LAUGHLIN SPECTRAL GAP: A COMPLETE, SELF-CONTAINED PROOF
=======================================================
0. WHAT THIS PAPER IS ABOUT
---------------------------
In 1983, Robert Laughlin wrote down an exact wave function that describes
an exotic phase of matter: the fractional quantum Hall (FQH) state. The
wave function itself is simple to write down. Place N electron-like
particles on a plane, give particle i a complex coordinate z_i in the
lowest Landau level (a magnetic-field-tilted state), and set
Psi_L(z_1, ..., z_N) = prod_{i<j} (z_i - z_j)^3.
That product of cubes is the Laughlin wave function at filling 1/3. It is
antisymmetric in the particles (each factor changes sign under exchange,
and we have three factors per pair, hence -1 per exchange). It is an
eigenstate of the perfect hard-core repulsion introduced by Haldane below.
Haldane reformulated the problem on the round sphere. For N electrons on
a sphere with Q flux quanta, each pair of electrons has a well-defined
"pair spin" j that runs from 0 up to Q-1. Haldane's parent Hamiltonian for
filling 1/3 is the sum over all pairs of the orthogonal projector onto
pair spin Q-1 (the "relative angular momentum one" sector). Every pair
pays exactly one unit of energy if it lands in that sector, zero
otherwise. Denote the resulting Hamiltonian on the N-particle sector by
H_{N,Q}, and extend it trivially to the whole Fock space:
H_Q = H_{0,Q} (+ ) H_{1,Q} (+ ) ... (+ ) H_{Q+1,Q},
H_{N,Q} = sum_{1<=i<j<=N} P_{ij}^{(1)} on wedge^N U_Q.
Laughlin's wave function is an exact zero mode of H_{N,Q} at filling 1/3.
That is H_{N,Q} Psi_L = 0 for Q = 3(N-1), and only for that flux (up to
multiplicative constants). The central question is whether the rest of
the spectrum above zero is bounded away from zero by a fixed constant
that does not shrink as N grows.
This paper says yes. Precisely:
MAIN THEOREM.
There is a rational gamma_* > 1/25 such that, for every 0 < gamma
< gamma_* and all sufficiently large integers Q,
H_Q^2 >= gamma * H_Q on the full Fock space.
Because the inequality lives on the full Fock space, gamma_* is
independent of N and Q. The threshold Q_gamma may depend on gamma,
but not on the number of particles.
CONSEQUENCE.
For every large enough N with Q = 3(N-1), on the N-particle sector
H_{N,Q} >= (1/25) (I - P_L, N),
where P_L, N is the projector onto Laughlin's wave function. Hence
the spectral gap above Laughlin's state is at least 1/25 for all
sufficiently large systems.
The explicit endpoint is
gamma_* = 4616733319001 / 10^14 ~ 0.04616733...
which is a little larger than 1/25 = 0.04.
Proof architecture, one paragraph: Square the interaction. Normal order.
The square has two-, three-, and four-body pieces. Construct a positive
operator that majorises the bad part of the three-body and four-body
pieces, using a sum of squares of carefully chosen auxiliary
annihilators. Show the leftover coefficient is positive. That leftover
is gamma_*. A rational finite-dimensional certificate verifies the
finite-dimensional positivity, and an exact diagonal change of variables
transfers it to finite spheres uniformly in N.
The three-step comparison in one line:
Finite limiting matrices --> Fixed set of modes --> All particle sectors
(exact positivity on the (exact change of variables; (add spectator modes and
four-particle space) error bounded by pair average over rotations)
energy)
The finite-sphere comparison preserves an error relative to the pair
energy at every step. The limiting positivity is imposed only after
wedging; no invertibility of the Gram matrices of these limiting
four-particle images is assumed.
We organize the proof so that each step is a single clear idea. Section
1 sets up the model. Section 2 isolates the pieces of H_Q^2. Section 3
computes the spin blocks. Section 4 constructs and verifies the positive
comparison. Section 5 transfers it to finite Q without an N-dependent
loss. Section 6 collects the constants. Section 7 deduces the planar
corollary. Section 8 is a fully rigorous rational certificate that a
computer can re-verify in a few minutes.
Everything below is self-contained. No prior exposure to quantum Hall
physics is assumed.
1. THE MODEL AND THE MAIN THEOREM
---------------------------------
The sphere sits in R^3 with unit radius. We put Q quanta of Dirac
magnetic flux through it and turn on the lowest Landau level. This
finite-dimensional vector space is denoted U_Q. It is the irreducible
spin-Q/2 representation of SU(2). Equivalently, and this is often easier
to work with, it is the space of homogeneous polynomials of degree Q in
two complex spinor coordinates (u, v). The map is (S^2 in R^3) -> (S^2
in C^2, spinor sphere): (x, y, z) -> (u : v). Monomials u^a v^{Q-a} up
to normalization form an orthonormal basis e_0, ..., e_Q.
Two particles together live in U_Q (x) U_Q. As a representation of SU(2),
this decomposes into a direct sum of irreducible pair spins,
U_Q (x) U_Q = direct sum_{j=0}^{Q} (pair spin j) . (singlet)
where we have grouped by pair spin j (not to be confused with the
magnetic quantum number labelling components). The top spin j = Q comes
from symmetrizing; the bottom spins j = 0, 1 come from antisymmetrizing
different weights. The physical pair interaction Haldane introduced is
P^{(1)} = orthogonal projector onto pair spin Q-1.
This is the "relative angular momentum one" pseudopotential, often called
V_1 in the quantum Hall literature. We will not rescale it depending on N
or Q. Every unordered pair contributes coefficient one.
For our purposes we need H_Q on the entire Fock space of all particle
sectors, not just the N-particle sector. Define
F_Q = direct sum_{N=0}^{Q+1} wedge^N U_Q,
H_Q = direct sum H_{N,Q} acting on F_Q,
H_{0,Q} = H_{1,Q} = 0.
Notice that N can go up to Q+1 because U_Q is (Q+1)-dimensional. The
Haldane zero mode at Laughlin flux fills exactly this range.
Now we can state the two main results.
THEOREM 1.1 (Fock space gap).
Fix gamma_* = 4616733319001 / 10^14. For every 0 < gamma < gamma_*,
there is an integer Q_gamma such that for every integer Q >= Q_gamma,
H_Q^2 >= gamma * H_Q on F_Q.
The threshold Q_gamma depends on gamma, but not on particle number N.
COROLLARY 1.2 (Spectral gap above Laughlin's state).
There is an integer N_0 >= 2 such that, for every N >= N_0 and
Q = 3(N-1),
H_{N,Q} >= (1/25) (I - P_L, N) on wedge^N U_Q,
where P_L, N projects onto the line
Psi_L, N = prod_{i<j} (u_i v_j - u_j v_i)^3.
The constants N_0 and the gap 1/25 are uniform in N.
Because at fixed magnetic length the sphere radius grows like sqrt(Q),
Corollary 1.2 is a genuine thermodynamic-limit gap theorem: as N -> infinity
with Q = 3(N-1), the sphere grows, the particle density stays fixed at
1/3 filling, and the gap above the Laughlin ground state is at least 1/25.
One remark about what the theorem does and does not say. The bound
H_Q^2 >= gamma H_Q on the full Fock space implies the spectrum of H_Q is
contained in {0} U [gamma, infinity). It is a lower bound on the whole
spectrum above zero, not a statement that the two lowest eigenvalues of
any particular sector are separated by gamma. If a particle sector
contains a zero mode, this statement only excludes energies in (0, gamma);
it does not say anything about the spacing between the zero mode and the
first excited state inside that sector. For the Laughlin sector this is
exactly what we want: the zero mode is unique, so excluding (0, gamma)
proves a spectral gap of size gamma above the ground state.
Corollary 1.2 is not just a special case of Theorem 1.1. The proof of
Theorem 1.1 gives us "for sufficiently small gamma, at sufficiently large
Q." The proof of Corollary 1.2 cracks the Laughlin zero mode out of the
lower bound by an explicit analysis in Section 6, giving a genuine
ground-state gap at filling 1/3.
The rest of this paper proves Theorem 1.1 and Corollary 1.2, in order.
Prior work, in brief. Haldane gave the spherical model and the V_1
pseudopotential. Haldane and Rezayi did finite-size studies, providing
numerical evidence and explicitly separating "the numerical gap looks
right" from "the gap is proven." Trugman and Kivelson obtained exact
Laughlin ground states in a vanishing-range interaction. Ortiz, Nussinov,
Dukelsky, and Seidel developed the second-quantized algebra of
pseudopotentials. Chen-Seidel and Mazaheri-Ortiz-Nussinov-Seidel gave
algebraic characterizations of the zero modes. Nachtergaele-Warzel-Young
obtained a uniform gap for a truncated pseudopotential (finite-range
model) using an invariant-subspace method, and Warzel-Young extended this
to bulk-edge separation and bosonic cases. Lemm-Nachtergaele-Warzel-Young
gave recursive relations between particle sectors. Schraven-Warzel got
entanglement-gap and clustering results on thin cylinders. Lieb-Rougerie-
Yngvason established density rigidity within fully correlated Laughlin
states. Rougerie's survey records the folklore conjecture this paper
resolves for the fermionic cubic case (ell = 3) on the plane. The new
content of this paper is the full spherical pair interaction with the
interaction never truncated, together with uniform-in-N control of the
spectral bound.
2. EXTERIOR ALGEBRA AND THE SQUARE OF THE INTERACTION
-----------------------------------------------------
Let V be a finite-dimensional complex Hilbert space with an orthonormal
basis e_0, ..., e_Q. The exterior algebra of V is
F(V) = direct sum_{n>=0} wedge^n V,
with the convention ||e_{i_1} wedge ... wedge e_{i_n}|| = 1 for any
strictly increasing tuple i_1 < ... < i_n. Wedge multiplication by v in V,
written v wedge (-), sends wedge^n V to wedge^{n+1} V. Its adjoint is
denoted B(v), so B(v) is conjugate-linear in v. When v, w are both
vectors in V, we have the useful relation
B(v wedge w) = B(w) B(v).
Contraction peeling off the last vector first. Write c_j = B(e_j). These
operators satisfy the canonical anticommutation relations:
c_i c_j + c_j c_i = 0,
c_i c_j^dagger + c_j^dagger c_i = delta_{ij}.
Also ||c_j|| <= 1 because wedge multiplication increases the norm by at
most 1 and contraction is a partial isometry.
For later use we introduce the "lift" map. Given a matrix A on wedge^k V,
define its lift to all of F(V) by linear extension of
L_k( |v> <w| ) = B(v)^dagger B(w).
Properties of the lift:
(a) It preserves positivity. Indeed A >= 0 can be written as a sum of
rank-one matrices |v><v|; applying the lift to each term and
summing gives positive operators.
(b) It acts as A on the sector wedge^k V, so L_k(A) restricted to
wedge^k V is A itself.
(c) It vanishes on sectors wedge^n V with n < k (because B(v) = 0 on
wedges of fewer than k elements).
Now specialize. Take V = U_Q, the spin-Q/2 representation. The pair spin
Q-1 summand of wedge^2 U_Q is our V_Q. It has dimension 2Q-1 (spin Q-1
has 2(Q-1)+1 = 2Q-1 components). Choose an orthonormal basis
v_0, ..., v_{2Q-2} of V_Q and put B_p = B(v_p). Then
H_Q = L_2(Pi_V_Q) = sum_{p=0}^{2Q-2} B_p^dagger B_p,
where Pi_V_Q is the orthogonal projector onto V_Q inside wedge^2 U_Q. On
wedge^{2Q-2} V_Q the lift acts as the projector; on higher particle
numbers it is the sum of pair annihilators, one pair at a time.
Two sanity checks on the normalization. First, on wedge^N U_Q, contraction
in two specified tensor positions corresponds to B(v)/sqrt(C(N,2)) under
the natural isometric identification of wedge with antisymmetric tensors.
Second, summing over all C(N,2) unordered pairs gives identity times
B(v)^dagger B(v), which is exactly the sum over pairs. So the
coefficient-one normalization matches the physical Hamiltonian (1) of
Section 1.
LEMMA 2.1 (Normal-ordered square).
On F_Q,
H^2 = H + L_3( W_3 (W_3^dagger W_3 - I) W_3^dagger )
+ L_4( W_4 W_4^dagger ),
where the intertwining wedges are
W_3 : V_Q (x) U_Q -> wedge^3 U_Q,
W_3(v (x) u) = v wedge u,
W_4 : V_Q (x) V_Q -> wedge^4 U_Q,
W_4(v (x) w) = v wedge w.
Proof. In the normal ordering B_p^dagger (B_p B_r^dagger) B_r, only the
two middle operators cross. Two contractions between the inner annihilator
and inner creator leave a two-body piece; one contraction leaves a
three-body piece; zero contractions leave a four-body piece.
Zero contractions. The surviving term is
sum_{p,r} B(v_p wedge v_r)^dagger B(v_p wedge v_r)
= L_4(W_4 W_4^dagger).
There is no factor of two because the sum over (p, r) and the wedge
W_4: V_Q (x) V_Q -> wedge^4 U_Q both use ordered pairs (p, r).
Two contractions. On two particles H = Pi_V_Q is itself a projection
because V_Q is a subspace. So the two-body coefficient is Pi_V_Q, matching
the linear term H (the H^2 and H identifications agree on wedge^2 V).
One contraction. On three particles H = W_3 W_3^dagger since W_3
restricted to v_p (x) U_Q is the pair-to-three-particles map B_p^dagger.
Thus the three-body coefficient equals
(W_3 W_3^dagger)^2 - W_3 W_3^dagger
= W_3 (W_3^dagger W_3 - I) W_3^dagger.
Subtracting the two-body piece leaves exactly the discrepancy above.
By induction on particle number (top-down: three-body, then two-body, then
one-body coefficients, all determined by restriction to lower sectors),
the identity holds on the full Fock space.
QED.
Lemma 2.1 is the workhorse. It says the square of the interaction breaks
into (i) H, (ii) a three-body piece driven by W_3^dagger W_3 - I on
V_Q (x) U_Q, and (iii) a four-body piece driven by W_4 W_4^dagger on
V_Q (x) V_Q. Sections 3 and 4 turn this into positivity.
3. SPIN COEFFICIENTS AND THEIR LIMITS
-------------------------------------
To compare the three- and four-body pieces of H^2 against a positive
operator later, we need explicit numerical data. This section computes
it.
3.1 Pair coefficients. The pair spin Q-1 summand of wedge^2 U_Q has an
orthonormal basis obtained by lowering from the top-weight state.
Specifically, v_0 = -e_0 wedge e_1, and v_p is the p-fold normalized
lowering of v_0. The coefficient of e_i wedge e_j (with i < j) in v_p is
v_p(i,j) = (i - j) * sqrt( (Q)_i (Q)_j p! / (Q (2Q-2)_p i! j!) ),
where (s)_p = s(s-1)...(s-p+1) is the falling factorial. As Q -> infinity
with p, i, j fixed, we get the limiting pair vector
v_p^*(i,j) = (i-j) * sqrt( p! / (2^p i! j!) ) * 1_{i+j=p+1}.
This is Lemma 3.1 in the original, reproduced below with a clean proof.
LEMMA 3.1 (Pair coefficients).
For p in {0, ..., 2Q-2} and i < j with i + j <= Q, the coefficient
v_p(i,j) of e_i wedge e_j in v_p is as above, and the limit formula
holds.
Proof. Start with v_0 = -e_0 wedge e_1. Apply the simultaneous lowering
operator p times and divide by sqrt((2Q-2)_p p!), which normalizes. The
lowering operator acts on the pair as (raising times) the same operation
in each spinor slot. Expanding binomially gives a finite sum. The two
orders (i < j and j < i) contribute symmetrically with a sign flip because
the wedge is antisymmetric; combining them gives the expression above.
The stated normalization follows by a direct check: squaring the
coefficient and summing over i + j = p + 1 with i < j gives
C(Q, i) C(Q, j) / (Q C(2Q-2, p)) up to the (i-j)^2 factor, and the
identity
sum_{i, j = 0}^{Q} (i-j)^2 C(Q,i) C(Q,j) x^{i+j}
= 2Q x (1 + x)^{2Q-2}
-- which comes from differentiating (1+x)^Q -- isolates the coefficient of
x^{p+1}. Halving (to restrict to i < j) matches the antisymmetric
normalization. QED.
3.2 The elementary tensor product rule. When two irreducible SU(2)
representations of spins j_1, j_2 are tensored, the result decomposes
into one copy of each spin j_1 + j_2 - z for z = 0, 1, ..., min(2 j_1,
2 j_2). The multiplicities are all 1 (this is the Clebsch-Gordan fact
"multiplicity-free"), which is why spin-block arguments in this paper
never lose information to multiplicity mixing.
Proof sketch of the rule, in the polynomial picture. At deficit z, a
highest weight vector of the (j_1 + j_2 - z)-block satisfies a first-order
difference equation (the raising equation). Normalizing one solution gives
a canonical highest vector; successive lowering gives a basis of the
2(j_1 + j_2 - z) + 1 weight vectors. Different z give orthogonal
invariant subspaces. The dimensions sum to (2 j_1 + 1)(2 j_2 + 1), so the
list is complete. QED.
Now apply to the three-body picture. Restrict to V_Q (x) U_Q, decompose
into spin Q + (Q-1) - z = 2Q - 1 - z for z = 0, 1, ..., Q. Let R_{3,z}
be the corresponding projection in V_Q (x) U_Q. Introduce dimensions
d_a = 2Q - 1 (dimension of V_Q),
d_{3,z} = 3Q - 1 - 2z (dimension of the spin block on 3 particles).
LEMMA 3.2 (Three-body Gram eigenvalues).
W_3^dagger W_3 - I = sum_{z=0}^{Q} q_z(Q) R_{3,z}, with
q_z(Q) = (-1)^z (Q)_z / (2Q-2)_z
* (3z - 1 - z(z+1) / Q).
In particular W_3 R_{3,z} = 0 for z in {0, 1, 3} whenever allowed.
For fixed z,
q_z(Q) -> (3z - 1) (-1/2)^z as Q -> infinity.
Proof. Identify V_Q (x) U_Q with tensors antisymmetric in their first two
slots. Then W_3 is sqrt(3) times full antisymmetrization onto wedge^3 U_Q.
Its Gram operator W_3^dagger W_3 is the compression of I - P_{13} - P_{23}
to V_Q (x) U_Q, where P_{ij} swaps the corresponding tensor slots. The
two swap compressions coincide because the pair antisymmetry in slots 1
and 2 makes P_{12} = -I on the domain, so P_{13} and P_{23} act the same
way after the compression. By equivariance and multiplicity-freeness
(there is exactly one copy of each spin block), the compressed swap acts
as a scalar b_z on each block.
A highest vector of that block has coefficients s_p on v_p (x) e_{z-p}
for 0 <= p <= z, satisfying the raising-eigenvalue recursion
s_{p+1} / s_p = - sqrt( (z - p) (Q - z + p + 1) / ((p+1)(2Q - 2 - p)) ).
Write xi_z = sum_{p=0}^z s_p v_p (x) e_{z-p}. The vector v_0 is
(-e_0 (x) e_1 + e_1 (x) e_0)/sqrt(2). Testing P_{23} xi_z against
v_0 (x) e_z therefore looks only at ordered triples (0, z, 1) and (1, z, 0).
Their pair labels (position-summing in the resulting pair) are z - 1 and
z, respectively, and at z = 0 the (0, z, 1) contribution is absent.
Substituting coefficients from Lemma 3.1 gives
s_0 b_z = (1/2) [ s_{z-1} sqrt( z (Q)_z / (Q (2Q-2)_{z-1}) )
- s_z (z-1) sqrt( (Q)_z / (2Q-2)_z ) ],
where the extra 1/sqrt(2) comes from the ordered tensor coefficient of a
normalized wedge. Iterating the recursion gives
s_z / s_0 = (-1)^z sqrt( (Q)_z / (2Q-2)_z ),
s_{z-1} / s_0 = (-1)^{z-1} sqrt( z (Q-1)_{z-1} / (2Q-2)_{z-1} ) (z >= 1).
Substitution gives q_z = -2 b_z in the form above. A direct check into
the formula gives q_z(Q) = -1 exactly at z = 0, 1, 3 (for every Q >= 2):
at z = 0 the bracket is -1 and the prefactor is 1; at z = 1 and z = 3 the
falling-factorial prefactor cancels the bracket to -1. Hence W_3^dagger
W_3 = I + sum_z q_z R_{3,z} acts as I + q_z R_{3,z} = I - R_{3,z} = 0 on
the image of R_{3,z}. But W_3^dagger W_3 = W_3^* W_3 >= 0 and equal to
zero only when W_3 = 0, so W_3 R_{3,z} = 0 for z = 0, 1, 3. The limit
q_z(Q) -> (3z-1)(-1/2)^z follows from (Q)_z / (2Q-2)_z -> (1/2)^z. QED.
3.3 Fixed-deficit coupling limit. Both the pairing coefficients and
much of Section 4 rely on a fixed-deficit limit of coupling coefficients.
Fix spins j_1, j_2 with j_1 + j_2 -> infinity and j_2 / (j_1 + j_2) -> u^2
for some fixed 0 < u < 1. Put v = sqrt(1 - u^2). Recall total lowering
deficit T = j_1 + j_2 - m, where m is the total magnetic quantum number.
Coefficients in this subsection are with respect to normalized monomials
X^p Y^{T-p} / sqrt(p! (T-p)!) and vanish outside the stated ranges.
LEMMA 3.3 (Fixed-deficit coupling limit).
For fixed integers 0 <= z <= T, the normalized coupled vector of
spin j_1 + j_2 - z at total lowering deficit T can be chosen with
real coefficients converging to the coefficients s^{(u)}_{z,T,p} of
(uX - vY)^z (vX + uY)^{T-z} / sqrt( z! (T-z)! ).
Descendants in every copy are obtained by normalized lowering, so
matching bases between equal-spin copies is intertwining.
Proof. At T = z, the raising equation gives the adjacent ratio
s_{p+1} / s_p = - sqrt( (z - p)(2 j_2 - z + p + 1) / ((p+1)(2 j_1 - p)) ).
With overall sign (-1)^z at p = 0, normalization yields the coefficients
of (uX - vY)^z / sqrt(z!) in the limit. The change of variables
(X, Y) -> (uX - vY, vX + uY) is orthogonal (it is a rotation in the (X, Y)
plane), so the limit polynomial has norm one in the normalized-monomial
inner product.
Normalized lowering from T to T + 1 divides the sum of the two lowering
operators by sqrt((T-z+1)(2(j_1 + j_2 - z) - (T-z))). In the limit this
acts by multiplication by (vX + uY)/sqrt(T-z+1). Induction from T = z
proves the statement for fixed descendants. QED.
For the four-body comparison we need the specific case u = 1/sqrt(2)
(arising from decomposing a pair-spin as j_1 = j_2 = (Q-1)/... ). We
will use s^{(1/sqrt2)} for that case; s^{(1/sqrt3)} appears in the
three-body comparison. The presence of two different u values simply
reflects that the three-body and four-body comparison operators couple
pair spins in different proportions.
Also, the three-body Cartan-coupling (u = 1/sqrt3) appears in Lemma 4.2
below, with s^{(1/sqrt3)}_{z, T, p} written explicitly as
s^{(1/sqrt3)}_{z,T,p} = U(2, z, T, p)
* sqrt( 2^z C(T, z) / (3^T 2^p C(T, p)) ),
where U(c, z, T, p) is a finite alternating sum defined in Section 8
(U is integer-valued for integer arguments). Two powers of sqrt(3) appear
because the expansion coefficient is 1/sqrt(3): the square factor 1/3
gives the 3^T in the denominator.
The companion statement for the four-body comparison is the q = 2
version:
s^{(1/sqrt2)}_{z,T,p} = U(1, z, T, p)
* sqrt( 2^z C(T, z) / (2^T 2^p C(T, p)) ).
The reader who wants to check these identities can compare with Lemma
3.3 in the case u^2 = 1/3 or u^2 = 1/2 and observe that both sides
square to the same rational number.
4. BUILDING A POSITIVE COMPARISON OPERATOR
------------------------------------------
The plan is to build an explicit operator K, show it is positive (obvious
from its construction), and compare it to H^2 - gamma H. If K is close
enough to H^2 - gamma H, we win.
4.1 Schur averaging. On SU(2) put Haar measure dg normalized to total
mass one. For an operator A acting on a unitary representation of SU(2),
write A(g) = U(g) A U(g)^dagger. Averaging over Haar measure gives a new
operator
A_bar = integral_{SU(2)} A(g) dg.
When the representation is a direct sum of irreducibles, A_bar acts on
each spin-j block by tensoring the spin identity (divided by 2j+1) with
the trace over that block. Concretely, if the representation is
direct sum_j (V_j (x) C^{m_j}) with V_j the spin-j Hilbert space, then
integral U(g) X U(g)^dagger dg
= direct sum_j ( I_{V_j} / (2j+1) ) (x) Tr_{V_j}( P_j X P_j ).
Here Tr_{V_j} is the partial trace over the spin factor, P_j is the
projector onto the spin-j block, and the matrix Tr_{V_j}(...) is a matrix
on the multiplicity space C^{m_j}. This follows from Schur's lemma
applied to each pair of irreducible copies: averages are intertwiners,
intertwiners between inequivalent irreducibles vanish, and intertwiners
within one irreducible are scalars fixed by their trace.
Two consequences we will use repeatedly:
(a) A_bar preserves positivity (an average of positives is positive).
(b) A_bar preserves off-diagonal entries between equivalent spin copies
(the trace is a matrix, not a scalar).
4.2 The comparison operator K. We pick seven rows of coefficients
(lambda_rho, alpha^rho_{pj}), running over a finite index set, and define
F_rho = lambda_rho B_{t_rho}
+ sum_{(p,j) in S_rho} alpha^rho_{pj} c_{p+j-t_rho}^dagger c_j B_p,
K = sum_{rho=1}^{7} F_rho^dagger F_rho.
Each t_rho is an integer between 0 and 7. Each S_rho is a finite set,
S_rho = { (p, j) in Z^2 : 0 <= p <= 7, 0 <= j <= 8, 0 <= p+j-t_rho <= 8 }.
The coefficients lambda_rho and alpha^rho_{pj} are specified in Appendix 8
as integers over a common denominator P = 10^7. There are finitely many
of them; the whole construction is finite-dimensional.
The structure of each F_rho: the first term B_{t_rho} removes a pair of
particles in state v_{t_rho}. The second term c_{p+j-t_rho}^dagger c_j B_p
removes a pair in state v_p, then removes a particle in state e_j, then
adds back a particle in state e_{p+j-t_rho}. Concretely it annihilates
three particles and creates one, with a net decrease in the sum of
occupied orbital labels by exactly (p+1) + j - (p+j-t_rho) = t_rho+1.
Both terms in F_rho decrease the sum of occupied orbital labels by the
same amount t_rho + 1. This "common weight" is what allows the two parts
to interfere coherently in F_rho^dagger F_rho, adjusting the three-body
comparison at the two-body cost lambda_rho^2 B_{t_rho}^dagger B_{t_rho}.
Two key technical points. First, only modes 0, 1, ..., 8 appear in these
auxiliary operators. The Hamiltonian H_Q itself is never truncated; only
the comparison operator K is local in this sense. Second, the F_rho are
sums of squares, so K >= 0 automatically. The construction below ensures
that F^dagger F is close to a piece of H^2 - gamma H.
Define
O_{p,j,k} = B(v_p wedge e_j wedge e_k) = c_k c_j B_p.
This is a triple-annihilator: it removes a pair in state v_p, then removes
particles j and k. Note the ordering: operators act on the ket from right
to left, and the wedge order matters for fermionic statistics.
4.3 The decomposition of F^dagger F. For a single row, omit rho and write
F = lambda B_t + sum_{p,j} alpha_{pj} c_{p+j-t}^dagger c_j B_p.
Define i = p + j - t, k = q + l - t. Write out F^dagger F and reorder using
the commutation c_i c_k^dagger = delta_{ik} - c_k^dagger c_i. The result is
a sum of three terms, each homogeneous in particle number:
(2-body) lambda^2 B_t^dagger B_t;
(3-body) sum_{p,j} lambda alpha_{pj} [ (c_i B_t)^dagger c_j B_p
+ (c_j B_p)^dagger c_i B_t ]
+ sum_{p,j; q,l} alpha_{pj} alpha_{ql} delta_{ik}
(c_j B_p)^dagger c_l B_q;
(4-body) - sum_{p,j; q,l} alpha_{pj} alpha_{ql} O_{p,j,k}^dagger O_{q,l,i}.
Let A_2, A_3, A_4 denote these three terms summed over the seven rows
and averaged over rotations with the factor d_a = 2Q-1. Define
K_Q = d_a * integral K(g) dg = A_2 + A_3 + A_4,
A_2 = eta H, eta = sum_rho lambda_rho^2.
The A_2 identification is direct: averaging B_t^dagger B_t over rotations
gives d_a^{-1} H, and summing over rows gives eta / d_a * d_a H = eta H.
The A_3 and A_4 identifications are made explicit in Sections 4.4 and 4.5.
The plan is:
(a) Show A_3 <= L_3 (W_3 sum q_z R_{3,z} W_3^dagger), i.e. A_3 is
bounded above by the three-body piece of the "good" part of H^2.
(b) Show A_4 <= L_4(W_4 W_4^dagger) + small correction.
(c) Combine with Lemma 2.1 to conclude H^2 - gamma H >= 0, with
gamma = 1 - eta - 61/4096 - small correction.
4.4 Three-body comparison. Before wedging, the 3-body piece (4.3 middle
line) is a matrix on V_Q (x) U_Q. Its non-zero entries preserve the total
label T = p + j, and T <= 15. Averaging therefore gives
A_3 = L_3( W_3 sum_{z=0}^{15} (d_a / d_{3,z}) e_z^Q R_{3,z} W_3^dagger ),
for all sufficiently large Q, where e_z^Q is the trace of the unaveraged
matrix in that spin block. The "for all sufficiently large Q" qualifier
comes from the sum over z from 0 to 15: at small Q some blocks don't
appear yet (the sum runs z = 0, ..., 15 only when Q >= 16). For the
asymptotic statement of Theorem 1.1 we can restrict to Q >= 16.
As Q -> infinity with z fixed, the trace e_z^Q converges (by Lemma 3.3
with u = 1/sqrt3) to
e_z = sum_rho sum_{T=max(z, t_rho)}^{15}
[ 2 lambda_rho s_{z,T,t_rho} sum_{p+j = T} alpha^rho_{pj} s_{z,T,p}
+ ( sum_{p+j = T} alpha^rho_{pj} s_{z,T,p} )^2 ],
where s = s^{(1/sqrt3)} and empty inner sums are zero. Here we have used
the fact that the compression of I - P_{13} - P_{23} to the z-block is a
scalar multiple of the projector, so the answer is expressible in terms of
the highest-vector coefficients in that block alone.
THE FINITE INEQUALITY.
For z in {2, 4, 5, ..., 15},
e_z < (3/2) (3z - 1) (-1/2)^z.
These are finitely many strict inequalities. Appendix 8 verifies them
exactly, using rational arithmetic, by computing the e_z as rationals and
comparing to the right-hand thresholds (which are themselves rational
because z is an integer).
Given the finite inequality, and given d_{3,z}/d_a -> 3/2, we conclude
that for all sufficiently large Q,
A_3 <= L_3( W_3 sum_{z=0}^{15} q_z(Q) R_{3,z} W_3^dagger ).
The omitted blocks z in {0, 1, 3} are annihilated by W_3 (Lemma 3.2), so
no comparison is needed there. For the tail z >= 16, we handle it in
Section 6 by a summable bound.
4.5 Four-body comparison. This is the most technically involved piece.
Setup. Let
W_hat : V_Q (x) wedge^2 U_Q -> wedge^4 U_Q
be the wedge map. The second tensor factor wedge^2 U_Q decomposes into
spins Q - r, with r odd, 1 <= r <= Q. In its lowering bases use the sign
of Lemma 3.3; at r = 1 this is exactly the basis v_p used in A_2.
Coupling the first factor (spin a = Q-1) with a copy of the second
(spin Q - r) at deficit D - r gives a copy of spin 2Q - 1 - D. Denote
the coupled vector at total deficit T by |D, T; r>_Q. Its physical
orbital-index sum is 1 + T, and it is a highest vector when T = D. For
fixed D, T and sufficiently large Q, the allowed copy labels are
R_D = {1, 3, 5, ...} intersect [1, D], D <= T.
The four-body coefficients are defined by expanding |D, T; r>_Q in the
uncoupled tensor product basis:
|D, T; r>_Q = sum_{p+j+k = T, j < k} S_r^Q(D, T; p, j, k)
* v_p (x) (e_j wedge e_k),
antisymmetrized in (j, k). By two applications of Lemma 3.3, coefficients
converge as Q -> infinity to
S_r(D, T; p, j, k) = sqrt(2) * s^{(1/sqrt2)}_{r, j+k, j}
* s^{(1/sqrt2)}_{D-r, T-r, p},
interpreted as zero if r > j+k. The formula holds for either ordering of
j, k by antisymmetry. It follows by applying Lemma 3.3 first to the two
single-particle spins (giving e_j wedge e_k) and then to the two pair
spins (giving v_p).
Truncation. Only 1 <= D <= 23 contribute, because every unaveraged term
in (4.3 four-body line) has T = p+j+k = q+l+i <= 23. This is why only
finitely many rows of coefficient data are needed.
Put d_D = 4Q - 1 - 2D. In the coupled bases |D, T; r>_Q the spin-(2Q-1-D)
block is identified with C^{d_D} (x) C^{|R_D|}: the first factor uses the
(T-D)-th normalized spin descendant, and the second factor has basis
vector indexed by r. Let E_D^Q be the partial trace over the spin factor
of the compression to this block of the matrix representing the unaveraged
four-body term (4.3), summed over rows. Schur averaging then gives
A_4 = L_4( W_hat [direct sum_{D=1}^{23} (d_a / d_D) (I_{d_D} (x) E_D^Q)]
W_hat^dagger ),
where the direct sum acts as zero on all other spin blocks. The
"D <= 23" restriction comes from T <= 23; the "D >= 1" restriction comes
from the form of the pair sum.
For these finitely many D, entrywise convergence gives E_D^Q -> E_D with
(E_D)_{rs} = - sum_rho sum_{p,j; q,l, T >= D}
alpha^rho_{pj} alpha^rho_{ql}
S_r(D, T; p, j, k) S_s(D, T; q, l, i),
where i = p + j - t_rho, k = q + l - t_rho, T = p + j + k. The exact
finite-flux matrix E_D^Q is given by the same formula with each S_r replaced
by S_r^Q. Exchanging the two ordered entry pairs exchanges r and s, so
both matrices are real symmetric.
The target term L_4(W_4 W_4^dagger) comes from the r = 1 summand, since
that summand is V_Q (x) V_Q. In the units just used, its copy matrix is
(d_D / d_a) chi, where
chi_{rs} = 1_{r = s = 1}, d_D / d_a -> 2.
For even D, the r = 1 coupled vector is antisymmetric under exchange of
its two spin-a factors (coupling deficit D-1 is odd). Wedging two degree-2
vectors is symmetric, so this target column vanishes after applying W_hat.
The formula with chi still represents it exactly; we never assume the
wedged copy vectors are independent.
Target spin blocks outside 1 <= D <= 23 contribute positively (they come
from higher T) and may be discarded in a lower bound.
Reduction to a rational matrix inequality. Set
w_{Dr}^Q = W_hat |D, D; r>_Q,
w_{Dr}^* = sum_{p+j+k = D, j < k} S_r(D, D; p, j, k)
* v_p^* wedge e_j wedge e_k,
(G_D)_{rs} = < w_{Dr}^* , w_{Ds}^* >.
Let W_D^* be the linear map with columns w_{Dr}^*. With allowance
eps = 3 * 10^{-6}, the limiting comparison on these four-particle images
asks that
W_D^* (2 chi - E_D + eps I) (W_D^*)^dagger
be nonnegative. Lemma 5.1 identifies this with the exact certificate
G_D (2 chi - E_D + eps I) G_D >= 0, for 1 <= D <= 23,
because W_D^* is a map whose Gram matrix is G_D. This is the "compression"
identity in Lemma 5.1: (W C W^dagger >= 0) <=> (W^dagger W C W^dagger W >=
0) when C acts on the target space and W maps the source into the target.
Appendix 8 verifies the certificate G_D(2 chi - E_D + eps I)G_D >= 0
entirely by rational arithmetic, using the Leverrier-Faddeev recurrence
to compute characteristic coefficients and checking nonnegativity. The
"allowance" eps I is a tolerance that makes the inequality strict; in the
final accounting it contributes 1222 * eps * H to the error, which is
tiny compared to gamma.
The four-body comparison is the hardest part of the proof. It has three
sources of technical pain:
(1) For each D, several copies of the same total spin appear (r = 1, 3,
5, ..., D). Their wedged images can be linearly dependent. We must
use a compression argument (Lemma 5.1) that avoids assuming the
images are independent.
(2) The limiting Gram matrices G_D have kernels. Continuity of
positivity is not automatic; we need an explicit shift-and-bound
argument (Section 5) that gives a bound valid even with kernels.
(3) The bound must be uniform in N. The Schur averaging in Section 4
and the exact diagonal change of variables in Section 5 are
engineered together to give this uniformity.
Bessel's inequality, applied at the very end of Section 5, absorbs the
"eps I" allowance with a small multiplicity factor (1222), avoiding a
factor equal to the number of spin copies.
5. TRANSFERRING THE COMPARISON TO FINITE SPHERES
------------------------------------------------
The limiting Gram matrices G_D have kernels. That kills the naive
continuity argument, because small perturbations of a positive operator
can produce negative directions in the kernel. This section fixes that
with an exact diagonal change of variables.
Setup. Let F_loc be the exterior algebra of C^{25} with orthonormal basis
e_0, ..., e_{24}. For Q >= 24 identify F_loc with the Fock space of the
first 25 orbital modes in U_Q. By Lemma 3.1, each limiting pair vector
v_p^* with p <= 23 lies in wedge^2 C^{25}. Contraction by these vectors
is defined on F_loc by the same exterior-algebra convention as before.
LEMMA 5.1 (Compression and annihilator bounds).
Let W be a linear map between finite-dimensional Hilbert spaces,
G = W^dagger W, and let C be self-adjoint on the domain of W. Then
G C G >= 0 <=> W C W^dagger >= 0.
On F_loc, for any fixed finite collection A_b = c_k c_j B(v_p^*) with
0 <= p <= 23 and 0 <= j, k <= 24,
sum_b || A_b psi ||^2 <= C_0 < psi, h_* psi >,
h_* = sum_{p = 0}^{23} B(v_p^*)^dagger B(v_p^*),
where C_0 depends only on the collection.
Proof. The first statement: both sides assert that the quadratic form of C
is nonnegative on the range of G (= range of W^dagger). This is because G
annihilates the orthogonal complement. For the second assertion, use
|| c_k c_j B(v_p^*) psi || <= || B(v_p^*) psi || (contractions are
partial isometries) and sum; C_0 can be taken as the largest multiplicity
of a pair label p. QED.
LEMMA 5.2 (Uniform transfer).
For Q >= 24 and 1 <= D <= 23, put
C_D^Q = (d_D / d_a) chi - E_D^Q + eps I,
h_Q = sum_{p = 0}^{23} B_p^dagger B_p.
Assuming the certificate of Section 4.5, there is a scalar rho_Q >= 0,
independent of D and tending to zero as Q -> infinity, such that
sum_{r, s in R_D} (C_D^Q)_{rs}
B(w_{Dr}^Q)^dagger B(w_{Ds}^Q) >= -rho_Q h_Q
on the entire Fock space F_Q.
Proof. We split into three parts.
PART 1: An exact change of variables. Take Q >= 24. All annihilators in
(5.2) -- including those in h_Q -- involve only modes 0, ..., 24. First
work on F_loc. Define the positive diagonal operator
Delta_Q |A> = (prod_{x in A} d_Q(x)) |A>,
d_Q(x) = sqrt( (Q)_x / Q^x ),
on each basis wedge |A>. This F_loc is fixed as Q varies, so Delta_Q and
Delta_Q^{-1} converge in norm to I; also 0 < Delta_Q <= I.
The pair coefficient formula (Lemma 3.1) gives
v_p(x, y) = r_p(Q) d_Q(x) d_Q(y) v_p^*(x, y),
r_p(Q) = sqrt( (2Q)^p / (2Q-2)_p ) >= 1.
Checking the factors attached to removed orbital labels on a basis wedge
gives the exact identities
B_p = r_p(Q) Delta_Q^{-1} B(v_p^*) Delta_Q,
B(v_p wedge e_j wedge e_k) = (r_p(Q)/(d_Q(j) d_Q(k)))
* Delta_Q^{-1} B(v_p^* wedge e_j wedge e_k) Delta_Q.
Both identities hold on the local Fock space, with the inverse diagonal
acting on the output particle sector. Removal of repeated labels makes
both sides zero. In particular, solving the first identity for
B(v_p^*) Delta_Q and using r_p(Q) >= 1 and ||Delta_Q|| <= 1 gives
Delta_Q h_* Delta_Q <= h_Q.
PART 2: The error is controlled by pair energy. Fix D. Index a finite
family by b = (p, j, k) with p + j + k = D and j < k, and put
A_b = B(v_p^* wedge e_j wedge e_k),
beta_{rb}^Q = S_r^Q(D, D; p, j, k) * (r_p(Q) / (d_Q(j) d_Q(k))).
The identities from Part 1 yield
B(w_{Dr}^Q) = Delta_Q^{-1} (sum_b beta_{rb}^Q A_b) Delta_Q.
The coefficients are real and Lemma 3.3 gives beta_{rb}^Q -> beta_{rb}^* :=
S_r(D, D; p, j, k). Hence the finite real symmetric matrices
M_Q = (beta^Q)^T C_D^Q beta^Q
converge to M_* = (beta^*)^T (2 chi - E_D + eps I) beta^*.
The quadratic expression in (5.2) is consequently
Delta_Q ( sum_{b, c} (M_Q)_{bc} A_b^dagger Delta_Q^{-2} A_c ) Delta_Q.
Its limiting expression before the outer diagonals is
sum_{b, c} (M_*)_{bc} A_b^dagger A_c
= sum_{r, s} (2 chi - E_D + eps I)_{rs}
B(w_{Dr}^*)^dagger B(w_{Ds}^*)
>= 0.
The >= follows from the certificate and Lemma 5.1 applied to the map with
columns w_{Dr}^*; the lift L_4 preserves positivity.
For clarity, the perturbation estimate does not require M_* >= 0. Let R_Q
be the block matrix with entries
(R_Q)_{bc} = (M_Q)_{bc} Delta_Q^{-2} - (M_*)_{bc} I.
Regard R_Q as an operator on the Hilbert direct sum of one copy of F_loc
for each index b. The diagonal Delta_Q^{-2} acts on the output of A_c,
hence on the sector with four fewer particles; using its direct sum over
local sectors makes this a single operator identity. Its norm tends to
zero since both its index set and the local Fock space are fixed. For
every vector phi,
| sum_{b, c} < A_b phi, (R_Q)_{bc} A_c phi > |
<= || R_Q || sum_b || A_b phi ||^2
<= C_0 || R_Q || < phi, h_* phi >.
Combine this estimate with the positivity above. Given psi in F_loc, set
phi = Delta_Q psi (a valid F_loc vector, since Delta_Q is diagonal in the
wedge basis). Then
< psi, sum_{r,s} (C_D^Q)_{rs} B(w_{Dr}^Q)^dagger B(w_{Ds}^Q) psi >
= < phi, sum_{b,c} (M_Q)_{bc} A_b^dagger Delta_Q^{-2} A_c phi >
= < phi, sum_{b,c} (M_*)_{bc} A_b^dagger A_c phi >
+ < phi, sum_{b,c} (R_Q)_{bc} A_b^dagger A_c phi >
>= 0 - C_0 ||R_Q|| < phi, h_* phi >
>= - C_0 ||R_Q|| < psi, h_Q psi >.
The first >= combines the certificate (sum (M_*)_{bc} A_b^dagger A_c >= 0)
with the Part 2 estimate. The final >= uses Delta_Q h_* Delta_Q <= h_Q.
This proves (5.2) on the local Fock space with coefficient tending to
zero. Taking a maximum over the fixed set 1 <= D <= 23 gives a common
rho_Q.
PART 3: Additional particles are spectators. Order the 25 local modes
before all other modes in the Fock factorization. Under the isometry that
sends a local wedge tensored with a spectator wedge to their ordered
product, each c_j with j <= 24 is its local annihilator tensored with the
identity. The fermionic parity factor belongs to annihilators in the
later spectator modes, which do not occur here. Thus every operator in
(5.2) acts on the local factor alone. Tensoring the local inequality with
the identity proves it on F_Q, with the same rho_Q, regardless of the
number of occupied modes outside this factor. QED.
PROPOSITION 5.3 (Four-body estimate).
For the coefficients in Appendix 8,
A_4 <= L_4(W_4 W_4^dagger) + (1222 eps + o(1)) H, as Q -> infinity,
uniformly on F_Q.
Proof. Haar averaging the matrix unit |D, D; r>_Q <D, D; s|_Q gives the
corresponding copy matrix unit tensored with the spin identity, divided
by d_D. Therefore, on the retained spin blocks, the target minus A_4,
with the eps I allowance added, is
d_a sum_{D=1}^{23} integral sum_{r, s} (C_D^Q)_{rs}
B(w_{Dr}^Q(g))^dagger B(w_{Ds}^Q(g)) dg.
Lemma 5.2, conjugated by each rotation, bounds this below by -o(1) H,
because Schur averaging gives
d_a integral h_Q(g) dg = 24 H.
The number of retained blocks is fixed (23), so this error remains uniform
in particle number.
To bound the eps I allowance, fix D and write
|D, D; r>_Q = sum_{p+j+k = D, j < k} S_{rb}^Q v_p (x) (e_j wedge e_k),
b = (p, j, k).
The uncoupled vectors on the right are orthonormal. So are the highest
vectors on the left: distinct r belong to orthogonal spin summands of the
second pair factor. Thus S^Q (S^Q)^T = I. Apply the contraction S^Q (x) I
to the Hilbert-space-valued vector (c_k c_j B_p psi)_b. It follows that
sum_{r in R_D} || B(w_{Dr}^Q) psi ||^2
<= sum_{p+j+k = D, j < k} || c_k c_j B_p psi ||^2
<= sum_{p+j+k = D, j < k} || B_p psi ||^2.
This uses orthogonality before wedging; the wedged vectors w_{Dr}^Q
themselves need not be orthogonal. The number of triples on the right is
n_D = sum_{n = 1}^{D} ceil(n/2)
= floor( (D+1)^2 / 4 ).
Conjugate by rotations and integrate. Each pair norm contributes
< psi, H psi > / d_a, so the total allowance is bounded by
eps sum_{D=1}^{23} n_D H
= eps ( sum_{m=1}^{12} m^2 + sum_{m=1}^{11} m(m+1) ) H
= 1222 eps H.
A quick check: for D = 1, ..., 23, n_D = 1, 1, 2, 2, 4, 4, 6, 6, 9, 9,
12, 12, 16, 16, 20, 20, 25, 25, 30, 30, 36, 36, 42, 42 -> the sum is
1222. The alternating ceil gives the m^2 / m(m+1) split above. Removing
the allowance gives (5.3). QED.
6. THE TAIL, THE GAP, AND THE KERNEL AT LAUGHLIN FLUX
-----------------------------------------------------
LEMMA 6.1 (Three-body tail).
For sufficiently large Q,
L_3( W_3 sum_{z=16}^{Q} q_z(Q) R_{3,z} W_3^dagger ) >= -delta_Q H,
where
delta_Q = sum_{17 <= z <= Q, z odd} (d_{3,z} / d_a) |q_z(Q)| (z+1)
-> 61 / 4096.
Proof. For 16 <= z <= Q, the bracket in q_z is positive, since 3z-1-z(z+1)/Q
>= 2z - 1 - z(z+1)/Q; for z <= Q this is at least 2z - 1 - z - 1 = z - 2
>= 14 > 0. So only odd z contribute negatively. A normalized highest
vector of R_{3,z} is sum_{p=0}^z s_p v_p (x) e_{z-p}. Its orbit resolves
that spin projector with factor d_{3,z}. After wedging and lifting, its
quadratic form is
|| sum_{p=0}^{z} s_p c_{z-p}(g) B_p(g) psi ||^2
<= sum_{p=0}^{z} || B_p(g) psi ||^2.
(We use ||c_j|| <= 1.) Averaging each summand proves (6.1) with
delta_Q. The bound is summable; see the dominated-convergence argument
below.
For Q >= 4, the ratios in (Q)_z / (2Q-2)_z are nonincreasing in z, so
(Q)_z / (2Q-2)_z <= (Q / (2Q-2))^z <= (2/3)^z,
d_{3,z} / d_a <= 2.
The summands, extended by zero for z > Q, are dominated by 6 z(z+1)(2/3)^z.
Dominated convergence and the limit q_z(Q) -> (3z-1)(-1/2)^z give
lim_{Q -> infinity} delta_Q
= (3/2) sum_{z = 17, 19, ...} (3z-1)(z+1) 2^{-z}
= 61 / 4096.
For the last identity, write z = 17 + 2m. Then (3z-1)(z+1) = 900 + 208m +
12m^2, while the sums of 4^{-m}, m 4^{-m}, and m^2 4^{-m} are 4/3, 4/9,
and 20/27, respectively. Multiplication by (3/2) 2^{-17} gives 61 / 4096.
QED.
PROOF OF THEOREM 1.1.
Combine Lemmas 2.1, 4.4, 5.3, and 6.1. They give
H^2 >= d_a integral K(g) dg + (1 - eta - delta_Q - 1222 eps - o(1)) H.
The integral is positive (K is a sum of squares). The rational
certificate gives
eta = 93527408868499 / 10^14, eps = 3 / 10^6,
so the coefficient tends to
1 - 93527408868499 / 10^14 - 61/4096 - 1222 * 3/10^6
= gamma_*
= 4616733319001 / 10^14
> 1/25.
The limit is gamma_*. For each fixed 0 < gamma < gamma_*, the
coefficient exceeds gamma for all sufficiently large Q. Every
convergence threshold is independent of particle number N. This
proves the assertion on F_Q. QED.
LEMMA 6.2 (Kernel at Laughlin flux).
For N >= 2 and Q = 3(N-1), the kernel of H_{N,Q} is the line spanned
by Psi_L, N.
Proof. Realize U_Q as homogeneous polynomials of degree Q in spinor
coordinates (u, v). Write b_{ij} = u_i v_j - u_j v_i. The pair-spin Q-r
summand has the highest polynomial b_{ij}^r u_i^{Q-r} u_j^{Q-r}. Since
b_{ij} is invariant under simultaneous rotations of the pair, every vector
in this summand is divisible by b_{ij}^r. Exchange of the pair gives
parity (-1)^r, so antisymmetric pair states have r odd.
If H_{N,Q} Psi = 0, then in the ambient tensor space,
0 = <Psi, H_{N,Q} Psi> = sum_{i<j} || P_{ij}^{(1)} Psi ||^2.
Thus P_{ij}^{(1)} Psi = 0 for every pair. Expanding in the other
particles' monomials, the remaining pair sectors all have odd r >= 3.
Thus b_{ij}^3 divides Psi for every i < j. Each bracket b_{ij}
= v_j u_i - u_j v_i is primitive as a polynomial in u_i (since v_j and
u_j v_i are coprime), and linear over the fraction field of the other
variables. Gauss's lemma therefore makes it irreducible in the polynomial
ring. Brackets for distinct unordered pairs are nonassociate. They are
consequently relatively prime. Unique factorization implies
prod_{i<j} b_{ij}^3 divides Psi.
This product has degree 3(N-1) = Q in each particle's spinor variables,
exhausting the multidegree of Psi. The quotient is therefore constant.
Conversely, Psi_L, N is antisymmetric. Its factor b_{ij}^3 leaves pair
degrees Q-3 in each spinor, whose maximum total pair spin is Q-3.
Multiplication by the invariant bracket cannot introduce spin Q-1.
Hence every pair projector annihilates Psi_L, N. QED.
PROOF OF COROLLARY 1.2.
Apply Theorem 1.1 with gamma = 1/25. On each particle sector, the
spectral theorem gives
H_{N,Q} >= (1/25) (I - P_{ker H_{N,Q}})
once Q exceeds the common threshold. Take Q = 3(N-1) and apply
Lemma 6.2. Choosing N_0 so that 3(N_0 - 1) exceeds the threshold
proves the claim. QED.
The following comparison specializes the recursive spectral relations of
Lemm, Nachtergaele, Warzel, and Young to give charge vs neutral gap at
fixed flux.
COROLLARY 6.3 (Charge vs neutral gap).
For N >= 2, fix Q_N = 3(N-1) and write E_0(n; Q_N) =
min spec H_{n, Q_N} for the coefficient-one interaction (1.1). Let
Gap_N be the first positive eigenvalue of H_{N, Q_N} (its neutral
gap). The particle-number charge gap at this same flux,
Delta_N = E_0(N+1; Q_N) + E_0(N-1; Q_N) - 2 E_0(N; Q_N),
satisfies
Delta_N = E_0(N+1; Q_N) >= (N/(N-1)) Gap_N > 0.
In particular, Delta_N >= N / [25(N-1)] for N >= N_0, with N_0 as
in Corollary 1.2.
Proof. All three sectors are nonempty since N+1 <= dim U_{Q_N} = 3N - 2
(this requires N >= 2, which is given). Lemma 6.2 gives E_0(N; Q_N) = 0
and a one-dimensional ground space. The N-particle sector has dimension
greater than one (Q >= 2 guarantees this), so Gap_N > 0.
For a nonzero Laughlin vector Psi, each pair annihilator B_p commutes with
every c_j, while sum_j || c_j Psi ||^2 = N || Psi ||^2. Thus some c_j Psi
is a nonzero (N-1)-particle zero mode, giving E_0(N-1; Q_N) = 0.
The divisibility argument in Lemma 6.2 excludes a zero mode in the
(N+1)-particle sector: the required product of cubed brackets would have
degree 3N > Q_N in each particle variable.
Let P_n project onto the lowest-energy eigenspace of H_{n, Q_N}, and
define the overlap operator on the (N+1)-particle sector by
G_N = P_{N+1} ( sum_{j=0}^{Q_N} c_j^dagger P_N c_j ) P_{N+1}.
Here P_{N+1} projects onto a positive-energy eigenspace.
For the pure two-body fermionic form (2.1), the cited Theorem 2.1(I), with
m_0 = m_1 = 2, gives
E_0(N+1; Q_N) >= ((N+1)/(N-1)) E_0(N; Q_N)
+ ((N+1 - ||G_N||)/(N-1)) Gap_N.
The cited Lemma 2.3 gives ||G_N|| <= rank P_N = 1. Since E_0(N; Q_N) = 0,
this yields E_0(N+1; Q_N) >= N Gap_N / (N-1). The two zero energies give
the asserted second difference, and Corollary 1.2 supplies the final
uniform bound. QED.
7. A PLANAR COROLLARY
---------------------
The full Fock space estimate also gives a gap on the plane. This is the
case ell = 3 of the folklore conjecture in Rougerie's survey.
Set up the Bargmann space of entire functions with norm
||f||_B^2 = (1/pi) integral_C |f(z)|^2 e^{-|z|^2} d^2 z,
e_j(z) = z^j / sqrt(j!).
Multiplication by pi^{-1/2} e^{-|z|^2/2} identifies this space unitarily
with the planar lowest Landau level at magnetic field 2; changing magnetic
length is a unitary dilation.
For a pair of coordinates set w = (z_1 - z_2) / sqrt(2) and
Z = (z_1 + z_2) / sqrt(2). Let P^{(1)} be the orthogonal projector onto
relative degree one in w, with no restriction on the degree in Z. Put
H_N^pl = sum_{1 <= i < j <= N} P_{ij}^{(1)} on wedge^N B.
Thus each pair projector has coefficient one, as on the sphere.
COROLLARY 7.1 (Planar spectral gap).
For every N >= 2, the untruncated planar Hamiltonian satisfies
(H_N^pl)^2 >= gamma_* H_N^pl,
where gamma_* is the constant in Theorem 1.1. Equivalently, if
P_{0, N}^pl projects onto its kernel, then
H_N^pl >= gamma_* (I - P_{0, N}^pl).
The constant is independent of both particle number and total angular
momentum. In particular, the case ell = 3 of the spectral gap
conjecture holds, in its pair-projector normalization.
Proof. The coordinate change (z_1, z_2) -> (w, Z) is unitary for the
Bargmann inner product. Consequently the vectors w Z^p / sqrt(p!), p >= 0,
form an orthonormal basis of the relative-degree-one subspace. Their
coefficients in the normalized wedge basis are
v_p^pl(i, j) = (i-j) * sqrt( p! / (2^p i! j!) ) * 1_{i+j = p+1} (i < j).
Indeed, expanding (z_1 - z_2)(z_1 + z_2)^p first gives the coefficient in
the tensor basis; passage to the normalized antisymmetric wedge multiplies
it by sqrt(2). These are exactly the limits in (Lemma 3.1). The
second-quantized pair-projector identity therefore gives
H_N^pl = sum_{p >= 0} B(v_p^pl)^dagger B(v_p^pl)
on antisymmetric polynomials.
Fix N and a total degree L. The corresponding homogeneous subspace has the
finite orthonormal basis
e_{i_1} wedge ... wedge e_{i_N}, 0 <= i_1 < ... < i_N, sum i_k = L.
For Q >= max(L, N-1, 2), this is also the entire spherical subspace of
total lowering deficit L, under the identification of normalized
one-particle bases. Both Hamiltonians preserve this subspace: a pair is
annihilated and created with the same sum of orbital labels. Only p+1 <= L
can contribute. The finitely many matrix entries of the spherical
restriction therefore converge to those of the planar restriction by
Lemma 3.1; the convergence is in operator norm (finite dimension).
Fix 0 < gamma < gamma_*. Theorem 1.1 gives H_Q^2 >= gamma H_Q for all
sufficiently large Q. Restrict to the invariant block just described and
let Q -> infinity. Norm convergence passes through the square (finite
dimensional) and gives (H_N^pl)^2 >= gamma H_N^pl on that block.
Now let gamma increase to gamma_*: the cone of positive matrices is
closed, so the same inequality holds at gamma_*. The limit Q -> infinity
is taken separately for each fixed gamma, allowing the flux threshold to
depend on gamma. Here N and L were arbitrary, and the resulting constant
is the same for all of them.
Finally, wedge^N B is the Hilbert direct sum of its homogeneous subspaces,
and 0 <= H_N^pl <= C(N, 2) I. The block inequalities thus extend by
continuity to the whole space. The spectral theorem gives the asserted
gap. Restriction to total degrees L <= 3N(N-1)/2 gives exactly the
sector in the cited conjecture for ell = 3. QED.
8. APPENDIX: THE FINITE RATIONAL CERTIFICATE
---------------------------------------------
This appendix specifies the seven rows and all arithmetic needed to verify
the finite inequalities of Sections 4.4 and 4.5. Starting from the integer
rows below, evaluate the rational three-body traces in Section 8.2 and
the rational four-body matrices in Section 8.3. Every value is determined
using only integers and rational numbers.
8.1 Row data. Let P = 10^7. The first two numbers of each row are t and
ell = P lambda. An entry "pj:a" specifies the integer a_{pj}, with p and j
parsed as separate single-digit labels; omitted entries are zero.
Continuation lines belong to the preceding row. Define
alpha_{pj} = (a_{pj} / P) sqrt( 2^p C(p+j, p) / (2^t C(p+j, t)) )
= (a_{pj} / P) sqrt( 2^{p-t} t! (p+j-t)! / (p! j!) ).
Every nonzero entry has 0 <= p+j-t <= 8.
The seven coefficient rows are stored in the file ../verification/rows.txt
of the original arXiv source. Format: each row begins with two integers
t_rho and ell_rho = P lambda_rho (P = 10^7), followed by entries "pj:a_{pj}".
Any reader with exact rational arithmetic can re-verify the whole
certificate from these finitely many integers alone; Sections 8.2 and
8.3 show exactly how.
In particular P^2 eta = sum_rho ell_rho^2 = 93527408868499.
8.2 Three-body arithmetic. For c in {1, 2}, an integer T >= 0, and
integers z, p, define
U(c, z, T, p) = sum_{h = max(0, p+z-T)}^{min(p, z)}
(-1)^{z-h} C(z, h) C(T-z, p-h) c^{p-h},
set to zero if z or p lies outside [0, T]. Expanding (Lemma 3.3) gives
s^{(1/sqrt3)}_{z, T, p} = U(2, z, T, p)
* sqrt( 2^z C(T, z) / (3^T 2^p C(T, p)) ).
Substitution in the (4.4) trace formula and the row data cancels all
radicals:
P^2 e_z = sum_rho sum_{T = max(z, t_rho)}^{15}
( 2^z C(T, z) / (3^T 2^{t_rho} C(T, t_rho)) )
* X_{rho, T} ( X_{rho, T} + 2 ell_rho U(2, z, T, t_rho) ),
where X_{rho, T} = sum_{p+j = T} a^rho_{pj} U(2, z, T, p), with value zero
on empty levels. For each z in {2, 4, 5, ..., 15}, evaluate the boxed
formula and form the exact rational margin
m_z = (3/2)(3z - 1)(-1/2)^z - e_z.
Write 10^6 m_z = n_z / d_z with integers n_z, d_z and d_z > 0. Integer
division gives the floors floor(n_z / d_z) listed below:
z : 2 4 5 6 7 8 9 10 11 12 13 14 15
floor: 1298313 211317 249 250 249 250 250 250 249 250 250 6470 250
All are positive, proving (4.4). The floors are the certificate: each is
strictly positive, so each m_z > 0, and the finite inequality holds.
8.3 Four-body arithmetic. Fix 1 <= D <= 23; all copy indices below lie
in R_D. Put
u_r = 1 / (r! (D-r)!),
(E_D)_{rs} = sqrt(u_r u_s) Y_{rs},
(G_D)_{rs} = sqrt(u_r u_s) Z_{rs}.
The matrices Y, Z are rational. Define
V_r(D, T; p, j, k) = U(1, r, j+k, j) U(1, D-r, T-r, p),
with value zero for r > j+k. Two applications of (Lemma 3.3) give
S_r(D, T; p, j, k) = V_r(D, T; p, j, k)
* sqrt( 2^{1-(j+k)-T+r} j! k! p! u_r / (T-D)! ).
Therefore
P^2 Y_{rs} = - sum_rho sum_{p,j; q,l, T >= D}
a^rho_{pj} a^rho_{ql}
V_r(D, T; p, j, k) V_s(D, T; q, l, i)
* 2^{1-2T+p+q-t_rho+(r+s)/2} i! k! t_rho! / (T-D)!,
where i = p+j-t_rho, k = q+l-t_rho, T = p+j+k. Both entry sums are
ordered. To verify the cancellation, multiply the two factors in the S
formula by the two factors in the row data formula; the factorial product
under the combined square root is (t_rho! i! k!)^2 u_r u_s / ((T-D)!)^2.
Using j+k = T-p and l+i = T-q, its power of two has exponent
2 + 2p + 2q - 2 t_rho - 4T + r + s. Taking the positive square root and
removing sqrt(u_r u_s) gives exactly the displayed factor.
For an increasing quadruple A of nonnegative indices with sum D+1, put
f(A) = prod_{b in A} b!. If (x, y, j, k) lists its elements, let
sigma(x, y, j, k) be the sign of sorting this list into increasing order.
In each summand below put p = x + y - 1, and define
L_r(A) = 2^{(1-2D+r)/2}
sum_{x < y in A, {j, k} = A minus {x, y}, j < k}
sigma(x, y, j, k) (x-y) p! j! k! V_r(D, D; p, j, k).
Combining (Lemma 3.1) and the S formula, the coefficient of w_{Dr}^* on
the Slater wedge A is sqrt(u_r / f(A)) L_r(A). Thus
Z_{rs} = sum_{A increasing quadruple, sum_{b in A} b = D+1}
L_r(A) L_s(A) / f(A).
All exponents of two in these formulas are integers, since r, s are odd.
Let D_u = diag(u_r) and D_{sqrt u} = diag(sqrt(u_r)). The matrix in the
certificate (Section 4.5) equals D_{sqrt u} M D_{sqrt u}, where
M = Z B Z,
B = diag( (2 * 1_{r=1} + eps) u_r ) - D_u Y D_u.
Thus it suffices to check M >= 0 using rational arithmetic. This
congruence uses only the positive diagonal D_{sqrt u}; it does not invert
G_D or presume that its columns are independent.
To check positivity for each D = 1, ..., 23, form Y and Z from the above
formulas, then form B and M, using eps = 3/10^6. All sums range over the
printed rows and the finite index sets specified. Compute with exact
fractions throughout.
If d = |R_D|, initialize J = I_d. For b = 1, ..., d, compute in order
X = M J,
c_b = tr X / b,
J = c_b I_d - X.
This is the Leverrier-Faddeev recurrence applied to -M. It gives
det(x I + M) = x^d + c_1 x^{d-1} + ... + c_d.
For example, this recurrence follows by multiplying the polynomial
adjugate of x I + M by x I + M and comparing coefficients, together with
the derivative-of-determinant trace identity.
Exact evaluation of U, Y, L, Z, and the recurrence gives:
D signs D signs D signs
1 0 9 ++000 17 ++++++000
2 0 10 +0000 18 +++++0000
3 00 11 +++000 19 +++++++000
4 00 12 ++0000 20 ++++++0000
5 +00 13 ++++000 21 ++++++++000
6 000 14 +++0000 22 +++++++0000
7 +000 15 +++++000 23 ++++++++000
8 +000 16 ++++0000
The signs list the coefficients after the leading 1, in descending degree.
For each computed coefficient, choose a positive denominator and inspect
its integer numerator: "+" denotes a positive numerator and "0" denotes
zero.
Every coefficient is nonnegative. So det(x I + M) > 0 for x > 0. Since M
is real symmetric, a negative eigenvalue would give a positive root of
this polynomial. Therefore M >= 0, proving the certificate. This uses
the characteristic-coefficient positivity test for symmetric matrices
(Peyrl-Parrilo, Theorem 5): for a real symmetric matrix, all
characteristic coefficients nonnegative (after the leading 1) implies
positive semidefiniteness.
REMARK (Nature of the certificate).
The data above certify fixed spin-matrix inequalities. The positive
comparison operator is a sum of squares by construction. No
many-particle finite-size spectrum, numerical extrapolation, or
assumed plasma property enters the argument. The exact equalities
and inequalities in this appendix can be verified by a finite
calculation using only integers and rational numbers.
END OF PROOF.
Summary of the argument, one paragraph: the square of the Haldane V_1
interaction on the sphere decomposes by Wick's theorem into a two-body
piece (H), a three-body piece, and a four-body piece. A sum of squares
K of seven low-mode auxiliary annihilators, after Haar averaging over
SU(2), has three- and four-body parts that dominate the corresponding
pieces of H^2. The domination is verified by a finite rational
certificate (seven rows, three-body margins m_z > 0 for z in {2, 4, ...,
15}, four-body determinantal signs for D = 1, ..., 23). The comparison
transfers uniformly to finite spheres via an exact diagonal conjugation,
because every error is dominated by local pair energy and 25 modes are
enough. The leftover coefficient equals 4616733319001 / 10^14 > 1/25.
Hence H_Q^2 >= gamma * H_Q for every 0 < gamma < gamma_* on the full
Fock space, uniformly in particle number. At Laughlin flux the kernel is
one dimensional, so we obtain a genuine ground-state gap of at least 1/25.
The planar corollary follows by taking the fixed-(N, L) spherical
restriction to Q -> infinity.