(i) / (1 + r(i)), (4.1)
where r(i) is the probability that, if we "pin" i (force it up by
an extra constraint), the loop through i returns to i at a strictly
positive elapsed time. Section 6 proves (4.1) as Proposition 6.3.
This converts the thermodynamic observable (magnetization) into a
purely probabilistic one (a return probability).
4.2 Expose the environment.
Under the pinned law, the loops that avoid all pins form a
"free" set. Conditioning on their shapes gives a "pinned
environment", and the return probability becomes the probability
that a "fresh walk" hits the same slot again, on this environment,
before hitting a pin. Section 7 makes this precise (Propositions
7.3 and 7.4).
So the problem becomes: understand a random walk in a random
periodic environment, on time scale up to Kβ.
4.3 Environments are almost always regular.
If we pin only a sparse set of points, the free loops will surround
them, and locally the walk sees an almost full cubic lattice. The
sparse-pin bootstrap (Proposition 11.1, Lemma 11.2, Proposition
11.3) makes this quantitative: the probability of a "large blocked
cluster" at an anchor z decays exponentially in the cluster radius,
uniformly in the pin set. Consequently almost every time-slice is
"regular" in a technical sense (section 8): no dense clustered
holes at z, and no long-range connected holes far from z.
4.4 On regular slices, the walk diffuses.
On regular slices, a Nash inequality (Proposition 9.1) gives
fast L^2 decay for non-concentrated walk distributions. Carefully
tracking this over many periods (Proposition 12.2) gives
||v(t)||_2 ≤ C (t)^{-3/4}, t ≥ β/8, (4.2)
for masses of size ≤ 1. This is diffusive smoothing in three
dimensions: The 3/4 is d/4 with d = 3, up to log corrections that
we suppress by careful choice of localization radius.
4.5 Sum over returns.
The finite return sum Lemma 13.1 expands r(i) as
r(i) ≤ Σ_{n=1}^{2K-1} (Q^n)_{ii} + remainder(K, β). (4.3)
Each (Q^n)_{ii} is the n-period return probability, and by (4.2)
it is bounded above and below by ~ κ (nβ)^{-3/2} plus small error,
via a comparison to an ideal walk (section 14). The remainder is
bounded by C K^{-1/2} β^{-3/2} (Proposition 13.2), with C
independent of K.
Summing over all ℓ slots at a site and dividing by ℓ, we get
(per-site deficit) ≈ κ Σ_{n=1}^{2K-1} n^{-3/2} β^{-3/2}
+ C K^{-1/2} β^{-3/2}
as β → ∞, then as K → ∞, giving κ ζ(3/2) β^{-3/2} exactly.
4.6 Pressure squeeze recovers the magnetization.
Section 16 turns a positive external field h > 0 into a random extra
pin set of density ~ β h. The magnetization is then a mixture of
down-probabilities with these extra pins. Taking h → 0, then
β → ∞, recovers m(β) from the per-site down-probability bounds.
The order of limits is crucial: first N → ∞ (thermodynamic limit),
then h → 0 (spontaneous symmetry breaking), then β → ∞ (zero
temperature). Section 16.3 keeps careful track of this.
Here is the reason the whole program works. Each step is proved in
isolation with explicit constants. Every error is bounded uniformly
in the model, β, and pin set. And the coefficients match the ideal
magnon gas exactly. There is no magic; just a small number of ideas
applied many times.
The technical burden is making the constants uniform. Because the
random environment is itself randomly chosen (with pin law
depending on T, h, volume), each estimate must be proved for
every realizable pin set. That is why sections 8-12 work so hard.
================================================================
5. SLOTS, SWAPS, AND THE LOOP REPRESENTATION
================================================================
The random interchange representation of Tóth and Nachtergaele is
the engine of the proof. It turns the spin-S partition function into
a sum over random "pictures": swaps of spin-1/2 slots on a time
circle, whose resulting cycles can be colored.
The output of this section is Proposition 5.2: the trace of the
partition function equals a sum over colorings of cycles, weighted
by the swaps that produced them. Everything after this is about
controlling that sum.
5.1 Slots and slot edges.
Fix S ∈ {1/2, 1, 3/2, ...} and set ℓ = 2S. Let Λ be a finite cube
and V_Λ = Λ × {1, ..., ℓ}. An element (x, α) of V_Λ is called a
slot at x. Each site x holds ℓ slots, and the physical spin-S
degree of freedom is the symmetric subspace of the ℓ slots. The
point of slots is this: each slot is just a spin-1/2 particle, and
spin-1/2 arithmetic is very simple. The symmetric combination
recovers the spin-S behavior.
We need a way to couple slots attached to different sites. For
each unordered lattice bond {x, y} with J(y-x) > 0, include ALL ℓ^2
slot pairs above it. Call the resulting pairs slot edges, and let
E_Λ be the set of slot edges. Put
c_e := J(y-x) / 2, for every slot edge e above {x,y},
w_Λ := Σ_{e ∈ E_Λ} J_e, J_e := J(y-x). (5.1)
Because there are ℓ^2 slot edges above every lattice bond, we
have w_Λ = ℓ^2 Σ_{x,y} J(y-x) (each bond counted twice, symmetric).
5.2 Pictures.
Fix β > 0 and view time on the circle T_β = R / β Z. Under the
base measure P_b, every slot edge e carries an independent Poisson
process of rate c_e on [0, β). Each event (a "mark") is a moment
in time at which the two slots at the two ends of e exchange their
spin state. Slots not involved in a mark just carry on unchanged.
At the end of the interval, we add one extra ingredient: at each
site x, apply an independent uniform random permutation of the ℓ
slots. This is the "seam completion". It joins time β to time 0,
so the whole object lives on the time circle and not on an interval.
A picture ω consists of its marks and its seam completions. Because
the slot edges connect sites, a mark traces a "strand" from one
slot to another. Following the strands through repeated copies of
the same picture (the time circle is periodically repeated for the
walk) gives directed cycles.
Key geometric fact: at time zero, every slot is on exactly one
cycle. If a_C is the number of time-zero slots on a cycle C, then
Σ_C a_C = ℓ |Λ|. (5.2)
A cycle coloring is a choice of sign σ_C ∈ {-1, +1} for each cycle
C. Its contribution to the total z-magnetization is
M(σ) := (1/2) Σ_C a_C σ_C. (5.3)
Up spins contribute +1/2, down spins -1/2, per time-zero slot on
the cycle. The formula sums, over all cycles, their weight in the
z-magnetization.
5.3 Statement of the loop representation.
LEMMA 5.1 (Lifted Hamiltonian).
Define the operator on (C^2)^{V_Λ}:
H_hat := - (1/2) Σ_{e ∈ E_Λ} J_e U_e + (1/4) w_Λ I,
M_hat := Σ_{(x,α) ∈ V_Λ} s^z_{(x,α)},
where U_e is the transposition of the two slots of edge e. Let
S_Λ be the product over x of the symmetrization projections
S_x := (1/ℓ!) Σ_{κ ∈ Sym(ℓ)} U_κ (on the slots at x).
Then on the range of S_Λ, H_hat acts as the spin-S Hamiltonian
H_{S,N} and M_hat acts as M_{S,N}.
Heuristic proof: On two spin-1/2 particles, s_i · s_j = (1/2) U_{ij}
- (1/4) I. Summing over the ℓ^2 slots above a bond {x,y} and
subtracting the extraneous (ℓ^2/4) I_dentity with the w_Λ term
gives back the spin-1/2 exchange for the full spin at x and y.
The symmetrization S_x is what identifies ℓ spin-1/2 particles
with one spin-S particle. Since H_hat is a sum over pairs of sites,
its action on the symmetric subspace is exactly H_{S,N}.
This is standard and is proved in Nachtergaele 1994.
PROPOSITION 5.2 (Loop representation).
For every finite cube, every β > 0, and every real h,
Z_Λ(h) = Tr_{sym}[ e^{-β(H_{S,N} - h M_{S,N})} ]
= e^{β w_Λ / 4} E_b [ Σ_σ e^{β h M(σ)} ]. (5.4)
Moreover, for every function φ on the finite spectrum of M_{S,N},
Tr_{sym}[ e^{-β H_{S,N}} φ(M_{S,N}) ] / Z_Λ(0)
= E_b [ Σ_σ φ(M(σ)) ] / E_b [ 2^{#cycles} ]. (5.5)
Here Tr_{sym} means the trace on the range of S_Λ.
Proof sketch. We use the key identity for spin-1/2 particles:
s_i · s_j = (1/2) U_{(ij)} - (1/4) I. (5.6)
Summing (5.6) over the ℓ^2 slot pairs above a bond {x,y}, and
using the constraint that H_hat involves the sum over all such pairs,
gives an identity of operators on (C^2)^{V_Λ}: H_hat equals (up to
a constant shift) the spin-S lift of H_{S,N}. The shift is
captured by w_Λ, as in (5.1). Lemma 5.1 is the clean version of this.
Now let φ be the mark-counting measure (the number of Poisson
marks on all slot edges). The expectation of U over the Poisson
process can be computed exactly, because independent Poisson
processes integrate to a matrix exponential:
E_b [ U_φ ] = exp[ β Σ_e c_e (U_e - I) ]
= e^{-β w_Λ/4} e^{-β H_hat}. (5.7)
The first equality holds because each Poisson mark is independent
and averages U over time; the second follows by elementary
computation once we recognize c_e = J_e/2 and w_Λ = Σ_e J_e. For
the precise derivation of (5.7), see Tóth 1993, or the sketch below.
Averaging the seam completions adds the symmetrization operator S_Λ,
because the uniform permutation at each site is exactly S_x = S_x².
So E_b[ U_φ ] on the symmetric range is exactly e^{-β H_hat} S_Λ.
Taking the trace on (C^2)^{V_Λ} and using cyclicity:
Tr( e^{-β H_hat} S_Λ e^{β h M_hat} ) = Tr( e^{-β H_hat} S_Λ e^{β h M_hat} S_Λ ).
The right side is the trace on the symmetric range. By Lemma 5.1,
this is Z_Λ(h).
Now the coloring calculation. In the up/down basis of each slot,
a permutation U_φ has diagonal entries (i.e., contributes to the
trace) exactly when the coloring at each slot-time point is
constant along its strand. Because the strands form cycles, the
condition is: the coloring is constant on each cycle.
The trace of a permutation matrix equals the number of its fixed
points. So
Tr( U_φ · (colored diagonal) ) = Σ_σ 1,
where σ runs over colorings that are constant on cycles. This
gives a weight 1 for each such coloring, and after multiplying by
e^{β h M(σ)}, summing over σ and averaging over the seam and the
Poisson marks, we get (5.4).
The spectral law (5.5) follows by putting φ(M_hat) instead of
e^{β h M_hat}; the same trace calculation applies.
5.4 The positivity trick.
Formula (5.4) is powerful but not immediately useful, because the
average over pictures has a sign-free weight e^{β h M(σ)} that can
grow exponentially in βh. To isolate the "zero-field" behavior,
define the pinned picture law. Actually, the trick is simpler:
factor out e^{βh SV}. Then
e^{-β h SV} Z_Λ(h) = e^{β w_Λ/4} E_b [ Σ_σ e^{-β h Σ_C a_C (1 - σ_C)/2} ].
Because the deficit SV - M(σ) = Σ_C a_C (1-σ_C)/2 (only DOWN cycles
contribute), and 0 ≤ (1-σ_C)/2 ≤ 1, the weight e^{-βh·(deficit)}
is between 0 and 1. So the field only DOWNWEIGHTS colorings in the
sum. This is why the field h acts like a "pin" in the limit h → ∞
(only the all-up coloring survives) and why h > 0 corresponds to
extra pins in the proof below. Section 16 makes this precise.
For the high-school reader, here is the takeaway. The trace has
become a sum over pictures and colorings, with weights in [0, 1]
coming from the field. The physical content is:
- each cycle contributes to magnetization by +a_C/2 (if up)
or -a_C/2 (if down)
- the field biases toward up.
The partition function is a weighted count of colored loops.
================================================================
6. PINS AND THE RETURN PROBABILITY
================================================================
The rest of the proof is a change of measure. We fix a "pin" - a
slot-time point that we force to be up - and ask how the picture
law changes. The answer (Proposition 6.5) is that the down
probability at a slot-time point i equals a return probability,
and the return probability is what a one-particle walk computes.
This converts an observable about the spin system into a
statement about random walks in a random environment.
6.1 The pin set D and the pinned law.
The boundary pin set is the entire boundary of the cube, over the
whole time circle:
D_0 := ∂Λ × {1, ..., ℓ} × T_β, (6.1)
where T_β = R/βZ is the time circle, and ∂Λ = {x ∈ Λ : the
sup-norm distance from x to Z^3 \ Λ is at most b_0}. Here
b_0 = max{|z|_∞ : J(z) > 0} is the bond length. This is the set
of sites that could have a bond to the exterior if the boundary
were open; it is the boundary strip of Λ.
An allowed pin set is D = D_0 ∪ F, where F is any finite
deterministic subset of V_Λ × T_β (slot-time points). For the rest
of the paper, "pin set" means "allowed pin set" unless stated
otherwise.
For a pin set D, define
n_D(ω) := #{cycles C of ω : C ∩ D = ∅},
B_D(ω) := ∪_{C : C ∩ D = ∅} C,
Z_D := E_b [ 2^{n_D(ω)} ]. (6.2)
The pinned measure is the change of measure
P_D(dω) := (1/Z_D) · 2^{n_D(ω)} P_b(dω). (6.3)
Because 0 ≤ n_D ≤ |V_Λ|, the weight 2^{n_D} is between 1 and
2^{|V_Λ|}, so Z_D ∈ [1, 2^{|V_Λ|}] and P_D is a probability measure.
6.2 The coloring interpretation of P_D.
Recall the coloring weight. In the loop representation (5.4), each
cycle C gets a color σ_C ∈ {-1, +1} contributing to M(σ). At zero
field h = 0, all colorings have equal weight, so the SUM over
colorings gives a factor of 2 per cycle, i.e., 2^{#cycles}.
Insisting that every cycle meeting D be up (σ_C = +1) removes the
factor of 2 for those cycles (they contribute 1 · e^0 = 1 weight
mathematically), while cycles avoiding D are still summed over with
σ_C ∈ {-1, +1}. If we divide the partition function by the free
factor 2^{#cycles}, the resulting measure gives a formula:
P_D = the law of the picture, conditioned on cycles meeting D
being up and cycles avoiding D independently fair.
Under this measure, define
d_D(i) := P_D( σ_{C(i)} = -1 ),
the probability that slot-time point i is "down" (i.e., on a down
cycle).
Key observation: if i is on a cycle C that MEETS D, then C is
forced up, so i is up (contributes 1 · 0 = 0 to d_D). If i is on
a cycle avoiding D, then C is fair, so i is down with probability
1/2. Hence
2 d_D(i) = P_D( i ∈ B_D ). (6.4)
This is the key formula connecting the magnetization (via d_D) to
the free-cycle density (via i ∈ B_D).
6.3 Boundary approximation.
The pin set D_0 is continuous (it contains every slot at every
time). Working directly with a continuous pin set is awkward, and
the safe thing is to approximate by finite dense cuts.
Let T_n ⊂ T_β be an increasing sequence of finite time sets
containing 0, with union dense in T_β. Set
D_{0,n} := ∂Λ × {1, ..., ℓ} × T_n,
D_n := F ∪ D_{0,n},
D := F ∪ D_0.
LEMMA 6.1 (Dense boundary cuts).
Almost surely, for all n sufficiently large, n_{D_n} = n_D and
B_{D_n} = B_D. Also P_{D_n} → P_D in total variation. And
bounded observables X_n → X almost surely converge in
expectation: E_{D_n}[X_n] → E_D[X].
Proof. Each cycle is a finite curve on the time circle; hitting the
boundary ∂Λ involves sitting on a boundary slot for a whole time
interval (the time between two swaps), which is strictly positive.
Since the union of T_n is dense, any such interval is eventually
detected. There are finitely many cycles, so one T_n detects all
boundary-touching cycles. Therefore n_{D_n} = n_D and B_{D_n} = B_D
for large n. Since 2^{n_D} ∈ [1, 2^{|V_Λ|}], bounded convergence on
the weights gives Z_{D_n} → Z_D and P_{D_n} → P_D in total
variation. ∎
From now on we work with the finite cut D_n when needed and pass
to the limit with Lemma 6.1.
6.4 Factorial domination.
Under the pinned measure P_D, Poisson marks are no longer
independent; the weight 2^{n_D} introduces correlations. We need a
uniform "path count" bound: if we look at lists of marks in
arbitrary subsets, the expected number (averaged over the pinned
measure, with the weight) is at most a constant multiple of the
base Poisson factorial moment.
Let (as before) ω_m = Σ_{u ∈ marks} δ_u be the counting measure of
all marks, on X_Λ = E_Λ × [0, β) with intensity
λ_b( {e} × dt ) = c_e dt.
LEMMA 6.2 (Factorial domination).
For any pin set D (finite or allowed), any integer j ≥ 1, and
any nonnegative measurable F on X_Λ^j,
E_D [ Σ_{(u_1, ..., u_j) ∈ (ω_m)^j_{≠}} F(u_1, ..., u_j) ]
≤ 2^j ∫ F dλ_b^{⊗j}. (6.5)
The bound also holds with any prescribed time ordering or with
time restrictions on the marks.
Proof sketch. Inserting a single mark (a transposition between two
slots) either splits one cycle into two or joins two cycles into
one. In the split case, one cycle becomes two, so n_D increases by
at most 1 (at most one of the two new cycles avoids D). In the
join case, two cycles become one, so n_D decreases (possibly by up
to 1). Either way, |n_D(after) - n_D(before)| ≤ 1, so 2^{n_D}
changes by a factor between 1/2 and 2.
Consider the insertion identity for a Poisson process with intensity
λ: for any nonnegative F on the space of ordered j-tuples of
distinct marks,
E_b [Σ_{(u_1,...,u_j) distinct in marks} F(u_1,...,u_j)]
= ∫ F(u_1,...,u_j) E_b [ Π_i dμ(d u_i) ] (no weight)
where dμ = dλ_b. This is the standard factorial moment identity.
Now weight by 2^{n_D(ω)}. By the pointwise bound 2^{n_D(ω with j
extra marks)} ≤ 2^j · 2^{n_D(ω)},
E_b [2^{n_D(ω)} Σ_{j-tuples} F(u_1,...,u_j)]
= ∫ F(u_1,...,u_j) E_b [2^{n_D(ω with j extra marks)}] dλ_b^{⊗j}
≤ 2^j Z_D · ∫ F · dλ_b^{⊗j}.
Dividing by Z_D proves (6.5). ∎
We will occasionally use the weaker constant 4^j instead of 2^j; the
factor doesn't matter.
For mixed lists that include both actual Poisson marks (from ω_m)
and independent fresh marks (from a separate Poisson family), the
same bound holds with each actual mark contributing a factor 2 (or
4, or 8 - whatever) and each fresh mark contributing a factor 1
(because there's no weighting for fresh marks). This is an
unconditional statement under P_D: it holds no matter what the
exposure law looks like.
6.5 The singleton pin identity.
Fix a pin set E and a slot-time point i ∉ E. Set P = E ∪ {i}.
Define
r(E, {i}) := P_P ( R_i(E, {i}) ) (6.6)
where R_i(E, {i}) is the event that the line (strand) from i returns
to i at a strictly positive elapsed time before visiting any point
of E.
PROPOSITION 6.3 (Singleton pin identity).
For i ∉ E:
d_E(i) = r(E, {i}) / (1 + r(E, {i})). (6.7)
Proof. Under P_P, the line from i returns to i; the "extra pin i"
forces the cycle through i to be up. But the LINE from i (not the
cycle as a whole) is a small segment: it returns to i at a positive
elapsed time, and the event A that it does so BEFORE hitting E is
just the event that the cycle through i avoids E.
Write r = r(E, {i}), A = {i ∈ B_E} = R_i(E, {i}) (as an event).
Under P_P, P_P(A) = r by definition.
Compare weights: on the event A, the cycle through i avoids E and
contains i (a pin). Under E, this cycle is treated as a free cycle
(every cycle avoiding E is free, including the one through i).
Under P = E ∪ {i}, the same cycle is PINNED (because it contains
i ∈ P). So 2^{n_E} = 2 · 2^{n_P} · 1_A + 2^{n_P} · 1_{A^c} =
2^{n_P}(1 + 1_A). Hence
Z_E = E_b [2^{n_E}] = E_b [2^{n_P} (1 + 1_A)] = Z_P(1 + r).
Also, P_E(A) = E_b [2^{n_E} 1_A] / Z_E = E_b [2^{n_P} · 2 · 1_A] /
(Z_P(1+r)) = 2 P_P(A) / (1+r) = 2r/(1+r).
Combining with (6.4): d_E(i) = (1/2) P_E(A) = r/(1+r). ∎
COROLLARY 6.4 (Uniform simplified form).
Since r ≥ 0 and r ≤ 1, we have d_E(i) ≤ r(E, {i}), and
|d_E(i) - r(E, {i})| ≤ r(E, {i})² ≤ r(E, {i}) · C β^{-3/2}
for some constant C once r(E, {i}) ≤ C β^{-3/2}.
6.6 The simultaneous pin inequality.
The singleton identity (6.7) is not enough. We need to know that
if we prescribe many pins, they are unlikely to ALL be on free
cycles - even though the singleton event is not rare. Negative
dependence is the technical tool.
DEFINITION 6.5 (Simultaneous return).
Let E be a pin set, T = {i_1, ..., i_{m_T}} a finite set of
distinct slot-time points at one deterministic time θ, with
T ∩ E = ∅. Put P = E ∪ T. Define
r(E, T) := (1/m_T) Σ_{i ∈ T} P_P ( R_i(E, T) ), (6.8)
where R_i(E, T) is the event that the line from i first visits
T (at some slot-time point of T other than i, at positive
elapsed time) before visiting any point of E.
PROPOSITION 6.6 (Simultaneous pin inequality).
For E, T as above:
P_E ( T ⊆ B_E ) ≤ ( 2 r(E, T) )^{m_T}. (6.9)
The factor 2 in front is an artifact; the important thing is that
the m_T-th power comes out. This is what turns a "single-slot"
bound into a "sparse set" bound.
Proof sketch. We use three finite-set facts from the stable
polynomial method of Borcea-Brändén-Liggett (see Ordering, Lemmas
3.2, 3.4, 3.6):
(a) Variable identification, partial derivatives, real
specialization, and coefficientwise limits of bounded-degree
polynomials preserve stability (no zeros when every variable
has positive imaginary part).
(b) If π is a permutation of a finite set V that is the product of
independent Bernoulli transpositions and deterministic
permutations, then the polynomial
E[ ∏_{j ∈ V} (x_j + y_{π(j)}) ]
is stable. The same holds for independent Poisson exchanges
with nonnegative rates, and for uniform permutations on
specified subsets.
(c) If a random subset A ⊆ V has stable generating polynomial
E[ ∏_{j ∈ A} y_j ], then for every B ⊆ V,
P(B ⊆ A) ≤ ∏_{j ∈ B} P(j ∈ A).
The proofs of (b) and (c) are in Ordering; we use them as black
boxes. The key calculation is as follows. Cut the time circle at
zero and at every pin time in P. Let V be the disjoint union of the
slot layers at these cuts. Let π be the permutation that sends each
cut vertex to the next cut vertex along the same strand. The
interval maps are independent and each factors as a product of
independent interval factors, so their stable symbols multiply to
give a stable polynomial
F_V(x, y) := E_b [ ∏_{j ∈ V} ( x_j + y_{π(j)} ) ].
Delete an unpinned vertex v ∈ V \ P by identifying x_v = y_v = t
and taking the t-linear coefficient. This "splices" the strands
through v.
After deleting all unpinned vertices, we get a stable "spliced"
symbol
F_P(x, y) = E_b [ 2^{n_P} ∏_{j ∈ P} ( x_j + y_{σ(j)} ) ],
where σ is the "next-pin" permutation on P: it sends each pin to
the next pin on the same cycle, going around once. The closure
facts preserve stability, so F_P is stable.
Differentiate F_P with respect to every x_j for j ∈ P \ T, set
every other x_j to zero, and divide by Z_P. What remains is
G_T(y) := E_P [ ∏_{i ∈ T} y_{σ(i)} ]. (6.10)
G_T is stable and nonzero (because G_T(1) = 1). It represents the
subset σ(T) ⊂ V_Λ.
By (c) with A = σ(T) and B = T (all four have σ(i) ∈ T):
P_P ( T ⊆ σ(T) ) ≤ ∏_{i ∈ T} P_P ( σ(i) ∈ T ).
The mean of |T ∩ σ(T)| gets computed as
m_T r(E, T) = E_P |T ∩ σ(T)| = Σ_{i ∈ T} P_P ( σ(i) ∈ T ),
so the AM-GM inequality gives
P_P ( T ⊆ σ(T) ) ≤ r(E, T)^{m_T}. (6.11)
The event T ⊆ σ(T) is equivalent to T ⊆ B_E. Indeed, |σ(T)| = |T|,
so T ⊆ σ(T) implies σ(T) = T; iterating σ on T then stays in T
forever, so cycles through T avoid E. Conversely, if T ⊆ B_E, the
cycle through each i ∈ T avoids E, so the next pin on that cycle
is also in T (or is i itself, if i is the only T point on its
cycle). Either way T ⊆ σ(T).
Finally, deleting m_T pins frees at most m_T cycles, so
2^{n_P} ≤ 2^{n_E} ≤ 2^{m_T} 2^{n_P}, and Z_E ≥ Z_P. Thus
P_E ( T ⊆ B_E ) = ( 1/Z_E ) E_b [ 2^{n_E} 1_{T ⊆ B_E} ]
≤ (2^{m_T} Z_P / Z_E) P_P ( T ⊆ B_E )
≤ 2^{m_T} · r(E, T)^{m_T}
= ( 2 r(E, T) )^{m_T}. ∎
For D_0-continuous pins, use Lemma 6.1 to pass from finite D_n to
D.
The power (2r)^{m_T} is what makes the sparse-pin bootstrap
(section 11) work: if each i_j has r(E, {i_j}) ≤ p, then the
probability that ALL of them are free is at most (2p)^{m_T}.
We've converted the down probability into a return probability, and
we've seen that sparse sets of pins are exponentially unlikely to
be on free cycles. What remains is to compute the return probability
of a single slot-time point as a Markov walk in a "punctured"
environment. That is the subject of section 7.
================================================================
7. EXPOSING THE FREE CYCLES AND THE FRESH WALK
================================================================
Pick a pin set E, a test set T, and put P = E ∪ T. The pinned picture
probability P_P describes the swaps in the full picture. We now
split the picture into two parts:
(a) the "exposed" cycles: cycles that avoid P (i.e., free cycles
of the original P);
(b) the "remaining" cycles: cycles that meet P.
The exposed cycles are removed from the picture, leaving a "hole"
at every slot-time point that belonged to an exposed cycle. The
remaining strands live on holes only. What remains is to characterize
the remaining picture's law - that is the conditioning Q_a, and the
crucial trick is that it equals the law of a Markov walk in the
"punctured" environment a.
7.1 What the exposure reveals.
Fix a pin set P = E ∪ T (for concrete: E a boundary-pin set, T a
finite test set). Under P_P, sample one picture ω. Then:
An "exposed" label is one whose cycle misses P entirely.
The set of exposed labels a is the collection of all these
cycles - their trajectories during one period, their seam
images, and their permutation data.
Concretely, a is a complete description of a family of interleaved
slot-time curves that never touch any pin. Its complement is what
we call the holes H_t = H_t(a), the slots not on exposed labels
at time t. The total hole count is constant in time. Because
exposed cycles avoid P by definition, every pin is a hole at its
own time. Also, every boundary slot is a hole at every time: boundary pins
are in P by definition, so the cycles through them are not exposed,
so boundary slots belong to the remaining picture.
7.2 The conditionally-remaining picture.
Given a, what is the conditional law of the REST of the picture
(the part on H_t)? Each Poisson edge e has some part in the exposed
picture, and some part in the remaining picture. Marks are
independent across edges. The marks on hole-only edges form a base
Poisson picture on holes. Additionally, the exposed picture may
"push" holes around (see below). The seam completions on exposed
slots are already revealed, and we have to complete the residual
permutation at each site by a uniform bijection of the remaining
slots.
The resulting conditional law Q_a on the remaining picture is:
- Independent Poisson exchanges on hole-only edges at rate
c_e, as in the base law, on [0, β).
- Forced transports: if an exposed mark touches an edge e at
time t between a hole and an exposed slot, we move the hole
in the opposite direction (into the just-vacated exposed
slot). This is a deterministic bijection on the holes.
- Across the seam, at each site x, a uniform completion of the
revealed mapping on the remaining slots.
Note that the boundary conditions and structure of the holes are
encoded in a; Q_a is a probability law on the remaining pictures
consistent with a.
7.3 The disintegration identity.
The remaining picture is not unconditioned: it must contain no
cycles missing P, because those would have been exposed (the
exposure is precisely the cycles missing P). Let
G_P := { every cycle of the remaining picture meets P },
g_P(a) := Q_a( G_P ).
The event G_P is the constraint that the exposure imposes on the
remaining picture.
PROPOSITION 7.1 (Exposure disintegration). Almost surely under
P_P, g_P(a) > 0 and
P_P ( remaining picture ∈ · | exposure = a )
= Q_a ( · | G_P ). (7.1)
More precisely, for every bounded measurable Φ of the exposure
and the remaining picture, we have the disintegration formula
E_P [ Φ (a, ω_rem) ]
= (1/Z_P) Σ_{A ⊆ V_Λ} ∫_{F_A} 2^{c_A(a)} ·
E_{Q_a} [ 1_{G_P} Φ (a, ω_rem) ] μ_A(da), (7.2)
where A is a deterministic initial label set, F_A is the event
that the A-strands avoid P and their endpoint set is A, and
c_A(a) is the number of cycles of A's endpoint permutation. μ_A
is the base law of A-strands. The marginal for the exposure on
branch A is
(1/Z_P) 1_{F_A} 2^{c_A(a)} g_P(a) μ_A(da). (7.3)
Proof sketch. Sample a full picture. Expose all labels that avoid
P. Given the exposed a, sample the remaining picture from Q_a
(unconditioned). But Q_a wouldn't give a picture whose cycles all
meet P; that's a further constraint. To weight by the constraint
correctly, split the base Poisson at each hole edge into two
independent "accepted" and "rejected" subfamilies. Accept a mark
from the first subfamily when at least one endpoint is exposed
(this is a forced transport), accept from the second when neither
endpoint is (this is a genuine hole-hole mark). If the base clock
has rate 2c_e on edge e, the "accepted" family has rate c_e on
each subfamily, so the accepted subfamily is exactly a Poisson
process of rate c_e on hole-only edges. This is the standard
"thinning" argument for Poisson processes.
Since each exposed cycle contributes a factor of 2 (from the fair
coloring) and each remaining cycle contributes a factor of 1 (it
meets P, hence is pinned up), we have 2^{n_P} = 2^{c_A(a)} on the
branch A. The conditioning on G_P is what selects the branches in
which "remaining cycles meet P"; without it, we could have a
picture whose cycles all avoid P, contradicting the choice of P.
Summing over all A-branches reproduces the total weight. Divide by
Z_P for the normalized law.∎
We don't need every detail of (7.2); the takeaway is that after
exposure, the remaining picture conditional on G_P is the Q_a law
conditioned on every remaining cycle hitting P.
7.4 Exterior powers and the next-pin kernel.
Under Q_a, the picture on holes has a transition matrix K from one
near-seam cut layer to the next. Specifically, at each one-period
cut, define K as the matrix K_{jk} = probability that a mass
starting at hole j ends at hole k after one period (with forced
transports built in). Rows and columns sum to at most 1 (contraction
on the killed space).
We want: given that every cycle meets some fixed subset I of the
hole set (analogous to "every remaining cycle meets P"), what is
the first-hit kernel of the K-chain to I?
The key algebraic input is the exterior-moments identity.
LEMMA 7.2 (Exterior moments).
Let Π be a random permutation matrix on a finite set V, which
is a distributional limit of products of independent Bernoulli
transpositions and deterministic permutations. Then
E[ ∧^k Π ] = ∧^k K, K := E[Π], for 0 ≤ k ≤ |V|. (7.4)
Here ∧^k denotes the k-th exterior power: a matrix acting on
alternating k-tensors, whose entries are the k × k minors of the
original matrix.
Proof. For a transposition matrix T_{ij} = I - (e_i - e_j)(e_i - e_j)^*,
compute: ∧^k(I + u(T_{ij} - I)) has every minor affine in u
(because the perturbation is rank-one, so two-column minors
vanish). This gives
∧^k ((1-u) I + u T_{ij}) = (1-u) ∧^k I + u ∧^k T_{ij}.
Multiplicativity of exterior powers over products and independence
over factors preserve the identity. Deterministic permutations
satisfy it directly. Continuity of minors passes it to
distributional limits.∎
PROPOSITION 7.3 (Next-pin kernel).
Let V = I ⊔ B be a finite set with I ≠ ∅. Suppose a random
permutation matrix Π on V satisfies (7.4). Let
G_I = { every cycle of Π meets I }. Then
P( G_I ) = det( I - K_{BB} ). (7.5)
Moreover, letting Σ be the column-forward "next-positive-visit"
permutation matrix mapping each i ∈ I to its first subsequent
visit to I (in positive cycle time),
E[ Σ | G_I ] = K_{II} + K_{IB} ( I - K_{BB} )^{-1} K_{BI}. (7.6)
The right side is the first-hit kernel of the Markov chain with
independent transitions according to K.
Proof sketch (following Ordering, Prop. 4.3). Fix a matrix D and
consider E[det(I - DΠ)]. By the principal-minor expansion and
Lemma 7.2,
E[det(I - DΠ)] = Σ_k (-1)^k Tr(∧^k D · ∧^k K) = det(I - DK).
Choose D = diag(W, I_B) with W an arbitrary complex |I| × |I|
matrix. If a cycle lies entirely in B, its indicator vector is
fixed by DΠ, so the determinant vanishes. On G_I, Π_{BB} is
nilpotent and block elimination gives det(I - DΠ) = det(I - WΣ),
where Σ is the next-visit matrix on I. Consequently
E[1_{G_I} det(I - WΣ)] = det of the 2 × 2 block matrix
[ I - W K_II -W K_IB ]
[ -K_BI I - K_BB ] (7.7)
Setting W = 0 gives (7.5) without any invertibility assumption.
When the RHS of (7.5) is positive, I - K_BB is invertible. Taking the
Schur complement in (7.7) and dividing by det(I - K_BB) gives
E[ det(I - WΣ) | G_I ] = det(I - WH), H := K_II + K_IB (I-K_BB)^{-1}K_BI.
Replace W by tW, compare linear coefficients, and use matrix units to
recover every entry of E[Σ | G_I] = H. ∎
Remark. The "next-pin kernel" formula (7.6) is a resolvent identity
for substochastic K. Reading it: the first hit of I either happens
immediately (K_II term) or after some number of B-step wanderings
(K_IB · (I-K_BB)^{-1} · K_BI). This is exactly the "return to I
before anything else" probability for a Markov chain with
transition matrix K. The right side is the first-hit kernel.
7.5 The fresh walk.
We now define a random walk that will compute the K-chain's first-hit
probability directly.
Given the exposure a, define the FRESH WALK as follows:
At time t on the time circle, the walk is at some slot in H_t
(the holes). It jumps across an available slot edge e at rate
c_e (available = both endpoints at holes at that time). It
follows the forced transports exactly as they are. At a seam,
it uses a fresh uniform residual completion (the same site-level
uniform bijection the Q_a uses, but sampled fresh on each
traversal rather than reusing the seam image).
The fresh walk uses the same environment as Q_a (same holes, same
forced transports, same rate c_e on hole edges), but it resamples
its transitions on each traversal. Its trajectories on disjoint
traversals are independent.
Denote by U the fresh-walk one-period kernel from a cutoff at
t = θ back to itself at t = θ + β (with the seam at the end).
The matrix U is the mean transition matrix K of Q_a in the sense
of section 7.4. It's a real, nonnegative, substochastic matrix
acting on mass functions on H_θ.
For i ∈ T (a hole at cutoff θ), let
q_i(a) := P^{walk}_{a,i} ( hit T at some strictly positive
time before hitting E ) (7.8)
where "hit T" means reaching any hole-time point in T; similarly
for E. If no hit occurs, the probability is 0.
PROPOSITION 7.4 (Fresh-walk identity).
For almost every exposure a under P_P, and every i ∈ T,
P_P ( R_i(E,T) | exposure = a ) = q_i(a). (7.9)
Consequently
m_T r(E, T) = E_P [ Σ_{i ∈ T} q_i(a) ]. (7.10)
Proof sketch. First suppose E, T, P are all finite pins. The
unconditioned layered permutation on holes under Q_a satisfies the
exterior-moments property by Lemma 7.2. Conditional on G_P (every
remaining cycle hits P), Proposition 7.3 identifies the
"next-positive-visit to P" first-hit kernel with the first-hit
kernel of the K-Markov chain. Those K-transitions are literally the
fresh-walk transitions between cut layers. Together with
Proposition 7.1, this proves the finite-pin case.
For continuous E = F ∪ D_0, approximate D_0 by finite cuts
D_{0,n} ⊂ D_0 and write E_n = F ∪ D_{0,n}, P_n = E_n ∪ T. Then
G_{P_n} ↑ G_P (each boundary visit is detected eventually, each
boundary visit is eventually included), so g_{P_n} → g_P > 0. For
large n, Prop 7.3 applies with pin layer P_n. The event R_i(E_n,T)
of returning to T before E_n decreases to R_i(E,T) (a finite
horizon since the line eventually returns to i on its finite cycle).
The fresh-walk probabilities q_i^{(n)}(a) likewise decrease to
q_i(a) (a finite-horizon argument, using that every boundary visit
takes positive time and so is detected by a dense cut). Bounded
convergence gives the result.∎
The good news: the fresh walk is a simpler object than the pinned
spin model. It's a random walk on a random periodic environment,
but the walk resamples its transitions each period. We can attack
it with classical tools (Nash, Dirichlet forms, flow arguments).
The bad news: the environment (the holes) is random and depends on
the exposure law, which depends on β, h, and the pin set. All of our
estimates must hold uniformly over the pin set, without ever
conditioning on properties of the environment.
Section 8 provides the geometric tools. Sections 9-12 use them to
show that on almost every exposure at almost every time-slice, the
fresh walk has good diffusion. Section 13 controls the return sum's
truncation. Section 14 compares to the ideal walk. Sections 15-17
complete the proof.
The conceptual picture:
- The exposure is a random subset a of the slot-time points.
- On the fresh walk with environment a, we want the probability of
hitting T before E.
- Averaging over a (under P_P) gives back the return probability
r(E, T) that we need.
- Each individual fresh-walk environment has a Nash inequality (or
fails to at some slices, which we can control probabilistically).
- On good environments, the fresh walk diffuses at (time)^{-3/4}.
- The return probability after n periods is thus roughly
(nβ)^{-3/2} (in 3D).
- Summing over n gives the Bloch coefficient ζ(3/2)/⋯.
The technical burden is making all of this rigorous. That is the
content of the following sections.
Before moving on, let me note one general convention used below:
the pin set E is kept FIXED throughout, and all our constants are
uniform in E, β, and the geometry. We never need to know the actual
shape of the exposure a; we only use its distribution through
Proposition 7.4.